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No. 35, Jalan 5/10B
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Selangor Darul Ehsan
CONTENTS
KANDUNGAN
1
THE STRUCTURE OF ATOMS
STRUKTUR ATOM
2
CHEMICAL FORMULA AND EQUATIONS
FORMULA DAN PERSAMAAN KIMIA
22
3
PERIODIC TABLE
JADUAL BERKALA
49
4
CHEMICAL BOND
IKATAN KIMIA
72
5
ELECTROCHEMISTRY
ELEKTROKIMIA
88
6
ACID AND BASES
ASID DAN BES
114
7
SALT
GARAM
139
8
MANUFACTURED SUBSTANCES IN INDUSTRY
BAHAN KIMIA DALAM INDUSTRI
168
Con-Chem F4 (B).indd 3
1
12/9/2011 6:00:06 PM
Chemistry Form 4 • MODULE
1
THE STRUCTURE OF ATOMS
STRUKTUR ATOM
MATTER / JIRIM
• PARTICLE THEORY OF MATTER / TEORI ZARAH JIRIM
– To state the particle theory of matter
Menyatakan teori zarah jirim
– To differentiate and draw the three types of particles i.e. atom, ion and molecule
Membezakan dan melukis tiga jenis zarah jirim iaitu atom, ion dan molekul
– To describe the laboratory activity to investigate the diffusion of particles in gas, a liquid and a solid. (To prove that matter is
made up of tiny and discrete particles)
Menghuraikan aktiviti makmal untuk mengkaji resapan zarah dalam gas, cecair dan pepejal (Untuk membuktikan bahawa jirim terdiri daripada
zarah-zarah yang halus dan diskrit)
• KINETIC THEORY OF MATTER / TEORI KINETIK JIRIM
– To state the kinetic theory of matter
Menyatakan teori kinetik jirim
– To relate the change of physical states of matters with energy change
Menghubungkaitkan perubahan keadaan jirim dengan perubahan tenaga
– To relate the change of energy in the particles with kinetic particle theory of matter
Menghubungkaitkan perubahan tenaga dalam zarah dengan perubahan tenaga kinetik zarah
THE STRUCTURE OF ATOMS / STRUKTUR ATOM
• HISTORY OF ATOMIC MODELS DEVELOPMENT / SEJARAH PERKEMBANGAN MODEL ATOM
– To state the contribution of scientists in the atomic structure model such as the scientists who discovered electron, proton,
nucleus, neutron and shell
Menyatakan sumbangan ahli sains kepada perkembangan model struktur atom dan ahli sains yang menemui elektron, proton, nukleus, neutron dan
petala
• SUBATOMIC PARTICLES / ZARAH-ZARAH SUBATOM
– To compare and differentiate subatomic particles i.e. proton, neutron and electron from the aspect of charge, relative mass
and location
Membanding dan membezakan zarah-zarah subatom iaitu proton, neutron dan elektron dari segi cas, jisim relatif dan kedudukan
– To state the meaning of proton number and nucleon number based on the subatomic particle
Menyatakan maksud nombor proton dan nombor nukleon berdasarkan zarah subatom
– To write the symbol of elements with proton number and nucleon number
Menulis simbol unsur yang mengandungi nombor proton dan nombor nukleon
• ISOTOPE / ISOTOP
– To state the meaning, examples and the use of isotopes
Menyatakan maksud isotop, contoh-contoh isotop dan kegunaan isotop
• ELECTRON ARRANGEMENT / SUSUNAN ELEKTRON
– To know the number of electron shells and number of electrons in the 1st, 2nd and 3rd shell
Mengetahui bilangan petala elektron serta bilangan elektron yang diisi dalam petala 1, 2 dan 3
– To write the electron arrangement of atoms based on proton number or number of electrons and state the number of valence
electron
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Menulis susunan elektron bagi suatu atom berdasarkan nombor proton atau bilangan elektron dan seterusnya menyatakan bilangan elektron valens
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MODULE • Chemistry Form 4
MATTER / JIRIM
Matter is any substance that has mass and occupies space.
Jirim adalah sebarang bahan yang mempunyai jisim dan memenuhi ruang.
The Particle Theory of Matter / Teori Zarah Jirim
atoms ,
Matter is made up of tiny and discrete particles. Three types of tiny particles are
1
atom
Jisim terdiri daripada zarah yang halus dan diskrit. Tiga jenis zarah tersebut ialah
,
ions
ion
and molecules .
molekul .
dan
Matter can be classified as element or compound. / Jirim boleh dikelaskan sebagai unsur atau sebatian.
Complete the following: / Lengkapkan yang berikut:
2
3
MATTER / JIRIM
ELEMENT / UNSUR
satu
A substance made from only
Bahan yang terdiri daripada
satu
type of atom.
jenis atom sahaja.
COMPOUND / SEBATIAN
two
or more
A substance made from
elements which are bonded together.
dua
Bahan yang terdiri daripada
atau
different
lebih
unsur berbeza yang terikat secara kimia.
Types of particles / Jenis zarah
Atom / Atom
The smallest neutral particle
of an element (Normally pure
metals, noble gases and a
few non-metal elements such
as carbon and silicon).
Zarah neutral yang paling kecil
bagi suatu unsur (Biasanya logam
tulen, gas adi dan beberapa unsur
bukan logam seperti karbon dan
silikon).
Example:
Types of particles / Jenis zarah
Molecule / Molekul
A neutral particle consists
of similar non-metal atoms
which are covalently-bonded.
Molecule / Molekul
A neutral particle consists
of different non-metal atoms
which are covalently-bonded.
Zarah neutral terdiri daripada
atom-atom bukan logam serupa
terikat secara ikatan kovalen.
Zarah neutral terdiri daripada
atom-atom bukan logam berlainan
terikat secara ikatan kovalen.
Example:
Example:
Contoh:
Contoh:
Oxygen gas, O2
Carbon dioxide gas, CO2
Gas oksigen, O2
Gas karbon dioksida, CO2
Contoh:
Sodium metal, Na
O O
Logam natrium, Na
O O
Na Na Na Na Na
Na Na Na Na Na Na
Na Na Na Na Na
O O
O
C
O
O
C
O
Ne
O
Example:
Contoh:
Na+ Cl – Na+ Cl – Na+
Air, H2O
H H
H H
Natrium klorida, NaCl
Water, H2O
Gas hidrogen, H2
H H
Ne
C
Zarah bercas positif atau negatif
terbentuk dari logam dan bukan
logam terikat secara ikatan ion.
Daya tarikan antara dua ion yang
berlawanan cas membentuk ikatan
ion.
Sodium chloride, NaCl
Hydrogen gas, H2
Neon gas, Ne
Gas Neon, Ne
O
Ion / Ion
Positively or negatively
charged particles, which
are formed from metal
atom and non-metal atom
respectively. The force of
attraction between the two
oppositely charged ions
forms an ionic bond.
H
H
O
O
Cl – Na+ Cl – Na + Cl –
H
H
O
H
H
Ne
Na+ Cl – Na+ Cl – Na+
Calcium oxide, CaO
Kalsium oksida, CaO
Ca2+ O 2– Ca2+ O 2– Ca2+
O 2– Ca2+ O 2– Ca2+ O 2–
Ca2+ O 2– Ca2+ O 2– Ca2+
–
Elements can be identified as metal or non-metal by referring to the Periodic Table.
–
Formation of molecule and ion will be studied in Chapter 4 (Chemical Bond).
Unsur boleh dikenal pasti sebagai logam atau bukan logam dengan merujuk kepada Jadual Berkala Unsur.
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Pembentukan molekul atau ion akan dipelajari dalam Tajuk 4 (Ikatan Kimia).
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Chemistry Form 4 • MODULE
4
Determine the type of particles in the following substances:
Tentukan jenis zarah bagi setiap bahan berikut:
Substances
Type of particle
Substances
Type of particle
Substances
Type of particle
Molecule
Sulphur dioxide
(SO2)
Sulfur dioksida
(SO2)
Molecule
Tetrachloromethane (CCl4)
Tetraklorometana (CCl4)
Molecule
Copper(II) sulphate
(CuSO4)
Kuprum(II) sulfat
(CuSO4 )
Ion
Iron (Fe)
Ferum (Fe)
Atom
Zink chloride (ZnCl2)
Zink klorida
(ZnCl2 )
Ion
Argon (Ar)
Argon (Ar)
Atom
Carbon (C)
Karbon (C)
Atom
Hydrogen peroxide (H2O2)
Hidrogen peroksida (H2O2)
Molecule
Bahan
Jenis zarah
Hydrogen gas (H2)
Gas hidrogen (H2)
5
Bahan
Jenis zarah
Bahan
Jenis zarah
Diffusion
Resapan
(a) The tiny and discrete particles that made up matter are constantly moving. In gases, these particles are very far apart
from each other, in liquids, the particles are closer together and in solids, they are arranged closely packed.
Jirim terdiri daripada zarah-zarah halus dan diskrit yang sentiasa bergerak. Dalam gas, susunan zarah-zarahnya adalah berjauhan
antara satu sama lain, dalam cecair, zarah-zarahnya disusun lebih rapat dan dalam pepejal, zarah-zarahnya disusun dengan sangat
padat dan teratur.
(b) Diffusion occurs when particles of a substance move between the particles of another substance.
Resapan berlaku apabila zarah-zarah suatu bahan bergerak di antara zarah-zarah bahan lain.
(c) Diffusion occurs in a solid, liquid and gas. Complete the following table:
Resapan berlaku dalam pepejal, cecair dan gas. Lengkapkan jadual berikut:
Diffusion in a gas
Resapan dalam gas
Experiment
Eksperimen
A few drops
of bromine
liquid
Beberapa titis
cecair bromin
After few
minutes
Selepas
beberapa
minit
Diffusion in a liquid
Resapan dalam cecair
Water
Air
After a
few hours
Selepas
beberapa jam
Potassium manganate(VII)
Kalium manganat(VII)
Observation
Pemerhatian
Explanation
Penerangan
The brown colour of bromine vapour,
far
Br2 spreads
throughout
the two jars.
The purple colour of solid potassium
manganate(VII), KMnO4 spreads
slowly
throughout the water.
Diffusion in a solid
Resapan dalam pepejal
Gel
Agar-agar
Copper(II)
sulphate
After a
day
Selepas
sehari
Kuprum(II) sulfat
The blue colour of copper(II) sulphate,
CuSO4 spreads very slowly
throughout the gel.
Warna perang wap bromin, Br2 merebak
cepat
memenuhi kedua-dua
dengan
balang gas.
Warna ungu pepejal kalium manganat(VII),
perlahan
KMnO4 merebak dengan
di dalam air.
Warna biru kuprum(II) sulfat,
sangat perlahan
CuSO4 merebak
di dalam agar-agar.
Bromine vapour, Br2 and air are made
molecules .
up of
Wap bromin, Br2 dan udara terdiri
molekul
daripada
.
molecules
diffuse
Bromine
quickly between
large
Potassium manganate(VII) is
made up of potassium ions and
ions
manganate(VII) ions. The
slowly
diffuse
between close
space of water particles which is in
liquid form.
Copper(II) sulphate, CuSO4 is made
ions
and
up of copper(II)
ions
ions
sulphate
. The
slow
diffuse very
between
closely packed space of gel particles
which is in solid form.
space of air particles which is in gas
form.
Kuprum(II) sulfat, CuSO4 terdiri daripada
ion
ion
kuprum(II) dan
Ion-ion
sulfat.
ini meresap dengan
perlahan
sangat
antara ruang
padat
zarah agar-agar yang
berbentuk pepejal.
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Molekul bromin meresap
pantas
besar
melalui ruang
antara zarahzarah udara yang berbentuk gas.
Kalium manganat(VII) terdiri daripada
ion kalium dan ion manganat(VII).
Ion-ion
perlahan
ini meresap
rapat
antara ruang
zarah air
yang berbentuk cecair.
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MODULE • Chemistry Form 4
(d) Conclusions:
Kesimpulan:
(i)
gas
than in liquid. There is
Diffusion occurs faster in
gas
gas
a
than a liquid. Particles in a
are
closer
are
together.
larger
space in between the particles of
further
apart. The particles in the liquid
gas
Resapan berlaku lebih cepat di dalam
berbanding di dalam cecair. Terdapat ruang yang
gas
gas
berbanding dengan cecair. Zarah-zarah
adalah
antara zarah-zarah
lebih
rapat
antara satu sama lain.
antara satu sama lain. Zarah-zarah cecair adalah
(ii)
liquid
than in solid. There is
Diffusion occurs faster in a
liquid
of a
than a solid. The particles in the solid are very
cecair
Resapan berlaku lebih cepat di dalam
cecair
antara zarah-zarah
dan
padat
larger
lebih besar
berjauhan
space in between the particles
close
together.
lebih besar
berbanding di dalam pepejal. Terdapat ruang yang
rapat
berbanding dengan pepejal. Zarah-zarah pepejal tersusun sangat
antara satu sama lain.
tiny
(iii) Bromine gas, potassium manganate(VII) and copper(II) sulphate are made up of
particles that are constantly moving/constant motion .
and
halus
Gas bromin, kalium manganat(VII) dan kuprum(II) sulfat terdiri daripada zarah-zarah
sentiasa bergerak
.
yang
discrete
diskrit
dan
The Kinetic Theory of Matter / Teori Kinetik Jirim
solid
Matter exists in three different states which are
1
pepejal
Jirim wujud dalam tiga keadaan iaitu
Matter that made up of
2
tiny
As the temperature increases, the
Apabila suhu meningkat, tenaga
discrete
and
halus
Jirim terdiri daripada zarah-zarah
3
,
kinetik
and
gas
gas
dan
.
.
moving
particles which are always in constantly
dan
kinetic
liquid
,
cecair
diskrit
yang sentiasa
bergerak
.
.
energy of particles increases and the particles move
zarah-zarah akan bertambah dan zarah-zarah akan bergerak dengan
faster
.
lebih cepat
.
Particles in different states of matter have different arrangement, strength of forces between them, movement and
energy content.
4
Zarah-zarah
dalam keadaan jirim yang berbeza mempunyai susunan, daya tarikan antara zarah, pergerakan dan kandungan
tenaga yang berbeza.
Complete the following table: / Lengkapkan jadual di bawah:
5
State of matter
Solid
Keadaan jirim
Liquid
Pepejal
Gas
Cecair
Gas
Draw the particles arrangement.
Each particle (atom/ ion/
molecule) is represented by
Lukis susunan zarah. Setiap zarah
(atom / ion / molekul) diwakili dengan
‘ ’
Particles arrangement
Susunan zarah
The particles are arranged
closely packed in
orderly
manner.
Zarah-zarah tersusun
teratur
dan
.
Particles movement
m
orderly manner
padat
.
Zarah-zarah tersusun
tidak teratur
tetapi
padat
.
The particles are very
widely separated
from
each other.
terpisah jauh
Zarah-zarah
antara satu sama lain.
Particles can only vibrate
rotate
about their
and
Particles can vibrate ,
rotate
move
and
Particles can vibrate ,
rotate
move
and
fixed position.
throughout the liquid.
freely.
Zarah bergetar dan berputar
pada kedudukan tetap.
Zarah bergetar , berputar dan
bergerak dalam cecair.
Zarah bergetar , berputar dan
bergerak bebas.
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Pergerakan zarah
The particles are arranged
closely packed but not in
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Chemistry Form 4 • MODULE
Strong
Attractive forces between the
particles
Daya tarikan antara zarah
forces between the
particles but weaker than
strong
forces
Very
between the particles.
the forces in the solid.
Daya tarikan yang sangat kuat
antara zarah-zarah.
kuat
Daya tarikan yang
antara zarah-zarah tetapi
lebih lemah
berbanding di
Weak
forces between
the perticles
lemah
Daya tarikan yang
antara zarah-zarah.
dalam pepejal.
Energy content of the particles
Kandungan tenaga zarah
6
low .
Energy content is very
Kandungan tenaga sangat
rendah
.
Energy content is higher
than solid but less than in a
gas.
very
Energy content is
high.
Kandungan tenaga lebih tinggi
daripada pepejal tetapi
lebih rendah
daripada gas.
Kandungan tenaga
tinggi.
sangat
Changes in the state of matter
Perubahan keadaan jirim
(a) Matter undergoes change of state when
heat
Jirim mengalami perubahan keadaan apabila tenaga
(i)
(ii)
energy is
haba
absorbed
or
serap
di
released/lose
atau di
When heat energy is absorbed by the matter (it is heated), the
increases and they vibrate faster.
kinetic
diserap
Apabila tenaga haba
oleh jirim (semasa dipanaskan), tenaga
dan zarah tersebut bergerak dengan lebih cepat.
kinetik
When matter releases heat energy (it is cooled), the
they vibrate less vigorously.
dibebaskan
Apabila tenaga haba
zarah tersebut bergerak kurang cergas.
kinetic
:
bebaskan
:
energy of the particles
zarah
bertambah
energy of the particles decreases and
oleh jirim (semasa disejukkan), tenaga kinetik zarah
berkurang
dan
(b) Inter - conversion of the states of matter:
Perubahan keadaan jirim:
Solid
Pepejal
7
Melting / Peleburan
Freezing / Pembekuan
Liquid
Cecair
Boiling/Evoporation / Pendidihan/Penyejatan
Condensation / Kondensasi
Gas
Gas
Determination of melting and freezing points of naphthalene
Penentuan takat lebur dan takat beku naftalena
Materials / Bahan: Naphthalene powder, water
Apparatus / Radas: Boiling tube, conical flask, beaker, retort stand, thermometer 0 – 100°C, stopwatch,
Bunsen burner and wire gauze
Procedure / Prosedur:
I. Heating of naphthalene / Pemanasan naftalena
Set-up of apparatus: / Susunan radas:
Thermometer / Termometer
Boiling tube / Tabung didih
Water / Air
Naphthalene / Naftalena
Heat
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MODULE • Chemistry Form 4
(a)
boiling tube
A
placed into it.
Tabung didih
3 - 5 cm
is filled
height with naphthalene powder and a
diisi dengan serbuk naftalena setinggi
3 – 5 cm
dan
thermometer
termometer
is
diletakkan
di dalamnya.
(b)
The boiling tube is immersed in a water bath as shown in the diagram so that the water level in the water bath
is higher than naphtalene powder in the boiling tube.
Tabung didih dimasukkan ke dalam kukus air seperti di dalam gambar rajah dan pastikan aras air dalam kukus air lebih tinggi
daripada aras naftalena dalam tabung didih.
(c)
heated
The water is
Air dipanaskan dan naftalena
(d)
and the naphthalene is
dikacau
perlahan-lahan dengan
slowly with
termometer
thermometer
.
.
60°C
, the stopwatch is started. The temperature of
When the temperature of naphthalene reaches
90°C
naphthalene is recorded at 30 seconds intervals until the temperature of naphthalene reaches
.
60°C
Apabila suhu naftalena mencapai
sehingga suhunya mencapai
II.
stirred
90°C
, mulakan jam randik. Suhu naftalena dicatat setiap
30 saat
.
Cooling of naphthalene / Penyejukan naftalena
Naphthalene
Naftalena
(a)
The boiling tube and its content is removed from the water bath and put into a
in the diagram.
conical flask
kelalang kon
Tabung didih dan kandungannya dikeluarkan daripada kukus air dan dipindahkan ke dalam
dalam gambar rajah.
(b)
as shown
seperti
stirred
constantly with thermometer throughout cooling
The content in the boiling tube is
supercooling
process to avoid
(the temperature of cooling liquid drops below freezing point, without
the appearance of a solid).
dikacau
Kandungan dalam tabung didih
perlahan-lahan dengan termometer sepanjang proses penyejukan untuk
penyejukan lampau
(Suhu cecair yang disejukkan turun melepasi takat beku tanpa pembentukan
mengelakkan
pepejal).
(c)
The temperature of naphthalene is recorded every
60°C
to
.
Suhu naftalena dicatat setiap
(d)
60°C
.
suhu
melawan
masa
dilukis untuk proses pemanasan dan penyejukan.
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sehingga suhunya mencapai
interval until the temperature drops
A graph of temperature against time is plotted for the heating and cooling process respectively.
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30 seconds
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Chemistry Form 4 • MODULE
The Explanation of the Heating Process of Matter / Penerangan Proses Pemanasan
1
The heating curve of naphthalene:
Lengkung pemanasan naftalena:
Temperature/°C
Suhu/°C
F
D
B
E
C
A
Time/s
Masa/s
2
faster
When a solid is heated, the particles absorb heat and move
absorbed
energy is
, the state of matter will change.
as its energy content increases. As the heat
lebih cepat
Apabila pepejal dipanaskan, zarah-zarah menyerap haba dan bergerak
diserap
menyebabkan perubahan keadaan jirim.
Tenaga haba
Point
Titik
A to B
A ke B
State of Matter
Explanation
Keadaan jirim
Penerangan
absorbed
Heat energy is
kinetic
Solid
energy to
diserap
pepejal
oleh zarah-zarah
lebih cepat
dan zarah bergetar dengan
absorbed
Heat energy
Solid and
Liquid
solid
by the particles in the
increase
and vibrate
naphthalene causing their
faster
. The temperature
increases.
Tenaga haba
bertambah
B to C
B ke C
disebabkan kandungan tenaga bertambah.
overcome
naftalena menyebabkan tenaga
meningkat
. Suhu semakin
by the particles in the
forces between particles so that the
remains constant
temperature
diserap
Tenaga haba yang
liquid
solid
C to D
C ke D
Liquid
increases
Tenaga haba
bertambah
Heat energy
overcome
D to E
D ke E
Liquid and
Gas
to form a
Tenaga haba
turn to
Gas
liquid
to
. The
pepejal
oleh zarah-zarah dalam
naftalena
pepejal
cecair
berubah menjadi
digunakan
absorbed
liquid
by the particles in the
increase
energy to
and move
untuk mengatasi
tetap
.
. Suhu adalah
naphthalene causing their
faster
. The temperature
.
diserap
oleh zarah-zarah
cecair
dan zarah-zarah bergerak dengan
absorbed
kinetik
naftalena menyebabkan tenaga
lebih cepat
meningkat
. Suhu semakin
.
by the particles in the
liquid
naphthalene is
the forces of attraction between particles. The particles begin to move
gas
. The temperature
diserap
oleh zarah-zarah dalam
remains constant
cecair
absorbed
energy to incerease and move
akan
used
to
freely
.
naftalena digunakan untuk mengatasi
bebas
gas
untuk membentuk
. Suhu
gas
by the particles in the
faster
. The temperature
causing their
increases
.
diserap
oleh zarah-zarah gas naftalena menyebabkan tenaga
Tenaga haba
lebih cepat
meningkat
dan zarah-zarah bergerak dengan
. Suhu semakin
kinetik
kinetic
akan bertambah
.
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E to F
E ke F
used
.
daya tarikan antara zarah-zarah. Zarah-zarah mula bergerak
tetap
adalah
.
Heat energy is
akan
.
naphthalene is
daya tarikan antara zarah-zarah supaya
Heat energy
kinetic
kinetik
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MODULE • Chemistry Form 4
completely changes to become a liquid is called the melting point .
absorbed
by the particles
During the melting process, the temperature remains unchanged because heat energy
used
liquid .
is
to overcome the forces between particles so that the solid change to turn into a
The constant temperature at which a
3
solid
takat lebur
berubah kepada keadaan cecair dipanggil
diserap
oleh zarah-zarah
Semasa proses peleburan, suhu tidak berubah kerana haba yang
mengatasi
cecair
daya tarikan antara zarah supaya pepejal berubah menjadi
.
Suhu tetap di mana suatu
pepejal
.
digunakan
untuk
completely changes to become a gas is called the boiling point .
absorbed
by the particles
During the boiling process, the temperature remains unchanged because heat energy
used
is
to overcome the forces between particles so that the liquid change to turn into a gas.
The constant temperature at which a
4
liquid
cecair
Suhu tetap di mana suatu bahan dalam keadaan
takat didih
berubah kepada keadaan gas dipanggil
.
diserap
digunakan
oleh zarah-zarah
untuk
Semasa proses pendidihan, suhu tidak berubah kerana haba yang
mengatasi
daya tarikan antara zarah supaya cecair berubah menjadi gas.
The Explanation for the Cooling Process of Matter: / Penerangan Proses Penyejukan Bahan:
The cooling curve of naphthalene:
1
Lengkung penyejukan naftalena:
Temperature/°C
Suhu/°C
P
Q
R
S
Time/s
Masa/s
slower
When the liquid is cooled, the particles in the liquid release energy and move
released
decreases. As the energy is
to the surrounding, the state of matter will change.
2
cecair
Apabila cecair disejukkan, zarah
membebaskan tenaga dan
dibebaskan
ke persekitaran.
berubah semasa tenaga
Point
Titik
State of matter
P ke Q
Liquid
Penerangan
released/given out
liquid
to the surrounding by the particles in the
liquid
kinetic
move
lose their
energy and
The particles in the
temperature decreases
Q ke R
R to S
m
Solid
released
by the
heat
The temperature
Haba
tenaga
slower. The
cecair
naftalena. Zarah-zarah dalam
semakin perlahan. Suhu semakin menurun
.
liquid
to the surrounding by the particles in
naphthalene is balanced
solid
energy released as the particles attract one another to form a
.
remains constant
.
dibebaskan
cecair
diimbangi
ke persekitaran oleh zarah-zarah dalam
naftalena
oleh
haba
terbebas
yang
apabila zarah-zarah tertarik antara satu sama lain untuk membentuk
pepejal
tetap
. Suhu adalah
.
The particles in the solid naphthalene
decreases .
Zarah-zarah dalam pepejal naftalena
menurun
Suhu semakin
.
releases
membebaskan
heat and vibrate
slower
tenaga dan bergetar dengan
. The temperature
lebih perlahan
.
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R ke S
Liquid and
Solid
The heat
naphthalene.
.
dibebaskan
ke persekitaran oleh zarah-zarah dalam
Haba
cecair
kinetik
kehilangan tenaga
dan bergerak
Q to R
semakin perlahan. Keadaan jirim
Explanation
Keadaan jirim
Heat is
P to Q
bergerak
as its energy content
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01-Chem F4 (3p).indd 8
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Chemistry Form 4 • MODULE
3
freezing point
changes to a solid is called
. During the freezing
released
process, the temperature remains unchanged because the heat
to the surrounding is balanced by the
The constant temperature at which a
liquid
heat released when the liquid particles rearrange themselves to become a
solid
.
takat beku
berubah kepada keadaan pepejal dipanggil
. Semasa proses
dibebaskan
diimbangi
ke persekitaran
oleh haba yang terbebas
pembekuan, suhu tidak berubah kerana haba yang
pepejal
.
apabila zarah-zarah cecair menyusun semula untuk membentuk
Suhu tetap di mana suatu
cecair
Keadaan Fizik Bahan pada Sebarang Suhu: / Physical State Of A Substance At Any Given Temperature:
1
A substance is in
solid
state if the temperature of the substance is below melting point
pepejal
Suatu bahan berada dalam keadaan
2
A substance is in
liquid
state if the temperature of the substance is between melting and boiling points.
cecair
Suatu bahan berada dalam keadaan
3
A substance is in
gas
jika suhu bahan tersebut lebih rendah daripada takat leburnya.
jika suhu bahan tersebut berada antara takat lebur dan takat didihnya.
state if the temperature of the substance is above boiling point.
Suatu bahan berada dalam keadaan
gas
jika suhu bahan tersebut lebih tinggi daripada takat didihnya.
EXERCISE / LATIHAN
1
The table below shows substances and their chemical formula.
Jadual di bawah menunjukkan bahan dan formula kimia masing-masing.
Substance / Bahan
Chemical formula / Formula kimia
Type of particle / Jenis zarah
Silver / Argentum
Ag
Atom
Potassium oxide / Kalium oksida
K2O
Ion
Ammonia / Ammonia
NH3
Molecule
Chlorine / Klorin
Cl2
Molecule
(a) State the type of particles that made up each substance in the table.
Nyatakan jenis zarah yang membentuk bahan dalam jadual di atas.
(b) Which of the substances are element? Explain your answer.
Yang manakah antara bahan tersebut merupakan suatu unsur? Jelaskan jawapan anda.
Silver and chlorine. Silver and chlorine are made up of one type of atom
(c) Which of the substance are compound? Explain your answer.
Yang manakah antara bahan tersebut merupakan suatu sebatian? Jelaskan jawapan anda.
Potassium oxide and ammonia. Potassium oxide and ammonia are made up of two different elements
2
The table below shows the melting and boiling points of substance P, Q and R.
Jadual di bawah menunjukkan takat lebur dan takat didih bagi bahan P, Q dan R.
Melting point / Takat lebur / °C
Boiling point / Takat didih / °C
P
–36
6
Q
–18
70
R
98
230
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Substance / Bahan
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MODULE • Chemistry Form 4
(a) (i)
What is meant by ‘melting point’?
Apakah yang dimaksudkan dengan ‘takat lebur’?
The constant temperature at which a solid charges to a liquid at particular pressure
(ii)
What is meant by ‘boiling point’?
Apakah yang dimaksudkan dengan ‘takat didih’?
The constant temperature at which a liquid changes to a gas at particular pressure
(b) Draw the particles arrangement of substances P, Q and R at room condition.
Lukis susunan zarah P, Q dan R pada keadaan bilik.
Substance P / Bahan P
(c)
(i)
Substance Q / Bahan Q
Substance R / Bahan R
What is the substance that exist in the form of liquid at 0°C.
Nyatakan bahan yang wujud dalam keadaan cecair pada suhu 0°C.
P, Q
(ii)
Give reason to your answer.
Jelaskan jawapan anda.
The temperature 0°C is above the melting point of Q and below the boiling point of Q
(d) (i)
Substance Q is heated from room temperature to 100°C. Sketch a graph of temperature against time for the
heating of substance Q.
Bahan Q dipanaskan dari suhu bilik hingga 100°C. Lakarkan graf suhu melawan masa bagi pemanasan bahan Q terhadap masa
untuk pemanasan bahan Q.
Temperature/°C
70
Time/s
(ii)
What is the state of matter of substance Q at 70°C?
Apakah keadaan fizik bahan Q pada 70°C?
Liquid and gas
(e) Compare the melting point of substances Q and R. Explain your answer.
Bandingkan takat lebur bahan Q dan R. Terangkan jawapan anda.
The melting point of substance R is higher than subtance Q. The attraction force between particles in substance R
m
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is stronger than Q. More heat is needed to overcome the force between particles in substance R.
d.
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Chemistry Form 4 • MODULE
3
The melting point of acetamide can be determined by heating solid acetamide until it melts as shown in the diagram
below. The temperature of acetemide is recorded every three minutes when it is left to cool down at room temperature.
Takat lebur asetamida boleh ditentukan dengan memanaskan pepejal asetamida sehingga lebur seperti dalam rajah di bawah. Suhu asetamida
dicatatkan setiap tiga minit semasa disejukkan pada suhu bilik.
Thermometer / Termometer
Boiling tube / Tabung didih
Water / Air
Acetamide / Asetamida
(a) What is the purpose of using water bath in the experiment?
Apakah tujuan menggunakan kukus air dalam eksperimen ini?
To ensure even heating of acetemide. Acetamide is easily combustible.
(b) State the name of another substance which its melting point can also be determined by using water bath as shown
in the above diagram.
Namakan satu bahan lain yang mana takat leburnya boleh ditentukan dengan menggunakan kukus air seperti rajah di atas.
Naphthalene
(c) Sodium nitrate has a melting point of 310°C. Can the melting point of sodium nitrate be determined by using the
water bath as shown in the diagram? Explain your answer.
Natrium nitrat mempunyai takat lebur 310°C. Bolehkah takat lebur natrium nitrat ditentukan dengan menggunakan kukus air seperti
yang ditunjukkan dalam rajah di atas? Jelaskan jawapan anda.
No, because the melting point of water is 100°C which is less than the melting point of sodium nitrate.
(d) Why do we need to stir the acetemide in the boiling tube in above experiment?
Mengapakah asetamida dalam tabung didih itu perlu dikacau sepanjang eksperimen?
To make sure the heat is distributed evenly
(e) The graph of temperature against time for the cooling of liquid acetamide is shown below.
Rajah di bawah menunjukkan graf suhu melawan masa untuk penyejukan cecair asetamida.
Temperature / Suhu/ °C
T3
T2
Q
R
T1
(i)
Time / Masa/s
What is the freezing point of acetamide?
Apakah takat beku asetamida?
T2°C
(ii)
The temperature between Q and R is constant. Explain.
Suhu antara titik Q dan R adalah tetap. Jelaskan.
The heat lost to the surrounding is balanced by the heat released when the liquid particles
rearrange themselves to become solid.
(f)
Acetemide exists as molecules. State the name of another compound that is made up of molecules.
Asetamida wujud sebagai molekul. Namakan sebatian lain yang terdiri daripada molekul.
Water/naphthalene
(g) What is the melting point of acetamide?
Apakah takat lebur asetamida?
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MODULE • Chemistry Form 4
The Atomic Structure / Struktur Atom
History of the development of atomic models:
1
Sejarah perkembangan model atom:
Scientist
Saintis
Atomic Model
Discovery
Model atom
Penemuan
(i)
Matter is made up of particles called
atoms
.
atom
.
Jirim terdiri daripada zarah-zarah dipanggil
(ii)
Dalton
created
Atoms cannot be
dicipta
Atom tidak boleh
,
destroyed
dimusnah
,
Positively charged sphere
Sfera bercas
(i)
positif
(ii)
Thomson
Electron charges negative
Elektron
(i)
Electron moves
outside the nucleus
Elektron
nukleus
Rutherford
Nukleus mengandungi
proton
, zarah subatom yang pertama.
positive
charge which embedded with
negatively charged particles called electrons .
nukleus
Proton
yang merupakan pusat bagi atom dan
.
is a part of the nucleus.
Proton
(iii)
yang mengandungi zarah
Discovered the nucleus as the centre of an atom and
positively charged
.
bercas positif
(ii)
mengandungi
Nucleus
that
contain proton
elektron
Atom is sphere of
Menjumpai
bergerak di luar
.
.
positif
Atom adalah sfera yang bercas
elektron
bercas negatif dipanggil
.
bercas negatif
.
.
electrons , the first subatomic particle.
Discovered the
Menjumpai
atau
sama
Atom daripada unsur sama adalah
dibahagi
identical
(iii) Atoms from the same element are
divided
or
adalah sebahagian daripada nukleus.
Electron
move outside the nucleus.
Elektron
bergerak di sekeliling nukleus.
(iv) Most of the mass of the atom found in the
Nukleus
nucleus
.
mempunyai hampir semua jisim atom.
Shell
Neils Bohr
Nucleus that
contain proton
(i)
Nukleus mengandungi
proton
(ii)
shells
Discovered the existence of electron
petala
Menjumpai kewujudan
Electrons move in the
Elektron
.
elektron.
shells
around the nucleus.
nukleus
bergerak di dalam petala mengelilingi
.
Electron
Shell
James
Chadwick
Nucleus that contain
proton and neutron
Nukleus mengandungi
proton dan neutron
Electron
(i)
Discovered the existence of
(ii)
proton
.
Nukleus mengandungi zarah-zarah neutral dipanggil
proton
zarah-zarah bercas positif dipanggil
.
neutron
dan
(iii) The mass of a neutron and proton is almost the same.
neutron
dan
proton
adalah hampir sama.
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.
.
Nucleus of an atom contains neutral particles called
neutron and positively charged particles called
Jisim
m
neutron
neutron
Menjumpai kewujudan
d.
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Chemistry Form 4 • MODULE
2
The structure of an atom: / Struktur Atom:
Shell / Petala
Nucleus that contain proton and neutron
Nukleus yang mengandungi proton dan neutron
Electron / Elektron
nucleus
(a) An atom has a central
Atom mempunyai
nucleus
(b) The
Nukleus
nukleus
shells
and electrons that move in the
around the nucleus.
petala
di tengahnya dan elektron bergerak di dalam
mengelilingi nukleus tersebut.
contains protons and neutrons.
mengandungi proton dan neutron.
+1 . Each electron has an electrical charge of
–1
. The neutron has no
(c) Each proton has charge of
charge
neutral
(it is
). An atom has the same number of protons and electrons, so the overall charge
zero
neutral
of atom is
. Atom is
. (If an atom loses or gains electrons it is called an ion – formation
of ion will be studied in Chapter 4)
(ianya adalah neutral ).
sifar
. Atom
Setiap atom mempunyai bilangan proton dan elektron yang sama, oleh itu cas keseluruhan bagi atom adalah
neutral
. (Suatu atom akan membentuk ion apabila ia kehilangan atau menerima elektron – pembentukan ion akan
adalah
Setiap proton bercas
+1
–1
. Setiap elektron bercas
. Neutron tidak mempunyai
cas
dipelajari dalam Tajuk 4.)
(d) The relative mass of a neutron and a proton which are in the nucleus is 1. The mass of an atom is obtained mainly
proton
and neutron .
from the number of
proton
Jisim relatif proton dan neutron di dalam nukleus ialah 1. Jisim suatu atom diperoleh daripada jumlah bilangan
neutron
dan bilangan
.
1
(e) The mass of an electron can be ignored as the mass of an electron is about
times the size of a proton or
1 840
neutron.
Jisim elektron boleh diabaikan kerana ia terlalu kecil iaitu
3
1 daripada jisim proton dan neutron.
1 840
Complete the following table:
Lengkapkan jadual di bawah:
Subatomic particles
Symbol
Charge
Relative atomic mass
Jisim atom relatif
Kedudukan
Electron/Elektron
e
– (negative)
1
=0
1 840
In the shells
Proton/Proton
p
+ (positive)
1
In the nucleus
Neutron/Neutron
n
neutral
1
In the nucleus
Zarah subatom
4
Simbol
Cas
Position
Atom is the smallest neutral particle of an element.
Atom adalah zarah neutral paling kecil dalam suatu unsur.
Complete the following diagram: / Lengkapkan yang berikut:
Na
Na
Na
Na
Sodium element
natrium
Unsur
natrium
Na
Na
Na
Na
Na
Na
Sodium element
Unsur
natrium
Na
Na
Na
Sodium
Atom
atom
natrium
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Unsur
Sodium element
Na
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MODULE • Chemistry Form 4
Proton number of an element (Refer to Periodic table of an element)
5
Nombor proton sesuatu unsur (Rujuk Jadual Berkala Unsur)
(a) Proton number of an
element
atom
is the number of proton in its
.
atom
Nombor proton sesuatu unsur adalah bilangan proton yang terdapat dalam
.
neutral .
(b) The number of proton of an atom is also equal to the number of electrons in the atom because atom is
Bilangan proton sesuatu atom adalah sama dengan bilangan elektron dalam atom kerana atom adalah
neutral
.
(c) Every element has its own proton number:
Setiap unsur mempunyai nombor protonnya tersendiri:
atom
– Proton number of potassium, K is 19. Potasium
in the shells.
Atom
Nombor proton untuk kalium, K ialah 19.
19 elektron
di dalam petala.
– Proton number of oxygen, O is 8. Oxygen
in the shells.
Nombor proton untuk oksigen, O ialah 8.
8 elektron
di dalam petala.
has 19 protons in the nucleus and 19 electrons
19 proton
kalium mempunyai
atom
Atom
8 protons
has
di dalam nukleus dan
in the nucleus and
8 proton
oksigen mempunyai
8 electrons
di dalam nukleus dan
Nucleon number of an element (Refer to Periodic table of an element)
6
Nombor nukleon sesuatu unsur (Rujuk Jadual Berkala Unsur)
(a) Nucleon number of an element is the total number of protons and neutrons in the nucleus of its
Nombor nukleon sesuatu unsur adalah jumlah bilangan proton dan neutron di dalam nukleus sesuatu
atom
atom
.
.
(b) Nucleon number is also known as a mass number.
Nombor nukleon juga dikenali sebagai nombor jisim.
(c) Nucleon number = number of proton + number of neutron.
Nombor nukleon = bilangan proton + bilangan neutron.
Symbol of Element And Standard Representation For An Atom of Element
Simbol Unsur dan Perwakilan Piawai bagi Atom Sesuatu Unsur
The symbol of an element is a short way of representing an element. If the symbol has only one letter, it must be a capital
letter. If it has two letters, the first is always a capital letter, while the second is always a small letter.
1
Simbol unsur adalah cara mudah untuk mewakilkan unsur. Jika simbol hanya terdiri daripada satu huruf, maka ia mesti ditulis dengan huruf
besar. Tetapi jika simbol terdiri daripada dua huruf, maka huruf pertama merupakan huruf besar dan huruf kedua merupakan huruf kecil.
Example: / Contoh:
Element
Unsur
Symbol
Element
Symbol
Element
Symbol
O
Nitrogen/Nitrogen
N
Calcium/Kalsium
Ca
Mg
Sodium/Natrium
Na
Copper/Kuprum
Cu
Potassium/Kalium
K
Chlorine/Klorin
Cl
Simbol
Oxygen/Oksigen
Magnesium/Magnesium
Hydrogen/Hidrogen
H
Unsur
Simbol
Unsur
Simbol
The first letter of each element is capitalised to show that it is a new element. This is helpful when writing a chemical formula.
For example KCl. There are two elements chemically bonded in KCl because there are two capital letters represent potassium and
chlorine.
Huruf yang pertama bagi setiap unsur ditulis dengan huruf besar untuk menunjukkan ia adalah unsur yang baru. Ini sangat berguna semasa menulis formula
kimia. Contohnya KCl. Terdapat dua unsur yang terikat secara kimia dalam KCl kerana adanya dua huruf besar yang mewakili kalium dan klorin.
Standard representation symbol represents
2
m
one atom
of an element. It can be written as:
sesuatu unsur. Ianya boleh ditulis sebagai:
Nucleon number/Nombor nukleon
A
Proton number/Nombor proton
Z
X
Symbol of an element/Simbol unsur
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Chemistry Form 4 • MODULE
Example: / Contoh:
27
A1
13
– The element is Aluminium.
Unsur itu adalah Aluminium.
27
– The nucleon number of Aluminium is
27
Nombor nukleon Aluminium adalah
.
13
– The proton number of Aluminium is
13
Nombor proton Aluminium adalah
– Aluminium has
13 protons
.
.
14 neutrons
13 proton
Atom Aluminium mempunyai
3
,
.
13
and
14 neutron
,
electrons.
13
dan
elektron.
Isotope / Isotop
(a) Isotopes are atoms of the same element with same number of protons but different number of neutrons.
Isotop ialah atom-atom unsur yang mempunyai bilangan proton yang sama tetapi bilangan neutron yang berbeza.
Or / Atau
Isotopes are atoms of the same element with same
proton
Isotop ialah atom-atom unsur yang mempunyai nombor
berbeza.
proton
number but different
nucleon
number.
nukleon
yang sama tetapi nombor
yang
Example: / Contoh:
1
1 H
2
1 H
Nucleon number/Nombor nukleon = 1
Nucleon number/Nombor nukleon = 2
Proton number/Nombor proton = 1
Proton number/Nombor proton = 1
Number of neutron/Bilangan neutron = 0
Number of neutron/Bilangan neutron = 1
– Hydrogen-1 and Hydrogen-2 are isotopes. Hydrogen-1 and Hydrogen-2 atoms have the same proton number or the same
number of protons
but
different
in nucleon number because of the difference in the number of
Atom Hidrogen-1 dan Hidrogen-2 mempunyai nombor proton atau
bilangan neutron
kerana perbezaan
.
– Isotopes have the same
arrangements.
Isotop mempunyai sifat
chemical
kimia
bilangan proton
properties but different
yang sama tetapi nombor nukleon yang
physical
neutron
.
berbeza
properties because they have the same electron
yang sama kerana mempunyai susunan elektron yang sama tetapi sifat
fizik
yang berbeza.
(b) Examples of the usage of isotopes:
Contoh kegunaan isotop:
i.
Medical field
Bidang perubatan
–
To detect brain cancer.
–
To detect thrombosis (blockage in blood vessel).
–
Sodium-24 is used to measure the rate of iodine absorption by thyroid gland.
–
Cobalt-60 is used to destroy cancer cells.
–
To kill microorganism in the sterilising process.
Untuk mengesan barah otak.
Untuk mengesan trombosis (saluran darah tersumbat).
Untuk mengukur kadar penyerapan iodin oleh kelenjar tiroid. Contoh: Natrium-24
Untuk memusnahkan sel barah. Contoh: Kobalt-60
Untuk membunuh mikroorganisma semasa proses pensterilan.
ii.
In the industrial field
Bidang industri
–
To detect wearing out in machines.
–
To detect any blockage in water, gas or oil pipes.
Untuk mengesan kehausan enjin.
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Untuk mengesan saluran paip air, gas atau minyak yang tersumbat.
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MODULE • Chemistry Form 4
–
To detect leakage of pipes underground.
–
To detect defects/cracks in the body of an aeroplane.
Untuk mengesan kebocoran paip bawah tanah.
Untuk mengesan keretakan atau kecacatan pada badan kapal terbang.
iii.
In the agriculture field
Bidang pertanian
–
To detect the rate of absorption of phosphate fertilizer in plants.
–
To sterile insect pests for plants.
Untuk mengesan kadar penyerapan baja fosfat oleh tumbuhan.
Untuk memandulkan serangga perosak tumbuhan.
iv.
In the archeology field
Bidang arkeologi
–
Carbon-14 can be used to estimate the age of artifacts.
Karbon-14 untuk menentukan usia sesuatu artifak.
Electron Arrangement
4
Susunan elektron
(a) The electrons are filled in specific shells. Every shell can be filled only with a certain number of electrons. For the
elements with atomic numbers 1-20:
Elektron diisi dalam petala tertentu. Setiap petala hanya boleh diisi dengan bilangan elektron tertentu. Bagi unsur-unsur yang
mempunyai nombor proton 1–20:
2
– First shell can be filled with a maximum of
electrons.
2
Petala pertama boleh diisi dengan bilangan maksimum
– Second shell can be filled with a maximum of
Petala kedua boleh diisi dengan bilangan maksimum
Petala ketiga boleh diisi dengan bilangan maksimum
electrons.
8
8
– Third shell can be filled with a maximum of
elektron.
8
elektron.
electrons.
8
elektron.
First shell is filled with 2 electrons (duplet)
Petala pertama diisi 2 elektron (duplet)
Second shell is filled with 8 electrons (octet)
Petala kedua diisi 8 elektron (oktet)
Third shell is filled with 8 electrons (octet)
Petala ketiga disi 8 elektron (oktet)
(b) Valence electrons are the electrons in the outermost shell of an atom.
Elektron valens: Elektron yang diisi dalam petala paling luar suatu atom.
Complete the following table:
5
Lengkapkan jadual berikut:
(a) Draw the electron arrangement and complete the description for each element:
Lukis susunan elektron bagi atom dan penerangan bagi setiap unsur berikut:
Standard
representation
of an element
Perwakilan piawai
unsur
Electron arrangement
of an atom
Lukiskan susunan elektron
bagi atom
Hydrogen Atom
Atom Hidrogen
m
Number of protons/Bilangan proton
1
Number of eletrons/Bilangan elektron
1
Number of neutrons/Bilangan neutron
0
Proton number/Nombor proton
1
Nucleon number/Nombor nukleon
1
Electron Arrangement/Susunan elektron
1
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Description
Penerangan
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Chemistry Form 4 • MODULE
Sodium Atom
Atom Natrium
23
Na
11
Na
Number of protons/Bilangan proton
11
Number of electrons/Bilangan elektron
11
Number of neutrons/Bilangan neutron
12
Proton number/Nombor proton
11
Nucleon number/Nombor nukleon
23
Electron Arrangement/Susunan elektron
2.8.1
(b) Choose the correct statement for the symbol of element X.
Pilih pernyataan yang betul bagi simbol unsur X.
23
Na
11
Statement
Pernyataan
Element X has 11 proton number.
Unsur X mempunyai 11 nombor proton.
The proton number of element X is 11.
Nombor proton unsur X ialah 11.
The proton number of atom X is 11.
Nombor proton atom X ialah 11.
The number of proton of element X is 11.
Bilangan proton unsur X ialah 11.
The number of proton of atom X is 11.
Bilangan proton atom X ialah 11.
Nucleon number of element X is 23.
Nombor nukleon unsur X ialah 23.
Nucleon number of atom X is 23.
Nombor nukleon atom X ialah 23.
Number of nucleon of element X is 23.
Bilangan nukleon unsur X ialah 23.
Atom X has 23 nucleon number.
Atom X mempunyai 23 nombor nukleon.
Neutron number of atom X is 12.
Nombor neutron atom X ialah 12.
Number of neutron of atom X is 12.
Bilangan neutron atom X ialah 12.
Number of neutron of element X is 12.
Tanda ( 3 / 7 )
7
3
3
7
3
3
3
7
7
7
3
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Bilangan neutron unsur X ialah 12.
Tick ( 3 / 7 )
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MODULE • Chemistry Form 4
EXERCISE / LATIHAN
Complete the following table:
1
Lengkapkan jadual berikut:
Element
Unsur
Hydrogen
Hidrogen
Helium
Helium
Boron
Boron
Carbon
Karbon
Nitrogen
Nitrogen
Neon
Neon
Sodium
Natrium
Magnesium
Magnesium
Calcium
Kalsium
Symbol of
element
Simbol unsur
Number of
proton
Bilangan
proton
Number of
electron
Bilangan
elektron
Number of
neutron
Bilangan
neutron
Proton
number
Nombor
proton
Nucleon
number
Nombor
nukleon
Electron
arrangement
Susunan
elektron atom
Number
of valence
electron
Bilangan
elektron valens
1
1 H
1
1
0
1
1
1
1
4
2 He
2
2
2
2
4
2
2
11
5 B
5
5
6
5
11
2.3
3
12
6 C
6
6
6
6
12
2.4
4
14
7 N
7
7
7
7
14
2.5
5
20
Ne
10
10
10
10
10
20
2.8
8
23
Na
11
11
11
12
11
23
2.8.1
1
24
Mg
12
12
12
12
12
24
2.8.2
2
40
Ca
20
20
20
20
20
40
2.8.8.2
2
The diagram below shows the symbol of atoms P, R and S.
2
Rajah di bawah menunjukkan simbol atom P, R dan S.
35
P
17
12
R
6
37
S
17
(a) What is meant by nucleon number / Apakah maksud nombor nukleon?
Nucleon number of an element is the total number of protons and neutrons in the nucleus of its atom
(b) What is the nucleon number of P / Apakah nombor nukleon atom P?
35
(c) State the number of neutron in atom P / Nyatakan bilangan neutron atom P.
18
(d) State number of proton in atom P / Nyatakan bilangan proton atom P.
17
(e)
(i)
What is meant by isotope / Apakah maksud isotop?
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Isotopes are atoms of the same element with same number of proton but different number of neutrons
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Chemistry Form 4 • MODULE
(ii)
State a pair of isotope in the diagram shown / Nyatakan sepasang isotop dalam rajah yang ditunjukkan.
P and S
(iii) Give reason for your answer in (e)(ii) / Berikan sebab bagi jawapan di (e)(ii).
Atom P and S have same proton number but different nucleon number//number of neutron
(f)
An isotope of R has 8 neutron. Write the symbol for the isotope R.
Isotop bagi atom R mempunyai 8 neutron. Tuliskan simbol bagi isotop R.
14
R
6
3
The table below shows the number of proton and neutron of atoms of elements P, Q and R.
Jadual di bawah menunjukkan bilangan proton dan neutron bagi atom unsur P, Q dan R.
Element
Unsur
Number of proton
Bilangan proton
Number of neutron
Bilangan neutron
P
1
0
Q
1
1
R
6
6
(a) Which of the atoms in the above table are isotope? Explain your answer.
Berdasarkan jadual di atas, atom yang manakah merupakan isotop? Terangkan jawapan anda.
P and Q. Atom P and Q have same number of proton but different number of neutron // nucleon number.
(b) (i)
Write the standard representation of element Q.
Tuliskan perwakilan piawai untuk unsur Q.
2
Q
1
(ii)
State three information that can be deduced from your answer in (b)(i).
Nyatakan tiga maklumat yang boleh didapati daripada jawapan anda di (b)(i).
The proton number of element Q is 1 // Number of proton of atom Q is 1
Nucleon number of element Q is 2 // Atomic mass of atom Q is 2
Number of neutron of atom Q is 1
Nucleus of atom Q contains 1p and 1n
(c)
(i)
Draw atomic structure for atom of element R.
Lukiskan struktur atom bagi atom unsur R.
6 protons + 6 neutrons
(ii)
Describe the atomic structure in (c)(i).
Huraikan struktur atom di (c)(i).
– The atom consists of 2 parts: the centre part called nucleus and the outer part called electron shell.
– The nucleus consists of 6 protons which are positively charged and 6 neutrons which are neutral.
– The electrons are in two shells, the first shell consists of two electrons and the second shell consists of
four electrons.
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– Electrons move around nucleus in the shells.
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MODULE • Chemistry Form 4
(d) Element R react with oxygen and to produce liquid Z at room temperature. The graph below shows the sketch of the
graph when liquid Z at room temperature, 27°C is cooled to –5°C.
Unsur R bertindak balas dengan oksigen dan menghasilkan cecair Z pada suhu bilik. Rajah di bawah menunjukkan lakaran graf
apabila cecair Z pada suhu bilik, 27°C disejukkan sehingga –5°C.
Temperature /°C
Suhu /°C
Time /s
0
t1
Masa /s
t2
−5
(i)
What is the state of matter of liquid Z from t1 to t2? Explain why is the temperature remain unchanged from
t1 to t2.
Apakah keadaan jirim Z daripada t1 hingga t2? Terangkan mengapa suhu tidak berubah daripada t1 hingga t2.
Liquid and solid. Heat lost to the surrounding is balanced by the heat released when the particles at 0 °C
(ii)
Draw the arrangement of particles of Z at 20°C.
Lukiskan susunan zarah-zarah Z pada suhu 20°C.
(iii) Describe the change in the particles movement when Z is cooled from room temperature to –5°C.
Nyatakan perubahan dalam pergerakan zarah-zarah apabila cecair Z disejukkan daripada suhu bilik ke –5°C.
The particles move slower
Objective Questions / Soalan Objektif
The diagram shows the arrangement of particles for a type
of matter that undergoes a change in physical state through
process X.
1
3
The diagram below shows the heating curve for substance X.
Rajah di bawah menunjukkan lengkung pemanasan bahan X.
Temperature / Suhu °C
Rajah di bawah menunjukkan susunan zarah sejenis bahan yang
mengalami perubahan keadaan fizik melalui proses X.
U
S
Q
X
T
R
P
Time (m)
Masa (m)
Which region of the graph does boiling process occur?
What is process X?
Bahagian manakah pada graf berlaku proses pendidihan?
Apakah proses X ?
A
B
Melting
Peleburan
Boiling
Pendidihan
C
D
Sublimation
Pemejalwapan
Antara bahan berikut, yang manakah mengalami pemejalwapan apabila
dipanaskan?
A
Sulphur
Sulfur
Ammonium chloride
Ammonium klorida
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C
D
Sodium chloride
Natrium klorida
C
D
ST
TU
Which of the following information is true?
Antara pernyataan berikut, yang manakah adalah betul?
Change of state
Perubahan keadaan
Process
Proses
Heat energy
Tenaga haba
A
Solid → Liquid
Melting
Released
Pepejal → Cecair
Peleburan
Dibebaskan
B
Liquid → Gas
Evaporation
Released
Cecair → Gas
Penyejatan
Dibebaskan
C
Gas → Solid
Sublimation
Released
Gas → Pepejal
Pemejalwapan
Dibebaskan
D
Gas → Liquid
Condensation
Absorbed
Gas → Cecair
Kondensasi
Diserap
Glucose
Glukosa
PQ
QR
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a
B
20
4
Which of the following substances can undergo sublimation
when heated?
2
m
A
B
Freezing
Pembekuan
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Chemistry Form 4 • MODULE
5
The diagram below shows the graph of temperature against
time when a liquid Y is cooled.
Rajah di bawah menunjukkan graf suhu melawan masa apabila cecair Y
disejukkan.
Substance
Bahan
Melting point/°C
Takat lebur/°C
Boiling point/°C
Takat didih/°C
S
–182
–162
T
–23
77
U
–97
65
V
41
182
W
132
290
Temperature / Suhu °C
t3
P
Q
t2
R
Which substance exists as liquid at room temperature?
t1
Bahan yang manakah wujud sebagai cecair pada suhu bilik?
S
A
Time (m)
Masa (m)
B
Which of the following statements are true about the curve?
Antara pernyataan berikut, yang manakah adalah betul tentang lengkung
itu?
I
At Q, liquid Y begins to freeze.
II
At PQ, particles in Y absorb heat from the surroundings.
III
Liquid Y freezes completely at S.
IV
The freezing point of Y is t2°C.
8
Pada Q, cecair Y mula membeku.
C
S only
S sahaja
D
S and T only
S dan T sahaja
Rajah di bawah menunjukkan perwakilan piawai atom kuprum.
64
Cu
29
Cecair Y membeku dengan lengkap pada S.
A
B
6
C
I and III only
I dan III sahaja
D
I and IV only
I dan IV sahaja
Which of the following is correct based on the symbol the
diagram?
II and III only
Antara berikut, yang manakah betul berdasarkan rajah di atas?
II dan III sahaja
Rajah di bawah menunjukkan graf suhu melawan masa apabila pepejal Z
dipanaskan.
Temperature / Suhu °C
9
0
1
2
3
4
5
6
7
8
9
29
64
29
B
35
29
64
C
64
35
29
D
29
64
35
The diagram below shows the standard representation of
beryllium atom.
Apakah bilangan elektron valens bagi atom berillium?
Masa (m)
Antara berikut, yang manakah adalah benar pada minit keempat?
7
A
What is the number of valence electrons of beryllium atom?
A
B
Which of the following is true during the fourth minute?
D
Number of electron
Bilangan elektron
9
Be
4
Time (m)
C
Nucleon number
Nombor nukleon
Rajah di bawah menunjukkan perwakilan piawai atom berillium.
80
B
Proton number
Nombor proton
II and IV only
II dan IV sahaja
The diagram below shows the graph of temperature against
time when solid Z is heated.
A
V and W only
V dan W sahaja
The diagram below shows standard representation of an atom
copper.
Pada PQ, zarah dalam Y menyerap haba dari persekitaran.
Takat beku bagi Y adalah t2°C.
T and U only
T dan U sahaja
All the molecules are in random motion.
Semua molekul bergerak secara rawak.
All the molecules are closely packed and in random
motion.
Semua molekul sangat rapat dan bergerak secara rawak.
All the molecules are vibrating at fixed positions.
Semua molekul bergetar pada kedudukan tetap.
Some of the molecules are vibrating at fixed positions but
some are in random motion.
Sebahagian molekul bergetar pada kedudukan tetap dan
sebahagian bergerak secara rawak.
The table shows the melting points and boiling points of
substances S, T, U, V and W.
C
D
4
7
The table below shows the proton number and the number of
neutrons for atoms of elements W, X, Y and Z.
Jadual di bawah menunjukkan nombor proton dan bilangan neutron bagi
atom unsur W, X, Y dan Z.
Element
Atom
Proton number
Nombor proton
Number of neutrons
Bilangan neutron
W
7
7
X
8
8
Y
8
9
Z
9
10
Which of the following pair of elements is isotope?
Antara pasangan berikut, yang manakah adalah isotop?
A
B
W and X
W dan X
W and Y
W dan Y
C
D
X and Y
X dan Y
Y and Z
Y dan Z
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Jadual di bawah menunjukkan takat lebur dan takat didih bahan S, T, U,
V dan W.
10
2
3
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MODULE • Chemistry Form 4
2
CHEMICAL FORMULA AND EQUATIONS
FORMULA DAN PERSAMAAN KIMIA
RELATIF MASS / JISIM RELATIF
• RELATIVE ATOMIC MASS / JISIM ATOM RELATIF (JAR)
– To state the meaning of relative mass and solve numerical problems
Menyatakan maksud jisim atom relatif dan menyelesaikan masalah pengiraan
• RELATIVE FORMULA MASS / JISIM FORMULA RELATIF (JFR)
– To state the meaning of RAM, RMM and RFM based on carbon-12 scale
Menyatakan maksud JAR, JMR dan JFR berdasarkan skala karbon-12
• RELATIVE MOLECULAR MASS / JISIM MOLEKUL RELATIF (JMR)
– To calculate RAM, RMM and RFM using the chemical formulae of various substances
Menghitung JAR, JMR dan JFR menggunakan formula kimia beberapa bahan
MOLE CONCEPT / KONSEP MOL
• MOLE AND THE NUMBER OF PARTICLES / MOL DAN BILANGAN ZARAH
– To solve numerical problems involving mole and the number of atoms/ ions/ molecules
Menyelesaikan masalah pengiraan melibatkan mol dan bilangan atom, ion dan molekul
• MOLE AND THE MASS OF SUBSTANCES / MOL DAN JISIM BAHAN
– To solve numerical problems involving mole and the mass of substances, number of particles and volume of gas using mole
concept
Menyelesaikan masalah pengiraan melibatkan mol, jisim bahan, bilangan zarah dan isipadu gas menggunakan konsep mol
• MOLE AND THE VOLUME OF GAS / MOL DAN ISIPADU GAS
– To solve numerical problems involving mole and the mass of substances, number of particles and volume of gas using mole
concept
Menyelesaikan masalah pengiraan melibatkan mol, jisim bahan, bilangan zarah dan isipadu gas menggunakan konsep mol
CHEMICAL FORMULA AND EQUATIONS / FORMULA DAN PERSAMAAN KIMIA
• EMPIRICAL FORMULA / FORMULA EMPIRIK
– Stating the purpose and describe the empirical formula laboratory activities to determine the formula empirical
Menyatakan maksud formula empirik dan menghuraikan aktiviti makmal untuk menentukan formula empirik
• MOLECULAR FORMULA / FORMULA MOLEKUL
– Solve calculation problems involving empirical formula
Menyelesaikan masalah pengiraan melibatkan formula empirik
• CHEMICAL FORMULAE / FORMULA KIMIA
– To write formula of anion and cation and to write chemical formula for ionic compounds
Menulis formula kation dan anion dan menulis formula kimia untuk sebatian ion
• CHEMICAL EQUATIONS / PERSAMAAN KIMIA
– Write a balanced chemical equation and solve problems arrangements involving the mole concept
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Menulis persamaan kimia seimbang dan menyelesaikan masalah pengiraan yang melibatkan konsep mol
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Chemistry Form 4 • MODULE
RELATIVE ATOMIC MASS (RAM) / JISIM ATOM RELATIF (JAR)
1
A single atom is too small and light and cannot be weighed directly.
Satu atom adalah terlalu ringan, kecil dan tidak dapat ditimbang secara langsung.
2
The best way to determine the mass of a single atom is to compare its mass to the mass of another atom of an element
that is used as a standard.
Cara yang paling sesuai untuk menentukan jisim satu atom ialah dengan membandingkan jisimnya dengan jisim suatu atom unsur lain
yang dianggap sebagai piawai.
3
Hydrogen was the first element to be chosen as the standard for comparing mass because the hydrogen atom is the
lightest atom with a mass of 1.0 a.m.u (atomic mass unit).
Hidrogen adalah unsur pertama dipilih sebagai piawai untuk membandingkan jisim kerana atom hidrogen adalah unsur yang paling
ringan dengan jisim 1.0 u.j.a (unit jisim atom).
Example:
Contoh:
• The mass of one helium atom is four times larger than one hydrogen atom.
Jisim satu atom Helium adalah 4 kali lebih besar daripada satu atom hidrogen.
• RAM for He is 4.
JAR untuk He ialah 4.
4
On the hydrogen scale, the relative atomic mass of an element means the mass of one atom of the element compared to
the mass of a single hydrogen atom:
Pada skala hidrogen, jisim atom relatif suatu unsur ditakrifkan sebagai jisim satu atom unsur berbanding jisim satu atom hidrogen:
Relative atomic mass of an element (RAM) / Jism atom relatif suatu unsur (JAR)
The average mass of one atom of the element / Jisim purata satu atom unsur
Mass of one hydrogen atom / Jisim satu atom hidrogen
=
• RAM has no unit.
JAR tiada unit.
• The new standard used today is the carbon-12 atom.
Piawai yang digunakan sekarang adalah berdasarkan atom karbon-12.
1
• RAM based on the carbon-12 scale is the mass of one atom of the element compared with
of the mass of an
12
atom of carbon-12:
JAR berdasarkan skala atom karbon-12 adalah jisim satu atom unsur berbanding dengan
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• Relative atomic mass of an element (RAM) / Jisim atom relatif suatu unsur (JAR)
The average mass on one atom of the element / Jisim purata satu atom unsur
=
1
× The mass of an atom of carbon-12 / Jisim satu atom karbon-12
12
1
jisim satu atom karbon-12:
12
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MODULE • Chemistry Form 4
RELATIVE MOLECULAR MASS (RMM) / RELATIVE FORMULA MASS (RFM)
JISIM MOLEKUL RELATIF (JMR) / JISIM FORMULA RELATIF (JFR)
RMM / JMR =
1
The average mass on one atom of the element / Jisim purata satu molekul
1
× The mass of an atom of carbon-12 / Jisim satu atom karbon-12
12
RMM is obtained by adding up the RAM of all the atoms that are present in the molecule.
2
JMR diperoleh dengan menambahkan JAR semua atom yang terdapat dalam satu molekul.
Molecular substance
Molecular formula
Relative molecular mass
O2
2 × 16 = 32
Water / Air
H2O
2 × 1 + 16 = 18
Carbon dioxide / Karbon dioksida
CO2
12 + 2 × 16 = 44
Ammonia / Ammonia
NH3
14 + 3 × 1 = 17
Bahan molekul
Formula molekul
Oxygen / Oksigen
Jisim molekul relatif
[Relative atomic mass / Jisim atom relatif : O = 16, H = 1, C = 12, N = 14]
For ionic substances, RMM is replaced with Relative Formula Mass (RFM).
3
Untuk sebatian ion, JMR digantikan dengan Jisim Formula Relatif (JFR).
Substance
Chemical formula
Relative molecular mass
Sodium chloride / Natrium klorida
NaCl
23 + 35.5 = 58.5
Potassium oxide / Kalium oksida
K2O
2 × 39 + 16 = 94
CuSO4
64 + 32 + 4 × 16 = 160
(NH4)2CO3
2 [14 + 4 × 1] + 12 + 3 × 16 = 96
Aluminium nitrate / Aluminium nitrat
Al(NO3)3
27 + 3 [14 + 3 × 16] = 213
Calcium hydroxide / Kalsium hidroksida
Ca(OH)2
40 + 2 [16 + 1] = 74
Lead(II) hydroxide / Plumbum(II) hidroksida
Pb(OH)2
207 + 2 [16 + 1] = 241
CuSO45H2O
64 + 32 + 4 × 16 + 5 [2 × 1 + 16] = 250
Bahan
Copper(II) sulphate / Kuprum(II) sulfat
Ammonium carbonate / Ammonium karbonat
Hydrated copper(II) sulphate / Kuprum(II) sulfat terhidrat
Formula kimia
Jisim formula relatif
[Relative atomic mass / Jisim atom relatif : Na = 23, Cl = 35.5, K = 39, O = 16, Cu = 64, S = 32, N = 14, H = 1, C = 12, Al = 27,
Ca = 40, Pb = 207]
(i)
The formula of metal oxide of M is M2O3. Its relative formula mass is 152. What is the relative atomic mass of
metal M?
Oksida logam M mempunyai formula M2O3. Jisim formula relatif ialah 152. Apakah jisim atom relatif logam M?
M = RAM for M
2M + 3 × 16 = 152
M = 52
(ii) Phosphorus forms a chloride with a formula PClx. Its relative molecular mass is 208.5. Calculate the value of x.
Fosforus membentuk sebatian klorida dengan formula PClx. Jisim molekul relatifnya adalah 208.5. Hitungkan nilai x.
[Relative atomic mass / Jisim atom relatif : P = 31, Cl = 35.5]
m
= 208.5
= 208.5 – 31
= 177.5
=5
Publica
n Sdn.
24
tio
Nil
a
31 + x × 35.5
35.5x
35.5x
x
d.
Bh
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Chemistry Form 4 • MODULE
MOLE CONCEPT / KONSEP MOL
Mole and the Number of Particles / Bilangan Mol dan Bilangan Zarah
1
2
To describe the amount of atoms, ions or molecules, mole is used.
Untuk menyatakan jumlah atom, ion atau molekul, unit mol digunakan.
A mole is an amount of substance that contains as many particles as the number of atoms in exactly 12 g of carbon-12.
Satu mol ialah jumlah bahan yang mengandungi bilangan zarah seperti mana yang terdapat dalam 12 g atom karbon-12.
3
A mole of a substance is the amount of substance which contains a constant number of particles (atoms, ions,
molecules), which is 6.02 × 1023.
Satu mol bahan adalah jumlah bahan yang mengandungi bilangan zarah yang tetap (atom, molekul, ion) iaitu 6.02 × 1023.
4
5
The number 6.02 × 1023 is called the Avogadro Constant or Avogadro Number (NA).
Nombor 6.02 × 1023 dikenali sebagai Pemalar Avogadro atau Nombor Avogadro (NA ).
For compounds that exist as molecules/ions, the number of atoms/ions in that compound must be known.
Bagi sebatian yang wujud dalam bentuk molekul/ion, bilangan atom/ion dalam sebatian itu mestilah diketahui.
6
7
The symbol of mole is mol.
Simbol untuk mol ialah mol.
Complete the following table:
Lengkapkan jadual berikut:
Substance
Formula
Bahan
Formula
Type of
particles
Model / Figure
Number of atom per
molecule/ Number of
positive and negative ion
Model / Rajah
Jenis zarah
Bilangan atom per molekul/
Bilangan ion positif dan negatif
8
Chlorine / Klorin
Cl2
Molecule
Cl Cl
Water / Air
H2O
Molecule
H O H
Ammonia / Ammonia
NH3
Molecule
H
H N H
Sulphur dioxide / Sulfur dioksida
SO2
Molecule
O S O
Magnesium chloride / Magnesium klorida
MgCl2
Ion
[Cl]– [Mg]2+ [Cl]–
Aluminium oxide / Aluminium oksida
Al2O3
Ion
[O]2– [A1]3+ [O]2– [A1]3+ [O]2–
Cl : 2
H: 2
O:1
N:1
H: 3
S:1
O:2
Mg2+ : 1
Cl–
:2
Al3+ : 2
O2–
:3
Relationship between number of moles and number of particles (atoms/ions/molecules):
Hubungan bilangan mol dan bilangan zarah (atom/ion/molekul):
Number of moles
Bilangan mol
9
× Avogadro Constant / Pemalar Avogadro
÷ Avogadro Constant / Pemalar Avogadro
Number of particles
Bilangan zarah
Complete the following: [Differentiate between “mole” dan “molecule”]
Lengkapkan yang berikut: [Bezakan antara “mol” dan “molekul”]
(a) 1 mol of Cl2
[Chlorine gas]
1 mol Cl2
[Gas klorin]
(b) 1 mol of NH3
[Ammonia gas]
molecules of chlorine, Cl2 / molekul klorin, Cl2
2 × 6.02 × 1023 atoms of chlorine, Cl / atom klorin, Cl
6.02 × 1023
4
molecules of ammonia, NH3 / molekul ammonia, NH3
1 mol of nitrogen atom, N / mol atom nitrogen, N
mol atoms / mol atom
3
mol of hydrogen atoms, H / mol atom hidrogen, H
n
io
Sdn. B
m
25
.
hd
Publicat
1 mol NH3
[Gas ammonia]
6.02 × 1023
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MODULE • Chemistry Form 4
0.25 × 6.02 × 1023
1
mol of NH3
4
[Ammonia gas]
(c)
1
mol NH3
4
[Gas ammonia]
1
mol of atoms
1
mol atom
molecules of ammonia, NH3 / molekul ammonia, NH3
0.25 mol of N atoms / mol atom N,
23
number of N atoms / bilangan atom N = 0.25 × 6.02 × 10
0.75 mol of H atoms / mol atom H,
number of H atoms / bilangan atom H =
2 mol of Mg2+ ions / mol ion Mg2+,
number of Mg2+ ions / bilangan ion Mg2+ =
(d) 2 mol of MgCl2
[Magnesium chloride]
2 mol MgCl2
[Magnesium klorida]
4 mol of Cl– ions / mol ion Cl–,
number of Cl- ions / bilangan ion Cl– =
0.75 × 6.02 × 1023
2 × 6.02 × 1023
4 × 6.02 × 1023
2 × 6.02 × 1023
(e) 2 mol of SO2
[Sulphur dioxide]
molecules of SO2 / molekul SO2
2 mol of S atoms / mol atom S,
number of S atoms / bilangan atom S =
3 × 2 = 6 mol of atoms
2 mol SO2
[Sulfur dioksida]
3×2=6
mol atom
2 × 6.02 × 1023
4 mol of O atoms / mol atom O,
number of O atoms / bilangan atom O =
4 × 6.02 × 1023
10 Complete the table below:
Lengkapkan jadual berikut:
Number of moles
Number of particles
Bilangan mol
0.5
0.5
Bilangan zarah
mole of carbon, C
3.01 × 1023 atoms of carbon
mol atom karbon, C
3.01 × 1023 atom karbon
0.2 moles of hydrogen gas, H2
(i)
0.2 mol gas hidrogen, H2
1
(ii)
molecules of hydrogen / molekul hidrogen
2 × 0.2 × 6.02 × 1023 atoms of hydrogen / atom hidrogen
6.02 × 1023 molecules of carbon dioxide contains:
mole of carbon dioxide molecules, CO2
1
0.2 × 6.02 × 1023
6.02 × 1023 molekul karbon dioksida mengandungi:
mol molekul karbon dioksida, CO2
6.02 × 1023
6.02 × 10
23
atoms of C and
atom C dan
2 × 6.02 × 1023
2 × 6.02 × 1023
atoms of O.
atom O.
NUMBER OF MOLES AND MASS OF SUBSTANCE / BILANGAN MOL DAN JISIM BAHAN
Molar mass / Jisim molar
(a) Molar mass is the mass of one mole of any substance / Jisim molar adalah jisim satu mol sebarang bahan.
(b) Molar Mass is the relative atomic mass, relative molecular mass and relative formula mass of a substance in
g mol–1.
1
(c)
Jisim molar adalah jisim atom relatif, jisim molekul relatif dan jisim formula relatif suatu bahan dalam g mol–1.
Molar mass of any substance is numerically equal to its relative mass (Relative atomic mass/ relative formula
mass/relative molecular mass).
Jisim molar sebarang bahan mempunyai nilai yang sama dengan jisim relatif (Jisim atom relatif/ jisim formula relatif/ jisim molekul
relatif).
Example / Contoh:
Molar mass of H2O = 18 g mol–1
2
Jisim molar H2O = 18 g mol–1
× RAM/ /RFM/RMM
Mass of 1 mol of H2O = 18 g
Jisim 1 mol H2O = 18 g
Mass of 2 mol of H2O = 2 mol × 18 g mol = 36 g
Jisim 2 mol H2O = 2 mol ×
Mass of
Publica
g mol–1 =
mol of H2O = 45 g
36
g
Bilangan mol
÷ RAM/ /RFM/RMM
Mass in gram
Jisim dalam gram
÷ JAR/JFR/JMR
mol H2O = 45 g
n Sdn.
26
2.5
18
× JAR/JFR/JMR
tio
Nil
a
Jisim
m
2.5
–1
Number of
moles
d.
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Chemistry Form 4 • MODULE
3
Complete the following table:
Lengkapkan jadual berikut:
Element/
Compound
Unsur/Sebatian
Copper
Chemical
formula
RAM/RMM/RFM
Formula kimia
Cu
RAM/JAR = 64
NaOH
RFM/JFR = 40
Penghitungan
–1
(a) Mass of 1 mol of Cu / Jisim 1 mol Cu : 1 mol × 64 g mol = 64 g
2 mol × 64 g mol–1 = 128 g
(b) Jisim 2 mol / Jisim 1 mol :
1
mol × 64 g mol–1 = 32 g
1
1
2
(c) Jisim
mol / Jisim
mol:
2
2
32 g
(d) Mass of 3.01 × 1023 Cu atoms / Jisim 3.01 × 1023 atom Cu:
Kuprum
Sodium hydroxide
Calculate
JAR/JMR/JFR
(a) Mass of 3 mol of sodium hydroxide:
Natrium hidroksida
Jisim 3 mol natrium hidroksida:
120 g
120 g
(b) Number of moles in 20 g sodium hydroxide:
Bilangan mol natrium hidroksida dalam 20 g:
Oxygen gas
Gas oksigen
O2
RMM/JMR =
32
(a) Mass of 2.5 mol of oxygen gas:
Jisim 2.5 mol gas oksigen:
0.5 mol
0.5 mol
2.5 mol × 32 g mol–1 = 80 g
2.5 mol × 32 g mol–1 = 80 g
(b) Number of moles is 1.5 mol oxygen gas:
Bilangan molekul dalam 1.5 mol gas oksigen:
1.5 mol × 6.02 × 1023
1
(c) Number of molecules in
mol of oxygen gas:
2
1
Bilangan molekul dalam
mol gas oksigen:
2
0.5 mol × 6.02 × 1023
(d) Number of atoms in 2 mol of oxygen gas:
Bilangan atom dalam 2 mol gas oksigen:
2 × 2 × 6.02 × 1023
Sodium chloride
NaCl
Natrium klorida
Zinc nitrate
Zink nitrat
Zn(NO3)2
RFM/JFR = 58.5
RFM/JFR =
189
Mass of 0.5 mol of NaCl / Jisim bagi 0.5 mol NaCl:
0.5 mol × 58.5 g mol–1 = 29.25 g
Number of moles in 37.8 g of zinc nitrate:
Bilangan mol dalam 37.8 g zink nitrat:
37.8 g/189 g mol–1 = 0.2 mol
[Relative atomic mass / Jisim atom relatif: Cu = 64, Na = 23, O = 16, H = 1, Cl = 35.5, Zn = 65, N = 14]
NUMBER OF MOLES AND VOLUME OF GAS / BILANGAN MOL DAN ISI PADU GAS
1
Molar volume of a gas: Volume occupied by one mole of any gas is 24 dm3 at room conditions and 22.4 dm3 at
standard temperature and pressure (STP).
Isi padu molar gas: Isipadu yang dipenuhi oleh satu mol sebarang gas iaitu 24 dm3 pada keadaan bilik dan 22.4 dm3 pada suhu dan
tekanan piawai (STP).
2
The molar volume of any gas is 24 dm3 at room conditions and 22.4 dm3 at STP.
Isi padu molar sebarang gas adalah 24 dm3 pada keadaan bilik dan 22.4 dm3 pada STP.
3
Generalisation: One mole of any gas always occupies the same volume under the same temperature and pressure:
Umumnya: satu mol sebarang jenis gas menempati isi padu yang sama pada suhu dan tekanan yang sama.
Example / Contoh:
(i) 1 mol of oxygen gas, 1 mol ammonia gas, 1 mol helium gas dan 1 mol sulphur dioxide gas occupy the same
volume of 24 dm3 at room conditions.
1 mol gas oksigen, 1 mol gas ammonia, 1 mol gas helium dan 1 mol gas sulfur dioksida menempati isi padu yang sama iaitu 24 dm3
pada keadaan bilik.
44.8
(ii) 2 mol of carbon dioxide gas occupies
44.8
dm3 pada STP.
n
io
Sdn. B
m
27
.
hd
Publicat
2 mol gas karbon dioksida menempati
dm3 pada STP.
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02-Chem F4 (3P).indd 27
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MODULE • Chemistry Form 4
(iii) 16 g of oxygen gas = 0.5 mol of oxygen gas. Therefore, 16 g of oxygen gas occupies a volume of
at room conditions [Relative atomic mass: O =16]
0.5
mol gas oksigen. Oleh itu, 16 g gas oksigen menempati isi padu
16 g gas oksigen =
[Jisim atom relatif; O = 16]
Number of moles of gas
Bilangan mol gas
× 24 dm3 mol–1/ 22.4 dm3 mol–1
12
12
dm3
dm3 pada keadaan bilik.
Volume of gas in dm2
Isi padu gas dalam dm3
÷ 24 dm3 mol–1/ 22.4 dm3 mol–1
Formula for conversion of unit:
Formula untuk penukaran unit:
Volume of gas in dm3
Isi padu gas dalam dm3
÷ 24 dm3 mol–1/ 22.4 dm3 mol–1
÷ (RAM/ /RFM/RMM) g mol–1
Mass in gram (g)
Jisim dalam gram (g)
× 24 dm3 mol–1/ 22.4 dm3 mol–1
÷ (JAR/JFR/JMR) g mol–1
Number of
moles
÷ (6.02 × 1023)
× (RAM/ /RFM/RMM) g mol–1
Bilangan
mol
× (6.02 × 1023)
× (JAR/JFR/JMR) g mol–1
Number of particles
Bilangan zarah
EXERCISE / LATIHAN
Relative atomic mass of calcium is 40 based on the carbon-12 scale.
1
Jisim atom relatif kalsium berdasarkan skala karbon-12 ialah 40.
(a) State the meaning of the statement above.
Nyatakan maksud penyataan di atas.
Mass of calcium atom is 4 times greater than
1
mass of carbon-12 atom.
12
(b) How many times is one calcium atom heavier than one oxygen atom? [Relative atomic mass: O = 16]
Berapa kalikah satu atom kalsium lebih berat daripada satu atom oksigen? [JAR: O = 16]
Relative atomic mass of calcium
40
=
= 2.5 times
Relative atomic mass of oxygen
16
(c) How many calcium atoms have the same mass as two atoms of bromine? [RAM Br = 80]
Berapakah bilangan atom kalsium yang mempunyai jisim yang sama dengan dua atom bromin? [Jisim atom relatif: Br = 80]
Number of calcium atom × 40 = 2 × 80
2 × 80
Number of calcium atom =
=4
40
A sampel of chlorine gas weighs 14.2 g. Calculate / Suatu sampel gas klorin berjisim 14.2 g. Hitungkan:
[Relative atomic mass / Jisim atom relatif : Cl = 35.5]
(a) Number of moles of chlorine atoms / Bilangan mol atom klorin.
14.2
Number of mol of chlorine atoms, Cl =
= 0.4 mol
35.5
2
(b) Number of moles of chlorine molecules (Cl2) / Bilangan mol molekul klorin (Cl2 ).
14.2
Number of mol of chlorine molecule, Cl2 =
= 0.2 mol
71
(c) Volume of chlorine gas at room conditions / Isi padu gas klorin pada keadaan bilik.
[Molar volume of gas = 24 dm3 mol–1 at room temperature and pressure]
[Isi padu molar gas = 24 dm3 mol–1 pada suhu dan tekanan piawai]
m
Publica
n Sdn.
28
tio
Nil
a
Volume of chlorine gas = 0.2 mol × 24 dm3 mol–1
= 4.8 dm3
d.
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02-Chem F4 (3P).indd 28
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Chemistry Form 4 • MODULE
3
(a) Calculate the number of atoms in the following substances / Hitungkan bilangan atom yang terdapat dalam bahan berikut:
[Relative atomic mass: N = 14; Zn = 65; Avogadro Constant = 6.02 × 1023]
[Jisim atom relatif: N = 14; Zn = 65; Pemalar Avogadro = 6.02 × 1023]
(i) 13 g of zinc / 13 g zink
13
= 0.2 mol
65
Number of zinc atom = 0.2 × 6.02 × 1023
= 1.204 × 1023
Number of mol of zinc atom =
(ii)
5.6 g of nitrogen gas / 5.6 g gas nitrogen
5.6
Number of mol of N atom =
= 0.4 mol
14
Number of N atom = 0.4 × 6.02 × 1023
= 2.408 × 1023
(b) Calculate the number of molecules in the following substances / Hitungkan bilangan molekul dalam bahan berikut:
[Relative atomic mass: N = 14, H = 1, Cl = 35.5, Avogadro Constant = 6.02 × 1023]
[Jisim atom relatif: N = 14, H = 1, Cl = 35.5, Pemalar Avogadro = 6.02 × 1023]
(i) 8.5 g of ammonia gas, NH3 / 8.5 g gas ammonia, NH3
8.5
× 6.02 × 1023
17
= 2.408 × 1023
(ii)
4
14.2 g of chlorine gas, Cl2 / 14.2 g gas klorin, Cl2
14.2
× 6.02 × 1023
71
= 1.2 × 1023
A gas jar contains 240 cm3 of carbon dioxide gas. Calculate:
Suatu balang gas berisi 240 cm3 gas karbon dioksida. Hitungkan:
[Relative atomic mass: C = 12, O = 16; Molar volume of gas = 24 dm3 mol–1 at room conditions]
[Jisim atom relatif: C = 12, O = 16; Isi padu molar gas: 24 dm3 mol–1 pada keadaan bilik]
(a) Number of moles of carbon dioxide gas / Bilangan mol gas karbon dioksida:
Number of moles of CO2 =
240
= 0.01 mol
24 000
(b) Number of molecules of carbon dioxide gas / Bilangan molekul gas karbon dioksida:
Number of molecules of CO2 = 0.01 × 6.02 × 1023
= 6.02 × 1021
(c) Mass of carbon dioxide gas / Jisim gas karbon dioksida:
Mass of CO2 = 0.01 mol × [12 + 2 × 16] g mol–1
= 0.44 g
5
What is the mass of chlorine molecules (Cl2) that contains twice as many molecules as that found in 3.6 g of water?
Berapakah jisim molekul klorin (Cl2 ) yang mengandungi dua kali ganda bilangan molekul yang terdapat dalam 3.6 g air?
[Relative atomic mass / Jisim atom relatif : H = 1, O = 16, Cl = 35.5]
Number of moles of chlorine molecule = 2 × no of mol in H2O
3.6
=2×
= 0.4 mol
18
Mass of Cl2 = 0.4 × 71= 28.4 g
6
Calculate the mass of carbon that has the same number of atoms as found in 4 g of magnesium.
Hitungkan jisim karbon yang mempunyai bilangan atom yang sama seperti yang terdapat dalam 4 g magnesium.
[Relative atomic mass / Jisim atom relatif : C = 12, Mg = 24]
n
io
Sdn. B
m
29
.
hd
Publicat
2g
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MODULE • Chemistry Form 4
Compare the number of molecule in 32 g of sulphur dioxide (SO2) with 7 g of nitrogen gas (N2). Explain your answer.
7
Bandingkan bilangan molekul dalam 32 g sulfur dioksida (SO2 ) dengan 7 g gas nitrogen (N2 ). Terangkan jawapan anda.
[Relative atomic mass / Jisim atom relatif : S = 32, O = 16, N = 14]
Number of moles of molecules in 32 g SO2 =
32
= 0.5 mol
64
7
= 0.25 mol
28
Number of molecule in 32 g SO2 is two times more than 7 g N2.
Number of mole in sulphur dioxide molecule is two times more than number of mole of nitrogen molecule.
Number of moles of molecules in 7 g N2 =
Compare number of atoms in 1.28 g of oxygen to the number of atoms in 1.3 g of zinc. Explain your answer.
8
Bandingkan bilangan atom dalam 1.28 g oksigen dengan bilangan atom dalam 1.3 g zink. Terangkan jawapan anda.
[Relative atomic mass / Jisim atom relatif : O = 16, Zn = 65]
1.28
= 0.08 mol
16
1.30
Number of mol of Zn atoms in 1.3 g Zn =
= 0.04 mol
65
Number of oxygen atoms in 1.28 g oxygen is 2 times more than number of zinc atoms in 1.3 g zinc.
Number of mol of oxygen atom is 2 times more than zinc atom.
Number of mol of O atoms in 1.28 g SO2 =
CHEMICAL FORMULAE AND CHEMICAL EQUATIONS / FORMULA KIMIA DAN PERSAMAAN KIMIA
Symbol of elements – use capital letters for the first alphabet and use small letters if there is a second alphabet.
1
Simbol unsur – gunakan huruf besar untuk huruf pertama dan huruf kecil jika ada huruf kedua.
Example / Contoh: Potassium / Kalium – K,
Calcium / Kalsium – Ca,
Iron / Ferum
– Fe,
Sodium / Natrium – Na
Nitrogen / Nitrogen – N
Fluorine / Fluorin – F
Chemical Formula – A set of chemical symbols for atoms of elements in whole numbers representing chemical
substances.
Formula kimia – Satu set simbol kimia bagi atom-atom unsur dengan gandaan nombor bulat yang mewakili bahan kimia.
Chemical substance
Bahan kimia
Water
Air
Ammonia
Ammonia
Propane
m
Notes
Formula kimia
H2O
NH3
C3H8
Catatan
2 atoms of H combines with 1 atom of O.
2 atom H bergabung dengan 1 atom O.
3 atoms of H combines with 1 atom of N.
3 atom H bergabung dengan 1 atom N.
3 atoms of C combines with 8 atoms of H.
3 atom C bergabung dengan 8 atom H.
2
Information that can be obtained from the chemical formula / Maklumat yang diperoleh daripada formula kimia:
(i) All the elements present in the compound / Jenis unsur yang terdapat dalam sebatian,
(ii) Number of atoms of each element in the compound / Bilangan atom setiap unsur yang terdapat dalam sebatian,
(iii) Calculation of RMM/RFM of the compound / Pengiraan JMR/JFR bagi sebatian.
3
Two types of chemical formula / Dua jenis formula kimia:
(i) Empirical formula / Formula empirik,
(ii) Molecular formula / Formula molekul.
Publica
n Sdn.
30
tio
Nil
a
Propana
Chemical formula
d.
Bh
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Chemistry Form 4 • MODULE
EMPIRICAL FORMULA / FORMULA EMPIRIK
1
2
A formula that shows the simplest whole number ratio of atoms of each element in a compound.
Formula yang menunjukkan nisbah nombor bulat teringkas bagi bilangan atom setiap unsur yang terdapat dalam sebatian.
The formula can be determined by calculating the simplest ratio of moles of atoms of each element in the compound.
Formula itu boleh ditentukan dengan menghitung nisbah bilangan mol atom bagi setiap unsur yang terdapat dalam sebatian.
3
Experiments to determine empirical formula of metal oxide / Formula empirik bagi oksida logam diperoleh dengan cara:
Empirical formula of magnesium oxide
Empirical formula of copper(II) oxide
Formula empirik magnesium oksida
Set-up of apparatus / Susunan radas:
Formula empirik kuprum(II) oksida
Set-up of apparatus / Susunan radas:
Copper(II) oxide
Kuprum(II) oksida
Magnesium
Magnesium
Hydrogen gas
Gas hidrogen
Heat
Heat
Panaskan
Panaskan
Reaction occurs / Tindak balas yang berlaku:
Reaction occurs / Tindak balas yang berlaku:
Magnesium dipanaskan dengan kuat di dalam mangkuk pijar untuk
bertindak balas dengan oksigen membentuk magnesium oksida.
Gas hidrogen dilalukan melalui kuprum(II) oksida yang dipanaskan.
Hidrogen menurunkan kuprum(II) oksida kepada kuprum dan air.
Balanced equation / Persamaan kimia seimbang:
Balanced equation / Persamaan kimia seimbang:
This method can also be used to determine the empirical
formulae of reactive metals such as aluminium oxide and zinc
oxide.
This method can also be used to determine the empirical
formulae of less reactive metals such as lead(II) oxide and
tin(II) oxide.
Kaedah ini boleh juga digunakan untuk menentukan formula empirik oksida
logam reaktif seperti aluminium oksida dan zink oksida.
Kaedah ini boleh juga digunakan untuk menentukan formula empirik oksida
logam kurang reaktif seperti plumbum(II) oksida and stanum(II) oksida.
Magnesium is burnt in a crucble to react with oxygen to form
magnesium oxide.
2Mg + O2 → 2MgO
4
Hydrogen gas is passed through heated copper(II) oxide.
Hydrogen reduces copper(II) oxide to form copper and water.
CuO + H2 → Cu + H2O
Experiment to Determine Empirical Formula of Magnesium Oxide
Eksperimen untuk Menentukan Formula Empirik Magnesium Oksida
In this experiment, magnesium reacts with oxygen in the air to form white fumes, magnesium oxide:
Semasa eksperimen ini, magnesium bertindak balas dengan oksigen dalam udara untuk membentuk asap putih, magnesium oksida:
Magnesium + Oxygen → Magnesium oxide
Magnesium + Oksigen → Magnesium oksida
Material / Bahan: Magnesium ribbon, sand paper
Apparatus / Radas:
Crucible with lid, tongs, Bunsen burner, tripod stand and balance
Set-up of apparatus / Susunan radas:
Magnesium ribbon
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MODULE • Chemistry Form 4
Procedure / Langkah:
crucible
(a) A
Mangkuk pijar
pita magnesium
Pita magnesium
sand paper
is cleaned with
coiled
is
gulung
di
crucible
(d) The
ditimbang.
..
loosely and placed in the crucible.
dan diletakkan dalam mangkuk pijar.
magnesium ribbon
together with the lid and
Mangkuk pijar
.
kertas pasir
dibersihkan dengan menggunakan
magnesium ribbon
(c) The
are weighed.
penutup
dengan
magnesium ribbon
(b) 10 cm of
10 cm
lid
and its
pita magnesium
bersama dengan penutup dan
are weighed again.
ditimbang.
(e) The apparatus is set up as shown in the diagram.
Radas disusun seperti dalam gambar rajah.
(f)
strongly
The crucible is heated
burn
Mangkuk pijar dipanaskan dengan
terbakar
, mangkuk pijar ditutup dengan
lid
pita magnesium
Apabila
kuat selama 2 minit lagi.
,
penutup
weighed again
, lid and its content are
is removed and the crucible is
dibuka dan mangkuk pijar dipanaskan dengan
heating
(k) The process of
cooling
,
suhu bilik
.
.
ditimbang sekali lagi
, penutup dan kandungannya
constant
lid
, the
, penutup dan kandungannya dibiarkan sejuk ke
Mangkuk pijar
.
repeated
and weighing are
until a
mass is obtained.
pemanasan
Proses
.
lid and its content are allowed to cool down to room temperature .
crucible
The
burning
terbakar
berhenti
crucible
Mangkuk pijar
(j)
. Apabila pita magnesium mula
dibuka sekali sekala dengan menggunakan penyepit.
(h) When the magnesium ribbon stops
heated strongly for another 2 minutes.
The
tanpa
penutup
.
penutup
of the crucible is lifted from time to time using a pair of tongs.
Penutup
(i)
. When the magnesium starts to
lid
, the crucible is covered with its
kuat
(g) The
lid
without its
tetap
penyejukan
,
dan penimbangan
diulang
beberapa kali sehingga jisim
diperoleh.
Observation / Pemerhatian:
Magnesium burns
brightly
Magnesium terbakar dengan
white fumes
to release
terang
membebaskan
and
wasap putih
white solid
is formed.
pepejal putih
dan kemudiannya membentuk
.
Inference / Inferens:
Magnesium is a
reactive
metal.
reaktif
Magnesium adalah logam yang
Magnesium reacts with
oxygen
m
in the air to form
oksigen
magnesium oxide
dalam udara membentuk
.
magnesium oksida
.
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Magnesium bertindak balas dengan
.
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Chemistry Form 4 • MODULE
Precaution steps / Langkah berjaga-jaga:
Step taken / Langkah yang diambil
sand paper .
Magnesium ribbon is cleaned with
Pita magnesium perlu digosok dengan
The
crucible lid
Purpose / Tujuan
kertas pasir
To remove the
ribbon.
.
is lifted from time to time.
crucible lid
replaced
then
on the surface of the magnesium
Untuk membuang lapisan oksida pada permukaan magnesium oksida.
oxygen
To allow
from the air to react with magnesium .
Untuk membenarkan oksigen masuk dan bertindak balas dengan magnesium .
Penutup mangkuk pijar dibuka sekali sekala.
The
oxide layer
To prevent fumes of
quickly.
magnesium oxide
from escaping.
Untuk mengelakkan wasap magnesium oksida dari terbebas.
Penutup mangkuk pijar kemudian ditutup semula dengan cepat.
The process of
heating , cooling and weighing are
repeated until a constant mass is obtained.
To ensure magnesium react
for magnesium oxide .
completely
oxygen
pemanasan ,
penyejukan
penimbang
dan
Proses
jisim tetap
diulang beberapa kali sehingga
diperoleh.
lengkap
Untuk memastikan semua magnesium telah bertindak balas
oksigen
dengan
untuk membentuk magnesium oksida .
with
to
Result / Keputusan:
Description / Penerangan
Mass (g) / Jisim (g)
Mass of crucible + lid
x
Jisim mangkuk pijar + penutup
Mass of crucible + lid + magnesium
y
Jisim mangkuk pijar + penutup + magnesium
Mass of crucible + lid + magnesium oxide
z
Jisim mangkuk pijar + penutup + magnesium oksida
Calculation / Pengiraan:
Element / Unsur
Mg
O
Mass (g) / Jisim (g)
y–x
z–y
Number of mole of atoms / Bilangan mol atom
y–x
24
z–y
16
Simplest ratio of moles / Nisbah mol teringkas
p
q
MgpOq
Empirical formula of magnesium oxide is
Formula empirik magnesium oksida ialah
5
Mg O
p
q
.
.
Experiment to Determine Empirical Formula of Copper(II) Oxide
Eksperimen untuk Menentukan Formula Empirik Kuprum(II) Oksida
Copper(II) Oxide + Hidrogen → Copper + Water
Kuprum(II) oksida + Hidrogen → Kuprum + Air
Set-up of apparatus / Susunan radas:
Copper(II) oxide
Burning of hydrogen gas
Hydrogen gas
Combustion tube
Heat
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Anhydrous calcium chloride, CaCl2
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MODULE • Chemistry Form 4
Observation / Pemerhatian:
The
black
Warna
colour of copper(II) oxide turns
hitam
brown
perang
kuprum(II) oksida menjadi
.
.
Inference / Inferens:
copper metal
Copper(II) oxide reacts with hydrogen to produce the brown
.
logam kuprum
Kuprum(II) oksida bertindak balas dengan hidrogen untuk menghasilkan
yang berwarna perang.
Precaution steps / Langkah berjaga-jaga:
Step taken / Langkah yang ambil
Purpose / Tujuan
Hydrogen gas is passed through anhydrous calcium
chloride.
hydrogen gas.
Gas hidrogen dialirkan melalui kalsium klorida kontang.
To
air
in the combustion tube.
air
explodes when lighted).
If the gas burns quietly without ‘pop’ sound , all the
has been removed from the combustion tube.
bunyi ‘pop’
Jika gas terbakar tanpa
daripada tabung pembakaran.
at mouth of the test tube.
Gas yang keluar daripada lubang kecil dikumpul dalam sebuah
tabung uji. Kayu uji menyala di letakkan di mulut tabung
uji.
continuous
air
dikeluarkan
, semua gas telah
To prevent hot copper from reacting with
copper(II) oxide
again.
Gas hidrogen dialirkan secara berterusan sepanjang eksperimen.
oxygen
to form
Untuk mengelakkan kuprum panas daripada bertindak balas dengan
oksigen dan membentuk kuprum(II) oksida .
heating , cooling and weighing are
constant
mass is obtained.
repeated until a
The process of
dan
tetap
all the
udara
dalam tabung pembakaran.
Untuk mengeluarkan semua
udara
(Campuran hidrogen dan
menghasilkan letupan apabila dinyalakan)
The gas that comes out from the small hole is collected
in the test tube. Then, a lighted wooden splinter is
pemanasan ,
penyejukan
Proses
diulang beberapa kali sehingga jisim
remove
gas hidrogen.
(The mixture of hydrogen gas and
kering
dialirkan melalui tabung pembakaran
Gas hidrogen
selama 5 hingga 10 minit.
The flow of hydrogen gas must be
throughout the experiment.
mengering
Kalsium klorida kontang menyerap wap air untuk
Dry
hydrogen is passed through the combustion
tube for 5 to 10 minutes.
placed
dry
Anhydrous calcium chloride absorb water vapour to
To ensure all copper(II) oxide has changed to
copper .
Untuk memastikan semua kuprum(II) oksida telah bertukar kepada
kuprum .
penimbang
diperoleh.
Result / Keputusan:
Description / Penerangan
Mass (g) / Jisim (g)
Mass of combustion tube + porcelain dish
x
Jisim tabung pembakaran + piring tanah liat
Mass of combustion tube + porcelain dish + copper(II) oxide
y
Jisim tabung pembakaran + piring tanah liat + kuprum(II) oksida
Mass of combustion tube + porcelain dish + copper
z
Jisim tabung pembakaran + piring tanah liat + kuprum
Calculation / Pengiraan:
Element / Unsur
Cu
O
Mass (g) / Jisim (g)
z–x
y–z
Number of mole of atoms / Bilangan mol atom
z–x
64
y–z
16
Simplest ratio of moles / Nisbah mol teringkas
p
q
Empirical formula of copper(II) oxide is
m
CupOq
.
.
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Formula empirik kuprum(II) oksida ialah
CupOq
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Chemistry Form 4 • MODULE
6
Explain why the set-up of apparatus to determine the empirical formula in both the experiments is different.
Terangkan mengapa susunan radas untuk menentukan formula empirik dalam kedua-dua eksperimen itu berbeza.
reactive
(a) Magnesium is
magnesium oxide
metal (above hidrogen in reactivity series). Magnesium
by
hydrogen
hydrogen gas
Kuprum di bawah
gas hidrogen
7
easily to form
.
reaktif
Magnesium adalah logam
membentuk magnesium oksida .
(b) Copper is below
reacts
teroksida
(terletak di atas hidrogen dalam siri kereaktifan. Magnesium mudah
in the metal reactivity series. Oxygen in copper(II) oxide can be
reduced/removed
to form copper and water.
hidrogen
dalam siri kereaktifan. Kuprum(II) okida boleh
diturunkan/disingkirkan
oleh
untuk membentuk kuprum dan air.
To calculate the empirical formula of a compound, use the following table:
Untuk menghitung formula empirik suatu sebatian, jadual di bawah boleh digunakan sebagai panduan:
Calculation steps / Langkah pengiraan:
Element / Unsur
(a) Calculate the mass of each element in the compound.
Hitungkan jisim setiap unsur dalam sebatian.
Mass of element (g) / Jisim unsur (g)
(b) Convert the mass of each element to number of mole of atom.
Number of mole of atom / Bilangan mol atom
Tukar jisim setiap unsur kepada bilangan mol atom.
(c) Calculate the simplest ratio of moles of atom of the elements.
Simplest ratio of moles / Nisbah mol teringkas
Hitungkan nisbah bilangan mol atom teringkas unsur-unsur tersebut.
EXERCISE / LATIHAN
1
When 11.95 g of metal X oxide is reduced by hydrogen, 10.35 g of metal X is produced. Calculate the empirical formula
of metal X oxide.
Apabila 11.95 g oksida logam X diturunkan oleh hidrogen, 10.35 g logam terhasil. Hitungkan formula empirik bagi oksida logam X.
[RAM / JAR: X = 207, O = 16]
X
O
Mass of element (g) / Jisim unsur (g)
10.35
1.6
Number of mole of atoms / Bilangan mol atom
0.05
0.1
Ratio of moles / Nisbah mol
1
2
Simplest ratio of moles / Nisbah mol teringkas
1
2
Element / Unsur
Empirical formula / Formula empirik:
2
XO2
.
A certain compound contains the following composition / Satu sebatian mengandungi komposisi unsur seperti berikut:
Na = 15.23%, Br = 52.98%, O = 31.79 % [Relative atomic mass / Jisim atom relatif: O = 16, Na = 23, Br = 80]
(Assume that 100 g of substance is used / Anggap 100 g bahan digunakan)
Element / Unsur
Na
Br
O
Mass of element (g) / Jisim unsur (g)
15.23
52.98
31.79
Number of mole of atoms / Bilangan mol atom
0.66
0.66
1.99
Ratio of moles / Nisbah mol
1
1
3.01
Simplest ratio of moles / Nisbah mol teringkas
1
1
3
NaBrO3
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Empirical formula / Formula empirik:
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MODULE • Chemistry Form 4
2.08 g of element X combines with 4.26 g of element Y to form a compound with formula XY3. Calculate the relative
atomic mass of element X. [RAM: Y = 35.5]
3
2.08 g unsur X bergabung dengan 4.26 g unsur Y untuk membentuk sebatian dengan formula XY3 . Hitung jisim atom relatif unsur X.
[JAR: Y = 35.5]
Element / Unsur
Mass of element (g)
Jisim unsur (g)
Number of mole of atoms
Bilangan mol atom
Simplest ratio of moles
Nisbah mol teringkas
X
Y
2.08
4.26
2.08
x
4.26 = 0.12
35.5
1
3
x = relative atomic mass of X
Mol X = 1
Mol Y
3
2.08
x
1
=
0.12
3
x = 52
2.07 g of element Z reacts with bromine to form 3.67 g of a compound with empirical formula ZBr2. Calculate the
relative atomic mass of element Z. [RAM: Br = 80]
4
2.07 g unsur Z bertindak balas dengan bromin membentuk 3.67 g sebatian dengan formula empirik ZBr2. Hitung jisim atom relatif bagi
unsur Z. [JAR: Br = 80]
Element / Unsur
Mass of element (g)
Jisim unsur (g)
Number of mole of atoms
Bilangan mol atom
Simplest ratio of moles
Nisbah mol teringkas
Z
Br
2.07
1.6
2.07
z
1.6 = 0.02
80
1
2
z = relative atomic mass of Z
Mol Z
Mol Br
2.08
z
0.02
= 1
2
1
2
z = 207
=
The statement below is about compound J / Pernyataan berikut adalah mengenai sebatian J.
5
• It is black solid / Merupakan pepejal hitam.
• Contains 12.8 g copper and 0.2 mol of oxygen / Mengandungi 12.8 g kuprum dan 0.2 mol oksigen.
[Relative atomic mass / Jisim atom relatif : Cu = 64]
(a) What is meant by empirical formula / Apakah maksud formula empirik?
A formula that shows the simplest whole number ratio of atoms of each element in a compound.
(b) (i)
Calculate the number of mol of copper atom / Hitung bilangan mol atom kuprum.
12.8
= 0.2 mol
64
(ii)
What is the empirical formula of compound J / Apakah formula empirik sebatian J ?
0.2 mol Cu : 0.2 mol O.
1 mol Cu : 1 mol O.
Empirical formula of Compound J is CuO.
(c) Compound J reacts completely with hydrogen to produce copper and compound Q.
Sebatian J bertindak balas lengkap dengan hidrogen menghasilkan kuprum dan sebatian Q.
(i)
State one observation for the reaction / Nyatakan satu pemerhatian daripada tindak balas tersebut.
Black solid change to brown
(ii)
Name two the substances that can be used to prepare hydrogen gas.
Namakan dua bahan yang digunakan untuk menyediakan gas hidrogen.
Zinc/magnesium and hydrochloric acid/nitric acid/sulphuric acid.
(iii) Name compound Q / Nama sebatian Q.
Water
(iv) Write a balanced equation for the reaction.
Tuliskan persamaan kimia yang seimbang bagi tindak balas tersebut.
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CuO + H2 → Cu + H2O
m
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Chemistry Form 4 • MODULE
(d) Draw a labelled diagram of the set-up of apparatus for the experiment.
Lukiskan gambar rajah berlabel susunan radas bagi tindak balas tersebut.
Gas hidrogen
Compound J
Heat
(e) (i)
Why is hydrogen gas passed through the combustion tube after heating has stpopped?
Mengapakah gas hidrogen dilalukan melalui tiub pembakaran selepas pemanasan dihentikan?
To avoid copper produced react with oxygen to form copper(II) oxide.
(ii)
State how to determine that the reaction between compound J and hydrogen has completed.
Nyatakan bagaimana menentukan tindak balas antara sebatian J dengan hidrogen telah lengkap.
By repeating the process of heating, cooling and weighing until constant mass is obtained.
(f)
(i)
Can the empirical formula of magnesium oxide be determined by the same method? Explain your answer.
Bolehkah formula empirik bagi magnesium oksida ditentukan dengan cara yang sama? Jelaskan jawapan anda.
Cannot. Magnesium is more reactive than hydrogen. Hydrogen cannot reduce magnesium oxide to form
magnesium.
(ii)
Magnesium can reduce copper oxide to copper. Explain why the empirical formula of the copper oxide cannot
be determined by heating the mixture of copper oxide and magnesium powder.
Magnesium boleh menurunkan kuprum oksida kepada kuprum. Terangkan mengapa formula empirik kuprum oksida tidak boleh
ditentukan dengan pemanasan campuran kuprum oksida dengan serbuk magnesium.
Magnesium oxide and copper produced are in solid form, copper cannot be separated from magnesium oxide.
The mass of copper cannot be weighed.
MOLECULAR FORMULA / FORMULA MOLEKUL
1
Molecular formula of a compound shows the actual number of atoms of each element that are present in a molecule of
the compound.
Formula molekul suatu sebatian menunjukkan bilangan sebenar atom bagi setiap unsur yang terdapat dalam satu molekul sebatian.
Molecular Formula = (empirical formula)n, where n is a integer.
Formula molekul = (Formula empirik)n, di mana n adalah integer.
2
Example / Contoh:
Compound
Molecular formula
Empirical formula
Value of n
Water / Air
H2O
H2O
1
Carbon dioxide / Karbon dioksida
CO2
CO2
1
H2SO4
H2SO4
1
Ethene / Etena
C2H4
CH2
2
Benzene / Benzena
C6H6
CH
6
Glucose / Glukosa
C6H12O6
CH2O
6
Sebatian
Sulphuric acid / Asid sulfurik
Formula molekul
Formula empirik
Nilai n
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The molecular formula and the empirical formula of a compound may be the same if the value of n = 1 but different if the
value is n > 1.
Formula molekul dan formula empirik suatu sebatian akan sama sekiranya nilai n = 1 tetapi akan berbeza sekiranya nilai n > 1.
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MODULE • Chemistry Form 4
EXERCISE / LATIHAN
The empirical formula of compound X is CH2 and relative molecular mass is 56. Determine the molecular formula of
compound X. [Relative atomic mass: H = 1; C = 12]
1
Formula empirik sebatian X adalah CH2 dan JMR adalah 56. Tentukan formula molekul sebatian X. [Jisim atom relatif: H = 1; C = 12]
(12 + 2)n = 56
56
n=
=4
14
Molecular formula = (CH2)4 = C4H8
2.58 g of a hydrocarbon contains 2.16 g of carbon. The relative molecular mass of the hydrocarbon is 86.
2
2.58 g suatu hidrokarbon mengandungi 2.16 g karbon. Jisim molekul relatif bagi hidrokarbon ini ialah 86.
[Relative atomic mass / Jisim atom relatif : H = 1; C = 12]
(i) Calculate the empirical formula of the hydrocarbon / Hitungkan formula empirik bagi hidrokarbon ini.
Element
C
H
Mass of element (g)
2.16
0.42
Number of mole of atoms
0.18
0.42
Ratio of moles
1
21 = 7
3
3
Simplest ratio of moles
3
7
Empirical formula = C3H7
(ii) Determine the molecular formula of the hydrocarbon / Tentukan formula molekul hidrokarbon tersebut.
(12 × 3 + 7 × 1)n = 86
86
n=
=2
43
Molecular formula = (C3H7)2 = C6H14
The diagram below shows the structural formula for benzene molecule.
3
Rajah di bawah menujukkan formula struktur bagi benzena.
H
H
C
H
C
C
C
C
H
C
H
H
(a) Name the element that make up benzene / Namakan unsur yang membentuk benzena.
Carbon and hydrogen
(b) What are the molecular formula and empirical formula for benzene?
Apakah formula molekul dan formula empirik bagi benzena?
Molecular formula / Formula molekul: C6H6
Empirical formula / Formula empirik: CH
(c) Compare and contrast the molecular formula and empirical formula for benzene.
Banding dan bezakan formula molekul dan formula empirik bagi benzena.
• Both empirical formula and molecular formula shows benzene is made up of
elements.
Kedua-dua fomula molekul dan formula empirik menunjukkan benzena terdiri dari unsur
actual
• Molecular formula shows the
molecule . Each benzene molecule
carbon
number of
consists of
sebenar
6 carbon
m
hidrogen
Publica
atom
dan
hydrogen
hidrogen
.
hydrogen
atoms in benzene
6
hydrogen
atoms and
atoms.
dan atom hidrogen dalam molekul
karbon
6
dan
atom
.
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Formula molekul menunjukkan bilangan
bagi atom
molekul
6
benzena terdiri daripada
benzena. Setiap
atoms and
carbon
d.
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Chemistry Form 4 • MODULE
• Empirical formula shows the simplest ratio of number carbon atoms to hydrogen atoms, the simplest
carbon
hydrogen
1:1
ratio of number of
atoms to
atoms in benzene is
.
Formula empirik benzena menunjukkan
nisbah paling ringkas
Nisbah paling ringkas bilangan atom
karbon
kepada
hidrogen
karbon
bilangan atoms
kepada atom
hidrogen
1
:
1
adalah
.
.
PERCENTAGE COMPOSITION BY MASS OF AN ELEMENT IN A COMPOUND
PERATUS KOMPOSISI UNSUR MENGIKUT JISIM DALAM SEBATIAN
Total RAM of the element in the compound × 100%
1
% composition by mass of an element =
% komposisi unsur mengikut jisim
2
Jumlah JAR unsur dalam suatu sebatian × 100%
RMM/RFM of compound/JMR/JFR sebatian
Example / Contoh:
Calculate the percentage composition by mass of nitrogen in the following compounds:
Hitungkan peratusan nitrogen mengikut jisim dalam sebatian berikut:
[Relative atomic mass / Jisim atom relatif : N = 14, H = 1, O = 16, S = 32, K = 39]
(i)
(NH4)2SO4
2 × 14
× 100%
132
= 21.2%
%N =
(ii) KNO3
14
× 100%
101
= 13.9%
%N =
CHEMICAL FORMULA FOR IONIC COMPOUNDS / FORMULA KIMIA BAGI SEBATIAN ION
1
Chemical formula of an ionic compound comprising of the ions Xm+ and Yn– is by exchanging the charges on each ion.
The formula obtained will be XmYn.
Formula kimia sebatian ion yang mengandungi ion X m+ dan Y n– boleh diperoleh melalui pertukaran bilangan cas setiap ion. Formula yang
diperoleh ialah XnYm.
2
Example / Contoh:
(i) Sodium oxide / Natrium oksida
Ion / Ion
Na+
O2–
+1
–2
Exchange of charges / Pertukaran bilangan cas
2
1
Smallest ratio / Nisbah teringkas
2
1
2 Na+
O2–
Charges / Bilangan cas
Number of combining ions / Bilangan ion yang bergabung
Formula / Formula
(ii) Copper(II) nitrate / Kuprum(II) nitrat
Cu2+
+2
2 (Ratio / Nisbah)
(iii) Zinc oxide / Zink oksida
Zn2+
+2
O2–
–2
2
2
1
⇒ ZnO
1 (Ratio / Nisbah)
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⇒ Cu(NO3)2
NO3–
–1
Na2O
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CuO
Copper(II) oxide
CaO
Calcium oxide
Ag2O
Silver oxide
Na2O
Sodium oxide
K2O
Potassium oxide
Ion aluminium
Al 3+
Aluminium ion
Ion plumbum(II)
Pb2+
Lead(II) ion
Ion zink
Zn2+
Zinc ion
PbCO3
Lead(II) carbonate
ZnCO3
Zinc carbonate
MgCO3
Magnesium
carbonate
CuCO3
Copper(II) carbonate
PbSO4
Lead(II) sulphate
ZnSO4
Zinc sulphate
MgSO4
Magnesium
sulphate
CuSO4
Copper(II) sulphate
CaSO4
Calcium sulphate
(NH4)2SO4
Ammonium
sulphate
(NH4)2CO3
Ammonium
carbonate
H2SO4
Sulphuric acid
Na2SO4
Sodium sulphate
Ag2SO4
Silver sulphate
CaCO3
Calcium carbonate
Ion klorida
Cl–,
Chloride ion
Ion bromida
Br–,
Bromide ion
CaBr2
Calcium bromide
NH4Br
Ammonium
bromide
AgBr
Silver bromide
HBr
Hydrobromic acid
NaBr
Sodium bromide
AlCl3
Aluminium
chloride
PbCl2
Lead(II) chloride
ZnCl2
Zinc chloride
MgCl2
Magnesium
chloride
PbI2
Lead(II) iodide
ZnI2
Zinc iodide
Ion nitrat
NO3–,
Nitrate ion
Ca(NO3 )2
Calcium nitrate
NH4NO3
Ammonium
nitrate
AgNO3
Silver nitrate
HNO3
Nitric acid
NaNO3
Sodium nitrate
Pb(OH)2
Lead(II) hydroxide
Zn(OH)2
Zinc hydroxide
Al(NO3)3
Aluminium nirate
Pb(NO3 )2
Lead(II) nitrate
Zn(NO3 )2
Zinc nitrate
Mg(NO3 )2
Magnesium nitrate
Cu(OH)2
Cu(NO3 )2
Copper(II) hydroxide Copper(II) nitrate
Ca(OH)2
Calcium hydroxide
AgOH
Silver hydroxide
NaOH
Sodium hydroxide
KOH
KNO3
Potassium hydroxide Potassium nitrate
Ion hidroksida
OH–,
Hydroxide ion
Mg(OH)2
MgI2
Magnesium
Magnesium iodide
hydroxide
CuI2
Copper(II) iodide
CaI2
Calcium iodide
NH4I
Ammonium
iodide
AgI
Silver iodide
HI
Hydroiodic acid
NaI
Sodium iodide
KI
Potassium iodide
Ion iodida
I–,
Iodide ion
Al(OH)3
AlBr3
AlI3
Aluminium
Aluminium bromide Aluminium iodide
hydroxide
PbBr2
Lead(II) bromide
ZnBr2
Zinc bromide
MgBr2
Magnesium
bromide
CuCl2
CuBr2
Copper(II) chloride Copper(II) bromide
CaCl2
Calcium chloride
NH4Cl
Ammonium
chloride
AgCl
Silver chloride
HCl
Hydrocloric acid
NaCl
Sodium chloride
K2SO4
KCl
KBr
Potassium sulphate Potassium chloride Potassium bromide
Ion sulfat
SO42–,
Sulphate ion
Ag2CO3
Silver carbonate
H2CO3
Carbonic acid
Na2CO3
Sodium carbonate
K2CO3
Potassium carbonate
Ion karbonat
CO32–,
Carbonat ion
Al2(SO4 )3
Al2O3
Al2(CO3 )3
Aluminium
Aluminium oxide Aluminium carbonate
sulphate
PbO
Lead(II) oxide
ZnO
Zinc oxide
MgO
Mg2+
Magnesium ion Magnesium
Ion magnesium
oxide
Ion kuprum(II)
Cu2+
Copper(II) ion
Ion kalsium
Ca2+
Calcium ion
Ion ammonium
NH4 +
Ammonium ion
Ion argentum
Ag+
Silver ion
Ion hidrogen
H+
Hydrogen ion
Ion natrium
Na+
Sodium ion
Ion kalium
K+
Potassium ion
Ion oksida
O2–,
Oxide ion
Aktiviti 1: TULIS FORMULA KIMIA DAN NAMA BAGI BAHAN KIMIA BERIKUT
ACTIVITY 1: WRITE THE CHEMICAL FORMULAE AND NAMES OF THE FOLLOWING COMMON COMPOUNDS
MODULE • Chemistry Form 4
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Ion aluminium
Aluminium ion
Ion plumbum(II)
Lead(II) ion
Ion zink
Zinc ion
Ion magnesium
Magnesium ion
Ion kuprum(II)
Copper(II) ion
Ion kalsium
Calcium ion
Ion ammonium
Ammonium ion
Ion argentum
Silver ion
Ion hidrogen
Hydrogen ion
Ion natrium
Sodium ion
Ion kalium
Potassium ion
Al2O3
PbO
ZnO
MgO
CuO
CaO
Ag2O
Na2O
K2O
Ion oksida
Oxide ion
Al2(CO3)3
PbCO3
ZnCO3
MgCO3
CuCO3
CaCO3
(NH4 )2CO3
Ag2CO3
H2CO3
Na2CO3
K2CO3
Ion karbonat
Carbonat ion
Al2(SO4 )3
PbSO4
ZnSO4
MgSO4
CuSO4
CaSO4
(NH4 )2SO4
Ag2SO4
H2SO4
Na2SO4
K2SO4
Ion sulfat
Sulphate ion
AlCl3
PbCl2
ZnCl2
MgCl2
CuCl2
CaCl2
NH4Cl
AgCl
HCl
NaCl
KCl
Ion klorida
Chloride ion
AlBr3
PbBr2
ZnBr2
MgBr2
CuBr2
CaBr2
NH4 Br
AgBr
HBr
NaBr
KBr
Ion bromida
Bromide ion
AlI3
PbI2
ZnI2
MgI2
CuI2
CaI2
NH4 I
AgI
HI
NaI
KI
Ion iodida
Iodide ion
Al(OH)3
Pb(OH)2
Zn(OH)2
Mg(OH)2
Cu(OH)2
Ca(OH)2
AgOH
NaOH
KOH
Ion hidroksida
Hydroxide ion
AKTIVITI 2: TANPA MERUJUK KEPADA JADUAL AKTIVITI 1, TULISKAN FORMULA KIMIA BAGI SEBATIAN BERIKUT
Al(NO3 )3
Pb(NO3 )2
Zn(NO3 )2
Mg(NO3 )2
Cu(NO3 )2
Ca(NO3 )2
NH4 NO3
AgNO3
HNO3
NaNO3
KNO3
Ion nitrat
Nitrate ion
ACTIVITY 2: WITHOUT REFERRING TO THE TABLE IN ACTIVITY 1, WRITE THE CHEMICAL FORMULAE OF THE FOLLOWING COMPOUNDS
Chemistry Form 4 • MODULE
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MODULE • Chemistry Form 4
ACTIVITY 3: WRITE THE CHEMICAL FORMULAE AND TYPE OF PARTICLES FOR THE FOLLOWING ELEMENT/COMPOUND
AKTIVITI 3: TULIS FORMULA KIMIA DAN JENIS ZARAH UNTUK UNSUR/SEBATIAN BERIKUT
Compound / Element
Sebatian/Unsur
Sodium sulphate
Natrium sulfat
Ammonium carbonate
Ammonium karbonat
Magnesium nitrate
Magnesium nitrat
Hyrochloric acid
Asid hidroklorik
Potassium oxide
Kalium oksida
Magnesium oxide
Magnesium oksida
Lead(II) carbonate
Plumbum(II) karbonat
Iron(III) sulphate
Ferum(III) sulfat
Magnesium chloride
Magnesium klorida
Zinc sulphate
Zink sulfat
Silver nitrate
Argentum nitrat
Ammonium sulphate
Ammonium sulfat
Zinc oxide
Zink oksida
Nitric acid
Asid nitrik
Ammonia gas
Gas ammonia
Magnesium
Magnesium
Zinc
Zink
Copper(II) sulphate
Kuprum(II) sulfat
Iodine
Iodin
Chlorine
m
Type of particles
Na2SO4
Ion
(NH4 )2CO3
Ion
Mg(NO3 )2
Ion
HCl
Ion
K2O
Ion
MgO
Ion
PbCO3
Ion
Fe2(SO4)3
Ion
MgCl2
Ion
ZnSO4
Ion
AgNO3
Ion
(NH4 )2SO4
Ion
ZnO
Ion
HNO3
Ion
NH3
Molecule
Mg
Atom
Zn
Atom
CuSO4
Ion
I2
Molecule
Cl2
Molecule
Formula
Jenis zarah
Compound / Element
Sebatian/Unsur
Zinc carbonate
Zink karbonat
Ammonium carbonate
Ammonium karbonat
Silver chloride
Argentum klorida
Sulphuric acid
Asid sulfurik
Copper(II) nitrate
Kuprum(II) nitrat
Hydrogen gas
Gas hidrogen
Carbon dioxide gas
Gas karbon dioksida
Oxygen gas
Gas oksigen
Aluminium sulphate
Aluminium sulfat
Lead(II) chloride
Plumbun(II) klorida
Potassium iodide
Kalium iodida
Copper(II) carbonate
Kuprum(II) karbonat
Potasium carbonate
Kalium karbonat
Sodium hydroxide
Natrium hidroksida
Aqueous ammonia
Ammonia akueus
Ammonium chloride
Ammonium klorida
Nitrogen dioxide gas
Gas nitrogen dioksida
Sodium chloride
Natrium klorida
Silver
Argentum
Bromine
Bromin
Formula
Type of particles
ZnCO3
Ion
(NH4 )2CO3
Ion
AgCl
Ion
H2SO4
Ion
Cu(NO3 )2
Ion
H2
Molecule
CO2
Molecule
O2
Molecule
Al2(SO4 )3
Ion
PbCl2
Ion
KI
Ion
CuCO3
Ion
K2CO3
Ion
NaOH
Ion
NH3(aq)
Ion and molecule
NH4Cl
Ion
NO2
Molecule
NaCl
Ion
Ag
Atom
Br2
Molecule
Formula
Jenis zarah
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Chemistry Form 4 • MODULE
CHEMICAL EQUATIONS / PERSAMAAN KIMIA
1
Two types of equation / Dua jenis persamaan:
• Equation in words / Persamaan perkataan
– using names of reactants and products / menggunakan nama bahan tindak balas dan hasil tindak balas;
• Equation using symbols / Persamaan menggunakan simbol
– reactants and products are represented by chemical formulae and have certain meanings
menggunakan formula kimia untuk mewakili bahan tindak balas dan hasil tindak balas serta menggunakan pelbagai jenis simbol yang
membawa makna tertentu.
Symbol / Simbol
+
2
Meaning / Maksud
Symbol / Simbol
Meaning / Maksud
Separating 2 reactants / products
(g)
Gaseous state
Mengasingkan 2 bahan / hasil
(g)
Keadaan gas
Produces
(aq)
Aqueous state
Menghasilkan
(ak)
Keadaan akueus
Reversible reaction
Gas released
Tindak balas berbalik
Gas terbebas
(s)
Solid state
Precipitation
(p)
Keadaan pepejal
Bahan termendap
(l)
Liquid state
(ce)
Keadaan cecair
Heating / Heat energy is given
∆
Pemanasan / Haba dibekalkan
Information obtained from chemical equation using symbols / Maklumat yang diperoleh daripada persamaan kimia bersimbol:
(a) Qualitative aspect / Aspek kualitatif : type of reactants and products involved in the chemical reaction and the state
of each reactant and product.
jenis bahan / hasil tindak balas yang terlibat dalam tindak balas dan keadaan fizikal bagi
setiap bahan / hasil tindak balas.
(b) Quantitative aspect / Aspek kuantitatif : number of moles of reactants and products involved in the chemical reaction
that is the coeffficients involved in a balanced equation of the formulae of
reactants and products.
bilangan mol yang bertindak balas dan hasil tindak balas yang terbentuk iaitu pekali bagi
setiap formula bahan dan hasil tindak balas.
Example / Contoh:
Zn (s) + 2HCl (aq)
ZnCl2 (aq) + H2 (g)
Zn (p) + 2HCl (ak)
ZnCl2 (ak) + H2 (g)
1 mol
2 mol
1 mol
1 mol
Interpretation / Tafsiran: 1 mol of zinc reacts with 2 mol of hydrochloric acid to produce 1 mol of zinc chloride and
1 mol of hydrogen.
1 mol zink bertindak balas dengan 2 mol asid hidroklorik menghasilkan 1 mol zink klorida dan 1 mol hidrogen.
3
Writing balanced chemical equations / Menulis persamaan kimia seimbang:
Step 1 / Langkah 1 : Write the correct chemical formulae for each reactant and product.
Tulis formula kimia bagi setiap bahan dan hasil tindak balas.
Step 2 / Langkah 2 : Detemine the number of atoms for each element / Tentukan bilangan atom setiap unsur.
Step 3 / Langkah 3 : Balance the number of atoms for each element by adjusting the coefficients in front of the chemical
formulae.
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MODULE • Chemistry Form 4
EXERCISE / LATIHAN
Write a balanced chemical equation for each of the following reactions:
Tulis persamaan kimia seimbang bagi setiap tindak balas yang berikut:
Zinc carbonate
1
ZnCO3
Zinc oxide + Carbon dioxide / Zink karbonat
Sulphuric acid + Sodium hydroxide
2
H2SO4 + 2NaOH
Sodium sulphate + Water / Asid sulfurik + Natrium hidroksida
Argentum nitrat + Natrium klorida
AgNO3 + NaCl AgCl + NaNO3
Silver chloride + Sodium nitrate
Argentum klorida + Natrium nitrat
Copper(II) oxide + Hydrochloric acid
4
Kuprum(II) oksida + Asid hidroklorik
CuO + 2HCl CuCl2 + H2O
Magnesium + Oxygen
5
2Mg + O2
2Na + 2H2O
K 2O + H 2O
Sodium hydroxide + Hydrogen / Natrium + Air
ZnO + 2HNO3
Lead(II) nitrate
9
Magnesium oksida
Natrium hidroksida + Hidrogen
Potassium hydroxide / Kalium oksida + Air
Kalium hidroksida
2KOH
Zinc oxide + Nitric acid
8
Magnesium oxide / Magnesium + Oksigen
2NaOH + H2
Potassium oxide + Water
7
Copper(II) chloride + Water
Kuprum(II) klorida + Air
2MgO
Sodium + Water
6
Natrium sulfat + Air
Na2SO4 + 2H2O
Silver nitrate + Sodium chloride
3
Zink oksida + Karbon dioksida
ZnO + CO2
Zinc nitrate + Water / Zink oksida + Asid nitrik
Zink nitrat + Air
Zn(NO3 )2 + H2O
Lead(II) oxide + Nitrogen dioxide + Oxygen
Plumbum(II) nitrat Plumbum (II) oksida + Nitrogen dioksida + Oksigen
2Pb(NO3 )2 2PbO + 4NO2 + O2
10 Aluminium nitrate
Aluminium oxide + Nitrogen dioxide + Oxygen
Aluminium nitrat Aluminium oksida + Nitrogen dioksida + Oksigen
4Al(NO3 )3 2Al2O3 + 12NO2 + 3O2
NUMERICAL PROBLEMS INVOLVING CHEMICAL EQUATIONS / PENGHITUNGAN BERKAITAN PERSAMAAN KIMIA
Calculation steps / Langkah perhitungan:
S1 / L1 : Write a balanced equation / Tulis persamaan kimia seimbang.
S2 / L2 : Write the information from the question above the equation / Tulis maklumat daripada soalan di atas persamaan.
S3 / L3 : Write the information from the chemical equation below the equation (information about the number of moles of
reactants/products).
Tulis maklumat daripada persamaan kimia di bawah persamaan (Maklumat perhubungan bilangan mol bahan/hasil tindak balas
terlibat).
S4 / L4 : Change the information in S2 into moles by using the method shown in the chart below.
Tukarkan maklumat L2 kepada mol menggunakan carta di bawah.
S5 / L5 : Use the relationship between number of moles of substance involved in S3 to find the answer.
Gunakan perhubungan bilangan mol bahan terlibat dalam L3 untuk mencari jawapan.
S6 / L6 : Change the information to the unit required using the chart below.
Tukar maklumat kepada unit yang dikehendaki dengan menggunakan carta di bawah.
Mass (g)
m
× (RAM/FRM/RMM) g mol–1
No. of moles (n)
Bilangan mol (n)
× 24 dm3 mol–1 / 22.4 dm3 mol–1
÷ 24 dm3 mol–1 / 22.4 dm3 mol–1
Volume of gas (dm3)
Isipadu gas (dm3)
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Chemistry Form 4 • MODULE
EXERCISE / LATIHAN
1
The equation shows the reaction between zinc and hydrochloric acid.
Persamaan menunjukkan tindak balas antara zink dengan asid hidroklorik.
Zn + 2HCl
ZnCl2 + H2
Calculate the mass of zinc required to react with excess hydrochloric acid to produce 6 dm3 of hydrogen gas at room
conditions. [Relative atomic mass: Zn = 65, Cl = 35.5, 1 mole of gas occupies 24 dm3 at room conditions]
Hitungkan jisim zink yang perlu ditindakbalaskan dengan asid hidroklorik berlebihan untuk menghasilkan 6 dm3 gas hidrogen pada
keadaan bilik. [Jisim atom relatif: Zn = 65, Cl = 35.5, 1 mol gas menempati 24 dm3 pada suhu bilik]
Mol of H2 =
6 dm3
= 0.25 mol
24 dm3 mol–1
From the equation,
1 mol of H2 : 1 mol of Zn
0.25 mol of H2 : 0.25 mol of Zn
Mass of Zn = 0.25 × 65 = 16.2 g
2
The equation shows the reaction between potassium and oxygen.
Persamaan berikut menunjukkan tindak balas antara kalium dengan oksigen.
4K + O2
2K2O
Calculate the mass of potassium required to produce 23.5 g of potassium oxide. [Relative atomic mass: K = 39, O = 16]
Hitungkan jisim kalium yang diperlukan untuk menghasilkan 23.5 g kalium oksida. [Jisim atom relatif: K = 39, O = 16]
Mol of K2O =
23.5
23.5
=
= 0.25 mol
(2 × 39 + 16)
94
From the equation,
2 mol of K2O : 4 mol of K
0.25 mol of K2O : 0.5 mol of K
Mass of K = 0.5 mol × 39 g mol–1 = 19.5 g
3
The equation shows the decomposition of hydrogen peroxide.
Persamaan menunjukkan penguraian hidrogen peroksida.
H2O2
H2O + O2
Balance the equation above. Calculate the number of moles of H2O2 that decomposes if 11.2 dm3 oxygen gas is collected
at STP. [Relative Atomic mass: H = 1, O = 16, molar volume of gas = 22.4 dm3 mol–1 at STP]
Seimbangkan persamaan di atas. Hitung bilangan mol H2O2 yang telah terurai sekiranya 11.2 dm3 gas oksigen dikumpulkan pada STP.
[Jisim atom relatif: H = 1, O = 16, isi padu molar gas = 22.4 dm3 mol–1 pada STP]
Mol of O2 =
11.2 dm3
= 0.5 mol
22.4 dm3 mol–1
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From the equation,
1 mol of O2 : 2 mol of H2O2
0.5 mol of O2 : 1.0 mol of H2O2
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MODULE • Chemistry Form 4
8.0 g of copper(II) oxide powder is added to excess dilute nitric acid and heated. Calculate the mass of copper(II) nitrate
produced. [Relative atomic mass: N = 14, O = 16, Cu = 64]
4
8.0 g serbuk kuprum(II) oksida dicampurkan kepada asid nitrik cair yang berlebihan dan dihangatkan. Hitungkan jisim kuprum(II) nitrat
yang terhasil. [Jisim atom relatif: N = 14, O = 16, Cu = 64]
CuO + 2HNO3
Mol of CuO =
Cu(NO3 )2 + H2O
8g
= 0.1 mol
(64 + 16)g mol–1
From the equation,
1 mol of CuO : 1 mol of Cu(NO3)2
0.1 of CuO : 0.1 mol of Cu(NO3)2
Mass of Cu(NO3)2 = 0.1 mol × 188 g mol–1 = 18.8 g
1.3 g of zinc reacts with excess dilute sulphuric acid. The products are zinc sulphate and hydrogen. Calculate the
volume of hidrogen gas released at STP. [Relative atomic mass: Zn = 65, 1 mol of gas occupies 22.4 dm3 mol–1at STP]
5
1.3 g zink bertindak balas dengan asid sulfurik cair yang berlebihan. Hasil tindak balas adalah zink sulfat dan hidrogen. Hitungkan isi padu
hidrogen yang terbebas pada STP. [Jisim atom relatif: Zn = 65, isipadu molar gas = 22.4 dm3 mol–1 pada STP]
Answer/Jawapan:
448 cm3
0.46 g of sodium burns completely in chlorine gas at room conditions to produce sodium chloride. Calculate the
volume of chlorine gas that has reacted. [Relative atomic mass: Na = 23, Molar volume of gas = 24 dm3 mol–1 at room
conditions]
6
0.46 g natrium terbakar lengkap dalam gas klorin pada keadaan bilik menghasilkan natrium klorida. Hitungkan isi padu klorin yang
diperlukan untuk bertindak balas lengkap. [Jisim atom relatif: Na = 23, isi padu molar gas = 24 dm3 mol–1 pada keadaan bilik]
Answer/Jawapan:
0.24 dm3
The equation shows the combustion of propane gas.
7
Persamaan menunjukkan pembakaran gas propana.
C3H8 + 5O2
3CO2 + 4H2O
720 cm3 of propane gas (C3H8) at room conditions burns in excess oxygen. Calculate the mass of carbon dioxide formed.
[Relative atomic mass: C = 12, O = 16, Molar volume of gas = 22.4 dm3 mol–1 at room conditions]
720 cm3 gas propana (C3H8) pada keadaan bilik terbakar dalam oksigen berlebihan. Hitungkan jisim karbon dioksida yang terbentuk.
[Jisim atom relatif: C = 12, O = 16, isi padu molar gas = 24 dm3 mol–1 pada keadaan bilik]
3.96 g
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Chemistry Form 4 • MODULE
Objective Questions / Soalan Objektif
1
2
The mass of one atom of element Y is two times more
than an atom of oxygen. What is the relative atomic
mass of element Y? [Relative atomic mass: O = 16]
Jisim satu atom unsur Y adalah dua kali lebih daripada
satu atom oksigen. Apakah jisim atom relatif bagi unsur Y?
[Jisim atom relatif: O = 16]
A 12
B 24
C 32
D 36
5
Jadual berikut menunujukkan jisim atom relatif bagi neon,
karbon, oksigen dan kalsium.
Element/Unsur Relative atomic mass/Jisim atom relatif
The chemical formula for butane is C4H10. Which of the
following statements are true about butane?
[Relative atomic mass: H = 1, C =12 and O =16,
Avogadro Constant = 6 × 1023 mol–1]
The empirical formula for butane is CH2.
II
Each butane molecule is made up of 4 carbon atoms
and 10 hydrogen atoms.
6
A I and II only
II, III and IV only
II dan III sahaja
D I, II, III and IV
I, II, III dan IV
7
A bottle contains 3.01 × 1023 of gas particles. What is
the number of moles of the gas in the bottle?
4
C 3.0 mol
D 6.0 mol
Which of the following gases contains 0.4 mol of atoms
at room temperature and pressure? [Molar volume of
gas = 24 dm3 mol–1 at room temperature and pressure]
Antara gas berikut, yang manakah mengandungi 0.4 mol atom
pada suhu dan tekanan bilik? [Isi padu molar gas = 24 dm3 mol–1
pada suhu dan tekanan bilik]
16 g oksigen mengandungi 6.02 × 1023 molekul oksigen
Mass of one oxygen atom is 16 times bigger than
one carbon atom
A bulb is filled with 1 800 cm3 of argon gas at room
conditions. What is the number of argon atom in the
bulb?
[Molar volume of gas = 24 dm3 mol–1 at room conditions,
Avogadro constant = 6.02 × 1023 mol–1]
C 4.8 dm3 CO2
D 4.8 dm3 NH3
C 8.03 × 1022
D 8.03 × 1021
What is the number of hydrogen atom in 0.1 mol of
water? [Avogadro constant: 6.02 × 1023 mol–1]
A 6.02 × 1022
B 60.2 × 1023
8
C
D
6.02 × 1023
3.01 × 1023
5 g of element X reacted with 8 g of element Y to form
a compound with the formula XY2. What is the relative
atomic mass of element X? [Relative atomic mass:
Y = 80]
5 g unsur X bertindak balas dengan 8 g unsur Y membentuk
sebatian dengan formula XY2. Apakah jisim atom relatif unsur X?
[Jisim atom relatif: Y = 80]
A 25
B 40
C 50
D 100
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A 4.8 dm3 Ne
B 4.8 dm3 O2
40
Berapakah bilangan atom oksigen dalam 0.1 mol air?
[Pemalar Avogadro = 6.02 × 1023 mol–1]
Sebuah botol mengandungi 3.01 × 1023 zarah gas. Berapakah
bilangan mol zarah gas dalam botol itu?
A 0.5 mol
B 1.0 mol
Calcium / Kalsium
A 4.515 × 1022
B 4.515 × 1023
II, III dan IV sahaja
3
16
Sebuah belon diisi dengan 1 800 cm3 gas argon pada keadaan
bilik. Berapakah bilangan atom argon dalam belon itu?
[Isipadu molar gas = 24 dm3 mol–1 pada keadaan bilik,
Pemalar Avogadro = 6.02 × 1023 mol–1]
I dan II sahaja
C
Oxygen / Oksigen
Jisim satu atom oksigen adalah 16 kali lebih besar daripada
satu atom karbon
Satu molekul butana mempunyai jisim 84 kali lebih daripada
jisim satu atom hidrogen.
II and III only
12
D
III 1 mol of butane contains a total of 8.4 × 1024
atoms.
B
Carbon / Karbon
molecule
Setiap molekul butana terdiri dari 4 atom karbon dan 10
atom hidrogen.
IV One butane molecule has a mass of 84 times higher
than the mass of 1 hydrogen atom.
20
Antara pernyataan berikut, yang manakah adalah benar?
[Pemalar Avogadro = 6.02 × 1023 mol–1]
A Mass of one calcium atom is 40 g
Jisim satu atom kalsium ialah 40 g
B Mass of 1 mol of neon is 20 g
Jisim 1 mol neon ialah 20 g
C 16 g of oxygen contains 6.02 × 1023 oxygen
Formula empirik butana ialah CH2.
Jumlah bilangan atom dalam 1 mol butana adalah 8.4 ×
1024.
Neon / Neon
Which of the following statements is true?
[Avogadro constant = 6.0 × 1023 mol–1]
Formula kimia bagi butana ialah C4H10. Antara pernyataan
berikut, yang manakah adalah benar tentang butana?
[Jisim atom relatif: H = 1, C = 12 dan O = 16, Pemalar Avogadro =
6 × 1023 mol–1]
I
The table below shows the relative atomic mass of neon,
carbon, oxygen and calcium.
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MODULE • Chemistry Form 4
9
The diagram below shows the set-up of apparatus to
determine the empirical formula of an oxide metal X.
11 The equation shows a decomposition of magnesium
nitrate when heated.
Rajah di bawah menunjukkan susunan radas bagi menentukan
formula empirik oksida logam X.
Persamaan di bawah menunjukkan penguraian nitrat apabila
dipanaskan.
Metal X
Logam X
Heat
Panaskan
Which of the following is metal X?
Antara berikut, yang manakah mungkin bagi logam X?
A Zinc
C
Zink
B
Tin
Stanum
Lead
Plumbum
D Copper
Kuprum
10 The following equation shows the decomposition reaction
of lead(II) nitrate when heated at room temperature and
pressure.
Persamaan tindak balas di bawah menunjukkan penguraian
plumbum(II) nitrat apabila dipanaskan pada suhu dan tekanan
bilik.
2Pb(NO3)2
Antara berikut, yang manakah adalah benar apabila 0.1 mol
plumbum(II) nitrat terurai?
[Jisim atom relatif: N = 14, O = 16, Pb = 207 dan 1 mol gas
menempati isipadu 24 dm3 pada suhu dan tekanan bilik]
I
66.2 g of lead(II) oxide is formed
II
22.3 g of lead(II) oxide is formed
66.2 g plumbum(II) oksida terbentuk
22.3 g plumbum(II) oksida terbentuk
III 2.4 dm3 of oxygen gases is given off
2.4 dm3 gas oksigen dibebaskan
IV 4 800 cm3 of nitrogen dioxide given off
4 800 cm3 nitrogen dioksida dibebaskan
I dan III sahaja
B
I and IV only
C
II and III only
I dan IV sahaja
II dan III sahaja
D II and IV only
m
Berapakah bilangan molekul oksigen apabila 7.4 g magnesium
nitrat terurai apabila dipanaskan?
[Jisim formula relatif Mg(NO3)2 = 148; Pemalar Avogadro =
6.02 × 1023 mol–1]
A
B
C
D
1.505 × 1022
3.010 × 1022
1.505 × 1023
3.010 × 1023
12 The equation below shows the chemical equation of the
combustion of ethanol in excess oxygen.
Persamaan di bawah menunjukkan persamaan kimia pembakaran
etanol dalam oksigen berlebihan.
2C2H5OH + 6O2
4CO2 + 6H2O
What is the volume of carbon dioxide gas released when
9.20 g ethanol burnt completely?
[Relative atomic mass of H = 1, C = 12, O = 16, 1 mol
of gas occupies 24 dm3 at room condition]
Apakah isi padu gas karbon dioksida dibebaskan apabila 9.20 g
etanol terbakar lengkap?
[Jisim atom relatif: H = 1, C = 12, O = 16, 1 mol gas menempati
24 dm3 pada keadaan bilik]
A
B
C
D
4.8 cm3
9.6 cm3
96.0 cm3
9 600 cm3
13 What is the percentage by mass of nitrogen content in
urea, CO(NH2)2? [Relative atomic mass: C = 12, N = 14,
H = 1 and O = 16]
Apakah peratus kandungan nitrogen mengikut jisim dalam urea,
CO(NH2)2? [Jisim atom relatif: C = 12, N = 14, H = 1 dan
O = 16]
A
B
C
D
23.3%
31.8%
46.7%
63.6%
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II dan IV sahaja
2MgO + 4NO2 + O2
What is the number of oxygen molecules is produced
when 7.4 g magnesium nitrate decomposed when
heated.
[Relative formula mass of Mg(NO3)2 = 148;
Avogadro constant = 6.02 × 1023 mol–1]
2PbO + 4NO2 + O2
Which of the following are true when 0.1 mol of lead(II)
nitrate is decomposed?
[Relative atomic mass: N = 14, O = 16, Pb = 207
and 1 mol gas occupies the volume of 24 dm3 at room
temperature and pressure]
A I and III only
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Chemistry Form 4 • MODULE
3
PERIODIC TABLE
JADUAL BERKALA
HISTORICAL DEVELOPMENT / SEJARAH PERKEMBANGAN
– To identify the contribution of scientists in the arrangement of elements in the Periodic Table.
Mengetahui sumbangan ahli sains untuk penyusunan unsur dalam Jadual Berkala.
– To get ideas on the arrangement of elements in the Periodic Table based on their proton numbers.
Mendapat idea penyusunan unsur dalam Jadual Berkala berdasarkan nombor proton.
ARRANGEMENT OF ELEMENT IN THE PERIODIC TABLE / PENYUSUNAN UNSUR DALAM JADUAL BERKALA
• GROUP / KUMPULAN
– To write the electron arrangement for atoms of elements with proton numbers 1 to 20.
Menulis susunan elektron bagi atom unsur dengan proton 1 hingga 20.
• PERIOD / KALA
– To determine the group and period based on the electron arrangement of atoms or otherwise.
Menentukan kumpulan dan kala berdasarkan susunan elektron atom dan sebaliknya.
PROPERTIES OF ELEMENTS IN THE PERIODIC TABLE / SIFAT-SIFAT UNSUR DALAM JADUAL BERKALA
• GROUP 18 / KUMPULAN 18
– To explain the existence of noble gases as monoatom and their uses.
Menerangkan kewujudan gas adi secara monoatom serta kegunaannya.
• GROUP 1 / KUMPULAN 1
– To explain physical properties, similar chemical properties (with water, oxygen and chlorine) and the different reactivities.
Menerangkan sifat fizik, sifat kimia yang sama (dengan air, oksigen dan dengan klorin) serta kereaktifan yang berbeza.
• GROUP 17 / KUMPULAN 17
– To explain physical properties, similar chemical properties (with water, sodium hydroxide and iron) and the different
reactivities.
Menerangkan sifat fizik, sifat kimia yang sama (dengan air, natrium hidroksida dan ferum) serta kereaktifan yang berbeza.
• PERIOD 3 / KALA 3
– To explain changes in atomic size, electronegativity, metallic properties as well as oxide properties across period 3 from left
to right.
Menerangkan perubahan saiz atom, keelektronegatifan, sifat kelogaman serta sifat oksida merentasi Kala 3 dari kiri ke kanan.
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• TRANSITION ELEMENTS / UNSUR PERALIHAN
– To state metallic properties of transition metals and their special characteristics.
Menyatakan sifat kelogaman unsur peralihan serta ciri-ciri istimewa unsur peralihan.
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MODULE • Chemistry Form 4
ADVANTAGES OF CLASSIFYING THE ELEMENTS IN THE PERIODIC TABLE
KEBAIKAN PENGELASAN UNSUR DALAM JADUAL BERKALA
1
Elements are arranged systematically in the Periodic Table in an increasing order of proton number which enables:
Unsur disusun secara sistematik dalam Jadual Berkala mengikut tertib pertambahan nombor proton yang membolehkan:
(a) Chemists to study, understand and remember the chemical and physical properties of all the elements and
compounds in an orderly manner,
Ahli kimia mempelajari, memahami dan mengingat sifat kimia dan sifat fizik semua unsur dan sebatian secara teratur.
(b) Properties of elements and their compounds to be predicted based on the position of elements in the Periodic Table,
Sifat unsur dan sebatiannya diramal berdasarkan kedudukan unsur dalam Jadual Berkala.
(c) Relationship between elements from different groups to be known.
Perhubungan unsur dari kumpulan yang berlainan diketahui.
CONTRIBUTION OF SCIENTIST TO THE HISTORICAL DEVELOPMENT OF THE PERIODIC TABLE
SUMBANGAN AHLI SAINS DALAM SEJARAH PERKEMBANGAN JADUAL BERKALA
Scientists / Saintis
Antoine Lavoisier
Discoveries / Penemuan
– Substances were classified into 4 groups with similar chemical properties.
Bahan dikelaskan kepada empat kumpulan dengan sifat kimia sama.
J.W Dobereiner
– Substances were arranged in 3 groups / Bahan disusun dalam tiga kumpulan.
– Groups with similar chemical properties were called Triads.
Kumpulan dengan sifat kimia sama dinamakan triad.
– Triad system was confined to some elements only / Sistem triad terhad kepada beberapa unsur sahaja.
John Newlands
– Elements were arranged in ascending atomic mass / Unsur disusun mengikut pertambahan jisim atom.
– Law of Octaves because similar chemical properties were repeated at every eighth element.
Hukum Oktaf kerana sifat sama berulang pada setiap unsur kelapan.
– This system was inaccurate because there were some elements with wrong mass numbers.
Sistem ini tidak tepat kerana ada unsur dengan nombor jisim salah.
Lothar Meyer
Mass of 1 mol (g) / Jisim 1 mol (g)
Density (g cm–3) / Ketumpatan (g cm–3)
– Plotted graph for the atomic volume against atomic mass / Memplotkan graf isi padu atom melawan jisim atom.
– Found that elements with similar chemical properties were positioned at equivalent places along the curve.
– The atomic volume / Isipadu atom =
Mendapati unsur dengan sifat kimia sama menduduki tempat setara dalam lengkungan.
Mendeleev
– Elements were arranged in ascending order of increasing atomic mass.
Unsur disusun mengikut pertambahan jisim atom.
– Elements with similar chemical properties were in the same group.
Unsur dengan sifat kimia sama berada dalam kumpulan sama.
– Empty spaces were allocated for elements yet to be discovered.
Ruang kosong disediakan untuk unsur yang belum ditemui.
– Contributor to the formation of the modern Periodic Table.
Penyumbang kepada pambentukan Jadual Berkala Moden.
Henry Moseley
– Classified concepts of proton number and elements in ascending order of increasing proton number.
Mengelaskan unsur berdasarkan konsep nombor proton dan menyusun unsur-unsur mengikut turutan nombor proton menaik.
– Contributor to the formation of the modern Periodic Table.
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Chemistry Form 4 • MODULE
THE ARRANGEMENT OF ELEMENTS IN THE MODERN PERIODIC TABLE
SUSUNAN UNSUR DALAM JADUAL BERKALA MODEN
1
Write the electron arrangement for each atom of element in the Periodic Table below.
Tuliskan susunan elektron untuk setiap atom unsur dalam Jadual Berkala di bawah.
Nucleon number / Nombor nukleon
Proton number / Nombor proton
A
Z
X
Symbol of an element / Simbol unsur
GROUP / KUMPULAN
1
1
P
E
R
I
O
D
/
K
A
L
A
2
3
1
3
Li
11
8
4
19
K
2.8.8.1
11
Be
2.1
Na
13
2
5
2.2
24
12
2.8.1
39
4
2
1
23
3
4
H*
1
7
2
18
LOGAM PERALIHAN
40
20
27
13
2.8.2
3
4
5
6
7
8
9
12
B
6
2.3
TRANSITION METALS
Mg
14
14
C
7
2.4
28
Al
16
16
N
8
2.5
31
Si
14
2.8.3
10 11 12
15
15
2.8.4
32
16
2.8.5
2
17
19
O
9
2.6
P
He
20
F
Ne
10
2.7
35
S
17
2.8.6
40
Cl
80
35
2.8.8.2
Elements in the Periodic Table are arranged horizontally in increasing order of
proton number
.
Unsur-unsur dalam Jadual Berkala disusun secara mendatar mengikut tertib pertambahan
nombor proton
.
Ar
18
2.8.7
Ca
2.8
2.8.8
Br
Two main components of the Periodic Table / Dua komponen utama Jadual Berkala:
(a) Group / Kumpulan
(b) Period / Kala
GROUP / KUMPULAN
1
The vertical column of elements in the Periodic Table arranged according to the number of valance electron in the
outermost shell of atoms is called groups.
Lajur
menegak
petala terluar
2
dalam Jadual Berkala yang disusun mengikut bilangan
elektron valens
yang terdapat pada
bagi atom dipanggil kumpulan.
There are 18 vertical columns, called Group 1, Group 2, and Group 3 until Group 18.
Terdapat 18 lajur disusun secara menegak disebut Kumpulan 1, Kumpulan 2, Kumpulan 3 hingga Kumpulan18.
Number of valence electrons
1
2
3
4
5
6
7
Group
1
2
13
14
15
16
17
Bilangan elektron valens
Kumpulan
8 (except Helium)
8 (kecuali Helium)
18
For atoms of elements with 3 to 8 valence electrons, the group
number is: 10 + number of valence electrons.
Bagi atom unsur dengan 3 hingga 8 elektron valens, nombor kumpulan ialah: 10 +
bilangan elektron valens.
Specific name of groups / Nama-nama khas kumpulan:
(a) Group 1: Alkali metals # / Kumpulan 1: Logam alkali #
(b) Group 2: Alkali-earth metals / Kumpulan 2: Logam alkali bumi
(c) Group 3 to 12: Transition elements # / Kumpulan 3 to 12: Unsur peralihan #
(d) Group 17: Halogens # / Kumpulan 17: Halogen #
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MODULE • Chemistry Form 4
(e) Group 18: Noble gases # / Kumpulan 18: Gas adi #
#The important groups that will be studied with respect to chemical and physical properties.
#
Kumpulan penting yang akan dipelajari dari segi sifat fizik dan sifat kimia.
4
Types of substances according to the groups / Jenis bahan mengikut kumpulan:
(a) Elements of group 1, 2 and 13 – atoms of each element have 1, 2 and 3 valence electrons respectively are metals.
Unsur Kumpulan 1, 2 dan 13 – atom setiap unsur mempunyai 1, 2 dan 3 elektron valens adalah logam.
(b) The elements of group 3 to 12 – transition elements are metals.
Unsur Kumpulan 3 hingga 12 – unsur peralihan yang merupakan logam.
(c) The elements of Group 14, 15, 16, 17 and 18 – atoms of each element have 4, 5, 6, 7 and 8 valence electrons
respectively are non-metals.
Unsur Kumpulan 14, 15, 16, 17 dan 18 – atom setiap unsur mempunyai 4, 5, 6, 7 dan 8 elektron valens adalah bukan logam.
PERIOD / KALA
The horizontal row of elements in the Periodic Table, consists of the same number of
electrons in an
atom
called period.
1
Baris unsur secara
atom
dalam
2
mendatar
dalam Jadual Berkala, mempunyai bilangan
petala
berisi
shell
occupied with
elektron
yang sama di
disebut sebagai kala.
There are seven horizontal rows of elements known as Period 1, 2, ....., 7 [Refer to the Periodic Table]
Terdapat tujuh baris unsur secara mendatar disebut Kala 1, 2, ....., 7 [Rujuk Jadual Berkala]
(a) Period 1 has 2 elements / Kala 1 mengandungi 2 unsur
(b) Period 2 and 3 have 8 elements # / Kala 2 dan 3 mengandungi 8 unsur #
(c) Period 4 and 5 have 18 elements / Kala 4 dan 5 mengandungi 18 unsur
(d) Period 6 has 32 elements / Kala 6 mengandungi 32 unsur
(e) Period 7 has 23 elements / Kala 7 mengandungi 23 unsur
Short periods, # Period 3 will be studied in
detail with respect to physical and chemical
properties / Kala pendek, # Kala 3 akan dipelajari
dengan terperinci dari segi sifat fizik dan sifat kimia
Long periods / Kala panjang
EXERCISE / LATIHAN
1
Complete the table below / Lengkapkan jadual berikut.
Element
Unsur
Proton number
Nombor proton
Electron arrangement
Susunan elektron
Number of
valence electrons
Kumpulan
Group
Number of shell
Period
Bilangan petala
Kala
m
H
1
1
1
1
1
1
He
2
2
2
18
1
1
Li
3
2.1
1
1
2
2
Be
4
2.2
2
2
2
2
B
5
2.3
3
13
2
2
C
6
2.4
4
14
2
2
N
7
2.5
5
15
2
2
O
8
2.6
6
16
2
2
F
9
2.7
7
17
2
2
Ne
10
2.8
8
18
2
2
Na
11
2.8.1
1
1
3
3
Mg
12
2.8.2
2
2
3
3
Al
13
2.8.3
3
13
3
3
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Chemistry Form 4 • MODULE
2
The diagram below shows the chemical symbols which represent elements X, Y and Z.
Rajah di bawah menunjukkan simbol kimia yang mewakili unsur X, Y dan Z.
23
11
X
12
6
Y
39
19
Z
(a) Explain how to determine the position of element X in the Periodic Table.
Terangkan bagaimana menentukan kedudukan unsur X dalam Jadual Berkala.
The proton number of element X is 11 and the number of proton in
electrons in atom X is 11 . The electron arrangement of atom X is
1
because
three shells
atom
2.8.1
atom
X has one valence electron . Element X is in period
occupied with electrons .
X is
11
. The number of
. Element X is located in Group
3 because atom X has
atom
X adalah 11 . Bilangan elektron dalam atom
Nombor proton unsur X adalah 11 dan bilangan proton dalam
11 . Susunan elektron bagi
atom
2.8.1 . Unsur X terletak dalam kumpulan
1
X adalah
kerana
X adalah
atom
satu
elektron
valens
3
atom
tiga
X mempunyai
. Unsur X berada dalam kala
kerana
X mempunyai
berisi
elektron
dengan
.
petala
(b) (i)
State the position of element Y in the Periodic Table. / Nyatakan kedudukan unsur Y dalam Jadual Berkala.
Element Y is located in Group 14 and Period 2.
(ii)
Explain how to determine the position of element Y in the Periodic Table.
Terangkan bagaimana anda menentukan kedudukan unsur Y dalam Jadual Berkala.
– The proton number of element Y is 6 and the number of proton in atom Y is 6.
– The electron arrangement of atom Y is 2.4.
– Element Y is located in Group 14 because atom Y has 4 valance electron.
– Element Y is in Period 2 because atom Y has 2 shells occupied/filled with electrons.
(c) Which of the above elements show the same chemical properties? Explain your answer.
Antara unsur di atas, yang manakah mempunyai sifat kimia yang sama? Terangkan jawapan anda.
– Element X and element Z.
– Electron arrangement of atom X is 2.8.1 and electron arrangement of atom Z is 2.8.8.1. Atoms X and Z have the
same number of valence electron.
GROUP 18 (NOBLE GASES) / KUMPULAN 18 (GAS ADI)
1
Consist of Helium (He), Neon (Ne), Argon (Ar), Krypton (Kr), Xenon (Xe) and Radon (Rn).
Terdiri dari Helium (He), Neon (Ne), Argon (Ar), Kripton (Kr), Xenon (Xe) dan Radon (Rn).
Elements / Unsur
Electron arrangement / Susunan elektron
Helium / Helium
Neon / Neon
Argon / Argon
Krypton / Kripton
2
2
2.8
2.8.8
2.8.18.8
Noble gases are chemically inert because the outermost shell of the atom has achieved duplet electron
arrangement for helium and octet electron arrangement for others.
Unsur Kumpulan 18 adalah lengai secara kimia kerana petala terluar atomnya telah mencapai susunan elektron duplet untuk helium
dan susunan elektron oktet untuk yang lain.
3
Noble gases do not react with other elements (the atom does not lose, gain or share electrons).
Unsur Kumpulan ini tidak bergabung dengan unsur lain (atomnya tidak akan menderma, menerima, atau berkongsi elektron).
4
These gases exist as single uncombined atoms and are said to be monatomic gases.
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Gas ini wujud sebagai atom tunggal iaitu sebagai gas monoatom.
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MODULE • Chemistry Form 4
5
Going down Group 18 / Menuruni Kumpulan 18:
(a) The atomic size is increasing because the number of
Saiz atom bertambah kerana bilangan
petala
shells
increases.
bertambah.
(b) The melting point and boiling points are very low because atoms of noble gases atoms are attracted by
weak
Van der Waals forces, less energy is required to overcome these forces. However, the melting and
boiling points increase going down the group because atomic size increases, causing the Van der Waal forces to
more
increase and
energy is required to overcome these forces.
Takat lebur dan takat didih sangat rendah kerana atom-atom gas adi ditarik oleh daya Van der Waals yang lemah , sedikit tenaga
diperlukan untuk mengatasi daya tersebut. Walau bagaimanapun, takat lebur dan takat didih bertambah menuruni kumpulan kerana
pertambahan saiz atom menyebabkan daya tarikan Van der Waals semakin bertambah, semakin banyak tenaga diperlukan untuk
mengatasinya.
(c) The density is low and increases gradually because the mass increases greatly compared to the volume going down
the group.
Ketumpatan rendah dan semakin meningkat kerana jisim bertambah dengan banyak berbanding dengan isi padu menuruni kumpulan.
6
All noble gases are insoluble in water and cannot conduct electricity in all conditions.
Semua gas adi tidak larut dalam air dan tidak dapat mengkonduksikan elektrik dalam semua keadaan.
7
Complete the uses of noble gases in the table below / Lengkapkan jadual kegunaan gas adi.
Noble gases / Gas adi
Uses / Kegunaan
Helium / Helium
To fill weather balloons and airship.
Neon / Neon
To fill neon light (for advertisement board).
Argon / Argon
To fill electrical bulb.
Krypton / Kripton
To fill photographic flash lamp.
Radon / Radon
To treat cancer.
GROUP 1 (ALKALI METALS) / KUMPULAN 1 (LOGAM ALKALI)
1
Consist of Lithium (Li), Sodium (Na), Potassium (K), Rubidium (Rb), Cesium (Cs) and Francium (Fr).
Terdiri dari Litium (Li), Natrium (Na), Kalium (K), Rubidium (Rb), Sesium (Cs) dan Fransium (Fr).
Elements
Symbol
Proton number
Electron arrangement
Number of shells
Lithium / Litium
Li
3
2.1
2
Sodium / Natrium
Na
11
2.8.1
3
Potassium / Kalium
K
19
2.8.8.1
4
Unsur
2
Simbol
Nombor proton
Susunan elektron
Bilangan petala
Physical properties / Sifat fizik:
(a) Grey solid with shiny surface / Pepejal kelabu dengan permukaan berkilat.
(b) Softer and the density is lower compared to other metals.
Lebih lembut dan ketumpatan yang lebih rendah berbanding dengan logam lain.
(c) Lower melting and boiling points compared to other metals.
Takat lebur dan takat didih lebih rendah berbanding dengan logam lain.
3
Changes in physical properties going down the group / Perubahan sifat fizik menuruni kumpulan:
(a) Atomic size increases because the number of shells increases / Saiz atom bertambah kerana bilangan petala bertambah.
(b) Density increases because mass increases faster than the increase in radius.
Ketumpatan bertambah kerana pertambahan jisim lebih cepat dari pertambahan jejari
(c) Melting and boiling points decrease because when the atomic size increases, the metal bonds get weaker.
Takat didih dan takat lebur berkurang kerana apabila saiz atom bertambah, ikatan logam semakin lemah.
4
Chemical properties of Group 1 elements / Sifat kimia unsur Kumpulan 1:
atoms
1
(a) All
of elements in Group 1 have
valence electron and achieve a stable duplet/octet electron
arrangement by releasing
m
Semua
melepaskan
Publica
electron to form
+1
charged ions:
1
unsur mempunyai
elektron valens dan mencapai susunan elektron oktet/duplet yang stabil dengan
satu
+1
elektron valens membentuk ion bercas
.
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Chemistry Form 4 • MODULE
Example / Contoh:
(i) Atom releases one electron to achieve stable duplet electron arrangement:
Atom litium melepaskan satu elektron untuk mencapai susunan elektron duplet yang stabil:
Li
+e
Electron arrangement / Susunan elektron : 2.1
Number of protons = 3, total charge: +3
Electron arrangement / Susunan elektron : 2
Number of protons = 3, total charge: +3
Bilangan proton = 3,
Bilangan proton = 3,
jumlah cas: +3
jumlah cas: +3
Number of electrons = 3, total charge: –3
Number of electrons = 2, total charge: –2
Bilangan elektron = 3,
Bilangan elektron = 2,
jumlah cas: –3
neutral
Lithium atom is
neutral
Atom litium adalah
(ii)
Li+
Positively
.
charges lithium ion, Li+ is formed.
positif
Ion litium bercas
.
jumlah cas: –2
, Li+ terbentuk.
Sodium atom releases one electron to achieve stable octet electron arrangement:
Atom natrium melepaskan satu elektron untuk mencapai susunan elektron oktet yang stabil:
Na
Na+
Electron arrangement / Susunan elektron : 2.8.1
Number of protons = 11, total charge: +11
Bilangan proton = 11,
Electron arrangement / Susunan elektron : 2.8
Number of protons = 11, total charge: +11
jumlah cas: +11
Bilangan proton = 11,
Number of electrons = 11, total charge: –11
Bilangan elektron = 11,
Atom natrium adalah
neutral
jumlah cas:
+11
–10
Number of electrons = 10, total charge:
jumlah cas: –11
neutral
Sodium atom is
+e
Bilangan elektron = 10,
.
Positively
.
jumlah cas:
–10
charges sodium ion, Na+ is formed.
Ion natrium bercas
positif
, Na+ terbentuk.
atoms
(b) All elements in Group 1 have similar chemical properties because all
in Group 1 have one valence
electron
releasing
electron and achieve the stable duplet/octet
arrangement by
its valence electron to
form a
positively
charged ions.
atom
unsur Kumpulan 1 mempunyai bilangan
Semua unsur Kumpulan 1 mempunyai sifat kimia yang sama kerana semua
elektron
yang stabil dengan melepaskan satu elektron valensnya
elektron valens yang sama iaitu satu dan mencapai susunan
untuk membentuk ion bercas
5
positif
.
The reactivity of alkali metals increases going down the Group 1:
Kereaktifan unsur logam alkali bertambah menuruni Kumpulan 1:
– Atoms of Group 1 metals achieve a stable duplet/octet electron arrangement
one
by releasing
valence electron to form +1 charged ion.
Menuruni Kumpulan 1, bilangan
petala
elektron valens pada petala terluar semakin
bertambah, saiz atom bertambah dan
jauh
dari nukleus.
– The strength of attraction from the proton in the nucleus to the valence
weaker .
elecron gets
Kekuatan tarikan nukleus kepada elektron valens semakin
– The valence electron is loosely held and it is
be released.
Li
Na
K
.
easier
for the electron to
senang
dilepaskan.
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Elektron valens ditarik dengan lemah dan ia makin
lemah
down Group 1
increases, the atomic size
further
increases and the valence electron in the outer most shell gets
away from the nucleus.
increases
shells
menurun Kumpulan 1
– Going down Group 1, the number of
Reactivity
Kereaktifan logam Kumpulan 1 bergantung pada kesenangan atom melepaskan elektron,
semakin senang elektron dilepaskan, kereaktifan logam semakin bertambah .
bertambah
– The reactivity of Group 1 metals depends on the tendency for atoms to
lose electrons; the easier it loses an electron, the reactivity of the metal
increases .
Kereaktifan
Atom logam Kumpulan 1 mencapai susunan elektron gas adi yang stabil dengan
satu
melepaskan
elektron valens membentuk ion bercas +1.
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MODULE • Chemistry Form 4
6
Chemical reactions of Group 1 elements / Tindak balas kimia unsur Kumpulan 1:
(a) Metal Group 1 reacts with water to produce alkali and hydrogen gas.
Logam Kumpulan 1 bertindak balas dengan air menghasilkan alkali dan gas hidrogen.
2X + 2H2O
2X + 2H2O
2XOH + H2, X is the metal of Group 1
2XOH + H2 , X adalah logam Kumpulan 1
Lithium / Litium
Water / Air
Procedure / Kaedah:
(i) Cut a small piece of lithium using a knife and forceps.
Potong sepotong litium menggunakan pisau dan forsep.
(ii)
Dry the oil on the surface of the lithium with filter paper.
Keringkan minyak pada permukaan litium menggunakan kertas turas.
(iii) Place the lithium slowly onto the water surface in a water trough.
Letakkan litium dengan perlahan di atas permukaan air di dalam bekas.
(iv) When the reactions stop, test the solution produced with red litmus paper.
Apabila tindak balas berhenti, uji larutan yang terhasil dengan kertas litmus merah.
(v) Record the observation / Catatkan semua pemerhatian.
(vi) Repeat steps (i) – (v) using sodium and potassium to replace lithium one by one.
Ulang langkah (i) – (v) dengan menggunakan natrium dan kalium menggantikan litium satu demi satu.
Observation / Pemerhatian:
Element
Observation
Li
Lithium moves slowly on the water
red
surface and produces
flame. The
colourless solution formed turns red litmus to
blue .
Unsur
Inference
Pemerhatian
Inferens
perlahan
quickly
Sodium moves
surface and produces
on the water
yellow flame. The
colourless solution formed turns red litmus to
blue .
Natrium bergerak cepat dengan nyalaan
kuning di atas permukaan air. Larutan tidak
berwarna menukarkan kertas litmus merah kepada
biru .
K
Potassium moves
very quickly
on the
yellow
flame.
water surface and produce
The colourless solution formed turns red litmus
blue .
to
sangat cepat
m
lithium hydroxide:
Litium adalah logam yang paling kurang reaktif
bertindak balas dengan air membentuk larutan
beralkali , litium hidroksida.
Balanced equation / Persamaan kimia seimbang:
2Li + 2H2O 2LiOH + H2
Sodium is reactive metal reacts with water
to produce alkaline solution, sodium
hydroxide.
reaktif
bertindak
Natrium adalah logam yang
balas dengan air membentuk larutan beralkali ,
natrium hidroksida.
Balanced equation / Persamaan kimia seimbang:
2Na + 2H2O 2NaOH + H2
the most reactive
metal
alkaline
reacts with water to produce
solution, potassium hydroxide.
Potassium is
Kalium adalah logam yang paling reaktif bertindak
balas dengan air membentuk larutan beralkali ,
kalium hidroksida.
Balanced equation / Persamaan kimia seimbang:
2K + 2H2O 2KOH + H2
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dengan nyalaan
Kalium bergerak
kuning di atas permukaan air. Larutan tidak
berwarna menukarkan kertas litmus merah kepada
biru .
Lithium is the least reactive metal reacts
with water to produce alkaline solution,
Reactivity increases down Group 1
Na
Kereaktifan
Kereaktifan bertambah menuruni Kumpulan 1
dengan nyalaan
Litium bergerak
merah di atas permukaan air. Larutan tidak
berwarna menukarkan kertas litmus merah kepada
biru .
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Chemistry Form 4 • MODULE
(b) Metal Group 1 reacts with oxygen or air to form metal oxide. The metal oxide dissolves in water to produce
alkaline solution.
Logam Kumpulan 1 bertindak balas dengan oksigen membentuk oksida logam. Oksida logam larut dalam air menghasilkan larutan
berakali.
X2O + H2O
X2O + H2O
4X + O2 2X2O
2XOH, X is a metal element of Group 1 (Li, Na and K)
2XOH, X adalah logam unsur Kumpulan 1 (Li, Na dan K)
Combustion spoon / Sudu pembakaran
Gas jar / Balang gas
Chlorine gas / Gas klorin
Burning lithium / Litium menyala
Procedure / Kaedah:
(i) Cut a small piece of lithium using a knife and forceps / Potong secebis kecil litium menggunakan pisau dan forsep.
(ii) Dry the oil on the surface of the lithium with filter paper.
Keringkan minyak pada permukaan litium dengan kertas turas.
(iii) Place the lithium in a combustion spoon and heat lithium until it start to burn.
Letakkan litium pada sudu pembakaraan dan panaskan litium dengan kuat hingga ia menyala.
(iv) Put the burning lithium into a gas jar of oxygen / Letakkan litium yang menyala dalam balang gas berisi oksigen.
(v) When the reaction stop, add water to dissolve the compound formed.
Apabila tindak balas berhenti, tambahkan air untuk melarutkan sebatian yang terbentuk.
(vi) Add a few drops of universal to the solution formed.
Tambahkan beberapa titis penunjuk universal kepada larutan yang terbentuk.
(vii) Record the observation / Catatkan pemerhatian.
(viii) Repeat steps (i) – (vii) using sodium and potassium to replace lithium one by one.
Ulang langkah (i) – (vii) menggunakan natrium dan kalium untuk menggantikan litium satu demi satu.
Observation / Pemerhatian:
Element
Observation
Unsur
Li
Inference
Pemerhatian
Litium adalah paling kurang reaktif terhadap
oksigen.
form
colourless
– Lithium reacts with oxygen to produce
lithium oxide .
soluble in water to
solution.
Litium bertindak balas dengan oksigen
membentuk litium oksida .
Pepejal putih larut dalam air membentuk
tidak berwarna .
larutan
– The solution turns green
indicator to purple .
Balanced equation / Persamaan kimia seimbang:
4Li + O2 2Li2O
universal
Larutan itu menukarkan warna penunjuk
hijau
kepada ungu
universal dari
.
– Lithium reacts with water to form
alkaline solution, lithium hydroxide.
Litium oksida bertindak balas dengan air
membentuk larutan beralkali, litium hidroksida
Reactivity increases down Group 1
– Lithium is the least reactive metal towards
oxygen.
Litium terbakar perlahan dengan nyalaan
merah menghasilkan pepejal putih .
white solid
Kereaktifan
Kereaktifan bertambah menuruni Kumpulan 1
– Lithium burns slowly with a red
flame to produce white solid .
– The
Reactivity
Inferen
.
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Balanced equation / Persamaan kimia seimbang:
Li2O + H2O 2LiOH
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MODULE • Chemistry Form 4
Na
– Sodium burns brightly with a yellow
flame to produce white solid .
terang
Natrium terbakar
kuning
menghasilkan
– The
form
white solid
dengan nyalaan
pepejal putih .
soluble in water to
colourless
solution.
Pepejal putih larut dalam air membentuk
tidak berwarna .
larutan
– The solution turns green
indicator to purple .
universal
Larutan itu menukarkan warna penunjuk
universal dari hijau kepada ungu
.
– Sodium is
reactive
Natrium adalah logam
metal towards oxygen.
reaktif
terhadap oksigen.
– Sodium reacts with oxygen to produce
sodium oxide .
Natrium bertindak balas dengan oksigen
membentuk natrium oksida .
Balanced equation / Persamaan kimia seimbang:
4Na + O2 2Na2O
– Sodium reacts with water to form
alkaline solution, sodium hydroxide.
Natrium bertindak balas dengan air membentuk
larutan beralkali , natrium hidroksida.
Balanced equation / Persamaan kimia seimbang:
Na2O + H2O 2NaOH
K
– Potassium burns very brightly
with a purple flame to produce
– Potassium is the most reactive
towards oxygen.
white solid .
Kalium terbakar sangat terang dengan nyalaan
ungu menghasilkan pepejal putih .
– The
form
white solid
soluble in water to
colourless
solution.
Pepejal putih larut dalam air membentuk
tidak berwarna .
larutan
– The solution turns green
indicator to purple .
universal
Larutan itu menukarkan warna penunjuk
universal dari hijau kepada ungu
.
Kalium adalah logam
oksigen.
paling reaktif
metal
terhadap
– Potassium reacts with oxygen to
produce potassium oxide .
Kalium bertindak balas dengan oksigen membentuk
kalium oksida .
Balanced equation / Persamaan kimia seimbang:
4K + O2 2K2O
– Potassium reacts with water to form
alkaline solution, potassium hydroxide.
Kalium oksida bertindak balas dengan air
membentuk larutan beralkali , kalium hidroksida.
Balanced equation / Persamaan kimia seimbang:
K2O + H2O 2KOH
(c) Metal Group 1 reacts with with chlorine to produce metal chloride.
Logam Kumpulan 1 bertindak balas dengan klorin menghasilkan logam klorida.
2X + Cl2
2X + Cl2 + 2H2O
X is a metal element of Group 1 (Li, Na and K)
2XCl , X adalah logam unsur Kumpulan 1 (Li, Na dan K)
Combustion spoon / Sudu pembakaran
Gas jar / Balang gas
Chlorine gas / Gas klorin
Burning of metal Group 1
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Chemistry Form 4 • MODULE
Observation / Pemerhatian:
Element
Observation
Unsur
Li
Inference
Pemerhatian
slowly
Lithium burns
flame to produce
red
with a
white
Reactivity
Inferen
solid.
Litium terbakar perlahan dengan nyalaan
merah
putih
menghasilkan pepejal
least reactive
– Lithium is
chlorine.
.
Litium adalah
klorin.
Kereaktifan
metal towards
paling kurang reaktif
terhadap
– Lithium reacts with chlorin to produce
lithium chloride .
Litium bertindak balas
litium klorida .
dengan klorin membentuk
Sodium burns brightly with a yellow
flame to produce white solid.
terang
– Sodium is
reactive
metal towards chlorine.
reaktif
Natrium adalah logam
terhadap klorin.
reacts
with chlorine to produce
– Sodium
sodium chloride .
dengan nyalaan
Natrium terbakar
kuning menghasilkan pepejal putih .
Natrium bertindak balas dengan klorin membentuk
natrium klorida .
Balanced equation / Persamaan kimia seimbang:
2Na + Cl2 2NaCl
K
very brightly
with
Potassium burns
a purple flame to produce white
– Potassium is the most reactive metal
towards chlorine.
dengan nyalaan
Kalium terbakar
ungu menghasilkan pepejal putih .
paling reaktif
Kalium adalah logam
klorin.
solid.
sangat terang
Reactivity increases down Group 1
Na
Kereaktifan bertambah menuruni Kumpulan 1
Balanced equation / Persamaan kimia seimbang:
2Li + Cl2 2LiCl
terhadap
– Potassium reacts with chlorine to produce
sodium chloride .
Kalium
bertindak balas
kalium klorida
dengan klorin membentuk
.
Balanced equation / Persamaan kimia seimbang:
2K + Cl2 2KCl
GROUP 17 (HALOGENS) / KUMPULAN 17 (HALOGEN)
1
Consist of Fluorine (F2), Chlorine (Cl2), Bromine (Br2), Iodine (I2) and Astatine (At2).
Terdiri dari Fluorin (F2 ), Klorin (Cl2 ), Bromin (Br2 ), Iodin (I2 ) dan Astatin (At2 ).
Elements
Symbol
Proton number
Fluorine / Fluorin
F2
Chlorine / Klorin
Cl2
Bromine / Bromin
Unsur
Iodine / Iodin
2
Electron arrangement
Number of shells
19
2.7
2
17
2.8.7
3
Br2
35
2.8.18.7
4
I2
53
2.8.18.18.7
5
Simbol
Nombor proton
Susunan elektron
Bilangan petala
Physical properties: Halogens cannot conduct heat and electricity in all state.
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Sifat fizik: Halogen tidak boleh mengkonduksi elektrik dan haba dalam semua keadaan.
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MODULE • Chemistry Form 4
3
Changes in physical properties going down the group / Perubahan sifat fizik menuruni kumpulan:
(a) The melting and boiling points are low because the molecules are attracted by weak Van der Waals forces, and small
amount of energy is required to overcome these forces. However the melting and boiling points increase going down
the group.
Takat didih dan takat lebur adalah rendah kerana molekul ditarik oleh tarikan Van der Waals yang lemah, sedikit tenaga diperlukan
untuk mengatasi daya itu. Walau bagaimanapun, takat lebur dan takat didih meningkat menuruni kumpulan.
Explanation / Penerangan:
–
The atomic size increases
molecules get larger.
bertambah
Saiz atom
–
shell
going down the Group 17 because of increasing in number of
menuruni kumpulan kerana dengan pertambahan bilangan
petala
, the size
, saiz molekul semakin besar.
The inter molecular forces of attraction (Van der Waals forces) between molecules become stronger.
Daya tarikan antara molekul (daya Van der Waals) antara molekul semakin kuat.
–
More heat is needed to overcome the stronger forces between molecules during melting or boiling.
Lebih banyak tenaga diperlukan untuk mengatasi daya antara molekul yang lebih kuat semasa peleburan atau pendidihan.
(b) Physical properties change from gas (fluorine and chlorine) to liquid (bromine) and to solid (iodine) at room
temperature due to increase in the strength of inter molecular forces from fluorine to iodine.
Keadaan fizik berubah dari gas (flourin dan klorin) kepada cecair (bromin) dan kepada pepejal (iodin) pada suhu bilik kerana
pertambahan kekuatan tarikan antara molekul dari flourin kepada iodin.
(c) The density is low and increases going down the group.
Ketumpatan adalah rendah dan semakin meningkat apabila menuruni kumpulan.
darker
(d) The colour of the elements becomes
going down the group: fluorine (light yellow), chlorine (greenish
yellow), bromine (brown) and iodine (purplish black).
Warna unsur semakin
iodin (ungu kehitaman).
4
gelap
menuruni kumpulan iaitu flourin (kuning muda), klorin (kuning kehijauan), bromin (perang) dan
Chemical properties of Group 17 elements / Sifat kimia unsur Kumpulan 17:
atoms
seven
(a) All
of elements in Group 17 have
valence electrons and achieve a stable octet electron
one
negatively
arrangement by accepting
electron to form
charged ions.
atom
Semua
menerima
satu
unsur Kumpulan 17 mempunyai
elektron membentuk ion bercas
Example / Contoh:
(i) Fluorine atom
Atom
elektron valens, mencapai susunan elektron oktet yang stabil dengan
negatif .
receives one electron to achieve stable duplet electron arrangement:
flourin menerima satu elektron untuk mencapai susunan elektron oktet yang stabil:
F
+e
F–
Electron arrangement / Susunan elektron : 2.7
Number of protons = 9,
total charge: +9
Electron arrangement / Susunan elektron : 2.8
Number of protons = 9,
total charge: +9
Bilangan proton = 9,
jumlah cas: +9
Bilangan proton = 9,
Number of electrons = 9,
total charge: –9
Number of electrons = 10, total charge: –10
Bilangan elektron = 9,
jumlah cas: –9
Bilangan elektron = 10,
neutral
Fluorine atom is
Atom flourin adalah
(ii)
tujuh
Negatively
.
neutral
.
jumlah cas: +9
jumlah cas: –10
charged fluoride ion, F– is formed.
Ion flourida, F– bercas
negatif
terbentuk.
Chlorine atom receives one electron to achieve stable octet electron arrangement:
Atom klorin menerima satu elektron untuk mencapai susunan elektron oktet yang stabil:
Cl
+e
Cl–
Electron arrangement / Susunan elektron : 2.8.7
Number of protons = 17, total charge: +17
Electron arrangement / Susunan elektron : 2.8.8
Number of protons = 17, total charge: +17
Bilangan proton = 17,
Bilangan proton = 17,
jumlah cas: +17
jumlah cas: +17
Number of electrons = 17, total charge: –17
Number of electrons = 18, total charge: –18
Bilangan elektron = 17,
Bilangan elektron = 18,
Chlorine atom is
Atom klorin adalah
jumlah cas: –17
neutral
neutral
.
.
Negatively
jumlah cas: –18
charged chloride ion, Cl– is formed.
Ion klorida, Cl– bercas
negatif
terbentuk.
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(b) All elements in Group 17 have similar chemical properties because atoms in Group 17 have seven valence
electron and achieve the stable octet electron arrangement by receiving one electron to form a negatively
charged ion.
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Chemistry Form 4 • MODULE
atom
tujuh
unsur Kumpulan 17 mempunyai
Semua unsur Kumpulan 17 mempunyai sifat kimia yang sama kerana
menerima
elektron valens sama dalam atom, mencapai susunan elektron oktet yang stabil dengan
satu elektron membentuk ion
negatif
.
bercas
5
Reactivity of halogens decreases going down the group / Kereaktifan halogen berkurang menuruni kumpulan:
– All the atoms of Group 17 have seven valence electrons and achieve a stable octet
electron arrangement by accepting one electron to form negatively charged ion.
– Going down Group 17, the number of
Apabila menuruni Kumpulan 17, bilangan
petala
atom menerima elektron.
increases, atomic
bertambah,
saiz
size
increases.
atom bertambah.
jauh
dari nukleus.
– The strength of attraction from the proton in the nucleus to attract one electron into the
outermost occupied shell becomes weaker .
Kekuatan tarikan daripada proton dalam nukleus untuk menarik satu elektron ke dalam petala luar
semakin lemah .
– The strength of a halogen atom to attract electron
astatine (electronegativity decreases).
Kekuatan atom halogen untuk menarik elektron
(keelektronegatifan berkurang).
6
decreases
berkurang
Cl
down Group 17
Petala luar semakin
menurun Kumpulan 17
– Outer shell becomes further from the nucleus.
F
decreases
shells
berkurang
kecenderungan
Kereaktifan unsur Kumpulan 17 bergantung pada
of the atom to receive
Reactivity
tendency
– The reactivity of a halogen atom depends on the
electron.
Kereaktifan
Semua atom unsur Kumpulan 17 mempunyai tujuh elektron valens dan mencapai susunan elektron
oktet yang stabil dengan menerima satu elektron membentuk ion bercas negatif .
Br
from fluorine to
dari fluorin ke astatin
Elements in group 17 exist as diatomic molecules. Two atoms of element sharing one pair of valence electrons to achieve
stable octet electron arrangement.
Unsur Kumpulan 17 wujud sebagai molekul dwiatom. Dua atom unsur berkongsi sepasang elektron valens untuk mencapai susunan elektron
oktet yang stabil.
Example: Two fluorine atoms share one pair of electron to form one fluorine molecule:
Contoh: Dua atom fluorin berkongsi sepasang elektron untuk membentuk molekul fluorin:
F
Share / Kongsi
kongsi
kongsi
Fluorine atom / Atom fluorin
F
F
F
Fluorine Molekul
moleculeflorin
/ Molekul fluorin
Fluorine atom / Atom fluorin
Chlorine, bromine and iodine exists as diatomic molecules. (Cl2, Br2 and I2)
Klorin, bromin dan iodin wujud sebagai molekul dwiatom (Cl2 , Br2 dan I2 )
7
Chemical properties reactions of Group 17 elements / Tindak balas kimia unsur Kumpulan 17:
(a) Halogen reacts with water with different reactivity / Halogen bertindak balas dengan air dengan kereaktifan berbeza:
X2 + H2O
HX + HOX, X is halogen. (Cl2, Br2 and I2) / X2 + H2O
Chlorine gas / Gas Klorin
HX + HOX, X adalah halogen. (Cl2 , Br2 dan I2 )
Bromine water / Air Bromin
Chlorine
Iodine crystals / Hablur Iodin
Bromine
Gas klorin
Bromin
Water
water
Water
air
Air
Procedure / Kaedah:
– Chlorine gas is passed through
water in a test tube.
Gas klorin dilalukan melalui air dalam
tabung uji.
– The solution produced tested with
blue litmus paper.
Larutan yang terhasil diuji dengan kertas
litmus biru.
Water / Air
Procedure / Kaedah:
– A few drops of bromine water are
added to water in a test tube.
Beberapa titis air bromin ditambah
kepada air dalam tabung uji.
– The test tube is shaken.
Tabung uji digoncang.
– The solution produced tested with
blue litmus paper.
Iodine cystals
Hablur iodin
Procedure / Kaedah:
– Some iodine crystals are added to
water in a test tube.
Sedikit hablur iodin ditambah kepada air
dalam tabung uji.
– The test tube is shaken.
Tabung uji digoncang.
– The solution produced tested with
blue litmus paper.
Larutan yang terhasil diuji dengan kertas
litmus biru.
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Larutan yang terhasil diuji dengan kertas
litmus biru.
Air
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MODULE • Chemistry Form 4
Observation / Pemerhatian:
– Chlorine dissolves rapidly in
water to form light yellow solution:
Klorin larut dengan cepat dalam air
menghasilkan larutan berwarna kuning
muda:
Cl2 + H2O
Observation / Pemerhatian:
– Bromine dissolves slowly in
water to form brown solution:
Bromin larut dengan perlahan dalam
air menghasilkan larutan berwarna
perang:
HCl + HOCl
– The solution changes
red
litmus paper to
Br2 + H2O
blue
qucikly decolourises it.
slowly decolourises it.
kepada merah dan
dengan cepat.
kepada merah dan
dengan perlahan.
Larutan menukarkan kertas litmus biru
blue
and
Larutan menukarkan kertas litmus biru
melunturkannya
sedikit
dalam
Iodin larut dengan
air menghasilkan larutan berwarna
perang:
HBr + HOBr
– The solution changes
red
litmus paper to
and
Observation / Pemerhatian:
– Iodine dissolves slightly in
water to form brown solution:
melunturkannya
I2 + H2O
HI + HOI
– The solution changes
red
litmus paper to
blue
. The
litmus paper is not decolourises .
Larutan menukarkan kertas litmus biru
kepada merah . Kertas litmus tidak
dilunturkan
.
Inference / Inferens:
– Chlorine, bromine and iodine react water to form acidic solution.
Klorin, bromin dan iodin bertindak balas dengan air membentuk larutan berasid.
– Solubility decreases from chlorine to iodine / Keterlarutan berkurang dari klorin kepada iodin.
(b) Halogens react with sodium hydroxide solution / Halogen bertindak balas dengan larutan natrium hidroksida:
X2 + 2NaOH
X2 + 2NaOH
NaX + NaOX + H2O, X2 is halogen. (Cl2, Br2 and I2)
NaX + NaOX + H2O, X2 adalah halogen. (Cl2 , Br2 dan I2 )
Complete the following / Lengkapkan yang berikut:
(i)
Cl2 + 2NaOH
(ii)
Br2 + 2NaOH
(iii) I2 + 2NaOH
NaCl + NaOCl + H2O
Reactivity decreases
NaBr + NaOBr + H2O
Kereaktifan berkurang
NaI + NaOI + H2O
(c) Halogens react with hot iron to form brown solid, iron(III) halide.
Halogen bertindak balas dengan besi panas membentuk pepejal perang, ferum(III) halida.
Iron wool / Wul besi
Iodine
Iodin
Chlorine or
Bromine
Klorin atau
Bromin
Iron wool
Heat
Wul besi
Haba
NaOH to absorb chlorine/bromine
NaOH untuk menyerap klorin/bromin
2Fe + 3X2
2Fe + 3X2
2FeX3, X2 represents any halogen. (Cl2, Br2 or I2)
2FeX3, X2 mewakili sebarang halogen. (Cl2 , Br2 atau I2 )
Halogen
Observation
Halogen
Chlorine
Klorin
Bromin
Iron wool burns
when cooled.
very brightly
Iodin
dan membentuk pepejal perang
dan membentuk pepejal
2Fe + 3Cl2
2FeCl3
2Fe + 3Br2
2FeBr3
with a dull glow and forms a
perlahan
dan membentuk pepejal perang
2Fe + 3I2
2FeI3
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and forms a brown solid
sangat terang
Iron wools glows slowly
brown solid when cooled.
Wul besi berbara dengan
apabila sejuk.
m
terang
Persamaan kimia
Iron wool burns brightly and forms a brown solid when
cooled.
Wul besi berbara dengan
perang apabila sejuk.
Iodine
Chemical equation
Pemerhatian
Wul besi terbakar dengan
apabila sejuk.
Bromine
Heat / Haba
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Chemistry Form 4 • MODULE
similar
Experiment (a), (b) dan (c) show that all halogens have
decreases going down the group:
chemical properties but their reactivity
sama
Eksperimen (a), (b) dan (c) menunjukkan semua halogen menunjukkan sifat kimia yang
berkurang apabila menuruni kumpulan.
Reactivity
decreases
tetapi kereaktifannya
berkurang
/ Kereaktifan
F2, Cl2, Br2 and I2 / F2 , Cl2 , Br2 dan I2
PERIOD / KALA
1
2
3
Horizontal rows in the periodic table / Baris mendatar dalam Jadual Berkala.
There are seven periods known as Period 1, 2, 3, 4, 5, 6, 7 / Terdapat 7 kala ditulis sebagai Kala 1, 2, 3, 4, 5, 6, 7.
The number of period of an element represents the number of shells occupy with electrons in each atom of element.
Nombor kala suatu unsur mewakili bilangan petala yang diisi oleh elektron di dalam setiap atom unsur.
Elements
Proton number
Unsur
4
Nombor proton
Number of shells
Period
Susunan elektron
Bilangan petala
Kala
Li
3
2.1
2
2
Na
11
2.8.1
3
3
K
19
2.8.8.1
4
4
Period 3 elements (complete the following table): / Unsur Kala 3 (lengkapkan jadual berikut):
Elements / Unsur
Na
Mg
Al
Si
P
S
Cl
Ar
11
12
13
14
15
16
17
18
2.8.1
2.8.2
2.8.3
2.8.4
2.8.5
2.8.6
2.8.7
2.8.8
3
3
3
3
3
3
3
3
+11
+12
+13
+14
+15
+16
+17
+18
0.191
0.160
0.130
0.118
0.110
0.102
0.099
0.095
Proton number
Nombor proton
Electron arrangement
Susunan elektron
Number of shells
Bilangan petala
Positive charge in the nucleus
Bilangan cas positif dalam nukleus
Radius (nm)
Jejari (nm)
5
Electron arrangement
Physical changes across the Period 3 (from left to right) / Perubahan fizik merentasi Kala 3(dari kiri ke kanan):
(a) Change in atomic radius across Period 3 / Perubahan jejari atom merentasi Kala 3:
The atomic radius of the atoms decreases from sodium to chlorine
berkurang dari natrium kepada klorin
Jejari atom
Na
Mg
Al
Si
S
P
Cl
16 p
Atom / Atom
Na
Mg
Al
Si
P
S
Cl
Number of proton / Bilangan proton
11 p
12 p
13 p
14 p
15 p
16 p
17 p
Positive charge / Cas positif
+11
+12
+13
+14
+15
+16
+17
Electron arrangement / Susunan elektron
2.8.1
2.8.2
2.8.3
2.8.4
2.8.5
2.8.6
2.8.7
–
All the atoms of elements have
3
shells
occupied with electrons
.
petala berisi elektron.
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MODULE • Chemistry Form 4
–
The proton number
increases
by one unit from sodium to chlorine.
Nombor proton bertambah satu unit dari natrium kepada klorin.
–
–
Increase in proton number causes the number of
positive
Pertambahan nombor proton menyebabkan bilangan cas
positif
increase
charge in the nucleus to
.
bertambah .
pada nukleus
increases .
The strength of attraction from the proton in the nucleus to the electrons in the shells
bertambah .
across period 3 / Jejari atom unsur berkurang
Daya tarikan proton dalam nukleus terhadap elektron dalam petala
– The size of atom decreases
(b) Change in electronegativity / Perubahan keelektronegatifan:
–
The atomic radius decreases
sodium to chlorine.
berkurang
Jejari atom
klorin.
–
elektron
berkurang
Keelektonegatifan
bertambah
dari natrium kepada
dari natrium kepada klorin.
increases
Tendency of atoms to attract electron to the outermost shells
The electronegativity
ke arah nukleusnya.
from sodium to chlorine.
bertambah
Kekuatan nukleus menarik elektron kepada petala paling luar
–
towards its nucleus.
due to the increasing nuclei attraction on the electrons in the shells from
kerana daya tarikan nukleus terhadap elektron dalam petala
decreases
The size of atom
Saiz atom
–
electron
Electronegativity: The strength of an atom in a molecule to attract
Kelektronegatifan: Kekuatan suatu atom dalam molekul menarik
–
merentasi Kala 3.
increases
bertambah
from sodium to chlorine.
dari natrium kepada klorin.
across Period 3 from sodium to chlorine.
merentasi Kala 3 dari natrium kepada klorin.
(c) Physical state / Keadaan fizik:
(i) The physical state of elements in a period changes from solid to gas from left to right.
Keadaan fizik unsur-unsur dalam suatu kala berubah dari pepejal kepada gas dari kiri ke kanan.
(ii)
Metals on the left are solid while non-metals on the right are usually gases.
Logam di sebelah kiri adalah pepejal dan bukan logam di sebelah kanan kebanyakannya adalah gas.
(d) Changes in metallic properties and electrical conductivity / Perubahan sifat kelogaman dan kekonduksian elektrik:
Na
Element / Unsur
Al
Si
P
S
Cl
Ar
Metallic properties
Semi metal
Non-metal
Sifat kelogaman
Logam
Separa logam
Bukan logam
Electrical conductivity
Good conductors
of electric.
Weak conductor of electric but it increases with
the presence of boron or phosphorous.
Cannot conduct electricity
Konduktor elektrik
yang baik.
Konduktor elektrik yang lemah tetapi bertambah dengan
kehadiran boron atau fosforus.
Uses: semi-conductor / Kegunaan: semi konduktor
Kekonduksian elektrik
6
Mg
Metal
Tidak boleh mengkonduksi
elektrik
Changes in properties of oxide of elements Period 3 / Sifat oksida unsur Kala 3:
Na
Mg
Basic oxide / Oksida bes
Basic oxide + Water
Oksida bes + Air
Alkali
Alkali
Oksida bes + Asid
Si
Amphoteric oxide + Acid
Oksida amfoterik + Asid
Oksida amfoterik + Alkali
Salt + Water
Garam + Air
Example / Contoh:
MgO + 2HCl MgCl2 + H2O
Salt + Water
Garam + Air
Amphoteric oxide + Alkali
Example / Contoh:
Na2O + H2O 2NaOH
Basic oxide + Acid
Al
Amphoteric oxide / Oksida amfoterik
P
Acidic oxide + Water
Oksida asid + Air
Salt + Water
Garam +Air
Example / Contoh:
Al2O3 + 6HNO3 2Al(NO3)3 +3H2O
Al2O3 + 2NaOH 2NaAlO2 + H2O
S
Cl
Acidic oxide / Oksida asid
Acid
Asid
Example / Contoh:
SO2 + H2O H2SO3
Acidic oxide + Alkali
Oksida asid + Alkali
Salt + Water
Garam + Air
Example / Contoh:
SiO2 + 2NaOH Na2SiO3 + H2O
(a) Elements in Period 3 can be classified as metals and non-metals based on basic and acidic properties of their
oxides / Unsur Kala 3 boleh dikelaskan sebagai logam dan bukan logam berdasarkan sifat kebesan dan keasidan oksidanya.
acid
salt
(i) Basic oxide is metal oxide that can react with
to form
and water .
m
asid
membentuk
garam
dan
air
.
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Oksida bes adalah oksida logam yang boleh bertindak balas dengan
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Chemistry Form 4 • MODULE
(ii)
Acidic oxide is non-metal oxide that can react with
alkali
to form
Oksida asid adalah oksida bukan logam yang boleh bertindak balas dengan
air
.
membentuk
to form
and
alkali
Oksida amfoterik adalah oksida yang boleh bertindak balas dengan
air
.
dan
asid
dan
water .
and
alkali
acid
(iii) Amphoteric oxide is oxide that can react with both
salt
alkali
garam
salt
dan
and
water .
garam
untuk membentuk
(b) Complete the following table / Lengkapkan jadual berikut:
(i) Reaction with water / Tindak balas dengan air:
Oxide
Solubility in water
Oksida
Sodium oxide, Na2O
pH
Type of oxide
pH larutan
Jenis oksida
14
Basic oxide
9
Basic oxide
–
–
–
–
3
Acidic oxide
3
Acidic oxide
White solid dissolve in water
Natrium oksida, Na2O
Pepejal putih larut dalam air
Magnesium oksida, MgO
White solid slightly dissolve in water
Magnesium oksida, MgO
Pepejal putih larut separa dalam air
Aluminium oxide, Al2O3
Insoluble
Aluminium oksida, Al2O3
Tidak larut
Silicon oxide, SiO2
Insoluble
Silikon oksida, SiO2
Tidak larut
Phosphorous oxide, P4O10
White solid dissolve in water
Fosforus oksida, P4O10
Pepejal putih larut dalam air
Sulphur dioxide, SO2
White solid dissolve in water
Sulfur dioksida, SO2
(ii)
Keterlarutan dalam air
Pepejal putih larut dalam air
Reaction between the oxide of elements Period 3 with nitric acid and sodium hydroxide solution:
Tindak balas antara oksida unsur Kala 3 dengan asid nitrik dan larutan natrium hidroksida:
Observation / Pemerhatian
Oxide
Oksida
Reaction with dilute nitric
acid
Reaction with sodium
hydroxide solution
The white solid dissolve to
form colourless solution.
No change. The white solid does
not dissolve.
Tindak balas dengan asid nitrik cair
Magnesium oxide, MgO
Magnesium oksida, MgO
Pepejal putih larut membentuk
larutan tanpa warna.
Aluminium oxide, Al2O3
Aluminium oksida, Al2O3
Silicon oxide, SiO2
Tiada perubahan. Pepejal putih tidak
larut.
No change. The white solid
does not dissolve.
Tiada perubahan. Pepejal putih tidak
larut.
Jenis oksida
Tindak balas dengan natrium hidroksida
Tiada perubahan. Pepejal putih tidak
larut.
The white solid dissolve to form
colourless solution.
Pepejal putih larut membentuk larutan
tanpa warna.
The white solid dissolve to form
colourless solution.
Pepejal putih larut membentuk larutan
tanpa warna.
Basic oxide
Amphoteric oxide
Acidic oxide
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Silikon oksida, SiO2
No change. The white solid
does not dissolve.
Type of oxide
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MODULE • Chemistry Form 4
7
Steps to compare and explain the change in atomic size/ radius/ electronegativity across Period 3, reactivity down
Group 1 and Group 17:
Langkah-langkah untuk membanding dan menerangkan perubahan saiz atom/ jejari/ keelekronegatifan merentasi Kala 3, kereaktifan
menuruni Kumpulan 1 dan Kumpulan 17:
16 p
Na
K
(a) To Compare Atomic Size/ Radius and Electronegativity Across Period 3:
Membanding Jejari/ Saiz Atom dan Keelektronegatifan Merentasi Kala 3:
(i) Compare number of shells in each atom.
Bandingkan bilangan petala dalam setiap atom.
(ii) Compare number of proton in the nucleus.
Bandingkan bilangan proton dalam nukleus.
(iii) Compare the strength of attraction from the nucleus to the electrons in the shells .
Bandingkan kekuatan tarikan dari proton dalam nukleus kepada elektron dalam petala .
(iv) Compare the atomic size/ Compare the electronegativity.
Bandingkan saiz atom/ Bandingkan keelektronegatifan.
(b) To Compare Reactivity Down Group 1 and Group 17:
Membanding Kereaktifan Menuruni Kumpulan 1 dan Kumpulan 17:
(i)
Compare number of shells in each atom.
Bandingkan bilangan petala dalam setiap atom.
(ii)
Compare the strength of proton in the nucleus to attract valence electron (Group 1)// to attract
electron to the outermost shells (Group 17).
Bandingkan kekuatan proton dalam nukleus menarik elektron valens (Kumpulan 1) // menarik
elektron ke petala paling luar (Kumpulan 17).
(iii) Compare tendency of the atom to release electron (Group 1)// receive electron (Group 17).
Bandingkan kecenderungan atom untuk melepaskan elektron (Kumpulan 1) // menerima
elektron (Kumpulan 17).
Reactivity decreases down Group 17/Kereaktifan berkurang menurun Kumpulan 17
Reactivity increases down Group 1/Kereaktifan bertambah menurun Kumpulan 1
Li
Atomic radius of the atoms decreases across Period 3 from sodium to chlorine.
Jejari atom berkurang merentasi Kala 3 dari natrium kepada klorin.
F
Cl
Br
TRANSITION ELEMENT / UNSUR PERALIHAN
1
Situated between Group 2 and 13. The examples of transition element are Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu and Zn.
Terletak antara Kumpulan 2 dan 13. Contoh unsur peralihan adalah Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu dan Zn.
2
Show metal properties: Shiny, conducts heat and electricity, malleable, high tensile strength, high melting point and
density.
Mempunyai sifat logam: Permukaan berkilat, konduktor haba dan elektrik, bersifat mulur, boleh ditempa, kekuatan tegangan yang tinggi,
takat lebur dan ketumpatan tinggi.
Special characteristics / Ciri istimewa:
(a) Form coloured compound / Membentuk sebatian berwarna.
Example / Contoh:
3
Iron(III) chloride is brown, iron(II) chloride is green and copper(II) sulphate is blue.
Ferum(III) klorida adalah perang, ferum(II) klorida adalah hijau dan kuprum(II) sulfat adalah biru.
(b) Form different oxidation numbers / Membentuk nombor pengoksidaan berbeza.
(c) Form complex ions: MnO4–, Cr2O72–, CrO42–, etc / Membentuk ion kompleks: MnO4–, Cr2O72–, CrO42–, dan sebagainya.
(d) Useful as a catalyst in industries / Berguna sebagai mangkin dalam industri.
Example / Contoh:
Iron: Haber Process in the manufacture of ammonia / Ferum: Proses Haber dalam penghasilan ammonia.
N2 + 3H2
Fe
2NH3
Vanadium(V) Oxide: Contact Process in the manufacture of sulphuric acid.
Vanadium(V) Oksida: Proses Sentuh dalam penghasilan asid sulfurik.
m
Platinum: Ostwald Process in the manufacture of nitric acid / Platinum: Proses Ostwald dalam penghasilan asid nitrik.
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Chemistry Form 4 • MODULE
EXERCISE / LATIHAN
1
The diagram below shows the electron arrangement for atoms P and Q.
Rajah di bawah menunjukkan susunan elektron bagi atom P dan Q.
P
Q
(a) Elements P and Q are placed in the same group in Periodic Table. State the group.
Unsur P dan Q terletak dalam kumpulan yang sama dalam Jadual Berkala. Nyatakan kumpulan itu.
Group 1
(b) How is elements P and Q kept in the laboratory? Give reason for your answer.
Bagaimanakah unsur P dan Q disimpan di dalam makmal? Berikan sebab bagi jawapan anda.
In paraffin oil. To prevent them from reacting with oxygen or water vapour in the atmosphere.
(c) (i)
Write chemical equation for the reaction between elements P with water.
Tuliskan persamaan kimia untuk tindak balas antara unsur P dengan air.
2P + 2H2O
(ii)
2POH + H2
What is the expected change of colour when a few drops of phenolphthalein are added into the aqueous
solution of the product? Explain your answer.
Apakah perubahan warna yang dijangkakan apabila beberapa titik fenolftalein dititiskan ke dalam larutan akueus yang
terhasil? Terangkan jawapan anda.
Colourless to purple/ pink. The solution formed is alkaline.
(iii) Between element P and element Q, which is more reactive in the reaction with water?
Antara unsur P dan Q, yang manakah lebih reaktif apabila bertindak balas dengan air?
Element Q is more reactive than P.
(iv) Explain your answer in (c)(iii) / Terangkan jawapan anda dalam (c)(iii).
The size of atom Q is larger than atom P.
The valence electron of atom Q is further away from the nucleus compared to atom P.
The attraction forces between proton in the nucleus to the valence electron of atom Q is weaker than atom P.
Atom Q is easier to release the valence electron compared to atom P.
(d) Name one element that has the same chemical properties as P and Q.
Namakan satu elemen yang mempunyai ciri-ciri kimia yang sama dengan P dan Q.
Potassium
2
The diagram below shows the information regarding elements W and X which are from the same group in the Periodic
Table.
Rajah di bawah menunjukkan maklumat mengenai unsur W dan X yang terletak di kumpulan yang sama dalam Jadual Berkala.
19
9
(a) (i)
(ii)
W
35
17
X
Write the electron arrangement of atom of elements W and X / Tuliskan susunan elektron bagi atom unsur W dan X.
2.7
2.8.7
Atom W / Atom W :
Atom X / Atom X :
State the position of elements W and X in the Periodic Table.
Nyatakan kedudukan unsur W dan X dalam Jadual Berkala.
Element W / Unsur W : Group 17, Period 2
Element X / Unsur X : Group 17, Period 3
(iii) Do elements W and X show the same chemical property? Explain your answer.
Adakah unsur W dan X menunjukkan sifat kimia yang serupa? Terangkan jawapan anda.
Elements W and X have the same chemical property. Atoms W and X have the same number of valence
electrons.
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(b) State the type of particles of substances W and X / Nyatakan jenis zarah yang terdapat pada W dan X.
Molecule
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MODULE • Chemistry Form 4
(c) (i)
Compare the boiling point of elements W and X. Explain the difference.
Bandingkan takat didih unsur W dan X. Terangkan perbezaan itu.
The boiling point of element X is higher than element W.
The size of molecule X2 is bigger than molecule W2 .
The forces of attraction between molecules X2 is stronger than molecule W2.
More heat energy is needed to overcome the stronger forces between molecules.
(d) (i)
Element X can react with sodium element to form a compound. Write the chemical equation for the reaction.
Unsur X boleh bertindak balas dengan unsur natrium untuk membentuk sebatian. Tulis persamaan kimia untuk tindak balas
tersebut.
X2 + 2Na
(ii)
2NaX
How does the reactivity of element W and element X differ? Explain your answer.
Bagaimanakah kereaktifan unsur W dan X berbeza? Terangkan jawapan anda.
Element W is more reactive than element X.
The size of atom W is smaller than atom X.
The outermost occupied shell of atom W is nearer to the nucleus compare to atom X.
The strength of the nucleus of atom W to attract electron to the outermost shell is stronger than atom X.
3
The table below shows the number of neutron and relative atomic mass of eight elements represented as P, Q, R, S, T, U
and W.
Jadual di bawah menunjukkan bilangan neutron dan jisim atom relatif bagi lapan unsur yang diwakili oleh huruf P, Q, R, S, T, U, V dan W.
Atom / Unsur
Number of neutron
Bilangan neutron dalam atom
Relative atomic mass
Jisim atom relatif
Number of proton
Bilangan proton dalam atom
Electron arrangement
Susunan elektron dalam atom
P
Q
R
S
T
U
V
W
12
12
14
14
16
16
18
22
23
24
27
28
31
32
35
40
11
12
13
14
15
16
17
18
2.8.1
2.8.2
2.8.3
2.8.4
2.8.5
2.8.6
2.8.7
2.8.8
(a) Complete the above table by writing the number of proton and electron arrangement for the atom of each element.
Lengkapkan jadual di atas dengan menulis bilangan proton dan susunan elektron bagi atom setiap unsur.
(b) (i)
State the period of element P – W in the Periodic Table. Explain your answer.
Nyatakan kala manakah unsur P – W terletak dalam Jadual Berkala? Terangkan jawapan anda.
Period 3 because P – W atoms have three shells occupied with electrons.
(ii)
What is the proton number of another element that is in the same group as P?
Nyatakan bilangan proton bagi unsur lain yang sama kumpulan dengan P.
3/19
(c) Write the standard representation for element Q / Tuliskan simbol perwakilan piawai untuk unsur Q.
24
Q
12
(d) Which element exist as / Unsur yang manakah wujud sebagai
W
monoatomic gas / gas monoatom?
(e) (i)
diatomic gas / gas dwiatom?
T/ U/ V
Which element can react vigorously with water to produce hydrogen gas?
Unsur yang manakah bertindak balas cergas dengan air untuk menghasilkan gas hidrogen?
P
m
Write the balanced equation for the reaction in (e)(i) / Tuliskan persamaan seimbang untuk tindak balas (e)(i).
2P + 2H2O
2POH + H2
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Chemistry Form 4 • MODULE
(f)
State the arrangement of elements T, U and V in the order of increasing atomic radius. Explain your answer.
Nyatakan susunan unsur T, U dan V dalam tertib pertambahan jejari atom. Terangkan jawapan anda.
V, U and T. Atoms of T, U, and V have three shells occupied with electrons.
The proton number // positive charges in the nucleus increases from T to V.
The forces of attraction between proton in the nucleus and the electrons in the shells increase from T to V.
The shells filled with electrons are pulled nearer to the nucleus from T to V.
4
The diagram below shows part of the Periodic Table of Elements. X, Y, A, B, D, E, F and G do not represent the actual
symbols.
Rajah di bawah menunjukkan sebahagian daripada Jadual Berkala Unsur. X, Y, A, B, D, E, F dan G tidak mewakili simbol yang sebenar.
X
Y
A
F
(a) (i)
(ii)
B
D
E
G
State the position of element B in the Periodic Table / Nyatakan kedudukan unsur B dalam Jadual Berkala.
Period 3, Group 13
Explain your answer in (a)(i) / Terangkan jawapan anda dalam (a)(i).
Electron arrangement atom B is 2.8.3. Atom B has three valence electrons, element B is in Group 13. Atom B
has three shells occupied with electrons, element B is in Period 3.
(b) (i)
(ii)
Which element is monatomic gas / Unsur yang manakah adalah gas monoatom?
Element Y
Explain your answer in (b)(i) / Terangkan jawapan anda dalam (b)(i).
Atom Y has achieved octet electron arrangement // has electron arrangement 2.8.
(c) Choose an element that / Pilih unsur yang:
(i)
exists in the form of molecule / wujud dalam bentuk molekul
(ii)
forms acidic oxide / membentuk oksida asid
(iii)
has atoms that have no neutron / atom yang tiada neutron
(iv)
is an alkali metal / logam alkali
(v)
forms amphoteric oxide / membentuk oksida amfoterik
B
(vi)
has a proton number of 15 / mempunyai nombor proton 15
C
(vii) is most electropositive / paling elektropositif
(viii) forms basic oxide / membentuk oksida bes
(ix)
forms coloured compound / membentuk sebatian berwarna
X/D/E
D/E
X
A/F
F
A/F
G
(d) Arrange Y, A, B, D and E according to the order of increasing atomic size.
Susun Y, A, B, D dan E mengikut tertib pertambahan saiz atom.
Y, E, D, B, A
(e) (i)
(ii)
Write the electron arrangement for an atom of element / Tulis susunan elektron bagi atom unsur:
2.8.5
2.8.7
D:
E:
Compare electronegativity of elements D and E / Bandingkan keelektronegatifan unsur D dan E.
Element E is more electronegative than element D.
(iii) Explain your answer in (e)(ii) / Terangkan jawapan anda dalam (e)(ii).
Atoms E and D have the same number of shells occupied with electrons. The number of proton in the nucleus
of atom E is more than atom D. The strength of proton in nucleus to attract electrons to the outermost shells
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in atom E is stronger than of atom D.
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MODULE • Chemistry Form 4
Objective Questions / Soalan Objektif
1
Proton number of element P is 8. What is the position of this
element in the Periodic Table of Elements?
Period / Kala
16
2
A
2
B
16
3
C
18
2
D
18
3
Potassium reacts with element Q from Group 17 in Periodic
Table. Which of the following chemical equations is correct?
Kalium bertindak balas dengan unsur Q dalam Kumpulan 17 dalam
Jadual Berkala Unsur. Antara persamaan kimia berikut, yang manakah
betul?
A K + Q KQ
B K+ + Q – KQ
3
C 2K + Q2 2KQ
D K + Q2 KQ2
Y oxide
Z oxide
A
Amphoteric
Acidic
Basic
Amfoterik
Asid
Bes
B
Amphoteric
Basic
Acidic
Amfoterik
Bes
Asid
C
Acidic
Amphoteric
Basic
Asid
Amfoterik
Bes
D
Acidic
Acidic
Basic
Asid
Asid
Bes
Oksida X
Nombor proton unsur P adalah 8. Apakah kedudukan unsur P dalam
Jadual Berkala Unsur?
Group/Kumpulan
X oxide
6
Jadual di bawah menunjukkan sifat oksida unsur X, Y dan Z yang berada
dalam Kala 3 Jadual Berkala Unsur.
X
Y
Z
Antara pernyataan berikut, yang manakah benar?
I
II
A
B
C
D
5
Calcium / Kalsium
III Potassium / Kalium
Sulphur / Sulfur
IV Nitrogen / Nitrogen
I and II only / I dan II sahaja
I and III only / I dan III sahaja
II and IV only / II dan IV sahaja
III and IV only / III dan IV sahaja
The diagram below shows the standard representation for
elements X, Y and Z.
– Oxide of Z reacts with sodium hydroxide solution.
Oksida Z bertindak balas dengan larutan natrium
hidroksida.
– Oxide of Z reacts with nitric acid.
A All the elements can conduct electricity.
Antara berikut, yang manakah dapat membentuk oksida asid?
Oksida Y bertindak balas dengan larutan natrium
hidroksida.
Oksida Y tidak bertindak balas dengan asid nitrik.
Z
Which of the following elements can form acidic oxide?
– Oxide of Y reacts with sodium hydroxide solution.
– Oxide of Y does not react with nitric acid
Which of the following statements is true?
4
Oksida X bertindak balas dengan asid nitrik.
natrium hidroksida.
Y
D
– Oxide of X reacts with nitric acid.
– Oxide of X does not react with sodium hydroxide
solution./Oksida X tidak bertindak balas dengan larutan
Rajah di bawah menunjukkan kedudukan unsur X, Y dan Z dalam Jadual
Berkala Unsur.
C
Property of the oxide formed
Sifat oksida yang terbentuk
Element
Unsur
The diagram below shows the position of elements X, Y and Z
in the Periodic Table.
B
Oksida Z
The table below shows the properties of the oxide of elements
X, Y and Z which are located in Period 3 of the Periodic Table.
X
Semua unsur boleh mengkonduksi elektrik.
All the elements exist as gas at room temperature.
Semua unsur wujud dalam bentuk gas pada suhu bilik.
The boiling points of the elements increase from
X
Y
Z.
Takat didih unsur bertambah dari X → Y → Z.
The density of the elements decreases going down from
X
Y
Z.
Ketumpatan unsur berkurang dari X → Y → Z.
Oksida Y
Oksida Z bertindak balas dengan asid nitrik.
What is the correct arrangement of elements X, Y and Z from
left to right in Period 3 of the Periodic Table?
Apakah susunan yang betul bagi unsur X, Y dan Z dari kiri ke kanan Kala
3 Jadual Berkala Unsur?
A Z, X, Y
B X, Z, Y
7
C X, Y, Z
D Y, Z, X
The following statements describe the characteristic of an
element:
Pernyataan berikut menerangkan sifat suatu unsur.
– Used as a catalyst / Digunakan sebagai mangkin.
– Forms coloured ions or compound.
Membentuk ion atau sebatian berwarna.
– Shows different oxidation number in its compound.
Menunjukkan numbor pengoksidaan yang berbeza.
Which of the following is the position of the element in the
Periodic Table of Element?
Antara berikut, yang manakah adalah kedudukan unsur tersebut dalam
Jadual Berkala Unsur?
Rajah di bawah menunjukkan simbol unsur X, Y dan Z.
27
13
X
32
16
Y
23
11
Z
What type of oxides are formed by X, Y and Z?
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A B
C
D
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Chemistry Form 4 • MODULE
8
The table below shows the proton number of elements in
Period 3 of the Periodic Table of Elements.
Jadual di bawah menunjukkan nombor proton unsur dalam Kala 3 Jadual
Berkala Unsur.
10 The table below shows the proton numbers of elements X and
Y.
Jadual di bawah menunjukkan nombor proton unsur X dan Y.
Elements / Unsur
Proton number / Nombor proton
Jejari (nm)
X
11
11
0.191
Y
19
Mg
12
0.160
Which statements are true about elements X and Y?
Al
13
0.130
Antara pernyataan berikut, yang manakah benar tentang unsur X dan
Y?
Si
14
0.118
I
P
15
0.110
S
16
0.102
Cl
17
0.099
Ar
18
0.095
Elements
Proton number
Radius (nm)
Na
Unsur
Nombor proton
II
III
IV
Why does the atomic radius of the atoms decrease from sodium
to argon in the period?
Mengapakah saiz atom berkurang dari natrium ke argon dalam kala?
A The number of valence electrons increases.
Bilangan elektron valens bertambah.
The electronegativity of the elements increases.
Keelektronegatifan unsur bertambah.
The properties of the elements change from metallic to
non-metallic.
Sifat unsur berubah dari logam kepada bukan logam.
The strength of attraction of the nucleus to the electrons
in the shells increases.
Kekuatan tarikan nukleus kepada elektron dalam petala
bertambah.
B
C
D
9
A
B
C
D
Atoms X and Y have one valence electron.
Atom X dan Y mempunyai satu elektron valens.
Elements X is more reactive than element Y.
Unsur X lebih reaktif daripada unsur Y.
Atom X has a bigger atomic size than atom Y.
Saiz atom X lebih besar daripada saiz atom Y.
Elements X and Y are in the same group in the Periodic
Table.
Unsur X dan Y berada dalam kumpulan sama dalam Jadual
Berkala.
I and III only / I dan III sahaja
I and IV only / I dan IV sahaja
II and III only / II dan III sahaja
II and IV only / II dan IV sahaja
The table below shows proton number for elements P, Q and
R.
Jadual di bawah menunjukkan nombor proton bagi unsur P, Q dan R.
Elements / Unsur
Proton number / Nombor proton
P
11
Q
17
R
19
Which of the following statements about these elements are
true?
Antara pernyataan berikut, yang manakah benar tentang unsur-unsur
tersebut?
I
II
III
IV
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A
B
C
D
P and R has the lowest number of valence electrons.
P dan R mempunyai bilangan elektron valens paling rendah.
P and R have similar chemical properties.
P dan R mempunyai sifat kimia yang sama.
Size of atom R is bigger than size of atom Q.
Saiz atom R lebih besar daripada saiz atom Q.
Element R is more electronegative than element Q.
Unsur R lebih elektronegatif daripada unsur Q.
I, II and III / I, II dan III
I, II dan IV / I, II dan IV
I, III dan IV / I, III dan IV
II, III dan IV / II, III dan IV
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MODULE • Chemistry Form 4
CHEMICAL BOND
4
IKATAN KIMIA
TYPE OF CHEMICAL BOND / JENIS IKATAN KIMIA
• IONIC BOND / IKATAN ION
– To predict the formulae of ionic compounds based on the electron arrangement.
Meramal formula sebatian ion berdasarkan susunan elektron.
– To describe the formation of ionic bond / Menghuraikan pembentukan ikatan ion.
– To draw the diagram of the formation of ionic bond / Melukis rajah pembentukan ikatan ion.
• COVALENT BONDS / IKATAN KOVALEN
– To predict the formulae of molecules of elements or covalent compounds as well as the types of covalent bond.
Meramal formula molekul unsur atau molekul sebatian kovalen serta jenis ikatan kovalen.
– To describe the formation of covalent bonds / Menghuraikan pembentukan ikatan kovalen.
– To draw the diagram of the formation of covalent bonds / Melukis rajah pembentukan ikatan kovalen.
PROPERTIES OF IONIC AND COVALENT COMPOUNDS / SIFAT SEBATIAN ION DAN KOVALEN
• IONIC COMPOUNDS / SEBATIAN ION
– To state and explain the properties from the aspect of melting point and electrical conductivity in solid and molten state.
Menyatakan dan menerangkan sifat dari segi takat lebur, kekonduksian elektrik dalam keadaan pepejal dan leburan.
• COVALENT COMPOUNDS / SEBATIAN KOVALEN
– To state the solubility in water and organic solvents / Menyatakan keterlarutan dalam air dan pelarut organik.
– To differentiate between ionic and covalent compounds / Membezakan sebatian ion dengan sebatian kovalen.
CHEMICAL BONDS BETWEEN ATOMS / IKATAN KIMIA ANTARA ATOM
Chemical bonds are formed when two or more atoms of elements bond together. Atoms form chemical bonds to achieve
a stable duplet or octet electron arrangement. There are two types of chemical bond, that is Ionic Bond and Covalent
Bond.
1
Ikatan kimia dibentuk apabila dua atau lebih atom-atom unsur berpadu. Atom-atom membentuk ikatan kimia untuk mencapai susunan
elektron yang stabil iaitu susunan elektron duplet atau oktet. Terdapat dua jenis ikatan kimia iaitu Ikatan Ion dan Ikatan Kovalen.
IONIC BOND / IKATAN ION
Ionic bond is formed between atoms of metal elements that release electrons to atoms of non-metal elements.
1
Ikatan ion terbentuk antara atom unsur logam yang melepaskan elektron kepada atom unsur bukan logam yang menerima elektron.
Atom of an element is neutral because the number of protons is equal to the number of electrons.
2
Atom suatu unsur adalah neutral kerana bilangan proton adalah sama dan dengan bilangan elektron.
Atoms of elements that release the electrons form positive ions and atoms that receive the electrons form negative ions
to achieve a stable octet or duplet electron arrangement:
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susunan elektron oktet atau duplet yang stabil.
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Chemistry Form 4 • MODULE
Complete the following table / Lengkapkan jadual di bawah:
Changes
Na
Na+ + e
Ca
Electron
arrangement
2.8.1
2.8
2.8.2
Total of positive
charges (From
number of proton)
+11
+11
Total of negative
charges (From
number of proton)
–11
Perubahan
Ca2+ + 2e
O + 2e
O2–
Cl + e
Cl–
2.8
2.6
2.8
2.8.7
2.8.8
+12
+12
+8
+8
+17
+17
–10
–12
–10
–8
–10
–17
–18
0
+1
0
+2
0
–2
0
–1
Sodium
atom
Sodium ion
Calcium
atom
Oxygen
atom
Oxide ion
Chlorine
atom
Susunan elektron
Jumlah cas positf
(Dari bilangan proton)
Jumlah cas negaitf
(Dari bilangan proton)
Total changes
Jumlah cas
Type of particles
Jenis zarah
Atom natrium
3
Atom kalsium
Calcium
ion
Atom oksigen
Atom klorin
Chlorine
ion
The positive ions and negative ions are attracted to one another with strong electrostatic forces. The electrostatic force
between the positive and negative ions forms ionic bond.
Ion positif dan ion negatif tertarik antara satu sama lain dengan daya elekrostatik yang kuat. Daya elektrostatik antara ion positif dan ion
negatif membentuk ikatan ion.
4
Ionic bond is usually formed between atoms from Groups 1, 2 and 13 (metal) with atoms from Groups 15, 16 and 17
(non-metal).
Ikatan ion biasanya dibentuk antara atom-atom daripada Kumpulan 1, 2 dan 13 (logam) dengan atom-atom dari Kumpulan 15, 16 dan 17
(bukan logam).
5
The maximum number of electrons transferred in the formation of ionic bond is usually three:
Bilangan maksimum elektron yang berpindah dalam pembentukan ikatan ion biasanya tiga.
(a) Atoms of elements in Groups 1, 2 and 13 release 1, 2 and 3 electrons respectively to form positively charged ions
(+1, +2 and +3).
Atom unsur Kumpulan 1, 2 dan 13 masing masing melepaskan 1, 2 dan 3 elektron membentuk ion bercas positif (+1, +2 dan +3).
(b) Atoms of elements in Groups 15, 16 and 17 receive 3, 2 and 1 electrons respectively to form negatively charged
ions (–3, –2 and –1)
Atom unsur Kumpulan 15, 16 dan 17 masing-masing menerima 3, 2 dan 1 elektron membentuk ion bercas negatif (–3, –2 dan –1).
6
Examples / Contoh-contoh:
(i) Sodium chloride / Natrium klorida
Predict the formula / Ramal formula:
Element
Proton number
Electron arrangement
Na
11
2.8.1
Cl
17
2.8.7
Unsur
Nombor proton
Susunan elektron
Na
Cl + e
Na+ + e
Cl–
Na+
Cl–
1
1
NaCl
Draw the electron arrangement of the compound formed.
Lukiskan susunan elektron bagi setiap sebatian yang terbentuk.
Transfer
Pindah
Na
Na
C1
Cl
Na
Na
C1
Cl
Sodium
atom, Na
Atom
natrium,
Na
Chlorine
atom,Cl
Cl
Atom klorin,
Sodium
ion, Na
Ion
natrium,
Na +
Chloride
ion,Cl
Cl–
Ion
klorida,
–
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MODULE • Chemistry Form 4
Explanation / Penerangan:
2.8.1
(a) Electron arrangement of sodium atom is
stable
Therefore sodium atom is not
one
. Sodium atom has
one
. Sodium atom releases
octet electron arrangement to form sodium ion , Na+ with electron arrangement
valence electron.
electron to achieve a stable
2.8
.
2.8.1 . Atom natrium mempunyai
satu
elektron valens. Dengan itu atom
Susunan elektron atom natrium ialah
stabil . Atom natrium melepaskan
satu
elektron ini untuk mencapai susunan elektron oktet yang
natrium tidak
stabil membentuk ion
natrium , Na+ dengan susunan elektron
2.8
.
2.8.7
seven
(b) Electron arrangement of chlorine atom is
. Chlorine atom has
valence electrons.
one
Chlorine atom receives
electron to achieve stable octet electron arrangement to form
chlorine
2.8.8
ion, Cl– with an octet arrangement of electron
2.8.7
Susunan elektron bagi atom klorin ialah
.
tujuh
. Atom klorin mempunyai
elektron valens. Atom klorin
satu
elektron membentuk ion klorida , Cl– dengan
mencapai susunan elektron oktet yang stabil dengan menerima
2.8.8 .
susunan elektron
(c)
Sodium ions , Na+ and chloride ions , Cl– ions are attracted with strong
bond formed is called ionic bond.
Ion natrium , Na+ dan
ikatan ion.
electrostatic
force. The
ion klorida , Cl– ditarik dengan daya elektrostastik yang kuat. Ikatan yang terbentuk dinamakan
(ii) Magnesium oxide / Magnesium oksida
Predict the formula / Ramal formula:
Element
Proton number
Electron arrangement
Mg
12
2.8.2
O
8
2.6
Unsur
Nombor proton
Susunan elektron
Mg
Mg+ + 2e
O + 2e
O2–
Draw the electron arrangement of the compound formed.
Mg2+
O2–
2
2
1
1
MgO
Lukiskan susunan elektron bagi setiap sebatian yang terbentuk.
2+
Mg
Mg
Pindah
Transfer
Magnesium
atom, Mg
Atom
magnesium,
Mg
O
O
2−
O
O
Mg
Mg
Oxygen
atom, OO
Atom oksigen,
Magnesium
ion, Mg
Mg 2+
Ion
magnesium,
2+
2–
2−
Oxide
ion, OO
Ion oksida,
Explanation / Penerangan:
2.8.2
(a) The electron arrangement of magnesium atom is
. Magnesium atom has
stable
electrons. Therefore magnesium atom is not
. Magnesium atom releases
electrons to achieve a stable octet electron arrangement to form
2.8
arrangement
.
magnesium ion , Mg
2+
two
valence
2
valence
with electron
2.8.2 . Atom magnesium mempunyai
dua
elektron di petala terluar. Maka atom
Susunan elektron atom magnesium
stabil
dua
. Atom magnesium melepaskan
elektron valens untuk mencapai susunan elektron oktet
magnesium tidak
yang stabil membentuk
ion magnesium , Mg2+ dengan susunan elektron
2.8
.
2.6
(b) The electron arrangement of oxygen atom is
. Oxygen atom is also unstable. Oxygen atom
receives two electrons to achieve a stable octet electron arrangement to form oxide ion , O2– with
electron arrangement
2.8
Susunan elektron atom oksigen ialah
.
2.6
. Atom oksigen juga tidak stabil, atom oksigen
m
ion oksida
menerima
, O dengan susunan elektron
dua elektron
2.8
.
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untuk mencapai susunan elektron oktet yang stabil membentuk
2–
d.
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Chemistry Form 4 • MODULE
(c)
Magnesium ion , Mg2+ and
formed is called ionic bond.
Ion magnesium , Mg2+ dan
dinamakan ikatan ion.
oxide ion , O2– are attracted by strong
ion oksida
electrostatic
force. The bond
, O2– ditarik dengan daya elektrostatik yang kuat. Ikatan yang terbentuk
(iii) Magnesium chloride /Magnesium klorida
Predict the formula / Ramal formula:
Element
Proton number
Electron arrangement
Mg
12
2.8.2
Cl
17
2.8.7
Unsur
Nombor proton
Mg
Cl + e
Susunan elektron
Mg2+ + 2e
Cl–
Mg2+
Cl–
1
2
MgCl2
Draw the electron arrangement of the compound formed.
Lukiskan susunan elektron bagi setiap sebatian yang terbentuk.
2+
Transfer
Pindah
Transfer
Pindah
Mg
C1
Chlorine
atom,
Atom klorin,
Cl Cl
C1
Magnesium
atom,Mg
Mg
Atom magnesium,
Mg
C1
Chlorine
atom,
Atom
klorin,
Cl Cl
C1
Chlorine
ion,ClCl– Ion
Magnesium
ion,Mg
Mg2+2+ Ion
Chlorine
ion,
Ion klorida,
magnesium,
klorida,
ClCl–
Explanation / Penerangan:
2.8.2
(a) The electron arrangement of magnesium atom is
2
. Magnesium atom has
stable
. Magnesium atom releases
in the outer shell. Therefore, magnesium atom is not
valence electrons to achieve a stable octet electron arrangement to form
2.8
electron arrangement
.
electrons
2
magnesium ion , Mg2+ with
2.8.2 . Atom magnesium mempunyai
2
elektron di petala terluar. Maka atom
Susunan elektron atom magnesium
stabil
2
. Atom magnesium melepaskan
elektron valens untuk mencapai susunan elektron oktet
magnesium tidak
ion magnesium , Mg2+ dengan susunan elektron
yang stabil membentuk
2.8
.
2.8.7
(b) The electron arrangement of chlorine atom is
. Chlorine atom is also unstable. Chlorine atom
receives one electron to achieve a stable octet electron arrangement to form chloride ion , Cl– with
electron arrangement
2.8.8
Susunan elektron atom klorin ialah
.
2.8.7
ion klorida , Cl dengan susunan elektron
mencapai susunan elektron oktet yang stabil membentuk
(c) As such,
Oleh itu,
one
satu
Daya
elektrostatik
–
magnesium atom releases
atom magnesium melepaskan
(d) Strong electrostatic
ionic bond.
menerima
. Atom klorin juga tidak stabil. Atom klorin
2
force is formed between
yang kuat terbentuk antara
2
2
electrons to
elektron kepada
2
2.8.8
.
chlorine atoms.
atom klorin.
magnesium ion , Mg
ion magnesium , Mg2+ dan
satu elektron untuk
2+
and
chloride ion , Cl– to form
ion klorida , Cl– membentuk ikatan ionik.
COVALENT BONDS / IKATAN KOVALEN
1
This bond is formed when two or more similar or different atoms share valence electrons between them, so that each
atom achieves the octet or duplet electron arrangement that is a stable electron arrangement for noble gases.
Ikatan ini terbentuk apabila dua atau lebih atom yang sama atau berlainan berkongsi elektron valens antara satu sama lain supaya setiap
atom mencapai susunan elektron oktet atau duplet iaitu susunan elektron gas adi yang stabil.
2
Normally, this bond is formed when similar or different non-metal atoms bond together. [Atoms from Groups 14, 15, 16
and 17]
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Ikatan ini biasanya terbentuk apabila atom-atom bukan logam berpadu. [Atom-atom dari Kumpulan 14, 15, 16 dan 17]
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MODULE • Chemistry Form 4
When atoms of non-metals share their valence electrons from their outermost shells to achieve stable duplet or octet
sharing
atoms
electron arrangement, covalent bonds are formed. The product of the
of electrons between
3
is called
molecule .
Apabila atom-atom bukan logam berkongsi elektron pada petala terluar untuk mencapai susunan elektron duplet atau oktet yang stabil,
ikatan kovalen terbentuk. Hasil daripada perkongsian elektron antara atom-atom ini membentuk
molekul
.
The molecules are neutral as there is no electron transfer involved. During the formation of covalent bonding ,
each atom contributes same number of electrons for sharing. The number of electrons shared can be one pair, two pairs
or three pairs.
4
neutral
kerana tidak melibatkan pemindahan elektron. Semasa pembentukan ikatan kovalen , setiap atom akan
Molekul adalah
menyumbang bilangan elektron yang sama untuk dikongsi. Bilangan elektron yang dikongsi boleh jadi sepasang, dua pasang atau tiga
pasang.
The forces that exist between molecules are Van der Waals forces that are weak. These forces become stronger when
the molecule size increases.
5
Daya yang wujud antara molekul adalah
daya Van der Waals
yang lemah. Daya ini semakin kuat apabila saiz molekul bertambah.
Examples / Contoh:
(i) Hydrogen molecule / Molekul hidrogen:
(a) Hydrogen atom has one electron in the first shell, with an electron arrangement of 1 needs one electron to
achieve a stable duplet electron arrangement.
6
Atom hidrogen mempunyai satu elektron pada petala pertama dengan susunan elektron 1 memerlukan satu elektron untuk
mencapai susunan elektron duplet yang stabil.
(b)
Two hydrogen atoms share a pair of electrons to form a hydrogen molecule.
Dua atom hidrogen berkongsi sepasang elektron membentuk satu molekul hidrogen.
(c)
Both hydrogen atoms achieve a stable duplet arrangement of electron.
Kedua-dua atom hidrogen mencapai susunan elektron duplet yang stabil.
Draw the electron arrangement of the compound formed / Lukiskan susunan elektron bagi sebatian yang terbentuk.
Share
Kongsi
Kongsi
Share
H
H
H
H
H
H
The number of electron pairs shared is
one
pair. Single covalent bond is formed.
Bilangan pasangan elektron dikongsi adalah
satu
pasang. Ikatan kovalen tunggal terbentuk.
(ii) Oxygen molecule / Molekul oksigen:
(a)
Oxygen atom with an electron arrangement 2.6 needs two electrons to achieve a stable
arrangement.
Atom oksigen dengan susunan elektron 2.6 memerlukan dua elektron untuk mencapai susunan elektron
(b)
Two oxygen atoms share
octet
oktet
electron
yang stabil.
two
pairs of electrons to achieve a stable octet arrangement of electron, form
an oxygen molecule. Each oxygen atom achieves stable octet electron arrangement.
dua
pasang elektron untuk mencapai susunan elektron oktet yang stabil, membentuk satu
oktet
yang stabil.
molekul oksigen. Setiap atom oksigen mencapai susunan elektron
Dua atom oksigen berkongsi
Draw the electron arrangement of the compound formed / Lukiskan susunan elektron bagi sebatian yang terbentuk.
O
O
Kongsi
Share
m
Oxygen
atom, OO
Atom
oksigen,
O
O
O
Oxygen molecule,
O22
Molekul
oksigen, O
The number of electron pairs shared is
2
pairs. Double covalent bond is formed.
Bilangan pasangan elektron dikongsi adalah
2
pasang. Ikatan kovalen ganda dua terbentuk.
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Oxygen
atom, OO
Atom
oksigen,
O
O
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Chemistry Form 4 • MODULE
(iii) Nitrogen molecule / Molekul nitrogen:
(a)
3
Atom nitrogen dengan susunan elektron 2.5 memerlukan
stabil.
(b)
3
Nitrogen atom with an electron arrangement 2.5 needs
arrangement.
electrons to achieve stable
elektron untuk mencapai susunan elektron
octet
oktet
yang
3
pairs of electrons to achieve a stable octet arrangement, form a
nitrogen molecule. Each nitrogen atom achieves stable octet electron arrangement.
Two nitrogen atoms share
3
oktet
pasang elektron untuk mencapai susunan elektron
oktet
yang stabil.
satu molekul nitrogen. Setiap atom nitrogen mencapai susunan elektron
Dua atom nitrogen berkongsi
yang stabil membentuk
Draw the electron arrangement of the compound formed / Lukiskan susunan elektron bagi sebatian yang terbentuk.
Share
Kongsi
N
Nitrogen
atom, N
N
Atom
nitrogen,
N
N
N
Nitrogen
atom, NN
Atom
nitrogen,
N
N
Nitrogen
Molekulmolecule,
nitrogen,NN2 2
The number of electron pairs shared is
3
pairs. Triple covalent bond is formed.
Bilangan pasangan elektron dikongsi adalah
3
pasang. Ikatan kovalen ganda tiga terbentuk.
(iv) Hydrogen chloride molecule /Molekul hidrogen klorida
Predict the formula / Ramal formula:
Element
Proton number
Electron arrangement
Susunan elektron
H
H
1
1
Cl
Cl
17
2.8.7
Unsur
Nombor proton
needs
1 electron
perlu
1 elektron
needs 1 electron
1 elektron
perlu
Cross the number of
electrons each atom needs:
HCl
Silangkan bilangan elektron yang
diperlukan oleh setiap atom: HCl
Draw the electron arrangement of the compound formed.
Lukiskan susunan elektron bagi setiap sebatian yang terbentuk.
H
Share
Kongsi
Hydrogen
atom, H
H
Atom
hidrogen,
H
H
Cl
C1
Chlorine
atom,
Atom
klorin,
ClCl
Cl
C1
Hydrogen
chloride molecule,
HCl
Molekul hidrogen
klorida, HCl
Explanation / Penerangan:
(a)
Hydrogen atom with an electron arrangement
duplet electron arrangement.
Atom
hidrogen dengan susunan elektron
1
1
memerlukan
needs
satu
one
electron to achieve a stable
elektron untuk mencapai susunan elektron
duplet yang stabil.
(b)
Chlorine atom with an electron arrangement 2.8.7 needs
electron arrangement.
Atom
klorin dengan susunan elektron 2.8.7 memerlukan
yang stabil.
(c)
One
chloride
Satu
one
chlorine atom share
molecule with the formula
atom klorin berkongsi
HCl
electron to achieve stable
elektron untuk mencapai susunan elektron
pair of electrons with
HCl .
pasang elektron dengan
one
satu
octet
oktet
hydrogen atom to form hydrogen
atom hidrogen membentuk
molekul
.
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hidrogen klorida dengan formula
satu
satu
one
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MODULE • Chemistry Form 4
(d)
One
chlorine atom contributes
electron for sharing.
Satu
satu
atom klorin menyumbang
dikongsi bersama.
One
(e)
Satu
(f)
Chlorine
duplet
Atom
one
chlorine atom forms
one
atom klorin membentuk
satu
atom
hydrogen atom contributes one
atom hidrogen menyumbang satu elektron untuk
single covalent bond with
one
hydrogen atom.
ikatan kovalen tunggal dengan
satu
atom hidrogen.
electron arrangement and hydrogen
atom
achieves stable
electron arrangement.
oktet
klorin mencapai susunan elektron
duplet
satu
elektron dan
octet
achieves stable
one
electron and
atom
yang stabil dan
hidrogen mencapai susunan elektron
yang stabil.
(v) Water molecule /Molekul air
Predict the formula / Ramal formula:
Element
Proton number
Electron arrangement
Susunan elektron
H
H
1
1
O
O
8
2.6
Unsur
Nombor proton
needs
1 electron
perlu
1 elektron
needs 2 electrons
2 elektron
perlu
Cross the number of
electrons each atom needs:
H2O
Silangkan bilangan elektron yang
diperlukan oleh setiap atom: H2O
Draw the electron arrangement of the compound formed.
Lukiskan susunan elektron bagi setiap sebatian yang terbentuk.
H
Kongsi
Share
Hydrogen
atom, H
Atom
hidrogen,
Kongsi
Share
O
O
Oxygen
atom, OO
Atom
oksigen,
H
H
H
H
Hydrogen
atom, H
Atom
hidrogen,
O
O
H
H
Water
molecule,
O
Molekul
air, HHO
2
2
Explanation / Penerangan:
(a)
Hydrogen atom with an electron arrangement
electron arrangement.
Atom
1
hidrogen dengan susunan elektron
1
memerlukan
needs electron to achieve a stable
satu
duplet
elektron untuk mencapai susunan elektron
duplet yang stabil.
(b)
(c)
Oxygen
octet
atom
with an electron arrangement
two
needs
electrons to achieve stable
electron arrangement.
Atom
oksigen dengan susunan elektron
oktet
yang stabil.
One
oxygen atom share
molecule with the formula
2.6
two
Satu
atom oksigen berkongsi
air dengan formula H2O .
(d)
2.6
memerlukan
dua
pairs of electrons with
H2O
.
dua
pasang elektron dengan
One
two
oxygen atom contributes
electron for sharing to form single
elektron untuk mencapai susunan elektron
two
dua
hydrogen atoms form water
atom hidrogen membentuk
molekul
electrons and each of the two hydrogen atoms contributes one
covalent bond.
Satu
dua
atom oksigen menyumbang
elektron dan setiap satu daripada dua atom hidrogen menyumbang satu
dikongsi
bersama membentuk ikatan kovalen tunggal.
elektron untuk
(e)
One
Satu
(f)
Oxygen
duplet
m
atom oksigen membentuk
atom
achieves stable
single covalent bonds with
dua
ikatan kovalen tunggal dengan
octet
two
dua
hydrogen atoms.
atom hidrogen.
electron arrangement and hydrogen
atom
achieves
electron arrangement.
oksigen mencapai susunan elektron
oktet
yang stabil dan
atom
hidrogen mencapai susunan elektron
duplet yang stabil.
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Atom
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Chemistry Form 4 • MODULE
(vi) The molecule formed between carbon and chlorine /Molekul yang terbentuk antara karbon dan klorin
Predict the formula / Ramal formula:
Element
Proton number
Electron arrangement
Susunan elektron
C
C
6
2.4
Cl
Cl
17
2.8.7
Unsur
Nombor proton
needs
Cross the number of electrons
each atom needs: CCl4
4 electrons
perlu
4 elektron
Silangkan bilangan elektron yang
diperlukan oleh setiap atom: CCl4
1 electron
needs
1 elektron
perlu
Draw the electron arrangement of the compound formed.
Lukiskan susunan elektron bagi setiap sebatian yang terbentuk.
Cl
Cl
Cl
Cl
Explanation / Penerangan:
(a)
(b)
(c)
Carbon
octet
atom
karbon dengan susunan elektron
oktet
yang stabil.
atom
2.4
empat
memerlukan
with an electron arrangement
klorin dengan susunan elektron
oktet
yang stabil.
2.8.7
2.8.7
memerlukan
satu
four
carbon atom share
pairs of electrons with
CCl4 .
tetrachloromethane molecule with the formula
atom karbon berkongsi empat
tetraklorometana berformula CCl4 .
four
elektron untuk mencapai susunan elektron
one
needs
One
One
electrons to achieve a stable
electron to achieve a stable
electron arrangement..
Atom
Satu
(d)
four
needs
electron arrangement.
Atom
Chlorine
octet
2.4
with an electron arrangement
pasang elektron dengan
elektron untuk mencapai susunan elektron
four
empat
four
carbon atom contributes
electrons and each of the
electron for sharing to form single covalent bond.
chlorine atoms to form
atom klorin membentuk
four
chlorine atoms contributes
Satu
satu
atom karbon menyumbang empat elektron dan setiap
daripada empat
satu
elektron untuk dikongsi bersama membentuk ikatan kovalen tunggal .
menyumbang
(e)
One
Satu
(f)
carbon atom forms
atom karbon membentuk
Carbon and chlorine
atoms
empat
single covalent bonds with
ikatan kovalen tunggal dengan
achieve stable
octet
karbon dan atom klorin mencapai susunan elektron
four
empat
atom klorin
chlorine atoms.
atom klorin.
electron arrangement.
oktet
yang stabil.
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Atom
four
molekul
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MODULE • Chemistry Form 4
7
Comparing the Formation of Ionic and Covalent Bonds / Perbandingan Pembentukan Ikatan Ion dan Kovalen
Ionic Bond / Ikatan Ion
Covalent Bond / Ikatan Kovalen
Type of
element
involved
Between metals (Groups 1, 2 and 13) and
non-metals (Groups 15, 16 and 17).
Electron
Electron is released by metal atoms and received
by non-metal atoms (electron transfer).
Jenis unsur
terlibat
Elektron
Jenis zarah
yang dihasilkan
How to
predict the
formulae
Antara bukan logam
14, 15, 16 dan 17).
logam
(Kumpulan 1, 2 dan 13) dengan
Antara
bukan logam (Kumpulan 15, 16 dan 17).
Elektron dilepaskan oleh atom logam dan
atom bukan logam (elektron berpindah).
Type of
particle
produced
Between non-metal
14, 15, 16 and 17).
diterima
oleh
Metal atom forms positive ion.
Non-metal atom forms negative ion.
positif .
Atom logam membentuk ion
negatif
Atom bukan logam membentuk ion
different
dengan
non-metals
(Groups
bukan logam
(Kumpulan
of electrons are shared
non-metal atoms.
Pasangan elektron dikongsi
sama atau berlainan.
by the same or
oleh atom-atom bukan logam
Neutral molecule .
Molekul
yang neutral.
.
Determine the coefficient of the charge of the ions and
criss cross.
Tentukan pekali cas pada ion dan silangkan.
Determine the number of electrons is needed to achieve
stable duplet or octet electron arrangement and criss
cross.
Tentukan bilangan elektron yang diperlukan untuk mencapai
susunan elektron duplet atau oktet yang stabil dan silangkan.
Bagaimana
meramal
formula
Example
of electron
arrangement
in the
particles
Pairs
and
+
A
2–
E
+
A
Contoh susunan
elektron dalam
zarah
Strong electrostatic forces between ions
Daya elektrostatik yang kuat antara ion
Strong covalent bond between atoms in the
molecules
Example of
ionic and
covalent
compounds
m
# Covalent bond is the shared pairs of electrons between
atoms in a molecule.
# Ikatan kovalen terhasil daripada perkongsian pasangan
elektron antara atom-atom dalam molekul.
Lead(II) bromide, PbBr2
Naphthalene, C8H10
Sodium chloride, NaCl
Acetamide, CH3CONH2
Copper(II) sulphate, CuSO4
Hexane, C6H14
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Contoh sebatian
ion dan kovalen
Ikatan kovalen yang kuat antara atom dalam molekul
# Ionic bond is the strong electrostatic force of attraction
between positively charged ion and negatively charged
ion.
# Ikatan ion terhasil daripada daya tarikan elektrostatik yang
kuat antara ion bercas positif dan ion bercas negatif.
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Chemistry Form 4 • MODULE
PHYSICAL PROPERTIES OF IONIC AND COVALENT COMPOUND
SIFAT FIZIK SEBATIAN ION DAN KOVALEN
Ionic compound / Sebatian ion
Example
Covalent compound / Sebatian kovalen
Sodium chloride, NaCl / Natrium klorida, NaCl
Carbon dioxide, CO2 / Karbon dioksida, CO2
Contoh
Weak Van
der Waals
forces
between
molecules
Daya Van der
Waals yang
lemah antara
molekul
Strong electrostatic forces between positive and
negative ions
Daya elektrostatik yang kuat antara ion
Strong covalent bond between atoms in the
molecules
Ikatan kovalen yang kuat antara atom dalam molekul
Type of forces
between
particles
Strong electrostatic force between ions.
Melting and
boiling points
–
Weak Van der walls forces (intermolecular force)
between molecule.
Daya elekrostatik yang kuat antara ion.
Daya Van der Waals yang lemah antara molekul.
Jenis daya antara
zarah
Takat lebur dan
takat didih
High
melting and boiling points because
positive ions and negative ions are attracted
by strong electrostatic force .
Takat lebur dan takat didih tinggi kerana ion positif
dan ion negatif ditarik oleh daya tarikan elektrostatik
yang kuat.
–
Large amount of energy is needed to
overcome it.
Banyak tenaga haba diperlukan untuk
– Low melting and boiling points because of
the weak “Van der Waals” force between molecules.
Takat lebur/takat didih rendah kerana daya
"Van der Waals" yang lemah antara molekul.
–
Small
amount of energy is needed to
overcome it.
Sedikit
mengatasinya .
tenaga haba diperlukan untuk
mengatasinya .
– Giant molecules such as silicon dioxide have very
high melting and boiling points.
Molekul raksaksa seperti silikon dioksida mempunyai takat
didih dan lebur yang amat tinggi.
Electrical
conductivity
Kekonduksian
elektrik
– Cannot conduct electricity when in
solid
but is able to conduct electricity when in
or aqueous form.
form
molten
Tidak boleh mengkonduksi elektrik dalam keadaan pepejal
tetapi boleh mengkonduksi elektrik dalam keadaan leburan
atau akueus .
– In solid form, the ions are not
Dalam bentuk pepejal, ion-ion tidak
bergerak .
free
bebas
to
move .
untuk
–
Cannot
conduct electricity in all state.
Tidak boleh
keadaan.
mengkonduksi elektrik dalam semua
– Covalent compound is made up of neutral
molecules .
Sebatian kovalen terdiri daripada
molekul
yang neutral.
– No free moving ions in molten or aqueous state.
Tidak ada ion bebas bergerak dalam keadaan leburan atau
akueus.
– In molten or aqueous state, the ions are free to
move to be attracted to the anode or cathode.
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MODULE • Chemistry Form 4
Ionic compound / Sebatian ion
Solubility
Keterlarutan
– Most are soluble
organic solvent*.
Kebanyakannya larut
dalam pelarut organik*
Covalent compound / Sebatian kovalen
in water and insoluble in
Insoluble in water but soluble in organic
solvents* (example: ether, alcohol, benzene,
tetrachloromethane and propanone). This is because
covalent molecules and organic solvents are both
held together by weak Van der Waals forces.
–
dalam air tetapi tidak larut
– This is because the polarisation of water molecule.
Water molecules have partially positive end (the
hydrogen end) and partially negative end (the
oxygen end).
Tidak larut dalam air tetapi larut dalam pelarut
organik* (contoh: eter, alkohol, benzena, tetraklorometana
dan propanon). Ini kerana molekul kovalen dan pelarut
organik ditarik oleh daya tarikan Van der Waals yang lemah.
* Organic solvents are covalent compounds that exist
as liquid at room temperature.
* Pelarut organik adalah sebatian kovalen yang wujud
dalam bentuk cecair pada suhu bilik.
Ini kerana air adalah molekul yang berkutub. Molekul
air mempunyai bahagian bercas separa positif (bahagian
hidrogen) dan bahagian bercas separa negatif (bahagian
oksigen).
EXERCISE / LATIHAN
The Table below shows the proton number of elements D, E, F, G, J and L.
1
Jadual di bawah menunjukkan nombor proton bagi unsur D, E, F, G, J dan L.
Element / Unsur
D
E
F
G
J
L
Proton number / Nombor proton
1
6
17
11
18
8
(a) Which element in the table are metal and non-metal / Unsur yang manakah merupakan logam dan bukan logam?
(i) Metal / Logam : G
(ii) Non-metals / Bukan logam : D, E, F, J, L
(b) State an element that exists as monoatomic gas. Explain your answer.
Nyatakan unsur yang wujud sebagai gas monoatom. Terangkan jawapan anda.
Element J, Atom J has 8 electrons in the outermost shell, the atom has achieved stable octet electron arrangement.
(c) Write the formula for the ion formed from an atom of element L.
Tuliskan formula ion yang terbentuk daripada atom unsur L.
L2–
(d) Element E reacts with element L to form a compound / Unsur E bertindak balas dengan unsur L untuk membentuk sebatian.
(i) State the type of bond present in this compound / Nyatakan jenis ikatan yang wujud dalam sebatian ini.
Covalent bond
(ii)
Write the formula of the compound formed / Tuliskan formula bagi sebatian yang terbentuk.
EL2
(iii) Explain how a compound is formed between element E and element L based on their electron arrangement.
Jelaskan dari segi susunan elektron bagaimana unsur E dan unsur L bergabung membentuk sebatian.
– E atom with electron arrangement 2.4 needs four electrons to achieve stable octet electron arrangement.
– L atom with an electron arrangement 2.6 needs two electrons to achieve octet electron arrangement.
– One E atom share four pairs of electrons with two L atoms to form a molecule with the formula EL2.
– One E atom contributes four electrons and each of the two L atoms contributes two electrons for sharing to
form double covalent bond.
– One E atom forms two double covalent bond with two L atoms.
– E atom and L atom achieve stable octet electron arrangement that is 2.8.
(e) (i)
Draw the electron arrangement of the compound formed / Lukiskan susunan elektron bagi sebatian yang terbentuk.
m
E
L
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Chemistry Form 4 • MODULE
(ii)
State one physical property of the compound / Nyatakan satu sifat fizik sebatian tersebut.
Low melting/boiling point // does not dissolve in water // dissolves in organic solvents // does not conduct
electricity in aqueous solution or molten state.
(f)
When element G is burnt in L gas, G burns rapidly and brightly with a yellow flame and produces white fumes.
Apabila unsur G dibakar dalam gas L, G terbakar cergas dengan nyalaan kuning terang dan menghasilkan wasap putih.
(i)
Write the equation for the reaction between element G and gas L.
Tuliskan persamaan kimia bagi tindak balas antara unsur G dan gas L.
4G + L2
(ii)
2G2L
.
Explain how a compound is formed between elements G and L based on their electron arrangement.
Jelaskan dari segi susunan elektron bagaimana unsur G dan L bergabung membentuk sebatian.
– The electron arrangement of G atom is 2.8.1. G atom is not stable. G atom releases one valence electron to
form G+ ion and achieve stable octet electron arrangement 2.8.
– The electron arrangement of L atom is 2.6. L atom is also unstable. L atom receives 2 electrons to form L2–
ion and achieves a stable octet electron arrangement 2.8.8.
– Therefore two G atoms release two electrons to one L atom, a strong electrostatic force is formed between
G+ and L2– ions.
(iii) Draw the electron arrangement of the compound formed / Lukiskan susunan elektron bagi sebatian yang terbentuk.
+
+
2–
G
L
G
(g) Compare the boiling point of the compounds formed in 1(d) and 1(e). Explain your answer.
Bandingkan takat didih sebatian yang terbentuk di 1(d) dan 1(e). Jelaskan jawapan anda.
– The boiling point of compound G2L is high and EL2 is low.
– The boiling point of compound G2L is high because positive ions and negative ions are attracted by strong
electrostatic force. Large amount of energy is needed to overcome it.
– The boiling point of EL2 is low because the molecules are attracted by weak Van der Waals forces, small amount
of energy is needed to overcome it.
2
The diagram below shows the electron arrangement of compound A. Compound A is formed from the reaction between
element X and element Y.
Rajah di bawah menunjukkan susunan elektron bagi sebatian A. Sebatian A terbentuk dari tindak balas antara unsur X dan unsur Y.
+
X
(a) (i)
(ii)
–
Y
Write the electron arrangement for atom of elements X and Y / Tuliskan susunan elektron bagi atom unsur X dan Y.
X: 2.8.1
Y: 2.8.7
Compare the size of atoms of elements X and Y. Explain your answer.
Bandingkan saiz atom unsur X dan unsur Y. Jelaskan jawapan anda.
– Atom Y is smaller than atom X.
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– Atom X and atom Y have the same number of shells occupied with electrons.
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MODULE • Chemistry Form 4
– The number of proton in the nucleus of atom Y is more than X.
– The strength nuclei attraction to the electrons in the shells of atom Y is stronger than X.
(b) How are X ion and Y ion formed from their respective atoms?
Bagaimana ion X dan ion Y terbentuk daripada atom masing-masing?
X ion / Ion X : Atom X releases one electron
Y ion / Ion Y : Atom Y receives one electron
(c) (i)
(ii)
Write the formula for compound A / Tuliskan formula sebatian A.
XY
Name type of bond in compound A / Namakan jenis ikatan dalam sebatian A.
Ionic compound
(iii) Write the chemical equation for the reaction between element X and element Y to form compound A.
Tuliskan persamaan kimia untuk tindak balas antara unsur X dan unsur Y untuk membentuk sebatian A.
2X + Y2
2XY
.
(d) Y can react with carbon to form a compound. Draw the electron arrangement for the compound formed.
[Given that proton number for carbon is 6]
Y bertindak balas dengan karbon untuk membentuk suatu sebatian. Lukiskan susunan elektron bagi sebatian yang terbentuk.
[Diberi nombor proton karbon ialah 6]
The table below shows the nucleon number, the number of neutrons and number of electrons in particles X, Y, Z, Q, R, T
and U.
3
Jadual di bawah menunjukkan nombor nukleon, bilangan neutron dan bilangan elektron bagi zarah X, Y, Z, Q, R, T dan U.
Particles / Zarah
X
Y
Z
Q
R
T
U
Nucleon number / Nombor nukleon
20
24
23
16
12
27
35
Number of proton / Bilangan proton
10
12
11
8
6
13
17
Number of neutron / Bilangan neutron
10
12
12
8
6
14
18
Number of electron / Bilangan elektron
10
10
11
10
6
10
17
(a) What is meant by nucleon number / Apakah maksud nombor nukleon?
The total number of proton and neutron in the nucleus of an atom.
(b) Complete the number of proton of the particles in the table above.
Lengkapkan bilangan proton bagi zarah dalam jadual di atas.
(c) State a particle which is / Nyatakan zarah yang merupakan
(i)
an atom of a non-metal / atom bukan logam
(ii)
an atom of a metal / atom logam
Y/T
(iv) a negative ion / ion negatif
Q
(v)
T
a positive ion with charge 3+ / ion positif dengan cas 3+
X
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(iii) a positive ion / ion positif
(vi) an atom of a noble gas / atom gas adi
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Chemistry Form 4 • MODULE
(d) Particle Y combines with particle Q to form a compound / Zarah Y bergabung dengan zarah Q untuk membentuk sebatian.
(i) State the type of compound formed / Nyatakan jenis sebatian yang terbentuk.
Ionic compound
(ii)
Write chemical formula for the compound formed / Tuliskan formula kimia bagi sebatian yang terbentuk.
YQ
(iii) Draw the electron arrangement of the compound formed / Lukiskan susunan elektron bagi sebatian yang terbentuk.
2+
2–
Q
Y
(e) Particle R combines with particle U to form a compound.
Zarah R bergabung dengan zarah U untuk menghasilkan suatu sebatian.
(i)
State the type of compound formed / Nyatakan jenis sebatian yang terbentuk.
Covalent compound
(ii)
(f)
Write a chemical formula for the compound formed / Tuliskan formula kimia bagi sebatian yang terbentuk.
RU4
Compare the electrical conductivity of the compounds formed in 3(d) and 3(e). Explain your answer.
Bandingkan kekonduksian elektrik bagi sebatian yang terbentuk di 3(d) dan di 3(e). Jelaskan jawapan anda.
– Compound in YQ cannot conduct electricity in solid state but can conduct electricity in molten or aqueous
solution. Compound RU4 cannot conduct electricity in molten and aqueous states.
– In solid form the ions in compound YQ are not free to move but in molten and aqueous state, the ions are free to
move to be attracted to the anode and cathode.
– Compound RU4 only consists of neutral molecules, there are no free moving ions in molten or aqueous state.
4
The table below shows the melting point and electrical conductivity of substances W, X, Y and Z.
Jadual di bawah menunjukkan takat lebur dan kekonduksian elektrik bagi bahan W, X , Y dan Z.
Substance
Bahan
Electrical conductivity / Kekonduksian elektrik
Melting point (°C)
Takat Lebur (°C)
V
–7
W
80
X
808
Y
1 080
Solid / Pepejal
Molten / Leburan
Cannot conduct electricity
Cannot conduct electricity
Tidak mengkonduksi elektrik
Tidak mengkonduksi elektrik
Cannot conduct electricity
Cannot conduct electricity
Tidak mengkonduksi elektrik
Tidak mengkonduksi elektrik
Cannot conduct electricity
Conduct electricity
Tidak mengkonduksi elektrik
Mengkonduksi elektrik
Conduct electricity
Conduct electricity
Mengkonduksi elektrik
Mengkonduksi elektrik
(a) Which of the substance is copper? Give reason for your answer.
Antara bahan di atas, yang manakah kuprum? Beri sebab bagi jawapan anda.
Y. It can conduct electricity in solid and molten state.
(b) (i)
State the type of particles in substances V and W / Nyatakan jenis zarah dalam bahan V dan W.
Molecule
(ii)
Explain why substances V and W cannot conduct electricity in solid and molten state.
Jelaskan mengapa bahan V dan W tidak boleh mengkonduksi elektrik dalam keadaan pepejal dan leburan.
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Substances V and W are made up of neutral molecules. No free moving ions in molten state.
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MODULE • Chemistry Form 4
(c) The boiling point of substance V is 59°C. What is the physical state of substance V at room temperature?
Takat didih bahan V adalah 59°C. Apakah keadaan fizikal bahan V pada suhu bilik?
Liquid
(d) Draw the arrangement of particle V at room temperature / Lukiskan susunan zarah V pada suhu bilik.
(e) Explain why the melting and boiling points of substances V and W are low?
Jelaskan mengapa takat lebur dan takat didih bahan V dan W rendah?
– Van der Waals / intermolecular forces between molecules are weak.
– Small amount of heat energy is required to overcome it.
(f)
(i)
(ii)
State the type of particle in substance X / Nyatakan jenis zarah dalam sebatian X.
Ion
.
Explain why substance X cannot conduct electricity in solid but can conduct electricity in molten state.
Jelaskan mengapa bahan X tidak boleh mengkonduksi elektrik dalam keadaan pepejal tetapi boleh mengkonduksi elektrik
dalam keadaan leburan.
Ions are not freely moving // ions are in a fixed position in solid state. Ion can move freely in molten state.
Objective Questions / Soalan Objektif
1
Which substance is an ionic compound?
Antara bahan berikut, yang manakah adalah sebatian ion?
A
B
C
D
2
4
Methane, CH4 / Metana, CH4
Carbon dioxide, CO2 / Karbon dioksida, CO2
Propanol, C3H7OH / Propanol, C3H7OH
Copper(II) oxide, CuO / Kuprum(II) oksida, CuO
12
Apakah susunan elektron bagi ion yang terbentuk dari atom T?
A
B
C
D
3
Volatile / Mudah meruap
Has a low melting point
Mempunyai takat lebur rendah
Insoluble in water / Tidak larut dalam air
Conducts electricity in the molten state
Mengalirkan arus elektrik dalam keadaan leburan
The diagram below shows the electron arrangement of a
compound formed between atoms X and Y.
5
Jadual di bawah menunjukkan susunan elektron atom P, Q, R dan S.
Y
X
Y
Y
Which of the following statements is true about the
compound?
Antara pernyataan berikut, yang manakah adalah benar tentang sebatian
itu?
A
B
C
m
Atom / Atom
Electron arrangement / Susunan elektron
P
2.4
Q
2.8.1
R
2.8.2
S
2.8.7
Which pair of atoms forms a compound by transferring of
electrons?
Antara pasangan berikut, yang manakah membentuk sebatian secara
perpindahan elektron?
A
B
C
D
P and S / P dan S
P and R / P dan R
Q and S / Q dan S
Q and R / Q dan R
Sebatian itu mempunyai takat lebur yang tinggi.
The compound conducts electricity.
Sebatian itu boleh mengkonduksi elektrik.
The compound is formed by sharing of electrons.
Sebatian terbentuk secara perkongsian elektron.
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It is an ionic compound / Ia adalah sebatian ion.
The compound has high melting point.
2.8
2.8.2
2.8.8
2.8.8.8
The table below shows the electron arrangements of atoms P,
Q, R and S.
Rajah di bawah menunjukkan susunan elektron dalam sebatian yang
terbentuk antara atom X and atom Y.
Y
T
What is the electron arrangement of ion formed by an atom of
T?
Which of the following is a property of zinc chloride?
C
D
Rajah menunjukkan simbol unsur T.
24
Antara berikut, yang manakah adalah sifat zink klorida?
A
B
The diagram shows symbol of an element T.
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Chemistry Form 4 • MODULE
6
The table below shows the proton number of four elements P,
Q, R and S.
Jadual di bawah menunjukkan nombor proton unsur P, Q, R dan S.
9
The diagram below shows the electron arrangement for an ion
of element Q.
Rajah di bawah menunjukkan susunan elektron ion unsur Q.
Element / Unsur
P
Q
R
S
Proton number / Nombor proton
6
8
17
20
2–
Which of the following pairs will form a compound with high
melting and boiling points?
Q
Antara pasangan berikut, yang manakah membentuk sebatian dengan
takat lebur dan takat didih yang tinggi?
A
B
7
C
D
P and Q / P dan Q
Q and S / Q dan S
P and R / P dan R
Q and R / Q dan R
The table below shows the proton number of elements X and
Y.
What are the number of protons and electrons in an atom of
element Q?
Apakah bilangan proton dan elektron dalam atom unsur Q?
Number of protons
Number of electrons
A
20
20
Jadual berikut menunjukkan nombor proton unsur X dan Y.
Element / Unsur
X
Y
Proton number / Nombor proton
6
8
Bilangan elektron
What type of bond and the chemical formula of the compound
formed between atoms X and Y?
B
20
18
C
16
16
Apakah jenis ikatan dan formula kimia bagi sebatian yang terbentuk
antara atom X dan Y ?
D
18
18
Type of bond
Chemical formula
A
Ion / Ion
YX2
B
Ion / Ion
XY2
Element / Unsur
P
Q
R
C
Covalent / Kovalen
XY2
Proton number / Nombor proton
10
11
12
D
Covalent / Kovalen
YX2
Jenis ikatan
8
Bilangan proton
Formula kimia
The diagram below shows the electron arrangement of ion
X +.
Rajah di bawah menunjukkan susunan elektron ion X +.
X
10 The table below shows the proton number of elements P, Q and
R.
Jadual di bawah menunjukkan nombor proton unsur P, Q dan R.
Which of the following particles contain 10 electrons?
Antara berikut, yang manakah adalah zarah yang mengandungi 10
elektron?
I
II
III
IV
A
B
C
Which of the following is the position of element X in the
Periodic Table?
D
Q
P
Q+
R2+
I, II and III only
I, II dan III sahaja
I, II and IV only
I, II dan IV sahaja
I, III and IV only
I, III dan IV sahaja
II, III and IV only
II, III dan IV sahaja
Antara berikut, yang manakah adalah kedudukan unsur X dalam Jadual
Berkala?
Group / Kumpulan
Period / Kala
A
1
3
B
18
3
1
4
D
18
4
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MODULE • Chemistry Form 4
5
ELECTROCHEMISTRY
ELEKTROKIMIA
ELECTROLYSIS / ELEKTROLISIS
• CONDUCTOR AND ELECTROLYTE / KONDUKTOR DAN ELEKTROLIT
– To differentiate between electrolyte and conductor with regard to electrical conductivity and any chemical changes that may occur.
Membezakan elektrolit dan konduktor dari segi kebolehan mengkonduksikan elektrik dan sebarang perubahan kimia yang berlaku.
– To list examples of substances which are classified as electrolytes and conductors.
Menyenaraikan contoh-contoh bahan yang dikelaskan sebagai elektrolit dan konduktor.
• ELECTROLYSIS CELL / SEL ELEKTROLISIS
– To draw and label the electrolytic cell / Melukis dan melabelkan sel elektrolisis.
– To identify anode and cathode in the electrolytic cell diagram / Mengenali anod dan katod dalam rajah sel elektrolisis.
• IONIC THEORY / TEORI ION
– To relate the existence of free moving ions in an electrolyte with the electron flow in an external circuit.
Mengaitkan kewujudan ion-ion yang bebas bergerak dalam elektrolit dengan proses pengaliran elektron dalam litar luar.
– To explain the electrolysis process / Menerangkan proses elektrolisis.
– To conclude that electrolysis process involve changes from electrical to chemical energy.
Membuat kesimpulan proses elektrolisis sebagai perubahan tenaga elektrik kepada tenaga kimia.
• FORMATION OF FREE MOVING IONS / PEMBENTUKAN ION BEBAS BERGERAK
– To differentiate molten and aqueous electrolytes / Membezakan elektrolit lebur dan akueus.
– To write the ionisation equation of molten and aqueous electrolytes.
Menulis persamaan pengionan untuk elektrolit lebur dan akueus.
• REACTION AT ELECTRODE / TINDAK BALAS DI ELEKTROD
– To write the discharge equation at the anode, where the anion releases electron. Focus on ions that are normally selected for
discharge, such as chloride, hydroxide and bromide ions.
Menulis persamaan di anod yang melibatkan anion melepaskan elektron. Fokus adalah kepada ion-ion yang biasa terpilih untuk nyahcas seperti ion
klorida, ion hidroksida dan ion bromida.
– To write the discharge equation at the cathode, where the cation receives electron. Focus on ions that are normally selected for
discharge, such as hydrogen, copper(II) and silver ions.
Menulis persamaan di katod yang melibatkan kation menerima elektron. Fokus adalah kepada ion yang biasa terpilih untuk nyahcas seperti ion
hidrogen, ion kuprum(II) dan ion argentum.
• FACTORS THAT AFFECT REACTIONS AT THE ELECTRODES
FAKTOR-FAKTOR YANG MEMPENGARUHI TINDAK BALAS DI ELEKTROD
(i)
The position of ions in the electrochemical series – for dilute solutions and inert electrodes.
Kedudukan ion dalam siri elektrokimia – bagi larutan cair dan elektrod lengai
(ii)
The concentration – for concentrated solutions and inert electrodes / Kepekatan – bagi larutan pekat dan elektrod lengai.
(iii) The types of electrode – for diluted solutions and reactive electrodes / Jenis elektrod – bagi larutan cair dan elektrod tak lengai.
• ELECTROLYSIS IN INDUSTRY / KEGUNAAN ELEKTROLISIS DALAM INDUSTRI
– Electrolysis in electroplating, purifying and extracting metals / Elektrolisis dalam penyaduran, penulenan dan pengekstrakan logam.
VOLTAIC CELL / SEL KIMIA
• ELECTROCHEMICAL SERIES / SIRI ELEKTROKIMIA
– To define and memorise the sequence of metal including hydrogen in the Electrochemical Series.
Menakrif dan menghafal siri logam termasuk hidrogen dalam Siri Elektrokimia.
• APPLICATION OF ELECTROCHEMICAL SERIES IN DISPLACEMENT OF METALS
APLIKASI SIRI ELEKTROKIMIA DALAM PENYESARAN LOGAM
– To predict the displacement of metal reactions based on the positions of metals in the Electrochemical Series.
Meramal tindak balas penyesaran logam berdasarkan kedudukan logam dalam Siri Elektrokimia.
– To write the equation of displacement reaction and to state the observations.
Menulis persamaan tindak balas penyesaran dan menyatakan pemerhatian.
– To describe the metal displacement experiment to construct the Electrochemical Series.
Menghuraikan eksperimen penyesaran logam bagi membina Siri Elektrokimia.
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• APPLICATION OF ELECTROCHEMICAL SERIES IN VOLTAIC CELL
APLIKASI SIRI ELEKTROKIMIA DALAM SEL KIMIA
– To determine the negative and positive terminals of a voltaic cell / Menentukan terminal negatif dan positif suatu sel kimia.
– To predict the voltage of voltaic cell / Meramal voltan sel kimia.
– To determine the direction of electron flow / Menentukan arah pengaliran elektron.
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Chemistry Form 4 • MODULE
ELECTROLYSIS / ELEKTROLISIS
1
Three types of substances that can be classified based on electrical conductivity.
Bahan boleh dibahagikan kepada tiga jenis berdasarkan kekonduksian elektrik.
Type of
substance
Definition
Conductor
Konduktor
Example
Definisi
Jenis bahan
Element that can conduct electricity
at solid or molten state without any
chemical changes , normally metals and
Contoh
Copper, lead, tin, silver and carbon
Kuprum, plumbum, stanum, argentum dan karbon
carbon.
Unsur yang boleh mengkonduksi arus elektrik
dalam keadaan pepejal atau leburan tanpa
perubahan kimia , biasanya logam dan karbon.
Electrolyte
Elektrolit
Compounds that can conduct electricity in
*molten state or *aqueous solution and
undergo chemical changes .
Sebatian yang boleh mengkonduksikan arus
elektrik dalam keadaan *lebur atau *akueus serta
mengalami perubahan kimia .
* Molten state: a solid that is heated until
it melts.
* Lebur: pepejal yang dipanaskan sehingga cair.
* Aqueous solution: a solid that is
dissolved in water.
* Akueus: pepejal yang larut di dalam air.
– Aqueous solution of ionic compound such as copper(II)
sulphate solution and sodium chloride solution.
Larutan akueus bagi sebatian ion contohnya larutan kuprum(II) sulfat
dan larutan natrium klorida.
– Aqueous solution of *acid or alkali such as hydrochloric acid
(HCl) and ammonia solution (NH3).
Larutan akueus *asid atau alkali contohnya asid hidroklorik (HCl) dan
larutan ammonia (NH3 ).
ionic
compounds such as molten lead(II)
– Molten
bromide, molten sodium chloride and molten aluminium
oxide.
ion
contohnya leburan plumbum(II)
Leburan sebatian
bromida, leburan natrium klorida dan leburan aluminium oksida.
* HCl and NH3 are covalent compounds, exist in form of
molecule without water but ionised in water. (Explanation
is in the next topic i.e acid and base)
* HCl dan NH3 adalah sebatian kovalen, yang terdiri daripada molekul
dalam keadaan tanpa air tetapi ianya terion dalam air (akan dijelaskan
dalam tajuk seterusnya iaitu dalam asid dan bes)
2
Nonelectrolyte
Compounds that cannot conduct electricity
in molten and aqueous solution.
Molten covalent compound such as naphthalene, molten
sulphur and liquid bromine.
Bukan elektrolit
Sebatian kimia yang tidak boleh mengkonduksikan
elektrik dalam keadaan lebur dan akueus.
Leburan sebatian
bromin.
process
Electrolysis is a
current passes
3
whereby an electrolyte is decomposed to its constituent elements when electric
penguraian elektrolit kepada unsur juzuknya apabila
Energy change in electrolysis process is electric energy to
chemical energy
Perubahan tenaga dalam proses elekrolisis adalah dari tenaga elektrik kepada
4
contohnya naftalena, sulfur lebur dan cecair
through it.
proses
Elektrolisis adalah
kovalen
arus elektrik
dialirkan melaluinya.
.
tenaga kimia
.
Conductor which is dipped into electrolyte which carries electric current in and out of electrolyte is called an
electrode
.
Electrode
is normally made up of
inert
substance such as carbon.
Konduktor yang dicelup dalam elektrolit yang mengalirkan arus elektrik ke dalam dan keluar daripada elektrolit dipanggil
Elektrod
lengai
biasanya terdiri daripada bahan
seperti karbon.
5
electrodes
An electrolytic cell is a set-up of apparatus that contains two
which are dipped in an
battery
and produce a chemical reaction when connected to a
(source of electricity).
elektrolit
.
electrolyte
dan menghasilkan
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elektrod
yang dicelup ke dalam
Sel elektrolisis adalah susunan radas yang terdiri daripada dua
bateri
. (sumber arus elektrik).
tindak balas kimia apabila disambungkan kepada
elektrod
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MODULE • Chemistry Form 4
Example of electrolytic cell / Contoh sel elektrolisis:
(i)
(ii)
(iii)
Electrolyte
A
Electrodes
Elektrod
Elektrolit
Electrode
Electrode
Electrolyte
Elektrod
Elektrod
Elektrolit
Electrodes
Electrolyte
Elektrolit
Heat
Panaskan
Electrolysis of aqueous electrolyte
(No gas released)
Electrolysis of molten electrolyte
Elektrolisis elektrolit lebur
Elektrod
A
Electrolysis of aqueous electrolyte
(Gas is released)
Elektrolisis elektrolit dalam bentuk akueus
(Tiada gas dibebaskan)
Elektrolisis elektrolit dalam bentuk larutan
(Gas dibebaskan)
Electric current from the battery flows into the electrolyte through the electrode. There are two types of electrode in the
electrolytic cell:
6
Arus elektrik dari bateri mengalir ke dalam elektrolit melalui elektrod. Terdapat dua jenis elektrod dalam sel elektrolisis:
(a) Anode: An electrode that is connected to the
Anod: Elektrod yang disambung kepada
positive terminal
terminal positif
bateri dalam sel elektrolisis.
negative terminal
(b) Cathode: An electrode that is connected to the
Katod: Elektrod yang disambung kepada
of the battery.
terminal negatif
of the battery.
bateri dalam sel elektrolisis.
An electrolyte consists of free moving ions because it is in a molten or aqueous state. Each ion moves to the opposite
charge electrode. There are two types of ions in electrolyte:
7
Dalam keadaan lebur atau akueus, elektrolit terdiri daripada ion-ion yang bergerak bebas. Setiap ion bergerak kepada elektrod yang
bertentangan cas. Terdapat dua jenis ion dalam elektrolit:
(a) Anions:
Anion: Ion
(b) Cations:
Kation: Ion
Negative
ions which are attracted and move to the
negatif
Positive
akan tertarik dan bergerak ke arah elektrod
positively
anod
yang bercas
positif
akan tertarik dan bergerak ke arah elektrod
katod
.
cathode
.
.
ions which are attracted and move to the negatively charged electrode,
positif
anode
charged electrode,
yang bercas
negatif
.
Electrolysis occurs at the electrode when electric current flows in the electrolytic cell. The stages in electrolysis process
are:
8
Proses elektrolisis berlaku di elektrod apabila arus elektrik mengalir melalui sel elektrolisis. Peringkat dalam proses elektrolisis adalah
seperti berikut:
(a) Anions (negative ions) are attracted and move to the
anode
of anode and become neutral atoms or molecule. The anions are
Anion (ion negatif) akan tertarik dan bergerak ke arah
menjadi atom/molekul. Anion dinyahcaskan pada anod.
(b) Electrons flow from the
Elektron mengalir dari
anode
anod
to the
ke
katod
anod
cathode
. The anions release electrons to the surface
discharged
at the anode.
. Anion melepaskan elektron pada permukaan anod dan
melalui wayar penyambung dalam
(c) Cations (positive ions) are attracted and move to the
cathode
of cathode and become neutral atoms or molecules. The cations are
katod
Kation (ion positif) akan tertarik dan bergerak ke arah
menjadi atom/molekul. Kation dinyahcaskan pada katod.
external circuit .
through the connecting wire in the
litar luar
.
. The cations receive electrons at the surface
discharged
at the cathode.
. Kation menerima elektron pada permukaan katod dan
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– Electrons flow through the external circuit / Elektron mengalir melalui litar luar.
– Chemical changes occur at the anode and cathode / Perubahan kimia berlaku di anod dan katod.
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Chemistry Form 4 • MODULE
FORMATION OF FREE MOVING IONS IN THE ELECTROLYTE
PEMBENTUKAN ION BERGERAK BEBAS DALAM ELEKTROLIT
1
Ionisation equation is an equation to determine the ions present in molten or aqueous electrolyte.
Persamaan pengionan adalah persamaan yang menunjukkan ion yang hadir dalam elektrolit sama ada dalam keadaan leburan atau
akueus.
(a) Example of ionisation of molten electrolyte (a compound that is heated until it melts)
Contoh pengionan elektrolit dalam keadaan leburan (sebatian yang dipanaskan sehingga lebur)
(i)
Molten sodium chloride / Natrium klorida lebur:
(ii)
Molten lead (II) bromide / Plumbum (II) bromida lebur: PbBr2 (s)
Na+(l) + Cl–(l)
NaCl (s)
(iii) Molten sodium oxide / Natrium oksida lebur:
Na2O (s)
(iv) Molten aluminium oxide / Aluminium oksida lebur:
Al2O3 (s)
Pb2+(l) + 2Br –(l)
2Na+(l) + O2–(l)
2Al3+(l) + 3O2–(l)
(b) Example of the ionisation on an aqueous electrolyte (a compound that is dissolved in water):
Contoh pengionan elektrolit dalam keadaan akueus (sebatian yang dilarutkan dalam air):
(i)
(ii)
Sodium chloride solution / Larutan natrium klorida:
NaCl(aq / ak )
Na+(aq) + Cl+(aq)
H2O
H+(aq) + OH–(aq)
Cu2+ + SO42–
Copper(II) sulphate solution / Larutan kuprum(II) sulfat: CuSO4(aq / ak )
H+ + OH–
H2O
(iii)
Sulphuric acid / Asid sulfurik:
2H+ + SO42–
H2SO4(aq / ak )
H+ + OH–
H2O
2
Ionisation of molten electrolyte produces cation and anion of the compound only. However the ionisation of an
aqueous electrolyte produces cation and anion from the ionisation of the compound and water.
Pengionan elektrolit dalam keadaan lebur hanya menghasilkan kation dan anion dari sebatian itu sahaja. Pengionan elektrolit dalam
keadaan akueus menghasilkan kation dan anion daripada sebatian dan air.
Example / Contoh:
(i)
molten
Ionisation of
Pengionan
leburan
sodium chloride produces Na+ and Cl– only.
natrium klorida menghasilkan Na+ dan Cl– sahaja.
aqueous
(ii) Ionisation of
sodium chloride produces Na+, H+, Cl– and OH–.
akueus
Pengionan larutan
natrium klorida menghasilkan Na+, H+, Cl– dan OH–.
REACTIONS AT THE ELECTRODES / TINDAK BALAS DI ELEKTROD
1
The process of cation gaining electron at the cathode or anion losing electrons at the anode is called discharged :
Proses apabila kation menerima elektron di katod atau anion melepaskan elektron di anod dipanggil
(a) A cation is
Kation
discharged
dinyahcaskan
dinyahcaskan
(c) When ions are
Apabila ion
2
apabila
discharged
(b) An anion is
Anion
when it
discharged
dinyahcaskan
menerima
when it
apabila
receives
elektron di katod.
electrons at the anode.
elektron di anod.
, they become neutral
, ianya akan menjadi
:
electrons at the cathode.
releases
melepaskan
nyahcas
atom
atom
atau
molecule .
or
molekul
The ionic equation that occurs at the anode and cathode to produce neutral
‘half equation’.
atom
atau
atom
molekul
or
molecule
is called
neutral dipanggil ‘persamaan
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Persamaan ion yang berlaku di anod dan di katod untuk menghasilkan
setengah’.
yang neutral.
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MODULE • Chemistry Form 4
Common half equation at the anode (anion/metal atom releases electrons):
3
Persamaan setengah yang biasa di anod (anion/atom logam melepaskan elektron):
Half equation
Explanation
Persamaan setengah
4OH–
2H2O + O2 + 4e
2Cl–
Cl2 + 2e
Cu
Cu2+ + 2e
Ag
Ag+ + e
release
Four hydroxide ions
molecule .
Empat ion hidroksida
melepaskan
Two chloride ions
release
melepaskan
Dua ion klorida
Br2 + 2e
2Br–
Penerangan
Two bromide ions
four electrons to form two water molecules and one oxygen
molekul
dua elektron membentuk satu
Dua ion bromida
dua elektron membentuk satu
Copper atom
releases
two electrons to form
melepaskan
Silver atom
releases
Atom argentum
melepaskan
molekul
.
molecule
.
bromin.
copper(II) ion
ion kuprum(II)
dua elektron membentuk
oksigen.
klorin.
two electrons to form one bromine
melepaskan
Atom kuprum
molecule
two electrons to form one chlorine
release
molekul
empat elektron membentuk dua molekul air dan satu
.
.
one electron to form
silver ion
satu elektron membentuk
ion argentum .
.
Common half equation at the cathode (cation receives electrons):
4
Persamaan setengah yang biasa di katod (kation menerima elektron):
Half equation
Explanation
Persamaan setengah
2H+ + 2e
Ag+ + e
Cu2+ + 2e
H2
Ag
Cu
Penerangan
Two hydrogen ions
Dua ion hidrogen
receive
menerima
molekul
dua elektron membentuk satu
Silver ion
receive
one electron to form one silver
Ion argentum
menerima
satu elektron membentuk satu
Copper(II) ion
Ion kuprum(II)
receives
menerima
molecule
two electrons to form one hydrogen
atom
atom
.
argentum.
two electrons to form one copper
atom
dua elektron membentuk satu
.
hidrogen.
atom
.
kuprum.
Write the equation of discharge of ion
5
Tuliskan persamaan setengah untuk nyahcas ion yang berikut:
(i)
Lead(II) ion to lead atom
:
Pb2+ + 2e
:
Ag++ e
:
2I–
Pb
Ion plumbum(II) kepada atom plumbum
(ii) Silver ion to silver atom /
Ag
Ion argentum kepada atom argentum
(iii) Iodide ion to iodine molecule
I2 + 2e
Ion iodida kepada molekul iodin
EXERCISE / LATIHAN
Using lead(II) bromide as an example, explain the electrolysis of molten lead(II) bromide. In your explanation, draw a labeled
diagram for the set up of apparatus and show the movement of particles by using arrows that occur in lead(II) bromide and
the direction of electron flow in the external circuit.
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Dengan menggunakan plumbum(II) bromida sebagai contoh, jelaskan elektrolisis leburan plumbum(II) bromida. Dalam penerangan anda,
lukiskan satu rajah susunan radas berlabel dan tunjukkan dengan anak panah pergerakan zarah yang berlaku dalam plumbum(II) bromida serta
arah aliran elektron dalam litar luar.
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Chemistry Form 4 • MODULE
Set-up of apparatus / Rajah susunan radas:
Carbon electrodes
Lead(II) bromide
Heat
Explanation / Penerangan:
– The ions present are lead(II) ions/ Pb2+ and bromide ions/ Br –.
– Bromide ion/ Br – move to the anode.
– Bromide ion/ Br – releases one electron to form bromine atom at the anode.
– Two bromine atoms combine to form bromine molecule.
– 2Br –
Br2 + 2e
– Lead(II) ions/Pb2+ move to the cathode.
– Lead(II) ions/Pb2+ receive two electrons to form lead atom at the cathode.
– Pb2+ + 2e
Pb
FACTOR THAT AFFECT THE ELECTROLYSIS OF AN AQUEOUS SOLUTION
FAKTOR YANG MEMPENGARUHI ELEKTROLISIS LARUTAN AKUEUS
1
When more than one type of ion are attracted towards the electrodes during electrolysis, only one type of ion is
selected to be discharged at each electrode. Selective discharge only occurs in aqueous solution because it usually
has more than one type of ion attracted to the anode or cathode.
Apabila lebih dari satu jenis ion bergerak ke elektrod semasa elektrolisis, hanya satu jenis ion sahaja yang akan dipilih untuk dinyahcas
pada setiap elektrod. Pemilihan nyahcas ion hanya berlaku di dalam larutan akueus sahaja kerana ia biasanya mempunyai lebih dari
satu jenis ion yang tertarik ke anod atau katod.
2
The selection of ion for discharge depends on three factors / Pemilihan ion untuk nyahcas bergantung pada tiga faktor:
(a) The position of ions in the electrochemical series (normally in dilute solution and inert electrode).
Kedudukan ion dalam siri elektrokimia (biasanya dalam larutan cair dan elektrod lengai).
(b) The concentration of electrolyte (normally in concentrated solution and inert electrode).
Kepekatan elektrolit (biasanya dalam larutan pekat dan elektrod lengai).
(c) The types of electrode (when reactive metal electrode is used).
Jenis elektrod (apabila elektrod logam reaktif digunakan).
3
The position of ions in the Electrochemical Series / Kedudukan ion dalam Siri Elektrokimia:
(a) When electrolysis is conducted on dilute solution and inert electrodes, the lower position of cation in the
Electrochemical Series, or anions in the lower position of the anion discharge series will be selected to be
discharged.
Apabila elektrolisis dijalankan ke atas larutan cair dan elektrod lengai, kation yang lebih rendah kedudukan dalam Siri Elektrokimia
atau anion yang lebih rendah kedudukan dalam siri discas anion akan dinyahcas.
Cation: K+, Na+, Ca2+, Mg2+, Al3+, Zn2+, Fe2+, Sn2+, Pb2+, H+, Cu2+, Ag+, and Au+
Kation: K+, Na+, Ca2+, Mg2+, Al3+, Zn2+, Fe2+, Sn2+, Pb2+, H+, Cu2+, Ag+, dan Au+
Increasing ease of discharge of ion from left to right
Ion semakin mudah dinyahcas dari kiri ke kanan
Anion: F , SO4 , NO3 , Cl–, Br –, I–, and OH–
–
2–
–
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Anion: F–, SO42–, NO3–, Cl–, Br –, I–, dan OH–
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MODULE • Chemistry Form 4
(b) Choose the ion to be discharged from the following pairs of ions. State the electrode where it occurs and write
the half equation for the discharge of ion:
Pilih ion yang akan dinyahcas dari pasangan ion berikut, nyatakan di elektrod mana ia berlaku dan tulis persamaan setengah untuk
nyahcas ion:
(i)
Ion hidroksida & ion sulfat
(ii)
4OH–
Hydroxide & sulphate ions : Half equation:
Hydroxide & nitrate ions
Ion hidroksida & ion nitrat
: Persamaan setengah:
: Half equation:
: Persamaan setengah:
(v)
Hydrogen & silver ions
Ion hidrogen & ion argentum
2H2O + O2 + 4e
2H2O + O2 + 4e
4OH
–
Cu
2+
2H2O + O2 + 4e
+ 2e
Cu
Cu2+ + 2e
: Persamaan setengah:
2H + 2e
: Persamaan setengah:
Ag + e
+
: Half equation:
: Persamaan setengah:
Ag
di
katod
.
.
cathode
at the
.
.
cathode
katod
.
.
cathode
katod
di
Ag
Ag + e
anod
di
.
.
anode
at the
at the
H2
+
anod
di
di
H2
2H+ + 2e
anode
at the
at the
Cu
+
(iv) Hydrogen & potassium ions : Half equation:
Ion hidrogen & ion kalium
4OH
4OH–
(iii) Hydrogen & copper(II) ions : Half equation:
Ion hidrogen & ion kuprum(II)
2H2O + O2 + 4e
–
.
.
(c) Complete the following table for the electrolysis of 0.1 mol dm–3 sodium nitrate solution using carbon electrode.
Lengkapkan jadual berikut bagi elektrolisis larutan natrium nitrat 0.1 mol dm–3 menggunakan elektrod karbon.
Set-up of apparatus
Susunan radas
Carbon electrodes
NaNO3
H2O
Equation of electrolyte ionisation
Persamaan pengionan elektrolit
Electrode / Elektrod
Ions that are attracted to the anode and
cathode
Ion yang ditarik ke anod dan katod
Half equation
Persamaan setengah
Name of the products
Anode / Anod
NO3–, OH–
4OH–
Cathode / Katod
Na+, H+
2H2O + O2 + 4e
2H+ + 2e
H2
Hydrogen
Pemerhatian
Gas bubbles are released.
Gas bubbles are released.
Confirmatory test (method and
observations)
– Insert a glowing wooden splinter into – When a lighted wooden splinter is
test tube.
placed near the mouth of the test tube.
– Glowing wooden splinter is lighted up. – A ‘pop’ sound is produced.
Observations
Ujian pengesahan (kaedah dan pemerhatian)
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Na+ + NO3–
H+ + OH–
Oxygen
Nama hasil
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Sodium nitrate
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Chemistry Form 4 • MODULE
(d) Complete the following table for the electrolysis of 0.1 mol dm–3 sulphuric acid using carbon electrodes.
Lengkapkan jadual berikut bagi elektrolisis asid sulfurik 0.1 mol dm–3 menggunakan elektrod karbon.
Set-up of apparatus
Susunan radas
Carbon electrodes
Sulphuric acid
H2SO4
Equation of electrolyte ionisation
H2O
Persamaan pengionan elektrolit
2H+ + SO42–
H+ + OH–
Electrode / Elektrod
Anode / Anod
Cathode / Katod
Ions that are attracted to the anode and
cathode
SO42–, OH–
H+
Ion yang ditarik ke anod dan katod
Half equation
4OH–
Persamaan setengah
Name of the products
Nama hasil
Observations
Pemerhatian
Confirmatory test (method and
observations)
Ujian pengesahan (kaedah dan pemerhatian)
2H2O + O2 + 4e
2H+ + 2e
H2
Oxygen
Hydrogen
Gas bubbles are released.
Gas bubbles are released.
– Insert a glowing wooden splinter into – When a lighted wooden splinter is
test tube.
placed near the mouth of the test tube.
– Glowing wooden splinter is lighted up. – A ‘pop’ sound is produced.
(e) Complete the following table for the electrolysis of 0.1 mol dm–3 copper(II) sulphate solution using carbon
electrodes.
Lengkapkan jadual berikut bagi elektrolisis larutan kuprum(II) sulfat 0.1 mol dm–3 menggunakan elektrod karbon.
Set-up of apparatus
Susunan radas
Carbon electrodes
Copper(II) sulphate
H2SO4
Equation of electrolyte ionisation
H2O
Persamaan pengionan elektrolit
Ion yang ditarik ke anod dan katod
Half equation
Persamaan setengah
Name of the products
Nama hasil
Observations
Pemerhatian
Confirmatory test (method and
observations)
H+ + OH–
Anode / Anod
Electrode / Elektrod
Ions that are attracted to the anode and
cathode
2H+ + SO42–
SO42–, OH–
4OH–
Cathode / Katod
Cu2+, H+
2H2O + O2 + 4e
Cu2+ + 2e
Cu
Oxygen
Copper
Gas bubbles are released.
Brown solid deposited
– Insert a glowing wooden splinter into test tube.
– Glowing wooden splinter is lighted up.
–
Ujian pengesahan (kaedah dan pemerhatian)
4
Concentration of electrolyte / Kepekatan elektrolit:
(a) When electrolysis is carried out using inert electrodes and concentrated solutions, ions that are more
concentrated will be discharged but this is only true for halide ions, which are Cl–, Br – and I–.
Apabila elektrolisis dijalankan menggunakan elektrod lengai dan larutan pekat, ion yang lebih pekat akan dinyahcas tetapi ia benar
untuk ion-ion halida sahaja iaitu Cl–, Br– dan I–.
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Nyatakan ion yang terpilih untuk dinyahcaskan di anod dan di katod bagi larutan pekat di bawah.
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(b) State the selected ions to be discharged at the anode and cathode for the following concentrated solutions.
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MODULE • Chemistry Form 4
(i)
Concentrated hydrochloric acid solution, using carbon electrodes
Larutan asid hidroklorik pekat menggunakan elektrod karbon
(ii)
Cl–
Anode / Anod:
Cathode / Katod:
Concentrated potassium iodide solution, using carbon electrodes
H+
Larutan kalium iodida pekat menggunakan elektrod karbon
l–
Anode / Anod:
Cathode / Katod:
(iii) Concentrated sodium chloride solution, using carbon electrodes
K+
Larutan natrium klorida pekat menggunakan elektrod karbon
Anode / Anod:
Cl–
H+
Cathode / Katod:
(c) Complete the following table for the electrolysis of 0.001 mol dm–3 hydrochloric acid and 2.0 mol dm–3 hydrochloric
acid, using carbon electrodes.
Lengkapkan jadual berikut bagi elektrolisis asid hidroklorik 0.001 mol dm–3 dan asid hidroklorik 2.0 mol dm–3 menggunakan elektrod
karbon.
Set-up of apparatus
Susunan radas
Carbon electrodes
Hydrochloric acid
HCl
Equation of electrolyte ionisation
H2O
Persamaan pengionan elektrolit
Half equation at the cathode
Persamaan setengah di katod
Observation at cathode
Pemerhatian di katod
Confirmatory test at cathode (method
and observations)
Ujian pengesahan (kaedah dan pemerhatian)
Name the products at the cathode
Nama hasil di katod
Ions that are attracted to the anode
Ion bergerak ke anod
Half equation at the anode
Persamaan setengah di anod
H+
H+
2H+ + 2e
H2
2H+ + 2e
H2
Gas bubbles are released.
Gas bubbles are released.
– Insert a burning wooden splinter into
the test tube.
– A ‘Pop’ sound is produced.
– Insert a burning wooden splinter into
the test tube.
– A ‘pop’ sound is produced.
Hydrogen gas
Hydrogen gas
Cl– , OH–
Cl– , OH–
4OH–
2H2O + O2 + 4e
2Cl–
Cl2 + 2e
Gas bubbles are released.
Confirmatory test at anode (method
and observations)
– Insert a glowing wooden splinter into – A damp blue litmus paper placed near
the test tube.
the mouth of the test tube.
– Glowing wooden splinter is lighted up. – The gas changed the damp blue litmus
paper to red and then bleached it.
Name the product at the anode
Nama hasil di anod
The concentration of hydrochloric acid
after a while and explanation
Kepekatan elektrolit selepas beberapa ketika
dan terangkan
Greenish yellow gas is released.
Oxygen gas
Chlorine gas
Concentration of hydrochloric acid
increases . Hydrogen gas is released at
Concentration of hydrochloric acid
decreases . Hydrogen gas released at
the cathode and oxygen gas is released
the cathode and chlorine gas released at
at the anode. Water decomposed to
the anode. Concentration of chloride
oxygen
gas and hydrogen gas. ions decreases.
bertambah .
Kepekatan asid hidroklorik
Gas hidrogen dibebaskan di katod dan gas
oksigen dibebaskan di anod. Air terurai kepada
gas oksigen dan gas hidrogen .
berkurang
Kepekatan asid hidroklorik
Gas hidrogen dibebaskan di katod dan gas
klorin dibebaskan di anod. Kepekatan ion
klorida berkurang.
.
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HCl 2.0 mol dm–3
Observations at anode / Pemerhatian
Ujian pengesahan (kaedah dan pemerhatian)
m
2.0 mol dm-3 of HCl
HCl 0.001 mol dm–3
Elektrolit
Ion bergerak ke katod
H+ + OH–
0.001 mol dm-3 of HCl
Electrolyte
Ions that are attracted to the cathode
H+ + Cl–
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Chemistry Form 4 • MODULE
(d) Complete the following table for the electrolysis of 2.0 mol dm–3 sodium iodide solution using carbon electrodes.
Lengkapkan jadual berikut bagi elektrolisis larutan natrium iodida 2.0 mol dm–3 menggunakan elektrod karbon.
Set-up of apparatus
Susunan radas
Carbon electrodes
Sodium iodide
NaI
H2O
Equation of electrolyte ionisation
Persamaan pengionan elektrolit
Na+ + I–
H+ + OH–
Electrode / Elektrod
Anode / Anod
Cathode / Katod
Ions that are attracted to the anode and
cathode
I–, OH–
Na+, H+
Ion yang ditarik ke anod dan katod
Half equation
2I–
Persamaan setengah
Name of the products
Nama hasil
Observations
Pemerhatian
Confirmatory test (method and
observations)
2H+ + 2e
H2
Iodine
Hydrogen
Brown solution is formed.
Gas bubbles are released.
– A few drops of starch solution added.
– Starch solution turns to dark blue.
– When a lighted wooden splinter is
placed near the mouth of the test tube.
– A ‘pop’ sound is produced.
Ujian pengesahan (kaedah dan pemerhatian)
5
I2 + 2e
Types of electrode
Jenis elektrod:
(a) There are two types of electrode
Terdapat dua jenis elektrod:
(i) Inert electrode – An electrode that acts as a conductor only and does not undergo any chemical changes.
Normally they are made of carbon or platinum.
(ii)
Elektrod lengai – Elektrod yang bertindak sebagai pengalir arus sahaja dan tidak mengalami perubahan kimia. Biasanya
diperbuat daripada karbon atau platinum.
Reactive electrode – An electrode that not only acts as a conductor but also undergoes chemical changes.
During the electrolysis, the metal atom at the anode releases electron to form metal ion, metal anode
becomes thinner while the less electropositive cation will be selected at the cathode which consist of metal
electrodes such as copper, silver and nickel.
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Elektrod reaktif – Elektrod yang bertindak bukan sahaja sebagai pengalir arus tetapi juga mengalami perubahan kimia.
Semasa proses elektrolisis berlaku, atom logam pada anod melepaskan elektron menjadi ion logam, anod logam menjadi
nipis manakala ion yang kurang elektropositif akan menyahcas di katod yang terdiri daripada logam seperti kuprum, argentum
dan nikel.
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MODULE • Chemistry Form 4
(b) Complete the following table for the electrolysis of 1 mol dm–3 copper(II) sulphate solution with carbon electrode
and copper electrode.
Lengkapkan jadual berikut bagi elektrolisis larutan kuprum(II) sulfat 1 mol dm–3 menggunakan elektrod karbon dan elektrod kuprum.
Set-up of apparatus
Susunan radas
Copper(II)
sulphate
Copper
electrodes
Carbon
electrodes
CuSO4 (aq / ak )
Cu2+ + SO42–
H2O
H+ + OH–
Equation of electrolyte ionisation
Persamaan pengionan elektrolit
Type of electrode
Carbon electrode
Jenis elektrod
The ions that move to the cathode
Ion bergerak ke katod
Half equation at the cathode
Persamaan setengah di katod
Name the product at the cathode
Nama hasil di katod
Observation at cathode
Pemerhatian di katod
The ions that move to the anode
Ion bergerak ke anod
Half equation at the anode
Persamaan setengah di anod
Name the product at anode
Nama hasil di anod
Observations at the anode
Pemerhatian di anod
Confirmatory test (method and
observations)
Ujian pengesahan (kaedah dan
pemerhatian )
The concentration of copper(II)
solution after a while and
explanation
m
Copper electrode
Elektrod karbon
Cu2+, H+
Elektrod kuprum
Cu2+, H+
Cu2+ + 2e
Cu
Cu2+ + 2e
Cu
Copper
Copper
Brown solid deposited
Brown solid deposited
SO42–, OH–
SO42–, OH–
4OH–
2H2O + O2 + 4e
Cu
Cu2+ + 2e
Oxygen gas
Copper(II) ion
– Gas bubbles are released.
– Intensity of blue colour decreases.
– Copper electrode becomes thinner.
– Intensity of blue colour remains
unchanged.
– Insert a glowing wooden splinter into
the test tube.
– Glowing wooden splinter is lighted up.
– Concentration of copper(II) sulphate
solution decreases.
– Copper(II) ions discharge as copper
atoms and deposited the cathode.
–
– Concentration of copper(II) sulphate
solution remains unchanged.
– The number of copper atoms form
copper(II) ions at the anode is equal to
the number of copper(II) ions form copper
atoms at the cathode.
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ketika dan terangkan
Copper(II)
sulphate
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Chemistry Form 4 • MODULE
EXERCISE / LATIHAN
1
Complete the table below / Lengkapkan jadual di bawah:
Electrolyte
Elektrolit
Electrode Factor that affects
Elektrod
electrolysis
Ions present
Ion yang hadir
Faktor yang
mempengaruhi
elektrolisis
Dilute
sulphuric acid
Carbon
H+, SO42–, OH–
Concentration of
electrolyte
H+, Cl–, OH–
Karbon
Position of ion in
the electrochemical
series
Ag+, NO3–, H+,
OH–
Silver
Type of electrode
Asid sulfurik cair
Concentrated
hydrochloric
acid
Persamaan setengah di anod dan
pemerhatian
Position of ion in
the electrochemical
series
Karbon
Carbon
Karbon
Carbon
Larutan
argentum nitrat
Silver nitrate
solution
4OH–
2H2O + O2 + 4e
Half equation at the
cathode and observation
Persamaan setengah di katod
dan pemerhatian
2H+ + 2e
H2
Gas bubbles are released.
Gas bubbles are released.
2Cl–
2H+ + 2e
Cl2 + 2e
H2
Greenish yellow gas is
released.
Gas bubbles are released.
4OH–
Gas bubbles are released.
Ag+ + e
Ag
Grey shiny solid deposited.
Ag+, NO3–, H+,
OH–
Ag
Ag+ + e
Anode becomes thinner.
Ag+ + e
Ag
Grey shiny solid deposited.
Position of ion in
the electrochemical
series
K+, I–, H+, OH–
4OH–
2H+ + 2e
Concentration of
electrolyte
K+, I–, H+, OH–
Position of ion in
the electrochemical
series
K+, SO42–, H+,
OH–
Asid hidroklorik
pekat
Silver nitrate
solution
Half equation at the
anode and observation
Argentum
2H2O + O2 + 4e
Larutan
argentum nitrat
Carbon
Dilute
Karbon
potassium
iodide solution
2H2O + O2 + 4e
H2
Gas bubbles are released.
Gas bubbles are released.
2I–
2H+ + 2e
Larutan kalium
iodida cair
Carbon
Concentrated
Karbon
potassium
iodide solution
I2 + 2e
H2
Brown solution formed.
Gas bubbles are released.
4OH–
2H+ + 2e
Larutan kalium
iodida pekat
Dilute
potassium
sulphate
solution
Carbon
Karbon
2H2O + O2 +4e
Gas bubbles are released.
H2
Gas bubbles are released.
Larutan kalium
sulfat cair
2
Electrolysis is carried out on a dilute potassium chloride solution using carbon electrodes. Explain how this electrolysis
occurs. Use a labelled diagram to explain your answer.
Proses elektrolisis dijalankan ke atas larutan kalium klorida cair menggunakan elektrod karbon. Jelaskan bagaimana proses elektrolisis ini
berlaku. Gunakan gambar rajah berlabel untuk menerangkan jawapan anda.
Set-up of apparatus / Susunan radas:
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Carbon electrode
Dilute potassium
chloride solution
Carbon electrode
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MODULE • Chemistry Form 4
Explanation / Penerangan:
K+, H+, Cl–
– Potassium chloride solution consist of
K , H , Cl
+
Larutan kalium klorida mengandungi ion
Cl–
–
ion and
Cl –
Ion
OH–
–
–
Ion OH
–
OH
dan
terletak di bawah ion
–
2H2O + O2 + 4e
is lower than
K+ ion
water
and
oksigen
dipilih untuk dinyahcaskan dengan melepaskan elektron membentuk molekul
molecule.
air
dan
.
.
Ion K +
dan
ion H +
bergerak ke katod.
in the electrochemical series.
terletak di bawah
H+ ion
is selectively discharged by receiving electrons to form
Ion H
oxygen
is selectively discharged by releasing electrons to form
Ion H +
+
yang bergerak bebas.
dalam siri elektrokimia.
– Half equation / Persamaan setengah:
K+ ion
H+ ion
–
and
move to the cathode /
–
ions that move freely.
–
ion in the electrochemical series.
Cl –
4OH–
H+ ion
OH–
and
bergerak ke anod.
Cl–
ion is lower than
OH– ion
–
ions move to the anode.
OH –
dan ion
OH –
Ion
OH–
+
ion K +
dalam siri elektrokimia
hydrogen
dipilih untuk dinyahcaskan dengan menerima elektron membentuk molekul
2H + + 2e
– Half equation / Persamaan setengah:
H2
molecules.
hidrogen
.
.
Describe an experiment to determine the product of electrolysis copper(II) sulphate solution with carbon electrode.
3
Your answer should include the observation, confirmatory test for the product at the anode and half equation at the
electrode.
Huraikan satu eksperimen untuk menentukan hasil elektrolisis larutan kuprum(II) sulfat menggunakan elektrod karbon. Dalam jawapan
anda perlu disertakan pemerhatian, ujian pengesahan untuk hasil yang terbentuk di anod dan persamaan setengah bagi tindak balas yang
berlaku di elektrod.
Answer / Jawapan:
Apparatus / Radas : Battery / power supply, carbon electrodes, wire, electrolytic cell, test tube, Ammeter [from a
labelled diagram]
Material / Bahan
–3
: 1 mol dm copper(II) sulphate solution
Carbon electrodes
Copper(II) sulphate
solution
Procedure / Langkah:
–3
(a) Pour 1 mol dm copper(II) sulphate
Masukkan
larutan
kuprum(II) sulfat
solution
1 mol dm–3
in the electrolytic cell until it is
ke dalam sel elektrolitik sehingga
(b) The apparatus is set up as shown in the diagram. Fill the
anode
invert the test tube on the
.
Radas disusunkan seperti dalam gambar rajah. Isi
anod
.
uji itu pada
tabung uji
test tube
dengan
half full
with copper(II) sulphate
larutan
.
separuh penuh
.
solution
and
kuprum(II) sulfat dan terbalikkan tabung
(c) Turn on the switch / Hidupkan suis.
(d) Collect the gas produced at the
anode
(e) Gas produced at the
m
anod
anod
/ Kumpulkan gas yang terhasil di
glowing wooden splinter
is tested with a
.
diuji dengan
kayu uji berbara
.
.
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Chemistry Form 4 • MODULE
Observation and half equation / Pemerhatian dan persamaan setengah:
Electrodes
Observation
Elektrod
4
Confirmatory test
Pemerhatian
Half equation
Ujian pengesahan
Cathode
Brown solid deposited
Anode
Gas bubbles are released
Persamaan setengah
Cu2+ + 2e
–
– Insert the glowing wooden splinter into the test
tube.
– The glowing wooden splinter is lighted up.
4OH–
Cu
2H2O + O2 + 4e
Copper(II) sulphate solution is electrolysed using copper electrodes.
Larutan kuprum(II) sulfat dielektrolisis dengan menggunakan elektrod kuprum.
(a) Write the formula of all the anions present in the solution / Tuliskan formula semua anion yang terdapat dalam larutan itu.
SO42–, OH–
(b) Write the half equation for the reaction at the / Tuliskan persamaan setengah untuk tindak balas di
Cu2+ + 2e
(i) anode / anod : Cu
(ii)
(c) (i)
2+
Cu
cathode / katod : Cu + 2e
From your observations, what happen to the intensity of the blue colour of the copper(II) sulphate solution
during electrolysis?
Daripada pemerhatian anda, nyatakan apakah yang berlaku ke atas keamatan warna biru larutan kuprum(II) sulfat semasa
proses elektrolisis?
The intensity of the blue colour of copper(II) sulphate remains unchanged.
(ii)
Explain your answer / Jelaskan jawapan anda.
The number of copper(II) ions become copper atoms at the cathode is equal to the number of copper atoms
become copper(II) ions at the anode.
(d) If the experiment is repeated with the copper electrodes being replaced by carbon electrodes, state the name of the
products formed at the
Jika eksperimen diulangi dengan menggantikan elektrod kuprum dengan elektrod karbon, namakan hasil yang terbentuk di
(i)
5
Oxygen
anode / anod:
(ii)
Copper
cathode / katod:
The diagram below shows the set-up of apparatus of an electrolytic cell.
Rajah di bawah menunjukkan susunan radas bagi sel elektrolisis.
Carbon electrode P
Carbon electrode Q
Elektrod karbon P
Elektrod karbon Q
Copper(II) nitrate solution
Larutan kuprum(II) nitrat
(a) Write the formula of all ions present in copper(II) nitrate solution.
Tuliskan formula semua ion yang hadir dalam larutan kuprum(II) nitrat.
Cu2+, NO3–, H+ and OH–
.
(b) Write half equation for the reaction at / Tuliskan persamaan setengah di:
2+
Cu
electrode P / elektrod P : Cu + 2e
–
2H2O + O2 + 4e
electrode Q / elektrod Q : 4OH
(c) (i) What is the colour of copper(II) nitrate / Apakah warna larutan kuprum(II) nitrat?
Blue
(ii)
What happens to the intensity of the colour of copper(II) nitrate solution? Explain your answer.
Apakah yang berlaku kepada keamatan warna larutan kuprum(II) nitrat? Jelaskan jawapan anda.
The intensity of the blue colour of copper(II) nitrate decreases. The concentration of Cu2+ decreases because
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copper(II) ions receive electrons to form copper atom at the cathode.
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MODULE • Chemistry Form 4
ELECTROLYSIS IN INDUSTRY / ELEKTROLISIS DALAM INDUSTRI
Three uses of electrolysis in industries are / Tiga kegunaan elektrolisis dalam industri ialah:
1
Application
Aplikasi
(a) Electroplating
Penyaduran logam
Example
Electrolyte
Contoh
Silver
electroplating
Elektrolit
Silver nitrate
solution
Penyaduran perak
(b) Purification of
metal
Penulenan logam
(c) Metal extraction
Pengekstrakan
logam
Purification of
copper
Molten
aluminium oxide
Pengekstrakan
aluminium
Cathode / Half equation
Anod / Persamaan setengah
Katod / Persamaan setengah
Anode / Anod:
Silver metal
Cathode / Katod:
Metal to be electroplated
Half equation / Persamaan setengah:
Ag Ag+ + e
Half equation / Persamaan setengah:
Ag+ + e Ag
Copper(II)
Anode / Anod:
sulphate solution Impure copper
Penulenan
kuprum
Extraction of
aluminium
Anode / Half equation
Cathode / Katod:
Pure copper
Half equation / Persamaan setengah:
Cu Cu2+ + 2e
Half equation / Persamaan setengah:
Cu2+ + 2e Cu
Anode / Anod:
Carbon
Cathode / Katod:
Carbon
Half equation / Persamaan setengah:
2O2– O2 + 4e
Half equation / Persamaan setengah:
Al3+ + 3e Al
The following diagram shows the aluminium extraction process.
2
Rajah di bawah menunjukkan proses pengekstrakan aluminium.
Substance Z / Bahan Z
Substance Y
Bahan Y
Substance X + cryolite
Substance W
Bahan X + kriolit
Bahan W
(a) State the name of the following substances / Nyatakan nama bahan-bahan berikut:
W : Liquid aluminium
X : Molten aluminium oxide
Y : Carbon
Z : Carbon
(b) Which substance acts as anode and cathode / Bahan yang manakah bertindak sebagai anod dan katod?
Anode / Anod : Z
Cathode / Katod : Y
(c) State the name of the product at anode and cathode / Namakan hasil yang diperoleh di anod dan katod.
Anode / Anod : Oxygen
Cathode / Katod : Aluminium
(d) Write the ionic equation for the reactions at / Tuliskan persamaan ion bagi tindak balas yang berlaku di
2–
3+
O2 + 4e
Al
anode / anod : 2O
cathode / katod : Al + 3e
(e) Why is cryolite added to X / Mengapakan kriolit ditambah ke dalam X ?
To lower down the melting point of aluminium oxide (from 2 045°C to 900°C ).
The diagram below shows the set-up of apparatus used in the purification of copper.
3
Rajah di bawah menunjukkan susunan radas yang digunakan untuk proses penulenan kuprum.
Electrode X
Electrode Y
Elektrod X
Elektrod Y
Electrode Z
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Chemistry Form 4 • MODULE
(a) State the name of the substance used as / Nyatakan nama bahan yang dijadikan sebagai:
electrode X / elektrod X : Impure copper
: Pure copper
electrolyte Z / elektrolit Z : Copper(II) sulphate solution
electrode Y / elektrod Y
(b) Write the half equation that occur at the / Tuliskan persamaan setengah yang berlaku di
Cu
Cu2+ + 2e
electrode X / elektrod X :
electrode Y / elektrod Y :
Cu2+ + 2e
Cu
(c) What are the observations at the / Apakah pemerhatian di
electrode X / elektrod X : Electrode becomes thinner
electrode Y / elektrod Y : Brown solid deposited
4
To purify metal an impure metal / Untuk menulenkan logam tak tulen:
Logam tak tulen
impure metal
(a) The
is used as the anode /
(b) The
pure metal
dijadikan sebagai anod.
Logam tulen
dijadikan sebagai katod.
is used as the cathode /
salt solution
(c) The electrolyte used is an
containing the ions of the purifying metal.
Elektrolit adalah
5
larutan garam
yang mengandungi ion logam yang hendak ditulenkan.
A student intends to electroplate an iron spoon with copper. Describe a laboratory experiment to electroplate the iron
ring. Your answer should involve the following:
Seorang pelajar bercadang untuk menyadurkan sebatang sudu besi dengan kuprum. Huraikan satu eksperimen di dalam makmal untuk
menyadur sebatang sudu besi. Jawapan anda perlu mengandungi:
– A labelled diagram showing the set-up of apparatus / Rajah berlabel menunjukkan susunan radas.
– Procedure / Kaedah.
– Half equation for the reactions at both electrodes / Persamaan setengah untuk tindak balas di kedua-dua elektrod.
– Observation at both electrodes / Pemerhatian di kedua-dua elektrod.
Answer / Jawapan:
Copper
Iron spoon
Copper(II) nitrate solution
Procedure / Kaedah:
(a) Copper plate and iron spoon are cleaned with
Kepingan kuprum dan sudu besi dibersihkan dengan
(b)
Copper(II) nitrate solution
Larutan kuprum(II) nitrat
(c)
sand paper
.
kertas pasir
.
beaker
is poured into a
dituangkan ke dalam bikar sehingga
until
half full
Iron spoon
is then connected to the negative terminal of battery while the copper plate is connected to the
positive terminal of the battery// Iron spoon is made as cathode while copper plate is made as anode.
Sudu besi
bateri//
disambungkan kepada terminal negatif bateri dan
Sudu besi dijadikan katod dan kepingan kuprum
(d) The iron spoon and the copper plate are dipped in the
Sudu besi dan plat kuprum
(e) The circuit is
completed
dicelup
kepingan kuprum
disambungkan kepada terminal positif
dijadikan anod.
copper(II) nitrate solution
larutan kuprum(II) nitrat
ke dalam
/ Litar dilengkapkan .
.
pepejal perang
Copper plate becomes thinner
terenap.
.
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Observation of the anode / Pemerhatian di anod :
as shown in the diagram.
seperti ditunjukkan dalam rajah.
Cu2+ + 2e Cu
Half equation at the cathode / Persamaan setengah di katod :
(g) Observation of the cathode: Brown solid is deposited / Pemerhatian di katod:
Cu Cu2+ + 2e
(h) Half equation at the anode / Persamaan setengah di anod :
(f)
(i)
.
separuh penuh .
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MODULE • Chemistry Form 4
To electroplate an object with metal / Untuk menyadur sesuatu objek dengan logam:
(a) The metal object to be electroplated is made to be cathode / Objek yang hendak disadur dijadikan
anod
anode
..
(b) The electroplating metal is made to be
/ Logam penyadur dijadikan
6
katod
..
(c) The electrolyte used is an aqueous salt solution containing the ions of the electroplating metal.
Elektrolit
yang digunakan adalah larutan akueus garam yang mengandungi ion logam penyadur.
ELECTROCHEMICAL SERIES / SIRI ELEKTROKIMIA
Electrochemical Series is an arrangement of
positive ion.
1
Siri Elektrokimia ialah susunan
logam
metals
according to their tendency to release electrons to form a
mengikut kecenderungan melepaskan elektron membentuk ion bercas
positif
.
The position of metal atoms in Electrochemical Series / Kedudukan atom logam dalam Siri Elektrokimia:
K, Na, Ca, Mg, Al, Zn , Fe, Sn ,Pb, Cu, Ag
2
Tendency of metal atom to release/donate electrons increases (electropositivity increases)
Kecenderungan untuk atom logam melepaskan/menderma elektron bertambah (keelektropositifan bertambah)
The position of metal ions (cation) in the Electrochemical Series / Kedudukan ion logam (kation) dalam Siri Elektrokimia:
K+, Na+, Ca2+, Mg2+, Al3+, Zn2+, Fe2+, Sn2+, Pb2+, *H+, Cu2+
3
Tendency of metal ion (cation) to receive/gain electrons increases
Kecenderungan untuk ion logam (kation) untuk menerima elektron bertambah
*H+ is also in the series of ion because it is present in aqueous solution of any electrolyte (salt solution/acid/alkali)
* H+ juga terdapat dalam siri ion kerana kehadiran ion H+ dalam elektrolit larutan akueus (larutan garam/asid/alkali)
METAL DISPLACEMENT REACTION / TINDAK BALAS PENYESARAN LOGAM
The metal which is situated at a higher position (higher tendency to release electron) in the Electrochemical Series is
metals
able to displace
below it from its salt solution .
1
Logam yang berada di kedudukan atas (kecenderungan melepaskan elektron yang tinggi) dalam Siri Elektrokimia dapat menyesarkan
logam
yang di bawahnya daripada larutan garam logam tersebut.
Example / Contoh:
2
Experiment / Eksperimen
Silver nitrate solution
Larutan argentum nitrat
Observation / Pemerhatian
– Copper strip becomes
thinner .
Kepingan kuprum
menipis .
grey
– A
deposited.
solid
Pepejal kelabu terenap.
– The colourless
solution turns blue.
Copper
Kuprum
Larutan tidak berwarna
bertukar menjadi biru.
Remark / Catatan
Inference / Inferens:
grey
solid is
– The
Pepejal
kelabu
adalah
.
argentum
.
copper(II) nitrate
– The blue solution is
Larutan biru adalah
silver
.
kuprum(II) nitrat .
Explanation / Penerangan:
Silver ion receives electrons to form
–
Ion
argentum
menerima elektron membentuk atom
– Copper atom releases electrons to form
Atom kuprum melepaskan elektron membentuk
silver
– Copper has displaced
Kuprum telah menyesarkan
Cu + 2AgNO3
– Copper is
above
silver
more
atom.
argentum .
copper(II) ion
ion kuprum(II)
.
.
from silver nitrate solution.
argentum
dari larutan argentum nitrat.
Cu(NO3)2 + 2Ag
.
electropositive than silver// Copper is
silver in the Electrochemical Series of metal.
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lebih
elektropositif daripada argentum //Kuprum
Kuprum adalah
di atas
terletak
argentum dalam Siri Elektrokimia logam.
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Chemistry Form 4 • MODULE
– Magnesium strip
becomes thinner .
Copper(II) sulphate
solution
Kepingan magnesium
menipis .
Larutan kuprum (II) sulfat
– The brown
deposited.
Inference / Inferens:
– The brown solid is
perang
Pepejal
solid
Pepejal perang terenap.
– The blue solution turn
colourless.
Larutan biru bertukar
menjadi tidak berwarna.
Magnesium
Magnesium
copper .
adalah
– The colourless solution is
kuprum
.
magnesium sulphate
magnesium sulfat
Larutan tidak berwarna adalah
.
.
Explanation / Penerangan:
Copper(II) ion receives electrons to form copper atom.
–
Ion
kuprum(II)
menerima elektron membentuk atom kuprum.
magnesium ion
– Magnesium atom releases electrons to form
ion magnesium
Atom magnesium melepaskan elektron membentuk
.
.
– Magnesium has displaced
solution.
copper
from copper(II) sulphate
Magnesium telah menyesarkan
kuprum
dari larutan kuprum(II) sulfat.
MgSO4 + Cu
Mg + Cu SO4
.
more
electropositive than copper// Magnesium
– Magnesium is
above
copper in the Electrochemical Series of metal.
is
Magnesium adalah
di atas
terletak
No observable changes.
Zinc nitrate solution
Tiada perubahan yang dapat
diperhatikan.
Larutan zink sulfat
lebih
elektropositif daripada kuprum// magnesium
kuprum dalam Siri Elektrokimia logam.
Inference / Inferens:
– No reaction occur.
Tiada tindak balas berlaku.
Explanation / Penerangan:
– Copper cannot displace
Kuprum
tidak boleh
zinc
menyesarkan
from zinc sulphate solution.
zink
daripada larutan zink sulfat.
more
electropositive than zinc// Copper is
– Copper is
zinc in the Electrochemical Series of metal.
below
Kuprum adalah kurang elektropositif daripada zink // kuprum terletak
di bawah zink dalam Siri Elektrokimia logam.
Copper / Kuprum
VOLTAIC CELL (CHEMICAL CELL) / SEL RINGKAS (SEL KIMIA)
1
A cell that produces electrical energy when chemical reactions occur in it.
Sel yang menghasilkan tenaga elektrik apabila berlaku tindak balas kimia di dalamnya.
2
electrical energy
Energy change in voltaic cell is chemical energy to
Perubahan tenaga dalam sel ringkas ialah dari tenaga kimia kepada
3
Produced when two
different
Terhasil apabila dua logam
4
5
.
metals are dipped in an electrolyte and are connected by an
berlainan
elektrolit
dicelup dalam
The voltage of chemical cell depends on the
distance
the further the distance between them, the
higher
Voltan sel kimia bergantung pada
tinggi
Elektrokimia, semakin
.
tenaga elektrik
jarak
dan disambung dengan
litar luar
external circuit
.
.
between the two metals in the Electrochemical Series, where
is the voltage.
antara dua logam dalam Siri Elektrokimia di mana semakin jauh dua logam dalam Siri
voltannya.
A more electropositive metal becomes the
positive terminal:
negative
negatif
sel. Logam yang kurang elektropositif akan menjadi terminal
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Logam yang lebih elektropositif akan menjadi terminal
positif
sel:
terminal of the cell. A less electropositive metal becomes the
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MODULE • Chemistry Form 4
Electrical current produced is detected by the galvanometer
Electrical energy)
(Chemical energy
Arus elektrik terhasil dikesan oleh galvanometer
(Tenaga kimia Tenaga elektrik)
Negative terminal / Terminal negatif
• More electropositive metal.
G
:
_
Positive terminal / Terminal positif
• Less electropositive metal.
+
Logam lebih elektropositif.
:
Logam kurang elektropositif.
• Metal atom will release electrons that will
flow through the external circuit. Metal atom
becomes metal ion (becomes thinner).
• The electrons that flow from the external
circuit are received by the positive ion in
the electrolyte through this terminal.
Atom logam akan melepaskan elektron yang akan
mengalir di litar luar. Atom logam menjadi ion logam
(semakin nipis).
Elektron yang akan mengalir dari litar luar diterima
oleh ion positif dalam elektrolit melalui terminal ini.
Example of simple voltaic cell / Contoh voltan sel ringkas:
6
V
Magnesium
Copper
Magnesium
Kuprum
Copper(II) sulphate solution
Larutan kuprum(II) sulfat
(a) Magnesium electrode is a
negative
negatif
Elektrod magnesium adalah terminal
–
Magnesium atom
Atom magnesium
releases
melepaskan
copper
:
kuprum :
daripada
elektron untuk membentuk ion magnesium, Mg2+.
Mg
Mg2+ + 2e
–
Magnesium electrode becomes
–
Electron flows through external circuit to the
thinner
positive
.
copper
kuprum
.
electrode.
.
electropositive
terminal because copper is less
positif
nipis
/ Elektrod magnesium menjadi
Elektron mengalir melalui litar luar ke elektrod
–
than
electrons to form magnesium ion, Mg2+.
Half equation / Persamaan setengah :
Elektrod kuprum adalah terminal
elektropositif
kerana magnesium lebih
–
(b) Copper electrode is a
electropositive
terminal because magnesium is more
kerana kuprum kurang
elektropositif
than
daripada
magnesium
:
magnesium
:
Electrons from magnesium flow through external circuit to copper electrode.
Elektron dari magnesium mengalir melalui litar luar ke elektrod kuprum.
–
Copper(II)
Ion
ion in the electrolyte
kuprum(II)
dalam elektrolit
receives
menerima
electron to form copper atom.
elektron untuk membentuk atom kuprum.
Cu + 2e Cu
Half equation / Persamaan setengah :
.
Brown solid is deposited on the surface of copper electrode.
–
+
–
Pepejal perang
terenap di permukaan elektrod kuprum.
(c) The concentration of copper(II) sulphate decreases because copper(II) ions discharged to copper atom at the
positive terminal. The intensity of blue colour of copper(II) sulphate decreases.
Kepekatan larutan kuprum(II) sulfat berkurang
warna biru larutan kuprum(II) sulfat berkurang.
kerana ion kuprum(II) dinyahcaskan kepada atom kuprum.
(d) If the magnesium metal is replaced with a zinc metal, the voltage reading
copper in the electrochemical series.
m
berkurang
because zinc is nearer to
kerana zink lebih dekat dengan kuprum
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Jika logam magnesium digantikan dengan logam zink, bacaan voltan akan
dalam siri elektrokimia.
decreases
Keamatan
d.
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Chemistry Form 4 • MODULE
7
Daniell cell / Sel Daniell
(a) It is an example of voltaic cell which consists of zinc electrode dipped in zinc sulphate solution, copper electrode
dipped in copper(II) sulphate solution and connected by a salt bridge or porous pot.
Merupakan satu contoh sel kimia yang terdiri daripada elektrod zink yang dicelup ke dalam larutan zink sulfat, elektrod kuprum
dicelupkan ke dalam larutan kuprum(II) sulfat dan dihubungkan dengan titian garam atau pasu berliang.
Zn / ZnSO4 // CuSO4 / Cu
(b) The function of porous pot or salt bridge is to allow the flow of ions through it so that the electric circuit is
completed.
Fungsi pasu berliang atau titian garam adalah untuk membenarkan ion-ion mengalir melaluinya dan melengkapkan litar.
(c) The diagram below shows the set-up of apparatus of Daniell cell.
Rajah di bawah menunjukkan susunan radas bagi sel Daniell.
Sulphuric acid
Asid sulfurik
Copper
Copper
Copper(II) sulphate
solution
(d) Zinc electrode is a
Zinc sulphate
Elektrod zink adalah terminal
–
releases
Zinc atom
Atom zink
melepaskan
than
daripada
copper
kuprum
:
:
electron to form zinc ion, Zn .
2+
elektron untuk membentuk ion zink, Zn2+.
Zn
Zn2+ + 2e
Half equation / Persamaan setengah :
–
Zinc electrode becomes
–
Electrons flow through external circuit to the
thinner
positive
Elektrod kuprum adalah terminal
.
nipis
/ Elektrod zink menjadi
copper
kuprum
Elektron mengalir melalui litar luar ke elektrod
–
elektropositif
kerana zink adalah lebih
–
(e) Copper electrode is a
Pasu berliang
electropositive
terminal because zinc is more
negatif
Porous pot
Larutan kuprum(II) sulfat
Zink sulfat
negative
Zink sulfat
Copper(II) sulphate
solution
Zink
Larutan kuprum(II) sulfat
Zinc / Zink
Zinc sulphate
Kuprum
Zinc
Kuprum
electrode.
.
electropositive
terminal because copper is less
positif
..
elektropositif
kerana kuprum kurang
than
daripada
zinc
:
zink
:
Electrons from zinc electrode flow through external circuit to copper electrode.
Elektron dari zink mengalir melalui litar luar ke elektrod kuprum.
–
Copper(II)
Ion
–
–
(f)
ion in the electrolyte
kuprum(II)
dalam elektrolit
receives
menerima
Half equation / Persamaan setengah :
electron to form copper atom.
elektron untuk membentuk atom kuprum.
Cu
2+
+ 2e
Cu
.
Brown solid
is deposited on the surface of copper electrode.
Pepejal perang
terenap di permukaan elektrod kuprum.
The concentration of copper(II) sulphate decreases because copper(II) ions are discharged to copper atoms. The
intensity of blue colour of copper(II) sulphate decreases.
Kepekatan larutan kuprum(II) sulfat berkurang
warna biru kuprum(II) sulfat berkurang.
kerana ion kuprum(II) telah dinyahcaskan kepada atom kuprum.
(g) If zinc metal is replaced with a magnesium metal, the voltage reading
further
from copper in the Electrochemical Series.
increases
Keamatan
because magnesium is
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Jika logam zink digantikan dengan logam magnesium, bacaan voltan bertambah kerana jarak antara magnesium dengan kuprum
jauh
daripada jarak antara zink dengan kuprum dalam Siri Elektrokimia.
lebih
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MODULE • Chemistry Form 4
Four main uses of the Electrochemical Series / Kegunaan utama Siri Elektrokimia:
(a) To predict the terminal of chemical cell / Untuk meramalkan terminal sel kimia
– The more electropositive metal is the negative terminal of the cell.
8
Logam yang lebih elektropositif ialah terminal negatif sel.
–
The less electropositive metal is the positive terminal of the cell.
Logam yang kurang elektropositif ialah terminal positif sel.
(b) To predict the voltage of chemical cell / Untuk meramalkan voltan sel kimia
– The further the distance between two metals in the Electrochemical Series, the higher is the voltage of the
chemical cell.
Semakin jauh jarak antara dua logam dalam Siri Elektrokimia, semakin tinggi bacaan voltan sel kimia.
(c) To predict the metal displacement reactions / Untuk meramalkan tindak balas penyesaran logam
– The more electropositive metal can displace a less electropositive metal from its salt solution.
Logam yang lebih elektropositif dapat menyesarkan logam yang kurang elektropositif daripada larutan garamnya.
(d) To predict the selected ion to be discharged at the electrode in an electrolysis
Untuk meramalkan pemilihan ion untuk dinyahcas di elektrod dalam proses elektrolisis
EXERCISE / LATIHAN
The table below shows the results of an experiment to construct the Electrochemical Series through the ability of metals
to displace other metals from their salt solution.
1
Jadual di bawah menunjukkan keputusan eksperimen untuk membina Siri Elektrokimia berdasarkan keupayaan suatu logam untuk
menyesarkan logam lain dari larutan garamnya.
Experiment I / Eksperimen I
Experiment II / Eksperimen II
P nitrate solution
R nitrate solution
Larutan P nitrat
Larutan R nitrat
Zinc / Zink
Zinc / Zink
Metal P is displaced, blue colour solution turn colourless.
Logam P disesarkan, larutan biru bertukar menjadi tanpa warna.
No reaction.
Tiada tindak balas.
(a) Based on the results in the table, arrange metal P, zinc and R in descending order of electropositivity.
Berdasarkan keputusan dalam jadual, susunkan logam P, zink dan R dalam tertib menurun keelektropositifan.
R, Zn, P
(b) Based on the observation in Experiment I / Berdasarkan pemerhatian dalam Eksperimen I,
(i) state the name the suitable metal P / namakan logam yang sesuai bagi P.
Copper
(ii)
.
zinc can displace metal P from P nitrate solution. Explain.
zink boleh menyesarkan logam P daripada larutan P nitrat. Terangkan.
Zinc is more electropositive than P.
.
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(iii) write the chemical equation for the reaction / tuliskan persamaan kimia untuk tindak balas.
Zn + Cu(NO3 )2
Zn(NO3 )2 + Cu
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Chemistry Form 4 • MODULE
2
The diagram below shows the set-up of the apparatus to arrange metals W, X, Y and Z based on the potential difference
of the metals.
Rajah di bawah menunjukkan susunan radas bagi eksperimen untuk menentukan kedudukan logam W, X, Y dan Z berdasarkan beza upaya
logam.
V
Metal electrode
Metal electrode
Elektrod logam
Elektrod logam
Electrolyte / Elektrolit
The table below shows the results of the experiment.
Jadual di bawah menunjukkan keputusan eksperimen.
Pair of metals
Pasangan logam
Potential difference (V)
Negative terminal
0.50
X
0.30
Y
1.10
Z
Beza keupayaan (V)
W and X
W dan X
X and Y
X dan Y
W and Z
W dan Z
Terminal negatif
(a) Arrange metals W, X, Y and Z in descending order of the electropositivity of metal.
Susunkan logam W, X, Y dan Z dalam tertib menurun keelektropositifan logam.
Z, Y, X, W
(b) (i)
.
Metals X and Z are used as electrodes in the diagram. State which metal acts as positive terminal.
Logam X dan Z digunakan sebagai terminal dalam rajah. Nyatakan logam yang manakah akan bertindak sebagai terminal
positif.
Metal X
Give reason for your answer in (b)(i) / Berikan sebab untuk jawapan anda di (b)(i).
(ii)
Metal X is less electropositive than metal Z.
.
(c) Predict the voltage of the cell in (b)(i) / Ramalkan nilai voltan dalam sel di (b)(i).
0.6 V
3
The diagram below shows the set-up of apparatus for two types of cell.
Rajah di bawah menunjukkan susunan radas untuk dua jenis sel.
Zinc
Copper
Copper
Zink
Kuprum
Kuprum
Copper(II) sulphate solution
Larutan kuprum(II) sulfat
Cell Y / Sel Y
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Cell X / Sel X
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MODULE • Chemistry Form 4
Complete the following table to compare cell X and cell Y :
Lengkapkan jadual berikut untuk membandingkan sel X dan sel Y :
Description
Cell X
Cell Y
Electrolytic cell
Chemical cell
Perkara
Sel X
Type of cell
Jenis sel
The energy change
Electrical energy
Perubahan tenaga
Ion presence in the
electrolyte
Sel Y
Chemical energy
Chemical energy
Cu2+, H+, SO42–, OH–
Ion hadir dalam
elektrolit
Electrical energy
Cu2+, H+, SO42–, OH–
Electrode
Anode / Anod: Copper
Negative terminal / Terminal negatif : Zinc
Elektrod
Cathode / Katod: Copper
Positive terminal / Terminal positif : Copper
Half equation
Anode / Anod: Cu
Negative terminal / Terminal negatif : Zn
Persamaan setengah
Observation
Pemerhatian
Cu2+ + 2e
2+
Cathode / Katod: Cu + 2e
Cu
Zn2+ + 2e
2+
Positive terminal / Terminal positif : Cu + 2e
Cu
Anode / Anod:
Negative terminal / Terminal negatif :
Copper electrode becomes thinner
Zinc electrode becomes thinner
Cathode / Katod:
Brown solid deposited
Positive terminal / Terminal positif :
Electrolyte / Elektrolit:
Electrolyte / Elektrolit:
Intensity blue colour of copper(II) sulphate
Intensity blue colour of copper(II) sulphate decreases
Brown solid deposited
remains unchanged
The diagram below shows the set-up of apparatus for an experiment.
4
Rajah di bawah menunjukkan susunan radas bagi suatu eksperimen.
V
–
+
Zinc / Zink
Anode
Cathode
Copper
Copper
Kuprum
Kuprum
Zinc sulphate solution
Copper(II) sulphate solution
Larutan zink sulfat
Larutan kuprum(II) sulfat
Porous pot
Copper(II) sulphate solution
Pasu berliang
Larutan kuprum(II) sulfat
Cell A / Set A
Cell B / Set B
(a) In the above diagram, label
Dalam gambar rajah di atas, label
(i)
the positive terminal and negative terminal Cell A,
(ii)
anode and cathode in Cell B.
anod dan katod bagi Sel B.
terminal positif dan terminal negatif bagi Sel A,
(b) What is the energy change in Cell A and Cell B?
Apakah perubahan tenaga dalam Sel A dan Sel B?
Cell A / Sel A : Chemical energy to electrical energy
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Cell B / Sel B : Electrical energy to chemical energy
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Chemistry Form 4 • MODULE
(c) What is the function of the porous pot in cell A?
Apakah fungsi pasu berliang dalam Sel A?
To allow the movement of ions through it.
(d) Referring to Cell A.
Merujuk kepada Sel A.
(i)
What is the observation at zinc electrode?
Apakah pemerhatian di elektrod zink?
Zinc electrode becomes thinner.
(ii)
Write the half equation for the reaction at zinc electrode.
Tuliskan persamaan setengah untuk tindak balas di elektrod zink.
Zn
Zn2+ + 2e
(iii) What is the observation at copper electrode / Apakah pemerhatian di elektrod kuprum?
Brown solid deposited.
(iv) Write the half equation for the reaction at copper electrode.
Tuliskan persamaan setengah untuk tindak balas di elektrod kuprum.
Cu2+ + 2e
(v)
Cu
.
After 30 minutes, what is the colour change of the copper(II) sulphate solution? Explain why.
Selepas 30 minit, apakah perubahan warna larutan kuprum(II) sulfat? Jelaskan mengapa.
– The intensity of blue colour decreases.
– Copper(II) ions are discharged to form copper atoms.
– Concentration of copper(II) ions in copper(II) sulphate decreases.
(e) Referring to Cell B.
Merujuk kepada Sel B.
(i)
What is the observation at the anode?
Apakah pemerhatian di anod?
Copper electrode becomes thinner.
(ii)
Write the half equation for the reaction at the anode.
Tuliskan persamaan setengah untuk tindak balas di anod.
Cu
Cu2+ + 2e
(iii) What is the observation at the cathode?
Apakah pemerhatian di katod?
Brown solid deposited.
(iv) Write the half equation for the reaction at copper electrode.
Tuliskan persamaan setengah untuk tindak balas di katod.
Cu2+ + 2e
(f)
Cu
The intensity of blue colour of copper(II) sulphate solution in the Cell B remains unchanged during the experiment.
Explain why.
Keamatan warna biru larutan kuprum(II) sulfat dalam Sel B tidak berubah semasa eksperimen. Jelaskan mengapa.
– The concentration of copper(II) sulphate remain unchanged.
– The rate of copper(II) ions discharged to copper atom at the cathode equals to the rate of copper atoms form
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copper(II) ions at the anode.
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MODULE • Chemistry Form 4
Objective Questions / Soalan Objektif
1
Which of the following is an electrolyte?
Antara berikut, yang manakah adalah elektrolit?
C
Copper electrode becomes
thicker
Copper electrode becomes
thinner
Elektrod kuprum semakin tebal
Elektrod kuprum semakin nipis
D
Gas bubbles are released
Copper electrode becomes
thicker
A Glacial ethanoic acid
Asid etanoik glasial
B
Molten naphthalene
C
Aqueous solution of zinc chloride
Naftalena lebur
Elektrod kuprum semakin tebal
Larutan akueus zink klorida
4
D Hydrogen chloride in methylbenzene
Hidrogen klorida dalam metilbenzena
2
Gelembung gas dibebaskan
The diagram below shows the set-up of apparatus of an
electrolysis process.
Rajah di bawah menunjukkan sususnan radas untuk proses elektrolisis.
The diagram below shows the set-up of apparatus used
to electrolyse substance X.
Electrolyte
Rajah di bawah menunjukkan susunan radas untuk elektrolisis
bahan X.
Elektrolit
Carbon electrode
P
Elektrod karbon
Q
Carbon electrode
Elektrod karbon
Carbon electrodes
Elektrod karbon
Which of the following electrolytes produce oxygen gas at
electrode Q?
Substance X
Antara elektrolit berikut, yang manakah membebaskan gas oksigen pada
elektrod Q?
Bahan X
I
Heat
Asid hidroklorik 1.0 mol dm–3
Panaskan
3
1.0 mol dm–3 hydrochloric acid
II
1.0 mol dm–3 sulphuric acid
Asid sulfurik 1.0 mol dm–3
Which of the following compounds can light up the bulb
when used as substance X?
III
Antara berikut, yang manakah boleh menyalakan mentol apabila
digunakan sebagai bahan X?
A Copper(II) nitrate / Kuprum(II) nitrat
B Lead(II) iodide / Plumbum(II) iodida
C Zinc carbonate / Zink karbonat
D Sodium carbonate / Natrium karbonat
IV
1.0 mol dm–3 nitric acid
A
B
C
D
Asid nitrik 1.0 mol dm–3
I and II only / I dan II sahaja
II and III only / II dan III sahaja
II and IV only / II dan IV sahaja
II, III and IV only / II, III dan IV sahaja
The diagram below shows the set-up of apparatus for
electrolysis of copper(II) sulphate solution.
Rajah di bawah menunjukkan susunan radas untuk elektrolisis larutan
kuprum(II) sulfat.
1.0 mol dm–3 potassium iodide solution
Larutan kalium iodida 1.0 mol dm–3
5
The table below shows the observation of electrolysis of a
substance Q using carbon electrode.
Jadual di bawah menunjukkan pemerhatian bagi elektrolisis bahan Q
menggunakan elektrod karbon.
Electrode
Observation
Elektrod
Copper
electrode X
Copper electrode Y
Elektrod kuprum Y
Elektrod
kuprum X
Copper(II)
sulphate solution
Anode
A greenish-yellow gas released
Anod
Gas kuning kehijauan terbebas
Cathode
A colorless gas which burns with a ‘pop’
sound is released
Katod
Gas tanpa warna terbakar dengan bunyi ‘pop’
dibebaskan
Larutan kuprum(II)
sulfat
What can be observed at the electrodes X and Y after
30 minutes?
Apakah yang dapat diperhatikan pada elektrod X dan Y selepas 30
minit?
A
B
X
Y
Copper electrode becomes
thinner
Copper electrode becomes
thicker
Elektrod kuprum semakin nipis
Elektrod kuprum semakin tebal
Copper electrode becomes
thinner
Gas bubbles are released
Pemerhatian
What is substance Q?
Apakah bahan Q?
A
B
C
D
1.0 mol dm–3 of hydrochloric acid.
Asid hidroklorik 1.0 mol dm–3.
1.0 mol dm–3 of potassium nitrate solution.
Larutan natrium nitrat 1.0 mol dm–3.
1.0 mol dm–3 of copper(II) chloride solution.
Larutan kuprum(II) klorida 1.0 mol dm–3.
1.0 mol dm–3 of magnesium bromide solution.
Larutan magnesium bromida 1.0 mol dm–3.
Gelembung gas dibebaskan
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Elektrod kuprum semakin nipis
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Chemistry Form 4 • MODULE
6
The diagram below shows the set-up of apparatus of a
chemical cell that shows the direction of electron flow from
zinc to metal Q.
8
The table below shows the information about three voltaic
cells.
Jadual di bawah menunjukkan maklumat tentang tiga sel kimia.
Rajah di bawah menunjukkan susunan radas sel kimia yang menunjukkan
arah pengaliran elektron ke logam Q.
Pair of metals
Pasangan logam
Q
Zinc
Zink
Larutan natrium
klorida cair
A
B
Apakah logam Q?
B
C
D
Copper
9
Iron
Beza upaya (V)
Terminal positif
W, Z
Z
3.1
X, Y
Y
0.3
W, X
X
1.8
Apakah beza upaya sel kimia apabila logam Y dipasangkan dengan logam
Z?
What is metal Q?
Kuprum
Potential difference (V)
What is the potential difference of the voltaic cell when metal
Y is paired with metal Z?
Dilute sodium
chloride solution
A
Positive
terminal
C
D
1.0 V
1.3 V
2.1 V
2.8 V
The diagram below shows the set-up of apparatus in a chemical
cell and electrolytic cell.
Rajah di bawah menunjukkan susunan radas bagi sel kimia dan sel
elekrolisis.
Besi
Aluminium
Aluminium
Magnesium
Magnesium
7
The diagram below shows the set-up of apparatus used to
purify impure copper by using electrolysis method.
Rajah di bawah menunjukkan susunan radas yang digunakan untuk
menulenkan kuprum tak tulen dengan menggunakan kaedah elektrolisis.
P
Zinc
Elektrolit Z
A
B
C
Antara berikut, yang manakah adalah kedudukan yang betul untuk
kuprum tulen dan kuprum tak tulen?
D
A
Elektrod Y
Impure copper
Pure copper
Kuprum tak tulen
Kuprum tulen
Electrolyte Z
Elektrolit Z
Copper(II) sulphate
solution
Larutan kuprum(II)
sulfat
B
Pure copper
Impure copper
Kuprum tulen
Kuprum tak tulen
Kuprum
Zinc
sulphate
solution
Which of the following shows the correct position of pure
copper and impure copper?
Electrode Y
Copper
Copper(II) sulphate solution
Larutan kuprum(II) sulfat
Antara berikut, yang manakah merupakan pemerhatian pada elektrod
R?
Electrolyte Z
Elektrod X
S
Which of the following is the observation at electrode R?
Y
Electrode X
R
Copper
Kuprum
Larutan
zink sulfat
X
Q
Zink
Pure copper
Sulphuric acid
Kuprum tak tulen
Kuprum tulen
Asid sulfurik
D
Pure copper
Impure copper
Sulphuric acid
Kuprum tulen
Kuprum tak tulen
Asid sulfurik
A colourless gas is released
Gas tanpa warna terbebas
A brown solid is deposited
Pepejal perang terenap
Jadual di bawah menunjukkan keputusan eksperimen untuk mengkaji
penyesaran logam daripada larutan garamnya menggunakan logam
lain.
Metal
Logam
Nitrate of Q
Nitrat bagi Q
Nitrate of S
Nitrat bagi S
P
Larutan kuprum(II)
sulfat
Impure copper
Electrode R becomes thicker
Elektrod R semakin tebal
10 The table below shows the results of an experiment to study the
displacement of metal from its solution using other metals.
Copper(II) sulphate
solution
C
Electrode R becomes thinner
Elektrod R semakin nipis
Q
–
– reaction occur / tindak balas berlaku
– no reaction / tiada tindak balas
Which of the following is the arrangement of metals P, Q and
R in ascending order of the tendency of the metals to form
ions?
Antara berikut, yang manakah adalah susunan logam P, Q dan R dalam
susunan menaik kecenderungan logam membentuk ion?
P, S, Q
Q, S, P
C
D
S, P, Q
S, Q, P
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B
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MODULE • Chemistry Form 4
6
ACID AND BASES
ASID DAN BES
ACID AND BASES / ACID DAN BES
• ACID / ASID
– To state the meaning of acid, give examples and write chemical equations and observations for the reaction of acids:
Menyatakan maksud asid, memberi contoh dan menulis persamaan tindak balas kimia dan pemerhatian bagi tindak balas asid:
(i) with carbonates / dengan karbonat
(ii) with metals / dengan logam
(iii) with bases / dengan bes
• BASICITY OF AN ACID / KEBESAN ASID
– To state the meaning of basicity of an acid and to write equations for the ionisation of monoprotic and diprotic acids.
Menyatakan maksud kebesan asid dan menulis persamaan pengionan asid monoprotik dan diprotik.
– To relate the basicity of acid/alkali with pH values / Mengaitkan kebesan asid /alkali dengan nilai pH.
• BASE / ALKALI / BES / ALKALI
– To state the meaning of base and to correlate base with alkali / Menyatakan maksud bes dan mengaitkan bes dengan alkali.
– To write chemical equations involving alkalis with acids and ammonium salts.
Menulis persamaan tindak balas kimia alkali dengan asid dan dengan garam ammonium.
ROLE OF WATER IN ACIDS AND ALKALI / PERANAN AIR DALAM ASID DAN ALKALI
– To explain why the acid and alkali properties are shown in the presence of water.
Menerangkan mengapa sifat asid dan alkali ditunjukkan dengan kehadiran air.
– To explain why the acid and alkali properties do not show in the absence of water or in non-water solvent.
Menerangkan mengapa sifat asid dan alkali tidak ditunjukkan tanpa kehadiran air atau dalam pelarut bukan air.
pH SCALE / SKALA pH
– To state the meaning of pH / Menyatakan maksud pH.
– To relate the pH value with the concentration of H+ ion for the acids and OH– ions for alkalis.
Mengaitkan nilai pH dengan kepekatan ion H+ bagi asid dan ion OH– bagi alkali.
• STRONG / WEAK ACID AND STRONG / WEAK ALKALI / ASID KUAT / LEMAH DAN ALKALI KUAT / LEMAH
– To list examples and equations for the ionisation of strong / weak acid and strong / weak alkali.
Menyenaraikan contoh dan menulis persamaan pengionan bagi asid kuat / lemah dan alkali kuat / lemah.
– To relate the pH value with the strength of acid / alkali / Mengaitkan nilai pH dengan kekuatan asid / alkali.
ACID AND ALKALI CONCENTRATION / KEPEKATAN ASID DAN ALKALI
– To state the meaning of concentration in g dm–3 and mol dm–3 / Menyatakan maksud kepekatan dalam unit g dm–3 dan mol dm–3.
– To state the meaning of standard solution and to describe the preparation of standard solution.
Menyatakan maksud larutan piawai dan menghuraikan eksperimen penyediaan larutan piawai.
MV
.
– To solve various problems with calculations related to the preparation of standard solution using n =
1 000
Menyelesaikan pelbagai masalah pengiraan berkaitan penyediaan larutan piawai menggunakan formula n = MV .
1 000
m
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• NEUTRALISATION OF ACID AND ALKALI / TINDAK BALAS PENEUTRALAN ASID DAN ALKALI
– To describe the titration of acid with alkali and to calculate acid / alkali concentrations if a standard solution are given.
Menghuraikan titratan asid dengan alkali dan menghitung kepekatan asid / alkali jika satu larutan piawai diberikan.
– To describe the type of indicators used and the colour changes at the end-point.
Menyatakan jenis penunjuk yang digunakan dan perubahan warna penunjuk pada takat akhir.
– To solve numerical problems involving neutralisation / Menyelesaikan masalah pengiraan berkaitan peneutralan.
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Chemistry Form 4 • MODULE
ACID / ASID
1
2
Acid is a chemical substance which ionises in water to produce hydrogen ion.
Asid ialah bahan kimia yang mengion dalam air menghasilkan ion hidrogen.
Acid tastes sour and turns moist blue litmus to red.
Asid mempunyai rasa yang masam dan menukar kertas litmus biru lembap menjadi merah.
3
Example of acid is hydrochloric acid / Contoh asid ialah asid hidroklorik :
(a) Hydrogen chloride gas is a *covalent compound exist in the form of molecule.
Gas hidrogen klorida ialah *sebatian kovalen wujud dalam bentuk molekul.
(b) As hydrogen chloride dissolves in water, hydrogen chloride molecule ionises to hydrogen ion and chloride ion in
aqueous solution. This aqueous solution is called hydrochloric acid.
Apabila hidrogen klorida melarut dalam air, molekul hidrogen klorida mengion kepada ion hidrogen dan ion klorida dalam larutan
akueus. Larutan akueus itu dipanggil asid hidroklorik.
HCl (aq / ak )
Hydrochloric acid / Asid hidroklorik
H+ (aq / ak )
Hydrogen ion / Ion hidrogen
+
Cl– (aq / ak )
Chloride ion / Ion klorida
(c) An aqueous hydrogen ion, H+(aq) is actually the hydrogen ion combined with water molecule to form
hydroxonium ion, H3O+. However this ion can be written as H+.
Ion hidrogen akueus, H+(ak) ialah ion hidrogen yang bergabung dengan molekul air membentuk ion hidroksonium, H3O+. Walau
bagaimanapun, ion ini boleh ditulis sebagai H+.
HCl (g)
+ H2O(l/ce)
H3O+ (aq/ak )
+ Cl– (aq/ak )
4
Hydrogen chloride
Ion hydroxonium
Ion klorida
Hidrogen klorida
Ion hidroksonium
Ion klorida
H3O+
Ion hydroxonium
H+(aq/ak )
Ion hidrogen
Ion hidroksonium
Ion hidrogen
The ionisation of hydrochloric acid is
represented as:
Pengionan asid hidroklorik diwakili oleh:
HCl (aq/ak)
H+ (aq/ak) + Cl– (aq/ak)
+ H2O
Basicity of an acid is the number of ionisable of hydrogen atom per molecule of an acid molecule in an aqueous
solution / Kebesan asid ialah bilangan atom hidrogen yang boleh mengion bagi setiap molekul asid dalam larutan akueus.
– Monoprotic: One acid molecule ionises to
one
hydrogen ion.
Monoprotik: Satu molekul asid mengion kepada
satu
ion hidrogen.
– Diprotic: One acid molecule ionises to
two
Diprotik: Satu molekul asid mengion kepada
dua
– Triprotic: One acid molecule ionises to
three
Triprotik: Satu molekul asid mengion kepada
tiga
hydrogen ion.
ion hidrogen.
hydrogen ion.
ion hidrogen.
Hydrochloric is monoprotic acid because one molecule of hydrochloric acid ionises to
Asid hidroklorik ialah sejenis asid monoprotik kerana satu molekul asid hidroklorik mengion kepada
5
one
satu
hydrogen ion.
ion hidrogen.
Examples of acid and their basicity / Contoh-contoh asid dan kebesannya:
Ionisation of acid
Pengionan asid
H+(aq)
+
Hydrogen ion
Nitrate ion
Asid nitrik
Ion hidrogen
Ion nitrat
+
Hydrogen ion
Sulphate ion
Asid sulfurik
Ion hidrogen
Ion sulfat
+
Hydrogen ion
Phosphate ion
Asid fosforik
Ion hidrogen
Ion fosfat
+
One
Monoprotic
Two
Diprotic
Three
Triprotic
One
Monoprotic
PO43–(aq)
H3PO4 (aq/ak )
Phosphoric acid
CH3COO–(aq)
Kebesan asid
SO42–(aq)
H2SO4 (aq/ak )
Sulphuric acid
3H+(aq)
Basicity of acid
Bilangan ion hidrogen dihasilkan bagi setiap
molekul asid
NO3–(aq)
HNO3 (aq/ak )
Nitric acid
2H+(aq)
Number of hydrogens ion produce
per molecule of acid
H+(aq)
*CH3COOH (aq/ak )
Ethanoic acid
Ethanoate ion
Hydrogen ion
Asid etanoik
Ion etanoat
Ion hidrogen
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*Not all hydrogen atoms in ethanoic acid are ionisable / *Bukan semua ion hidrogen dalam asid etanoik boleh mengion
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MODULE • Chemistry Form 4
BASES / BES
Bases is a chemical substance that reacts with acid to produce salt and water only. For example,
Bes ialah sejenis bahan kimia yang bertindak balas dengan asid menghasilkan garam dan air sahaja. Contohnya,
1
(a) Copper(II) oxide (a base) reacts with sulphuric acid to produce copper(II) sulphate (a salt) and water.
Kuprum(II) oksida (bes) bertindak balas dengan asid sulfurik menghasilkan kuprum(II) sulfat (garam) dan air.
CuSO4
CuO + H2SO4
+
H2O
(b) Zinc hydroxide (a base) reacts with hydrochloric acid to produce zinc chloride (a salt) and water.
Zink hidroksida (bes) bertindak balas dengan asid hidroklorik menghasilkan zink klorida (garam) dan air.
ZnCl2
H2O
Zn(OH)2 + 2HCl
+
Most bases are metal oxide or metal hydroxide which are ionic compound. Example of bases are magnesium oxide, zinc
oxide, sodium hydroxide and potassium hydroxide.
2
Kebanyakan bes ialah oksida logam atau hidroksida logam yang merupakan sebatian ion. Contoh-contoh bes ialah magnesium oksida, zink
oksida, natrium hidroksida dan kalium hidroksida.
The bases that can dissolve in water (soluble bases) are known as alkali.
3
Bes yang boleh melarut dalam air (bes larut) dikenali sebagai alkali.
Sodium hydroxide and potassium hydroxide are soluble in water and they are called alkali whereas magnesium oxide
and zinc oxide are called bases as they are insoluble in water.
4
Natrium hidroksida dan kalium hidroksida larut dalam air dan dipanggil sebagai alkali manakala magnesium oksida dan zink oksida
dipanggil sebagai bes kerana tidak terlarut dalam air.
Alkali is a base that is soluble in water and ionises to hydroxide ion. For example,
Alkali ialah bes yang larut dalam air dan mengion kepada ion hidroksida. Contohnya,
5
(a) Sodium hydroxide dissolves in water and ionises to hydroxide ion.
Natrium hidroksida terlarut dalam air dan mengion kepada ion hidroksida.
NaOH (aq/ak )
Na+ (aq/ak ) + OH– (aq/ak )
(b) Ammonia solution is obtained by dissolving ammonia molecule in water, ionisation occur to produce a hydroxide
ion, OH–.
Larutan ammonia diperoleh dengan melarutkan molekul ammonia dalam air, pengionan berlaku menghasilkan ion hidroksida, OH–.
–
NH3 (g) + H2O (l/ce )
NH+
4 (aq/ak ) + OH (aq/ak )
(c) Other examples of alkalis are barium hydroxide and calcium hydroxide.
Contoh alkali lain adalah barium hidroksida dan kalsium hidroksida.
Alkali tastes bitter, slippery and turns moist red litmus to blue.
6
Alkali mempunyai rasa yang pahit, licin dan menukarkan kertas litmus merah lembap kepada biru.
EXERCISE / LATIHAN
Complete the following table / Lengkapkan jadual berikut :
Soluble base (alkali) / Bes terlarut (alkali)
Name / Nama
Sodium oxide
Natrium oksida
Potassium oxide
Kalium oksida
Ammonia
Ammonia
Sodium hydroxide
Natrium hidroksida
Potassium hydroxide
Kalium hidroksida
Barium hydroxide
Barium hidroksida
Formula / Formula
Ionisation equation / Persamaan pengionan
Na2O
Na2O(s) + H2O
2NaOH(aq)
NaOH(aq)
Na+ (aq) + OH– (aq)
K2O
K2O(s) + H2O
2KOH(aq)
KOH(aq)
K+ (aq) + OH– (aq)
NH3
NH3(g)+ H2O
NaOH
NaOH(aq)
KOH
KOH(aq)
Ba(OH)2
Ba(OH)2(aq)
NH4+(aq) + OH–(aq)
Na+ (aq) + OH– (aq)
K+ (aq) + OH– (aq)
Ba2+(aq) + 2OH– (aq)
Insoluble base / Bes tak terlarut
Name / Nama
Formula / Formula
Copper(II) oxide
Kuprum(II) oksida
Copper(II) hydroxide
Kuprum(II) hidroksida
Zinc hydroxide
Zink hidroksida
Aluminium oxide
Aluminium oksida
Lead(II) hydroxide
Plumbum(II) hidroksida
Magnesium hydroxide
Magnesium hidroksida
CuO
Cu(OH)2
Zn(OH)2
Al2O3
Pb(OH)2
Mg(OH)2
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Bases that can dissolve in water (soluble bases) are known as alkali / Bes yang larut dalam air (bes larut) dipanggil alkali
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Chemistry Form 4 • MODULE
CHEMICAL PROPERTIES OF ACID / SIFAT-SIFAT KIMIA ASID
1
Acid react with metal, base / alkali and metal carbonate / Asid bertindak balas dengan logam, bes/alkali dan karbonat logam:
Chemical properties
Sifat-sifat kimia
1 Acid + Metal
Asid + Logam
Salt + Hydrogen
Garam + Hidrogen
* Acid react with the metals that are
more electropositive than hydrogen
in electrochemical series, acids do not
react with copper and silver (type of
reaction is displacement, the metals
that are placed above hydrogen in
Electrochemical Series can displace
hydrogen from acid)
* Asid bertindak balas dengan logam-logam
yang lebih elektropositif daripada hidrogen
dalam Siri Elektrokimia, asid tidak bertindak
balas dengan kuprum dan argentum (jenis
tindak balas ialah penyesaran, logam-logam
di atas hidrogen dalam Siri Elektrokimia
boleh menyesarkan hidrogen daripada asid)
* Application of the reaction:
* Aplikasi tindak balas:
– Preparation of soluble salt (Topic
Salt)
Penyediaan garam terlarut (Tajuk
Garam)
– Preparation of hydrogen gas in
determination of the empirical
formula of copper(II) oxide (Topic
Chemical Formula and Equation)
Penyediaan gas hidrogen dalam
menentukan formula empirik kuprum(II)
oksida (Tajuk Formula dan Persamaan
Kimia)
2 Acid + Metal carbonate
Water + Carbon dioxide
Asid + Karbonat logam
Karbon dioksida
Salt +
Garam + Air +
Example of experiment
Contoh eksperimen
Zinc + Hydrochloric acid
Zink + Asid hidroklorik
Lighted wooden
splinter
Kayu uji menyala
Hydrochloric acid
Asid hidroklorik
Magnesium powder
Serbuk magnesium
(a) About 5 cm3 of dilute
hydrochloric acid is poured
into a test tube.
Sebanyak 5 cm3 asid hidroklorik
cair dimasukkan ke dalam tabung
uji.
(b) One spatula of magnesium
powder is added to the acid.
Satu spatula serbuk magnesium
ditambah kepada asid.
(c) A burning wooden splinter is
placed at the mouth of the test
tube.
Chemical equation:
Persamaan kimia:
Pepejal kelabu
terlarut.
– Gas bubbles
are released.
When a
burning
wooden
splinter is
placed at the
mouth of the
test tube,
‘pop sound’ is
produced.
Mg + 2HCl
MgCl2 + H2
Inference / Inferens :
– Magnesium reacts with
hydrochloric acid.
Magnesium bertindak balas
dengan asid hidroklorik.
–
Hydrogen gas is
released.
Gas
hidrogen
terbebas.
Gelembung gas
dibebaskan.
Apabila kayu
uji menyala
didekatkan pada
mulut tabung
uji, bunyi ‘pop’
dihasilkan.
Semua pemerhatian direkodkan.
Calcium carbonate + Nitric acid
Kalsium karbonat + Asid nitrik
Hydrochloric acid
Asid hidroklorik
Lime water
Air kapur
– Preparation of soluble salt (Topic
Salt)
Ujian pengesahan bagi ion karbonat
dalam analisis kualitatif garam (Tajuk
Garam)
– The grey solid
dissolves.
Catatan
(d) The observations are recorded.
*Aplikasi tindak balas:
– Confirmatory test for anion
carbonate ion in qualitative
analysis of salt (Topic Salt)
Remark
Pemerhatian
Kayu uji menyala diletakkan pada
mulut tabung uji.
*Application of the reaction:
Penyediaan garam terlarut (Tajuk
Garam)
Observation
Calcium carbonate / Kalsium karbonat
(a) About 5 cm3 of dilute
hydrochloric acid is poured
into a test tube.
Sebanyak 5 cm3 asid hidroklorik
cair dimasukkan ke dalam tabung
uji.
(b) One spatula of calcium
carbonate powder is added
into the test tube.
Satu spatula serbuk kalsium
karbonat dimasukkan ke dalam asid.
– The white
solid
dissolves.
Pepejal putih
terlarut.
Chemical equation:
Persamaan kimia:
CaCO3 + 2HCl
CaCl2 + H2O + CO2
– Gas bubbles
are released.
Inference / Inferens :
When the
– Calcium carbonate
gas passed
reacts with nitric acid.
through lime
Kalsium
karbonat
water, the lime
bertindak balas dengan
water turns
asid hidroklorik.
chalky.
Carbon dioxide gas
Gelembung gas
–
terbebas. Apabila
is released.
gas tersebut
dilalukan melalui
air kapur, air
kapur menjadi
keruh.
Gas karbon dioksida
terbebas.
(c) The gas released is passed
through lime water as shown
in the diagram.
Gas yang dibebaskan dilalukan
melalui air kapur seperti
ditunjukkan dalam rajah.
(d) The observations are recorded.
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Semua pemerhatian direkodkan.
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MODULE • Chemistry Form 4
3 Acid + Base / Alkali
Asid + Bes / Alkali
Salt + Water
Garam + Air
*Acid neutralises base/alkali
Copper(II) oxide + Sulphuric acid
Kuprum(II) oksida + Asid sulfurik
Sulphuric acid / Asid sulfurik
– The black
solid
dissolves.
Persamaan kimia:
CuO + H2SO4
Pepejal hitam
terlarut.
* Asid meneutralkan bes/alkali
*Application of the reaction:
– The colourless
solution turns
blue.
*Aplikasi tindak balas:
– Preparation of soluble salt (Topic
Salt)
Penyediaan garam terlarut (Tajuk
Garam)
Chemical equation:
Copper(II) oxide / Kuprum(II) oksida
(a) Dilute hydrochloric acid is
poured into a beaker until half
full.
Larutan tanpa
warna bertukar
menjadi biru.
Asid hidroklorik cair dimasukkan
dalam bikar hingga separuh penuh.
(b) The acid is warmed gently.
Asid dihangatkan.
(c) One spatula of copper(II) oxide
powders added to the acid.
CuSO4 + H2O
Inference / Inferens :
– Copper(II) oxide reacts
with sulphuric acid.
Kuprum(II) oksida
bertindak balas dengan
asid sulfurik.
– The blue solution is
copper(II) sulphate .
Larutan biru tersebut ialah
kuprum(II) sulfat .
Satu spatula serbuk kuprum(II)
oksida ditambahkan kepda asid
tersebut.
(d) The mixture is stirred with a
glass rod.
Campuran dikacau dengan rod kaca.
(e) The observations are recorded.
Semua pemerhatian direkodkan.
Write the chemical formulae for the following compounds / Tuliskan formula kimia bagi sebatian berikut:
2
Compound / Sebatian
Hydrochloric acid
Asid hidroklorik
Nitric acid
Asid nitrik
Sulphuric acid
Asid sulfurik
Ethanoic acid
Asid etanoik
Sodium hydroxide
Natrium hidroksida
Potassium hydroxide
Kalium hidroksida
Calcium hydroxide
Kalsium hidroksida
Sodium carbonate
Natrium karbonat
Magnesium hydroxide
Magnesium hidroksida
Ammonium sulphate
Ammonium sulfat
Hydroxide ion
Ion hidroksida
Sodium sulphate
Natrium sulfat
Carbon dioxide
Karbon dioksida
Copper(II) carbonate
Kuprum(II) karbonat
Water
m
HCl
HNO3
H2 SO4
CH3COOH
NaOH
KOH
Ca(OH)2
Na2CO3
Mg(OH)2
(NH4 )2SO4
OH–
Na2 SO4
CO2
CuCO3
H2O
Compound / Sebatian
Magnesium oxide
Magnesium oksida
Calcium oxide
Kalsium oksida
Copper(II) oxide
Kuprum(II) oksida
Lead(II) oxide
Plumbum(II) oksida
Sodium nitrate
Natrium nitrat
Potassium sulphate
Kalium sulfat
Barium hydroxide
Barium hidroksida
Sodium chloride
Natrium klorida
Magnesium
Magnesium
Zinc
Zink
Sodium
Natrium
Calcium carbonate
Kalsium karbonat
Hydrogen gas
Gas hidrogen
Sodium oxide
Natrium oksida
Magnesium nitrate
Magnesium nitrat
Chemical formulae / Formula kimia
MgO
CaO
CuO
PbO
NaNO3
K2 SO4
Ba(OH)2
NaCl
Mg
Zn
Na
CaCO3
H2
Na2O
Mg(NO3 )2
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Air
Chemical formulae / Formula kimia
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Chemistry Form 4 • MODULE
3
Ionic equation / Persamaan ion :
Ionic equation shows particles that change during chemical reaction.
Persamaan ion menunjukkan zarah yang berubah semasa tindak balas kimia.
Example / Contoh :
(i) Reaction between sulphuric acid and sodium hydroxide solution:
Tindak balas antara asid sulfurik dengan larutan natrium hidroksida:
Write balanced equation / Tulis persamaan seimbang :
H2SO4 + 2NaOH
Na2SO4 + 2H2O
Write the formula of all the particles in the reactants and products:
Tulis formula bagi semua zarah dalam bahan tindak balas dan hasil tindak balas:
2H+ + SO42– + 2Na+ + 2OH–
2Na+ + SO42– + 2H2O
Remove all the particles in the reactants and products which remain unchanged:
Keluarkan semua zarah dalam bahan dan hasil tindak balas yang tidak berubah:
2H+ + SO42– + 2Na+ + 2OH–
2Na+ + SO42– + 2H2O
Ionic equation / Persamaan ion :
2H2O ⇒ H+ + OH–
2H+ + 2OH–
H2O
(ii) Reaction between zinc oxide and hydrochloric acid / Tindak balas antara zink dengan asid hidroklorik :
Write balanced equation / Tulis persamaan seimbang :
2HCl + Zn
ZnCl2 + H2
Write the formula of all the particles in the reactants and products:
Tulis formula bagi semua zarah dalam bahan tindak balas dan hasil tindak balas:
2H+ + 2Cl– + Zn
Zn2+ + 2Cl– + H2
Remove all the particles in the reactants and products which remain unchanged:
Keluarkan semua zarah dalam bahan dan hasil tindak balas yang tidak berubah:
2H+ + 2Cl– + Zn
Zn2+ + 2Cl– + H2
Ionic equation / Persamaan ion :
2H+ + Zn
4
Zn2+ + H2
Write the chemical equations and ionic equation for the following reactions:
Tulis persamaan kimia dan persamaan ion untuk tindak balas berikut:
Reactant / Bahan tindak balas
Hydrochloric acid and #magnesium oxide
Asid hidroklorik dan #magnesium oksida
Hydrochloric acid and sodium hydroxide
Asid hidroklorik dan natrium hidroksida
Hydrochloric acid and magnesium
Asid hidroklorik dan magnesium
Hydrochloric acid and #calcium carbonate
Asid hidroklorik dan #kalsium karbonat
Sulphuric acid and zinc
Asid sulfurik dan zink
Sulphuric acid and #zinc oxide
Asid sulfurik dan #zink oksida
Sulphuric acid and #zinc carbonate
Asid sulfurik dan #zink karbonat
Nitric acid and #copper(II) oxide
Asid nitrik dan #kuprum(II) oksida
Nitric acid and sodium hydroxide
Asid nitrik dan natrium hidroksida
Chemical equations / Persamaan kimia
MgO + 2HCl
MgCl2 + H2O
HCl + NaOH
NaCl + H2O
2HCl + Mg
2HCl + CaCO3
H2SO4 + Zn
H2SO4 + ZnO
H2SO4 +ZnCO3
2HNO3 + CuO
HNO3 + NaOH
MgCl2 + H2
CaCl2 + CO2 + H2O
ZnSO4 + H2
ZnSO4 +H2O
ZnSO4 + CO2 + H2O
Cu(NO3)2 + H2O
NaNO3 + H2O
Ionic equation / Persamaan ion
2H+ + MgO
Mg2+ + H2O
H+ + OH–
2H+ + Mg
2H+ + CaCO3
Mg2+ + H2
Ca2+ + CO2 + H2O
2H+ + Zn
Zn2+ + H2
2H+ + ZnO
2H+ + ZnCO3
H2O
Zn2+ + H2O
Zn2+ + CO2 + H2O
2H+ + CuO
H+ + OH–
Cu2+ + H2O
H2O
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# Ions in magnesium oxide, calcium carbonate, zinc oxide, zinc carbonate and copper(II) oxide cannot be separated because
the compounds are insoluble in water and the ions do not ionise.
#
Ion dalam magnesium oksida, kasium karbonat, zink oksida, zink karbonat dan kuprum(II) oksida tidak boleh diasingkan kerana sebatian tersebut
tidak larut dalam air dan ion-ionnya tidak mengion.
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MODULE • Chemistry Form 4
CHEMICAL PROPERTIES OF ALKALIS / SIFAT KIMIA ALKALI
Chemical properties
Sifat-sifat kimia
1 Alkali + Acid
Alkali + Asid
Salt + Water
Garam + Air
*Alkali neutralises acid / Alkali meneutralkan asid.
*Application of the reaction / Aplikasi tindak balas :
– Preparation of soluble salt (Topic Salt)
Penyediaan garam terlarut (Tajuk Garam)
2 Alkali + Ammonium salt
Alkali + Garam Ammonium
Salt + Water + Ammonia gas
Garam + Air + Gas ammonia
*Ammonia gas is released when alkali is heated with
ammonium salt. Ammonia gas has pungent smell and turn
moist red litmus paper to blue.
*Gas ammonia dibebaskan apabila alkali dipanaskan dengan garam
ammonium. Gas ammonia mempunyai bau yang sengit dan menukar
kertas litmus merah lembap kepada biru.
*Application of the reaction / Aplikasi tindak balas :
Write the balance chemical equation for the reaction
Tuliskan persamaan kimia seimbang bagi tindak balas
(a) Potassium hydroxide and sulphuric acid
Kalium hidroksida dan asid sulfurik :
H2SO4 + 2KOH
K2SO4 + 2H2O
(b) Barium hydroxide and hydrochloric acid:
Barium hidroksida dan asid hidroklorik:
2HCl + Ba(OH)2
BaCl2 + H2O
(c) Ammonium chloride and potassium hydroxide:
Ammonium klorida dan kalium hidroksida:
KOH + NH4Cl
KCl + H2O + NH3
(d) Ammonium sulphate and sodium hydroxide:
Ammonium sulfat dan natrium hidroksida:
2NaOH + (NH4)2SO4
Na2SO4 + 2H2O + 2NH3
– Confirmatory test for cations ammonium in qualitative
analysis of salt (Topic Salt)
Ujian pengesahan kation ammonium dalam analisis kualitatif garam
(Tajuk Garam)
3 Alkali + Metal ion
Alkali + Ion logam
Insoluble metal hydroxide
Logam hidroksida tak larut
(e) 2OH–(aq/ak) + Mg2+(aq/ak)
Magnesium hydroxide
(white precipitate)
*Most of the metal hydroxides are insoluble.
*Kebanyakan logam hidroksida tak terlarut.
Magnesium hidroksida
(mendakan putih)
*Hydroxides of transition element metals are coloured.
*Hidroksida bagi logam peralihan adalah berwarna.
*Application of the reaction / Aplikasi tindak balas :
Mg(OH)2(p)
(f) 2OH–(aq/ak) + Cu2+(aq/ak)
– Confirmatory test for cations in qualitative analysis of salt
(Topic Salt)
Ujian pengesahan bagi kation dalam analisis kualitatif garam (Tajuk
Garam)
Cu(OH)2(p)
Copper(II) hydroxide
(blue precipitate)
Kuprum(II) hidroksida
(mendakan biru)
ROLE OF WATER AND THE PROPERTIES OF ACID / PERANAN AIR DAN SIFAT ASID
An acid shows its acidic properties when it is dissolved in water.
1
Asid menunjukkan sifat keasidannya apabila terlarut dalam air.
Acid molecules ionise in aqueous solution to form hydrogen ions. The presence of hydrogen ions is needed for the acid
to show its acidic properties.
2
Molekul asid mengion dalam larutan akueus membentuk ion hidrogen. Kehadiran ion hidrogen diperlukan oleh asid untuk menunjukkan
sifat keasidannya.
Acid will remain in the form of molecules in two conditions / Asid akan kekal dalam bentuk molekul dengan dua keadaan:
(a) Without the presence of water for example dry hydrogen chloride gas and *glacial ethanoic acid.
3
Tanpa kehadiran air seperti gas hidrogen klorida kering dan *asid etanoik glasial
(b) Acid is dissolved in *organic solvent for example solution of hydrogen chloride in methylbenzene and ethanoic
acid in propanone.
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Asid dilarutkan dalam *pelarut organik seperti larutan hidrogen klorida dalam metilbenzena dan asid etanoik dalam propanon.
* Glacial ethanoic acid is pure ethanoic acid / Asid etanoik glasial ialah asid etanoik tulen.
* Organic solvent is covalent compound that exist as liquid at room temperature such as propanone,
methylbenzene and trichloromethane.
* Pelarut organik ialah sebatian kovalen yang wujud dalam bentuk cecair pada suhu bilik seperti propanon, metilbenzena dan
triklorometana.
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Chemistry Form 4 • MODULE
4
Example / Contoh :
Glacial ethanoic acid
Solution of hydrogen chloride in
methylbenzene
Asid etanoik glasial
Larutan hidrogen klorida dalam metilbenzena
HCl
CH3COOH
CH3COOH
HCl
CH3COOH
HCl
CH3COOH
CH3COOH
HCl
HCl
CH3COOH
CH3COOH
Larutan hidrogen klorida dalam air
(asid hidroklorik)
HCl
H+
Cl-
HCl
HCl
HCl
Solution of hydrogen chloride in
water (hydrochloric acid)
Molekul asid etanoik glasial tidak mengion .
• Glacial ethanoic exist as molecule
only, no hydrogen ions present.
Etanoik glasial hanya terdiri daripada
molekul CH COOH sahaja, tiada ion
3
hidrogen hadir.
H+
H+
Cl-
Cl-
HCl
• Hydrogen chloride molecules in
methylbenzene do not ionise
H+
Water / Air
.
Molekul hidrogen klorida dalam metilbenzena
tidak mengion .
• Hydrogen chloride exist as molecule
only, there are no hydrogen ions
present.
Hidrogen klorida wujud sebagai
sahaja, tiada ion hidrogen hadir.
Cl-
H+
Methylbenzene / Metilbenzena
• Glacial ethanoic acid molecules do not
ionise .
Cl-
molekul
• Hydrogen chloride molecule in water
ionises :
Molekul hidrogen klorida dalam air
mengion :
HCl (aq/ak)
H+ (aq/ak) + Cl– (aq/ak)
• Hydrogen ions and chloride ions
present.
Ion
hidrogen dan
ion
klorida hadir.
• Hydrogen chloride in water
• Glacial ethanoic acid and hydrogen chloride in methylbenzene does not show acidic
(hydrochloric acid) shows acidic
properties:
properties:
Asid etanoik glasial dan hidrogen klorida dalam metilbenzena tidak menunjukkan sifat asid:
(i) They
do not
Sebatian tersebut
tidak
do not
turn
Sebatian tersebut
tidak
(ii) They
Hidrogen klorida dalam air (asid hidroklorik)
menunjukkan sifat asid:
react with metal, base or metal carbonate.
bertindak balas dengan logam, bes dan karbonat logam.
blue
litmus paper to
menukarkan warna kertas litmus
red
biru
.
kepada
react
with
(i) Hydrochloric acid
metal, base or metal carbonate.
merah .
• There are no free moving ions, hydrogen chloride in methylbenzene and glacial
ethanoic acid cannot conduct electricity (non-electrolyte).
Asid hidroklorik bertindak balas
dengan logam, bes dan karbonat logam.
blue
(ii) Hydrogen ions turn
litmus paper to
Tidak wujud ion bebas bergerak , asid etanoik glasial dan hidrogen klorida dalam metilbenzena
tidak dapat mengkonduksikan elektrik (bukan elektrolit).
red
.
Ion hidrogen menukarkan warna kertas
biru
kepada merah .
litmus
• There are free moving ions,
can
conduct
hydrochloric acid
electricity (electrolyte).
Terdapat ion yang bebas bergerak , asid
dapat
hidroklorik
mengkonduksikan
elektrik (elektrolit).
ROLE OF WATER AND THE PROPERTIES OF ALKALI / PERANAN AIR DAN SIFAT ALKALI
1
In the presence of water, an alkali dissolves and ionises to produce hydroxide ions. For example potassium hydroxide
solution and ammonia solution.
Dengan kehadiran air, alkali melarut dan mengion menghasilkan ion hidroksida. Contohnya larutan kalium hidroksida dan larutan
ammonia:
KOH(aq/ak )
K+(aq/ak ) + OH–(aq/ak )
NH3(g) + H2O(l/ce)
NH4+(aq/ak ) + OH–(aq/ak )
2
Without water or in organic solvents, no hydroxide ions are produced, so the alkaline properties are not shown.
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Tanpa air atau dalam pelarut organik, tiada ion hidroksida yang dihasilkan, maka sifat-sifat alkali tidak ditunjukkan.
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MODULE • Chemistry Form 4
EXERCISE / LATIHAN
The diagram below shows the apparatus set-up to investigate the role of water and other solvent in showing the
properties of acid and the observations made from the investigation.
1
Rajah di bawah menunjukkan susunan radas untuk mengkaji peranan air atau pelarut lain dalam menunjukkan sifat asid serta pemerhatian
yang dibuat.
Experiment / Eksperimen
I
II
Set-up of apparatus
Susunan radas
Asid hidroklorik dalam air
Hydrochloric acid in
tetrachloromethane
Magnesium ribbon
Magnesium ribbon
Hydrochloric acid in water
Asid hidroklorik dalam tetraklorometana
Pita magnesium
Observation
• Bubbles of gas are released
Pemerhatian
Gelembung gas dibebaskan
Pita magnesium
• No bubble of gas
Tiada gelembung gas
• Magnesium ribbon dissolves
Pita magnesium larut
(a) What is meant by acid / Apakah yang dimaksudkan dengan asid ?
Acid is a chemical substance which ionises in water to produce hydrogen ion.
(b) (i)
Name the bubble of gas released in Experiment I / Namakan gas yang terbebas dalam Eksperimen I.
Hydrogen gas
(ii)
Write the chemical equation for the formation of the bubbles in Experiment I.
Tulis persamaan kimia untuk pembentukan gelembung gas dalam Eksperimen I.
Mg + 2HCl
MgCl2 + H2
(iii) Write the ionic equation for the chemical equation in (b)(ii).
Tulis persamaan ion untuk persamaan kimia dalam (b)(ii).
Mg + 2H+
Mg2+ + H2
(c) Compare observation in Experiment I and Experiment II. Explain your answer.
Bandingkan pemerhatian dalam Eksperimen I dan Eksperimen II. Terangkan jawapan anda.
–
Hydrochloric acid in water in Experiment I
Asid hidroklorik dalam air dalam Eksperimen I
–
reacts
with magnesium.
bertindak balas
dengan magnesium.
Hydrochloric acid in tetrachloromethane in Experiment I do not react with magnesium.
Asid hidroklorik dalam tetraklorometana dalam Eksperimen II
–
Hydrochloric acid in water
–
H+ ions react with
ionises
magnesium atom
tidak bertindak balas
to H+ / Asid hidroklorik dalam air
HCl H+ + Cl–
dengan magnesium.
mengion
kepada ion H+:
to produce hydrogen molecule:
Ion H+ bertindak balas dengan atom magnesium untuk menghasilkan molekul hidrogen:
Mg + 2H+
–
Mg2+ + H2
Hydrochloric acid in tetrachloromethane remains in the form of
m
. Tiada ion
hydrogen
hidrogen
ion present.
hadir.
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Asid hidroklorik dalam tetraklorometana kekal dalam bentuk
molekul
molecule . No
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Chemistry Form 4 • MODULE
2
The diagram below shows the set-up of apparatus to prepare two solutions of ammonia in solvent X and solvent Y. A
piece of red litmus paper is dropped into each beaker.
Gambar rajah di bawah menunjukkan susunan radas bagi menyediakan dua larutan ammonia dalam pelarut X dan pelarut Y. Sekeping
kertas litmus merah dimasukkan ke dalam setiap bikar.
Ammonia
Ammonia
Ammonia
Ammonia
Solvent X
Solvent Y
Pelarut X
Pelarut Y
Beaker A / Bikar A
Beaker B / Bikar B
The table below shows the observation on the red litmus paper in solvent X and solvent Y.
Jadual di bawah menunjukkan pemerhatian ke atas kertas litmus merah dalam pelarut X dan pelarut Y.
Solution / Larutan
Observation / Pemerhatian
Ammonia in solvent X in beaker A
The red litmus paper turns blue.
Ammonia dalam pelarut X dalam bikar A
Kertas litmus merah bertukar menjadi biru.
Ammonia in solvent Y in beaker B
No visible change in the colour of red litmus paper.
Ammonia dalam pelarut Y dalam bikar B
Tiada perubahan yang nyata pada warna kertas litmus merah.
(a) Name possible substances that can be solvent X and solvent Y.
Namakan bahan-bahan yang mungkin bagi pelarut X dan pelarut Y.
Solvent X / Pelarut X : Water
Solvent Y / Pelarut Y : Propanone / methylbenzene / trichloromethane
(b) Explain the difference in the observation on the beakers A and B.
Terangkan perbezaan antara pemerhatian dalam bikar A dengan bikar B.
–
Ammonia gas in beaker A is
hydroxide ions:
dissolved
Gas ammonia dalam bikar A
ion hidroksida :
larut
in water, ammonia molecules
dalam air, molekul ammonia
–
The presence of
hydroxide
Kehadiran ion-ion
hidroksida
Ammonia gas in beaker B is
molecules do not ionise .
Gas ammonia dalam bikar B
mengion .
–
No
hydroxide
Tiada ion
(c) (i)
mengion
to ammonium ion and
kepada ion
ammonium
dan
NH 4+ (ak) + OH– (ak)
NH3 (g) + H2O (l/ce)
–
ionise
ions change the red litmus paper to blue.
menukar kertas litmus merah kepada biru.
dissolved
larut
propanone / methylbenzene / trichloromethane
in
propanon / metilbenzena / triklorometana
dalam
, ammonia
, molekul ammonia tidak
ions present, the red litmus paper remains unchanged.
hidroksida
, warna merah kertas litmus tidak berubah.
Between solution in beakers A and B, which one is an electrolyte and non-electrolyte? Explain your answer.
Antara larutan dalam bikar A dangan bikar B, yang manakah elektrolit dan bukan elektrolit? Terangkan jawapan anda.
–
an electrolyte
Solution in beaker A is
, it contains
ionisation
of ammonia molecules in water.
free moving ions
elektrolit
, ia mengandungi ion-ion yang
Larutan dalam bikar A ialah
pengionan
molekul ammonia dalam air.
–
a non-electrolyte , ammonia molecules
Solution in beaker B is
propanone / methylbenzene / trichloromethane .
, molekul ammonia
bebas bergerak
do not ionise
tidak mengion
daripada
in
dalam
.
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bukan elektrolit
Larutan dalam bikar B
propanon / metilbenzena / triklorometana
from the
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MODULE • Chemistry Form 4
(ii)
Draw a labelled diagram to show the set-up of apparatus used to show the electrical conductivity of an
electrolyte.
Lukiskan gambar rajah berlabel yang menunjukkan susunan radas yang digunakan untuk menunjukkan kekonduksian arus
elektrik bagi sesuatu elektrolit.
Carbon electrode
Elektrod
karbon
Carbon
Elektrodelectrode
karbon
Elektrolit
Electrolyte
THE pH SCALE / SKALA pH
The pH is a scale of numbers to measure the degree of acidity and alkalinity of an aqueous solution based on the
concentration of hydrogen ions, H+ or hydroxide ions, OH–.
1
Skala pH ialah skala bernombor untuk mengukur darjah keasidan dan kealkalian suatu larutan akueus berdasarkan kepekatan ion
hidrogen, H+ atau ion hidroksida, OH–.
The pH scale has the range of number from 0 to 14 / Skala pH bernombor dari 0 hingga 14 :
2
pH
0
1
2
3
4
5
6
pH < 7:
• Acidic solution / Larutan berasid.
• The lower the pH value, the higher is the
concentration of hydrogen ion, H+.
7
pH = 7
Neutral
Neutral
8
9
10
11
12
13
14
pH > 7:
• Alkaline solution / Larutan beralkali.
• The higher the pH value, the higher is the
concentration of hydroxide ion, OH–.
Semakin rendah nilai pH, semakin tinggi
kepekatan ion hidrogen, H+.
Semakin tinggi nilai pH, semakin tinggi
kepekatan ion hidroksida, OH–.
The pH of an aqueous solution can be measured by / Nilai pH bagi sesuatu larutan akueus boleh diukur dengan menggunakan:
(a) pH meter / Meter pH
(b) Acid-base indicator / Penunjuk asid-bes
Complete the following table / Lengkapkan jadual berikut :
3
Colour / Warna
Indicator
Penunjuk
Acid / Asid
Neutral / Neutral
Alkali / Alkali
Litmus solution / Larutan litmus
Red
Purple
Blue
Methyl orange / Metil jingga
Red
Orange
Yellow
Colourless
Colourless
Pink
Red
Green
Purple
Phenolphthalein / Fenolftalein
Universal indicator / Penunjuk universal
THE STRENGTH OF ACID AND ALKALI / KEKUATAN ASID DAN ALKALI
The strength of acid and alkali depend on the degree of ionisation or dissociation of the acid and alkali in water.
1
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Kekuatan asid dan alkali bergantung pada darjah pengionan asid dan alkali dalam air.
(a) A strong acid is an acid that ionises completely in water to produce high concentration of hydrogen ion, H+.
Asid kuat ialah asid yang mengion sepenuhnya dalam air menghasilkan kepekatan ion hidrogen, H+ yang tinggi.
(b) A weak acid is an acid that partially ionises in water to produce low concentration of hydrogen ion, H+.
Asid lemah ialah asid yang mengion separa dalam air menghasilkan kepekatan ion hidrogen, H+ yang rendah.
(c) A strong alkali is an alkali that ionises completely in water to produce high concentration of hydroxide ion,
OH–.
Alkali kuat ialah alkali yang mengion sepenuhnya dalam air menghasilkan kepekatan ion hidroksida, OH– yang tinggi.
(d) A weak alkali is an alkali that partially ionises completely in water to produce low concentration of hydroxide
ion, OH–.
Alkali lemah ialah alkali yang mengion separa dalam air menghasilkan kepekatan ion hidroksida, OH– yang rendah.
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Chemistry Form 4 • MODULE
2
Example of different strength of acid and alkali / Contoh asid dan alkali dengan kekuatan yang berbeza.
Acid / Alkali
Asid / Alkali
Strong acid
Asid kuat
Example
Contoh
Ionisation equation
Persamaan ion
Hydrochloric HCl (aq/ak )
acid, HCl
H+ (aq/ak ) + Cl– (aq/ak )
Asid
hidroklorik,
HCl
Nitric acid,
HNO3
Asid nitrik,
HNO3
Sulphuric
acid, H2SO4
Asid sulfurik,
H2SO4
Weak acid
Asid lemah
Ethanoic
acid,
CH3COOH
Asid etanoik,
CH3COOH
Carbonic
acid, H2CO3
Asid karbonik,
H2CO3
Strong alkali
Alkali kuat
Sodium
hydroxide,
NaOH
HNO3(aq/ak )
H+ (aq/ak ) + NO3– (aq/ak )
H2SO4 (aq/ak )
2H+ (aq/ak ) + SO42– (aq/ak )
Alkali lemah
Ammonia
solution,
NH3(aq)
Larutan
ammonia,
NH3(ak)
All hydrogen chloride molecules that
H+ and Cl–
+
–
dissolve in water ionises completely into H dan Cl
hydrogen
chloride
ions and
ions.
H+ and NO3–
All nitric acid ionises completely in
water into hydrogen ions and nitrate
H+ dan NO3–
ions.
H+ and SO42–
All sulphuric acid ionises completely
into hydrogen ions and sulphate ions.
H+ dan SO42–
Semua asid sulfurik mengion sepenuhnya dalam
air kepada ion hidrogen dan ion sulfat .
CH3COOH (aq/ak )
Ethanoic acid partially ionises in water
CH3COO (aq/ak ) + H (aq/ak ) into etahnoate ions and hydrogen
ions. Some remain in the form of CH3COOH
molecules
.
–
+
separa
kepada ion
Asid etanoik mengion
etanoat dan ion hidrogen . Sebahagian lagi
kekal dalam bentuk molekul CH3COOH.
H2CO3 (aq/ak )
2H+ (aq/ak ) + CO32– (aq/ak )
Carbonic acid partially ionises in water
into carbonate ions and hydrogen ion. Some
molecules .
remain in the form of H CO
2
3
Sebahagian asid karbonik mengion dalam air
kepada ion karbonat dan ion hidrogen. Sebahagian
lagi kekal dalam bentuk molekul H2CO3.
NaOH (aq/ak )
Na (aq) + OH (aq)
+
–
CH3COOH,
CH3COO– and
H+
CH3COOH,
CH3COO– dan H+
H2CO3, H+ and
CO32–
H2CO3, H+ dan
CO32–
Sodium hydroxide ionises completely in
water into sodium ions and hydroxide
ions.
Na+ and OH–
Potassium hydroxide ionises completely
potassium
ions and
in water into
hydroxide
ions.
K+ and OH–
Na+
dan
OH–
Natrium hidroksida mengion sepenuhnya dalam
air kepada ion natrium dan ion hidroksida .
KOH (aq/ak )
K+ (aq) + OH– (aq)
K+
dan
OH–
Kalium hidroksida mengion sepenuhnya dalam air
kalium
dan ion hidroksida .
kepada ion
Ba(OH)2 (aq/ak )
Ba (aq) + 2OH (aq)
2+
–
Ba2+ and OH–
Barium hydroxide ionises completely in
water into barium ions and hydroxide
Ba2+ dan OH–
ions.
Barium
hidroksida,
Ba(OH)2
Weak alkali
Zarah-zarah
yang hadir
Semua asid nitrik mengion sepenuhnya dalam
air kepada ion hidrogen dan ion nitrat .
Kalium
hidroksida,
KOH
Barium
hydroxide,
Ba(OH)2
Penerangan
Semua molekul hidrogen klorida melarut dalam
air dan mengion sepenuhnya kepada ion
hidrogen dan ion klorida .
Natrium
hidroksida,
NaOH
Potassium
hydroxide,
KOH
Particles
present
Explanation
Barium hidroksida mengion sepenuhnya dalam
air kepada ion barium dan ion hidroksida .
NH3 (g)+ H2 O(l/ce)
+
NH4 (aq/ak )
+ OH–(aq/ak )
NH3, NH4+ and
OH–
Ammonia partially ionises in water into
ammonium ions and hydroxide ions,
some remain in the form of NH3 molecules .
separa
dalam air kepada
Ammonia mengion
ion ammonium dan ion hidroksida ,
sebahagian lagi kekal dalam bentuk molekul NH .
NH3, NH4+ dan
OH–
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MODULE • Chemistry Form 4
CONCENTRATION OF ACID AND ALKALI / KEPEKATAN ASID DAN ALKALI
A solution is a homogeneous mixture formed when a solute is dissolved in a solvent. For example copper(II) sulphate
solution is prepared by dissolving copper(II) sulphate powder (solute) in water (solvent).
1
Larutan adalah campuran homogen yang terbentuk apabila bahan larut dilarutkan dalam pelarut. Contohnya larutan kuprum(II) sulfat
disediakan dengan melarutkan serbuk kuprum(II) sulfat (bahan larut) di dalam air (pelarut).
2
Concentration of a solution the quantity of solute in a given volume of solution which is usually 1 dm3 of solution.
Kepekatan sesuatu larutan ialah kuantiti bahan terlarut dalam isi padu larutan yang tertentu, biasanya isi padu 1 dm3 larutan.
3
Concentration can be expressed in two ways / Kepekatan boleh diwakili dengan dua cara :
(a) Mass of solute in gram per 1 dm3 solution, g dm–3/ Jisim bahan larut dalam gram bagi setiap 1 dm3 larutan, g dm–3.
Concentration of solution (g dm–3) =
Mass of solute in gram (g) / Jisim bahan larut dalam gram (g)
Kepekatan larutan (g dm–3)
Volume of solution (dm3) / Isi padu larutan (dm3)
(b) Number of moles of solute in 1 dm3 solution, mol dm–3 / Bilangan mol bahan larut dalam 1 dm3 larutan, mol dm–3.
Concentration of solution (mol dm–3) =
Kepekatan larutan (mol dm–3)
Number of mole of solute (mol) / Bilangan mol bahan larut (mol)
Volume of solution (dm3) / Isi padu larutan (dm3)
The concentration in mol dm–3 is called molarity or molar concentration. The unit mol dm–3 can be represented by ‘M’.
4
Kepekatan dalam mol dm–3 dipanggil sebagai kemolaran atau kepekatan molar. Unit mol dm–3 boleh diwakili dengan‘M’.
Molarity =
Kemolaran
Number of mole of solute (mol) / Bilangan mol bahan larut (mol)
Bilangan mol bahan terlarut
M = Concentration in mol dm–3
(molarity)
Volume of solution (dm3) / Isi padu larutan (dm3)
Number of mole of solute (mol) = Molarity × Volume (dm3)
Bilangan mol bahan larut (mol)
n = Number of moles of solute
Kepekatan dalam mol dm–3
(kemolaran)
Kemolaran × Isi padu (dm3)
n = MV
Mv
n =
1 000
V = Volume of solution in dm3
Isi padu larutan dalam dm3
v = Volume of solution in cm3
Isi padu larutan dalam cm3
The concentration of a solution can be converted from mol dm to g dm and vice versa.
–3
5
–3
Kepekatan larutan boleh ditukar daripada mol dm–3 kepada g dm–3 dan sebaliknya.
× molar mass of the solute / jisim molar bahan terlarut
mol dm–3
g dm–3
÷ molar mass of the solute / jisim molar bahan terlarut
The pH value of an acid or an alkali depends on the concentration of hydrogen ions or hydroxide ions:
6
Nilai pH bagi asid atau alkali bergantung pada kepekatan ion hidrogen atau ion hidroksida:
The higher the concentration of hydrogen ions in acidic solution, the lower the pH value.
Semakin tinggi kepekatan ion hidrogen dalam larutan berasid, semakin rendah nilai pH.
The higher the concentration of hydroxide ions in alkaline solution, the higher the pH value.
Semakin tinggi kepekatan ion hidroksida dalam larutan beralkali, semakin tinggi nilai pH.
The pH value of an acid or an alkali is depends on / Nilai pH bagi asid atau alkali bergantung pada:
(a) The strength of acid or alkali / Kekuatan asid atau alkali
– the degree of ionisation or dissociation of the acid and alkali in water / darjah pengionan asid atau alkali dalam air.
(b) Molarity of acid or alkali / Kemolaran asid atau alkali
– the concentration of acid or alkali in mol dm–3 / kepekatan bahan terlarut dalam mol dm–3.
(c) Basicity of an acid / Kebesan asid
– the number ionisable of hydrogen atom per molecule of an acid molecule in an aqueous solution.
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bilangan atom hidrogen per molekul asid yang terion dalam larutan akueus.
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Bandingkan
kepekatan ion
hidrogen dan
nilai pH
Compare
concentration
of H+ and pH
value
Bacaan pH
meter
pH meter
reading
Eksperimen
Experiment
0.01 mol dm–3 HCl
2.98
H+
0.1 mol dm–3
+ Cl–
H+
0.01 mol dm–3
–3
+ Cl–
Nilai pH bagi asid hidroklorik 0.1 mol dm–3 lebih
rendah
daripada asid hidroklorik 0.01 mol dm–3.
hydrochloric acid.
– The pH value of 0.1 mol dm–3 of hydrochloric
acid is lower than 0.01 mol dm–3 of
Kepekatan ion hidrogen dalam asid hidroklorik
tinggi
daripada asid
0.1 mol dm–3 lebih
hidroklorik 0.01 mol dm–3.
–3
0.01 mol dm of hydrochloric acid.
– Concentration hydrogen ion in 0.1 mol dm of
hydrochloric acid is higher than
0.01 mol dm–3
HCl
Asid hidroklorik 0.01 mol dm–3 mengion kepada
0.01 mol dm–3 ion hidrogen:
– 0.01 mol dm–3 of hydrochloric acid ionises to
–3
form 0.01 mol dm hydrogen ion:
0.1 mol dm–3
HCl
Asid hidroklorik 0.1 mol dm–3 mengion kepada
0.1 mol dm–3 ion hidrogen:
– 0.1 mol dm–3 of hydrochloric acid ionises to
–3
form 0.1 mol dm hydrogen ion:
kuat
yang
Asid hidroklorik adalah asid
mengion lengkap dalam air kepada ion hidrogen.
– Hydrochloric acid is a strong acid ionises
completely in water to hydrogen ion.
0.1 mol dm–3 HCl
1.21
I
2.25
diprotik
.
acid.
–3
2H+
0.1 mol dm–3
+ SO42–
H+
+ Cl–
–3
0.05 mol dm
Nilai pH bagi asid sulfurik 0.05 mol dm–3 lebih rendah
daripada asid hidroklorik 0.05 mol dm–3.
hydrochloric acid.
– The pH value of 0.05 mol dm–3 of sulphuric
lower than 0.05 mol dm–3 of
acid is
Kepekatan ion hidrogen dalam asid sulfurik
0.05 mol dm–3 adalah dua kali ganda (lebih tinggi)
daripada asid hidroklorik 0.05 mol dm–3.
0.05 mol dm–3 of hydrochloric acid.
– Concentration hydrogen ion in 0.05 mol dm–3
of sulphuric acid is double of (higher than)
0.05 mol dm–3
HCl
Asid hidroklorik 0.05 mol dm–3 mengion lengkap dalam
–3
air menghasilkan 0.05 mol dm
ion hidrogen:
– 0.05 mol dm–3 of ionises completely in water
–3
to form 0.05 mol dm hydrogen ion:
Asid hidroklorik adalah asid kuat monoprotik .
– Hydrochloric acid is a strong monoprotic
acid.
0.05 mol dm
H2SO4
Asid sulfurik 0.05 mol dm–3 mengion lengkap kepada
0.1 mol dm–3 ion hidrogen:
hydrogen ion:
– 0.05 mol dm-3 of sulphuric acid ionises
–3
completely in water to form 0.1 mol dm
Asid sulfurik adalah asid kuat
diprotic
0.05 mol dm–3 HCl
– Sulphuric acid is a strong
0.05 mol dm–3 H2SO4
1.15
II
0.1 mol dm–3 CH3COOH
3.45
kuat yang mengion
lengkap dalam
–3
H+
0.1 mol dm–3
+ Cl–
H+
+ CH3COO–(aq/ak )
less than/kurang dari
0.1 mol dm–3
0.1 mol dm–3
lower
Nilai pH bagi asid hidroklorik 0.1 mol dm–3 lebih rendah daripada
asid etanoik 0.1 mol dm–3.
– The pH value of 0.1 mol dm–3 of hydrochloric acid
than of 0.1 mol dm–3 of ethanoic acid.
Kepekatan ion hidrogen dalam asid hidroklorik 0.1 mol dm–3 lebih
tinggi daripada asid etanoik 0.1 mol dm–3.
ethanoic acid.
– Concentration hydrogen ion in 0.1 mol dm–3 of
hydrochloric acid is higher than of 0.1 mol dm–3 of
0.1 mol dm–3
CH3COOH(aq/ak )
Asid etanoik 0.1 mol dm–3 mengion kurang daripada
ion hidrogen:
– 0.1 mol dm–3 of ethanoic acid ionises to less than
0.1 mol dm–3 hydrogen ion:
Asid etanoik adalah asid lemah mengion separa dalam air
menghasilkan kepekatan ion hidrogen yang lebih rendah .
weak
acid ionises partially in
– Ethanoic acid is a
water to produce lower concentration hydrogen ion.
0.1 mol dm
HCl
–3
Asid hidroklorik 0.1 mol dm–3 mengion lengkap kepada 0.1 mol dm
ion hidrogen:
– 0.1 mol dm–3 of hydrochloric acid ionises to form
0.1 mol dm–3 hydrogen ion:
Asid hidroklorik adalah asid
air kepada ion hidrogen.
– Hydrochloric acid is a strong acid ionises completely
in water to hydrogen ion.
0.1 mol dm–3 HCl
1.21
III
Rajah di bawah menunjukkan bacaan pH meter untuk pelbagai jenis dan kepekatan asid. Tujuan eksperimen adalah untuk mengkaji hubungan antara kepekatan ion hidrogen dengan nilai pH.
Bandingkan kepekatan ion hidrogen dan nilai pH untuk asid-asid yang berikut. Terangkan jawapan anda.
The diagram below shows the reading of pH meter for different types and concentration of acids. The aim of the experiment is to investigate the relationship between
concentration of hydrogen ions with the pH value. Compare the concentration of hydrogen ions and the pH value of the following acids. Explain your answer.
Example / Contoh:
Chemistry Form 4 • MODULE
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MODULE • Chemistry Form 4
PREPARATION OF STANDARD SOLUTION / PENYEDIAAN LARUTAN PIAWAI
Standard solution is a solution in which its concentration is accurately known.
1
Larutan piawai ialah larutan yang kepekatannya diketahui dengan tepat.
The steps taken in preparing a standard solution are:
2
Langkah-langkah yang diambil dalam menyediakan larutan piawai adalah:
(a) Calculate the mass of solute needed to give the required volume and molarity.
Hitung jisim bahan larut yang diperlukan untuk menghasilkan isi padu dan kemolaran yang dikehendaki.
(b) The solute is weighed / Bahan larut ditimbang.
(c) The solute is completely dissolved in distilled water and then transferred to a volumetric flask partially filled with
distilled water.
Bahan larut dilarutkan sepenuhnya dalam air suling dan dipindahkan kepada kelalang volumetrik yang sebahagiannya sudah diisi
dengan air suling.
(d) Distilled water is added to the calibration mark of the volumetric flask and the flask is inverted to make sure
thorough mixing.
Air suling ditambah ke dalam kelalang volumetrik hingga tanda senggatan dan kelalang volumetrik ditelangkupkan beberapa kali
untuk memastikan campuran sekata.
PREPARATION OF A SOLUTION BY DILUTION / PENYEDIAAN LARUTAN SECARA PENCAIRAN
Adding water to the standard solution lowered the concentration of the solution. Since no solute is added, the amount of
solute in the solution before and after dilution remains unchanged:
Penambahan air kepada larutan piawai merendahkan kepekatan larutan tersebut. Oleh kerana tiada bahan terlarut yang ditambah, kandungan
bahan terlarut dalam larutan sebelum dan selepas pencairan tidak berubah:
Number of mol of solute before dilution = Number of mole of solute after dilution
Bilangan mol bahan terlarut sebelum pencairan
M1V1
1 000
Therefore / Oleh itu,
Bilangan mol bahan terlarut selepas pencairan
=
M2V2
1 000
M1V1 = M2V2
M1 = Initial concentration of the solute / Kemolaran larutan awal
V1 = Initial volume of the solution in cm3 / Isipadu larutan awal dalam cm3
M2 = Final concentration of the solute / Kemolaran larutan akhir
V2 = Final volume of the solution in cm3 / Isipadu larutan akhir dalam cm3
Example / Contoh :
(a) What is meant by a standard solution / Apakah yang dimaksudkan dengan larutan piawai ?
1
Standard solution is a solution in which its concentration is accurately known.
(b) (i)
You are given solid sodium hydroxide. Describe the procedure to prepare 500 cm3 of 1.0 mol dm–3 sodium
hydroxide solution. [Relative atomic mass: H = 1; O = 16; Na = 23]
Anda diberi pepejal natrium hidroksida. Huraikan kaedah untuk menyediakan 500 cm3 larutan natrium hidroksida
1.0 mol dm–3. [Jisim atom relatif: H = 1, O = 16, Na = 23]
Calculate the mass of sodium hydroxide / Hitung jisim natrium hidroksida :
–
–
–
Molar mass of NaOH / Jisim molar NaOH = 23 + 16 + 1 = 40 g mol
Mol NaOH / Bilangan mol NaOH = 500 × 1.0/1 000 = 0.5 mol
–1
Mass of NaOH / Jisim NaOH = mol / Bilangan mol × molar mass / Jisim molar
–1
= 0.5 mol × 40 g mol = 20.0 g
Preparation of 500 cm3 1.0 mol dm–3 sodium hydroxide:
Penyediaan 500 cm3 larutan natrium hidroksida 1.0 mol dm–3:
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20.0 g
20.0 g
of NaOH accurately using a
NaOH dengan tepat menggunakan
weighing bottle .
botol penimbang
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Chemistry Form 4 • MODULE
–
–
Dissolve
20.0 g
Larutkan
20.0 g
distilled water
of NaOH in
air suling
NaOH dalam
Transfer the content into a 500 cm
in a beaker.
di dalam bikar.
volumetric flask .
3
Pindahkan kandungan ke dalam kelalang volumetrik 500 cm3.
–
–
Rinse
the beaker with distilled water and transfer all the contents into the volumetric flask .
Bilas
bikar dengan
Distilled water
air suling
is added to the
dan pindahkan semua kandungan ke dalam kelalang volumetrik .
volumetric flask
Air suling ditambah kepada kelalang volumetrik hingga
–
The volumetric flask is closed tightly with
homogeneous solution.
Kelalang volumetrik ditutup dengan
penutup
tanda senggatan
stopper
dan
calibration mark .
until the
.
inverted
and
ditelangkupkan
a few times to get
beberapa kali untuk mendapatkan larutan
yang homogen.
(ii)
Describe how you would prepare 250 cm3 of 0.1 mol dm–3 sodium hydroxide from the above solution.
Huraikan bagaimana anda menyediakan 250 cm3 larutan natrium hidroksida 0.1 mol dm–3 daripada larutan di atas.
Calculate the volume of 1 mol dm–3 sodium hydroxide used:
Hitung isi padu natrium hidroksida 1 mol dm–3 yang digunakan:
–
M1 × V1 = M2 × V2
–
V1 =
M2 × V2
= 0.1 × 250 = 25 cm3
M1
1
Preparation of 250 cm3 1.0 mol dm–3 sodium hydroxide solution:
Penyediaan 250 cm3 larutan natrium hidroksida 1.0 mol dm–3:
–
–
–
A pipette is filled with
25 cm3
of 1.0 mol dm–3 sodium hydroxide solution.
Sebuah pipet diisi dengan
25 cm3
larutan natrium hidroksida 1.0 mol dm–3.
25 cm3
of 1.0 mol dm–3 NaOH is transferred into 250 cm3
25 cm3
NaOH 1.0 mol dm dipindahkan kepada kelalang volumetrik 250 cm3.
Distilled water
is added to the
volumetric flask
Air suling ditambah kepada kelalang volumetrik hingga
–
volumetric flask .
–3
The volumetric flask is closed tightly with
homogeneous solution.
Kelalang volumetrik ditutup dengan
penutup
until the
stopper
dan
calibration mark .
tanda senggatan
and
ditelangkupkan
.
inverted
a few times to get
beberapa kali untuk mendapatkan larutan
yang homogen.
EXERCISE / LATIHAN
1
The table below shows the pH value of a few substances / Jadual di bawah menunjukkan nilai pH bagi beberapa bahan.
Substance / Bahan
pH value / Nilai pH
Ethanoic acid 0.1 mol dm / Asid etanoik 0.1 mol dm
3
Hydrochloric acid 0.1 mol dm–3 / Asid hidroklorik 0.1 mol dm–3
1
Glacial ethanoic acid / Asid etanoik glasial
7
–3
(a) (i)
–3
What is meant by weak acid and strong acid / Apakah yang dimaksudkan dengan asid lemah dan asid kuat ?
Weak acid / Asid lemah : An acid that partially ionises in water to produce low concentration of hydrogen ion,
H+.
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Strong acid / Asid kuat : An acid that completely ionises in water to produce high concentration of hydrogen
ion, H+.
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MODULE • Chemistry Form 4
(ii)
Between ethanoic acid and hydrochloric acid, which acid has the higher concentration of H+ ion? Explain
your answer.
Antara asid etanoik dengan asid hidroklorik, asid manakah mempunyai kepekatan ion H+ yang lebih tinggi? Terangkan
jawapan anda.
–
higher
Hydrochloric acid has
concentration of H+ than ethanoic acid.
tinggi
Asid hidroklorik mempunyai kepekatan ion H+ yang lebih
–
Hydrochloric acid is a strong acid which ionises completely in water to produce
concentration of H+:
asid kuat
Asid hidroklorik ialah
tinggi
yang lebih
:
HCl (aq/ak )
–
berbanding dengan asid etanoik.
sepenuhnya
yang mengion
lemah
:
CH3COOH (aq/ak )
dalam air untuk menghasilkan kepekatan ion H+
H+(aq/ak ) + Cl–(aq/ak )
Ethanoic acid is a weak acid which ionises
H+:
Asid etanoik ialah asid
rendah
yang lebih
higher
partially
in water to produce lower concentration of
separa
yang mengion
dalam air untuk menghasilkan kepekatan ion H+
CH3COO– (aq/ak ) + H+ (aq/ak )
(iii) Why do ethanoic acid and hydrochloric acid have different pH value?
Mengapakah asid etanoik dan asid hidroklorik mempunyai nilai pH yang berbeza?
–
The concentration H+ in hydrochloric acid is
tinggi
Kepekatan H dalam asid hidroklorik
+
–
The concentration H in ehanoic acid is
+
Kepekatan H dalam asid etanoik
+
rendah
higher
, the pH value is
rendah
, nilai pH lebih
lower
, nilai pH lebih
.
.
, the pH value is
tinggi
lower
higher
.
.
(b) Glacial ethanoic acid has a pH value of 7 but a solution of ethanoic acid has a pH value less than 7. Explain the
observation.
Asid etanoik glasial mempunyai nilai pH 7 tetapi asid etanoik mempunyai nilai pH yang kurang daripada 7. Terangkan pemerhatian
tersebut.
–
Glacial ethanoic acid molecules do not ionise . Glacial ethanoic acid consists of only CH3COOH
molecules . The CH COOH molecules are
neutral . No hydrogen ions present. The pH value of
3
glacial ethanoic acid is 7.
mengion . Asid etanoik glasial hanya terdiri daripada
molekul
CH3COOH
Molekul asid etanoik glasial tidak
Molekul
hidrogen
CH COOH adalah neutral. Tiada ion
hadir. Nilai pH asid etanoik glasial adalah 7.
sahaja.
3
– Ethanoic acid ionises partially in water to produce ethanoate ions and
solution to have acidic property. The pH value of the solution is less than 7.
hydrogen
etanoat
hidrogen
dan ion
Asid etanoik mengion separa dalam air untuk menghasilkan ion
asid
. Nilai pH bagi larutan tersebut adalah kurang daripada 7.
larutan mempunyai sifat
ions causes the
yang menyebabkan
The table shows the pH value of a few solution / Jadual berikut menunjukkan nilai pH bagi beberapa jenis larutan berbeza.
2
Solution / Larutan
P
Q
R
S
T
U
pH
1
3
5
7
11
14
(a) (i)
Which solution has the highest concentration of hydrogen ion?
Larutan manakah yang mempunyai kepekatan ion hidrogen yang paling tinggi?
Solution P
(ii)
Which solution has the highest concentration of hydroxide ion?
Larutan yang manakah mempunyai kepekatan ion hidroksida yang paling tinggi?
Solution U
m
(i)
0.01 mol dm–3 of hydrochloric acid / 0.01 mol dm–3 asid hidroklorik ?
Q
(ii)
0.01 mol dm–3 of ethanoic acid / 0.01 mol dm–3 asid etanoik ?
R
(iii) 0.1 mol dm–3 ammonia aqueous / 0.1 mol dm–3 larutan ammonia ?
T
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(b) Which is the following solution could be / Larutan manakah yang mungkin
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Chemistry Form 4 • MODULE
P
(iv) 1 mol dm–3 of hydrochloric acid / 1 mol dm–3 asid hidroklorik ?
(v)
U
1 mol dm–3 sodium hydroxide solution / 1 mol dm–3 larutan natrium hidroksida ?
S
(vi) 1 mol dm–3 potassium sulphate solution / 1 mol dm–3 larutan kalium sulfat ?
(c) (i)
State two solutions which react to form neutral solution.
Nyatakan dua larutan yang bertindak balas untuk membentuk larutan neutral.
P/Q/R and T/U // Hydrochloric acid/ethanoic acid with ammonia aqueous/sodium hydroxide solution.
(ii)
Which solutions will produce carbon dioxide gas when calcium carbonate powder is added?
Larutan manakah menghasilkan gas karbon dioksida apabila ditambah serbuk kalsium karbonat?
P/Q // Hydrochloric acid/ethanoic acid
3
The molarity of sodium hydroxide solution is 2 mol dm–3. What is the concentration of the solution in g dm–3?
[RAM: Na = 23, 0 = 16, H = 1]
Kemolaran larutan natrium hidroksida ialah 2 mol dm–3. Apakah kepekatan larutan tersebut dalam g dm–3?
[JAR: Na = 23, O = 16, H = 1]
Answer / Jawapan :
4
80 g dm–3
Calculate the molarity of the solution obtained when 14 g potassium hydroxide is dissolved in distilled water to make
up 500 cm3 solution. [RAM: K = 39, H = 1, O = 16]
Hitung kemolaran larutan yang diperoleh apabila 14 g kalium hidroksida dilarutkan dalam air suling untuk menyediakan larutan yang
berisi padu 500 cm3. [JAR: K = 39, H = 1, O = 16]
Answer / Jawapan :
5
0.5 mol dm–3
Calculate the molarity of a solution which is prepared by dissolving 0.5 mol of hydrogen chloride, HCl in distilled water
to make up 250 cm3 solution.
Hitung kemolaran larutan yang disediakan dengan melarutkan 0.5 mol hidrogen klorida, HCl dalam air suling untuk menyediakan larutan
yang berisi padu 250 cm3.
Answer / Jawapan :
6
2 mol dm–3
How much of sodium hydroxide in gram should be dissolved in water to prepare 500 cm3 of 0.5 mol dm–3 sodium
hydroxide solution? [RAM: Na = 23, O = 16, H = 1]
Berapakah jisim natrium hidroksida dalam gram yang patut dilarutkan dalam air untuk menyediakan 500 cm3 larutan natrium hidroksida
0.5 mol dm–3? [JAR: Na = 23, O = 16, H = 1]
10 g
Answer / Jawapan :
7
300 cm3 water is added to 200 cm3 hydrochloric acid, 1 mol dm–3. What is the resulting molarity of the solution?
Jika 300 cm3 air ditambah kepada 200 cm3 asid hidroklorik 1 mol dm–3, apakah kemolaran bagi larutan yang dihasilkan?
0.4 mol dm–3
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Answer / Jawapan :
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MODULE • Chemistry Form 4
Calculate the volume of nitric acid, 1 mol dm–3 needed to be diluted by distilled water to obtain 500 cm3 of nitric acid,
0.1 mol dm–3.
8
Hitung isi padu asid nitrik 1 mol dm–3 yang diperlukan untuk dilarutkan oleh air suling bagi menghasilkan 500 cm3 asid nitrik 0.1 mol dm–3.
Answer / Jawapan :
50 cm3
(a) Compare the number of mol of H+ ions which are present in 50 cm3 of 1 mol dm–3 of sulphuric acid and 50 cm3 of
1 mol dm–3 of hydrochloric acid. Explain your answer.
9
Bandingkan bilangan mol ion H+ yang hadir dalam 50 cm3 asid sulfurik 1 mol dm–3 dengan 50 cm3 asid hidroklorik 1 mol dm–3.
Terangkan jawapan anda.
Acid
Asid
Calculate number of
hydrogen ion, H+
50 cm3 of 1 mol dm–3 of sulphuric acid
50 cm3 asid sulfurik 1 mol dm–3
50 × 1
1 000
= 0.05 mol
50 cm3 of 1 mol dm–3 of hydrochloric acid
50 cm3 asid hidroklorik 1 mol dm–3
50 × 1
1 000
= 0.05 mol
Number of mol of sulphuric acid =
Number of mol of hydrochloric acid =
H2SO4
HCl
Hitung bilangan mol
ion hidrogen, H+
2H+ + SO42–
From the equation,
1 mol of H2SO4 : 2 mol of H
H+ + Cl–
From the equation,
1 mol of HCl : 1 mol of H+
+
0.05 mol of H2SO4 : 0.1 mol of H+
0.05 mol of HCl : 0.05 mol of H+
Compare the number The number of H+ in 50 cm3 of 1 mol dm–3 of sulphuric acid is twice of the number of H+ in
of hydrogen ions
50 cm3 of 1 mol dm–3 of hydrochloric acid.
Bandingkan bilangan
ion hidrogen
Explanation
Penerangan
Sulphuric acid is diprotic acid whereas hydrochloric acid is monoprotic acid.
1 mol of sulphuric ionises to 2 mol of H+ whereas 1 mol of hydrochloric acid ionises to 1 mol of H+.
The number of H+ in the same volume and concentration of both acids is doubled in sulphuric acid
compared to hydrochloric acid.
(b) Suggest the volume of 1 mol dm–3 of hydrochloric acid that has the same number of H+ with 50 cm3 of 1 mol dm–3
of sulphuric acid.
Cadangkan isi padu asid hidroklorik 1 mol dm–3 yang mempunyai bilangan ion H+ yang sama dengan 50 cm3 asid sulfurik 1 mol dm–3.
100 cm3
NEUTRALISATION / PENEUTRALAN
Neutralisation is the reaction between an acid and a base to form only salt and water:
Peneutralan ialah tindak balas antara asid dan bes untuk membentuk garam dan air sahaja:
1
Acid / Asid + Base / Bes
Salt / Garam + Water / Air
Example / Contoh :
HCl (aq/ak ) + NaOH (aq/ak )
2HNO3 (aq/ak ) + MgO (s/p)
NaCl (aq/ak ) + H2O (l/ce)
Mg(NO3)2 (aq/ak ) + H2O (l/ce)
In neutralisation, the acidity of an acid is neutralised by an alkali. At the same time the alkalinity of an alkali is
neutralised by an acid. The hydrogen ions in the acid react with hydroxide ions in the alkali to produce water:
2
Dalam peneutralan, keasidan asid dineutralkan oleh alkali. Pada masa yang sama, kealkalian alkali dineutralkan oleh asid. Ion hidrogen
dalam asid bertindak balas dengan ion hidroksida dalam alkali:
m
H2O (l/ce)
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H+ (aq/ak ) + OH– (aq/ak )
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Chemistry Form 4 • MODULE
3
Application of neutralisation in daily life / Aplikasi peneutralan dalam kehidupan seharian :
Application
Example
Aplikasi
Agriculture
Agrikultur
Contoh
1 Acidic soil is treated with powdered
CaCO3)or ashes of burnt wood.
soda lime
(calcium oxide, CaO),
limestone
(calcium carbonate,
Tanah berasid dirawat dengan serbuk kapur tohor (kalsium oksida, CaO), batu kapur (kalsium karbonat, CaCO3) atau abu
daripada kayu api.
2 Basic soil is treated with compost. The
alkalis in basic soil.
Tanah berbes dirawat dengan kompos. Gas
dalam tanah berbes.
acidic
berasid
gas from the decomposition of compost
yang terbebas daripada penguraian kompos
3 The acidity of water farming is controlled by adding
soda lime
neutralises
meneutralkan
the
alkali
, (calcium oxide, CaO).
Keasidan air dalam pertanian dikawal dengan menambah kapur tohor (kalsium oksida, CaO).
Industries
Industri
soda lime
1 Acidic gases emitted by industries are neutralised with
are released into air.
, (calcium oxide, CaO) before the gases
Gas-gas berasid yang dibebaskan oleh kilang dineutralkan dengan kapur tohor (kalsium oksida, CaO), sebelum gas-gas tersebut
dibebaskan ke udara.
2 Organic acid produced by bacteria in latex neutralises by
ammonia
and prevents coagulation.
Ammonia meneutralkan asid organik yang dihasilkan oleh bakteria dalam lateks dan mencegah penggumpalan.
Health
Kesihatan
1 Excess acid in the stomach is neutralised with its anti-acids that contain bases such as aluminium hydroxide ,
calcium carbonate
and magnesium hydroxide .
Anti-asid mengandungi bes seperti aluminium hidroksida ,
meneutralkan asid berlebihan dalam perut.
2 Toothpastes contain
mouth.
Ubat gigi mengandungi
mulut.
3
Baking powder
Serbuk penaik
4
Vinegar
Cuka
4
bases
bes
kalsium karbonat
dan
magnesium hidroksida
untuk
(such as magnesium hydroxide) to neutralise the acid produced by bacteria in
(seperti magnesium hidroksida) untuk meneutralkan asid yang dihasilkan oleh bakteria dalam
(sodium hydrogen carbonate) is used to cure alkaline bee stings.
(natrium hidrogen karbonat) digunakan untuk merawat sengatan lebah yang beralkali.
(Ethanoic acid) is used to cure acidic wasp sting.
(asid etanoik) digunakan untuk merawat sengatan tebuan yang berasid.
An acid-base titration / Pentitratan asid-bes :
(a) It is a technique used to determine the volume of an acid required to neutralise a fixed volume of an alkali with
the help of acid-base indicator.
Ianya adalah teknik yang digunakan untuk menentukan isi padu asid yang diperlukan untuk meneutralkan isi padu tertentu alkali
dengan bantuan penunjuk asid-bes. Penunjuk yang biasa digunakan adalah fenolftalein dan metil jingga.
(b) Steps taken are / Langkah-langkah yang diambil adalah :
(i)
An exact volume of alkali is measured with a pipette and poured into a conical flask.
Isi padu alkali yang tepat diukur dengan pipet dan dituang ke dalam kelalang kon.
A few drops of indicator are added to the alkali / Beberapa titik penunjuk ditambah kepada alkali.
(ii)
(iii) A burette is filled with an acid. An acid is added drop by drop into the alkali in the conical flask until the
indicator changes colour, indicating the pH of neutral solution produced.
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Buret diisi dengan asid. Asid ditambah setitik demi setitik kepada alkali dalam kelalang kon sehingga warna penunjuk bertukar,
menunjukkan pH larutan neutral telah dihasilkan.
(c) When the acid has completely neutralised the given volume of an alkali, the titration has reached the end point.
Apabila asid sudah lengkap meneutralkan isi padu alkali yang diberi, pentitratan telah mencapai takat akhir.
(d) The end point is the point in the titration at which the indicator changes colour.
Takat akhir ialah takat dalam pentitratan di mana penunjuk bertukar warna.
(e) The commonly used indicators are phenolphthalein and methyl orange.
Penunjuk yang biasa digunakan adalah fenolftalein dan metil jingga.
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MODULE • Chemistry Form 4
The general steps used in any calculation involving neutralisation:
5
Langkah umum dalam penghitungan yang melibatkan peneutralan:
Step / Langkah 1 : Write the balanced equation / Tulis persamaan yang seimbang.
Step / Langkah 2 : Write the information from the question above the equation.
Tulis maklumat daripada soalan di atas persamaan.
Step / Langkah 3 : Write the information from the chemical equation below the equation (number of moles of substance
involved).
Tulis maklumat daripada persamaan kimia di bawah persamaan (bilangan mol bahan yang terlibat).
Step / Langkah 4 : Change the information to mole / Tukar maklumat kepada mol.
Step / Langkah 5 : Use the relationship between the number of moles of the substances in Step 3.
Guna hubungan di antara bilangan mol bahan-bahan dalam Langkah 3.
Step / Langkah 6 : Convert the number of mol to the required unit with the formula:
Tukar bilangan mol kepada unit yang diperlukan dengan menggunakan formula:
n=
n
M
V
v
Mv
1 000
atau n = MV
= Number of moles of solute / Bilangan mol bahan terlarut
= Concentration in mol dm–3 (molarity) / Kepekatan dalam mol dm–3 (kemolaran)
= Volume of solution in dm3 / Isi padu larutan dalam dm3
= Volume of solution in cm3 / Isi padu larutan dalam cm3
EXERCISE / LATIHAN
50 cm3 of 1 mol dm–3 sodium hydroxide solution is neutralised by 25 cm3 of sulphuric acid. Calculate the concentration
of sulphuric acid in mol dm–3 and g dm–3. [RAM: H = 1, S = 32, O = 16]
1
50 cm3 larutan natrium hidroksida 1 mol dm–3 dineutralkan oleh 25 cm3 asid sulfurik. Hitung kepekatan asid sulfurik dalam mol dm–3 dan
g dm–3. [JAR: H = 1, S = 32, O = 16]
M = 1 mol dm–3
V = 50 cm3
M=?
V = 25 cm3
2NaOH + H2SO4
Na2SO4 + 2H2O
Number of mol of NaOH = 1 ×
From the equation,
n mol
V dm3
0.025 mol
=
= 1 mol dm–3
25
dm3
1 000
Concentration of H2SO4 =
50
= 0.05 mol
1 000
2 mol of NaOH : 1 mol of H2SO4
0.05 mol of NaOH : 0.025 mol of H2SO4
Concentration of H2SO4
= 1 mol dm–3 × (2 × 1 + 32 + 16 × 4) g mol–1
= 98 g dm–3
Calculate the volume of 2 mol dm–3 sodium hydroxide needed to neutralise 100 cm3 of 1 mol dm–3 hydrochloric acid.
2
Hitung isi padu larutan natrium hidroksida 2 mol dm–3 yang diperlukan untuk meneutralkan 100 cm3 asid hidroklorik 1 mol dm–3.
M = 2 mol dm–3
V = ? cm3
M = 1 mol dm–3
V = 100 cm3
NaOH
HCl
+
Number of mol of HCl = 1 ×
NaCl + H2O
100
= 0.1 mol
1 000
1 mol of HCl : 1 mol of mol NaOH
0.1 mol of HCl : 0.1 mol of mol NaOH
n mol
Volume of NaOH =
M mol dm–3
0.1 mol
=
2 mol dm–3
= 0.05 dm–3
= 50 cm3
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From the equation,
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Chemistry Form 4 • MODULE
3
Experiment I / Eksperimen I
1 mol dm–3 of nitric acid is used to neutralise 100 cm3 of 1 mol dm–3 sodium hydroxide solution.
Asid nitrik 1 mol dm–3 digunakan untuk meneutralkan 100 cm3 larutan natrium hidroksida 1 mol dm–3.
Experiment II / Eksperimen II
1 mol dm–3 of sulphuric acid is used to neutralise 100 cm3 of 1 mol dm–3 sodium hydroxide solution.
Asid sulfurik 1 mol dm–3 digunakan untuk meneutralkan 100 cm3 larutan natrium hidroksida 1 mol dm–3.
Compare the volume of acids needed to neutralise 100 cm3 of 1 mol dm–3 sodium hydroxide solution in Experiment I
and Experiment II. Explain your answer.
Bandingkan isi padu asid yang diperlukan untuk meneutralkan 100 cm3 larutan natrium hidroksida 1 mol dm–3 dalam Eksperimen I dan
Eksperimen II. Terangkan jawapan anda.
Answer / Jawapan:
Experiment
Experiment I
Eksperimen
Balanced equation
NaOH + HNO3
Persamaan kimia
Calculation
Pengiraan
Experiment II
Eksperimen I
Eksperimen II
NaNO3 + H2O
2NaOH + H2SO4
100
1 000
= 0.1 mol
From the equation / Daripada persamaan :
1 mol NaOH :
1 mol HNO
100
1 000
= 0.1 mol
From the equation / Daripada persamaan :
2 mol NaOH / NaOH :
1 mol H SO
Mol of KOH / Bilangan mol NaOH = 1 ×
0.1
Mol of KOH / Bilangan mol NaOH = 1 ×
3
2
4
0.1 mol NaOH / NaOH : 0.05 mol H SO
mol HNO3
2
4
Mv
Mv
Mol of HCl / Bilangan mol HNO3 =
Mol of H2SO4 / Bilangan mol H2SO4 =
1 000
1 000
M = Concentration of HNO3 / Kepekatan HNO3
M = Concentration of H2SO4 / Kepekatan H2SO4
v = Volume of HNO3 in cm3 / Isi padu HNO3 dalam cm3 v = Volume of H2SO4 in cm3 / Isi padu H2SO4 dalam cm3
1 mol dm–3 × v
1 mol dm–3 × v
= 0.1 mol
= 0.1 mol
1 000
1 000
3
3
v = 100 cm
v = 50 cm
mol NaOH :
0.1
Na2SO4 + 2H2O
Comparison and
explanation
– The volume of acid needed in Experiment II is doubled of Experiment I.
Perbandingan dan
penerangan
– Sulphuric acid is
Isi padu asid nitrik yang diperlukan adalah dua kali ganda dalam Eksperimen I dibandingkan dengan Eksperimen II.
diprotic
Asid sulfurik adalah asid
acid while nitric acid is
diprotik
– One mol of sulphuric ionises
monoprotic .
manakala asid nitrik adalah asid
two
Satu mol asid sulfurik mengion kepada
ion H+.
dua
monoprotik
.
mol of H , one mol nitric acid ionises to
+
one
mol of H+.
mol ion H manakala satu mol asid nitrik mengion kepada
+
satu
mol
– The number of H+ in the same volume and concentration of both acids is doubled in sulphuric acid
compared to hydrochloric acid.
Bilangan ion H+ dalam isi padu dan kepekatan yang sama bagi kedua-dua asid adalah dua kali ganda dalam asid
sulfurik dibandingkan dengan asid nitrik.
4
Diagram below shows the apparatus set-up for the titration of potassium hydroxide solution with sulphuric acid.
Gambar rajah di bawah menunjukkan susunan radas bagi pentitratan larutan kalium hidroksida dengan asid sulfurik.
0.5 mol dm–3 sulphuric acid
Asid sulfurik 0.5 mol dm–3
50 cm3 of 1 mol dm3 potassium hydroxide solution + methyl orange
50 cm3 larutan kalium hidroksida 1 mol dm3 + metil jingga
0.5 mol dm–3 sulphuric acid is titrated to 50 cm3 of 1 mol dm3 potassium hydroxide solution and methyl orange is used
as indicator.
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Asid sulfurik 0.5 mol dm–3 ditambahkan kepada 50 cm3 larutan kalium hidroksida 1 mol dm3 dan metil jingga digunakan sebagai penunjuk.
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MODULE • Chemistry Form 4
(a) (i)
Name the reaction between sulphuric acid and potassium hydroxide.
Namakan tindak balas antara asid sulfurik dengan kalium hidroksida.
Neutralisation
(ii)
Name the salt formed in the reaction / Namakan garam yang terbentuk dalam tindak balas tersebut.
Potassium sulphate
(b) Suggest an apparatus that can be used to measure 25.0 cm3 of potassium hydroxide solution accurately.
Cadangkan radas yang boleh digunakan untuk mengukur 25.0 cm3 larutan kalum hidroksida dengan tepat.
Pipette
(c) What is the colour of methyl orange / Apakah warna metil jingga dalam
(i) in potassium hydroxide solution / larutan kalium hidroksida?
Red
(ii)
in sulphuric acid / asid sulfurik?
Yellow
(iii) at the end point of the titration / pada titik akhir pentitratan?
Orange
(d) (i)
Write a balanced equation for the reaction that occurs / Tuliskan persamaan seimbang bagi tindak balas yang berlaku.
2KOH + H2SO4
(ii)
K2SO4 + 2H2O
Calculate the volume of the 0.1 mol dm–3 sulphuric acid needed to completely react with 50 cm3 of 1 mol dm3
potassium hydroxide.
Hitung isi padu asid sulfurik 0.1 mol dm–3 yang diperlukan untuk bertindak balas dengan lengkap dengan 50 cm3 larutan
kalium hidroksida 1 mol dm–3.
Number of mol KOH = 1 ×
From the equation,
50
= 0.05 mol
1 000
2 mol of KOH : 1 mol of H2SO4
0.05 mol of KOH : 0.025 mol of H2SO4
n mol
M mol dm–3
0.025 mol
=
1 mol dm–3
= 0.025 dm3
= 25 cm3
Volume of H2SO4 =
(e) (i)
The experiment is repeated with 0.1 mol dm–3 hydrochloric acid to replace sulphuric acid. Predict the volume
of hydrochloric acid needed to neutralise 50.0 cm3 potassium hydroxide solution.
Eksperimen diulang dengan menggunakan asid hidroklorik 0.1 mol dm–3 untuk menggantikan asid sulfurik. Ramalkan isipadu asid
hidroklorik yang diperlukan untuk meneutralkan 50.0 cm3 larutan kalium hidroksida.
50 cm3 // double the volume of sulphuric acid
(ii)
Explain your answer in (e)(i) / Terangkan jawapan anda di (e)(i).
– Hydrochloric acid is a monoprotic acid whereas sulphuric acid is a
Asid hidroklorik ialah asid
–
monoprotik
manakala asid sulfurik ialah asid
diprotic
diprotik
The same volume and concentration of both acids, hydrochloric acid contains
of mole of H+ ions as in sulphuric acid.
m
half
separuh
the number
bilangan
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Pada isi padu dan kepekatan yang sama untuk kedua-dua asid, asid hidroklorik mengandungi
mol ion H+ daripada asid sulfurik.
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Chemistry Form 4 • MODULE
Objective Questions / Soalan objektif
1
Which of the following substances changes blue litmus
paper to red when dissolved in water?
5
The table below shows the concentration of hydrochloric
acid and ethanoic acid.
Jadual di bawah menunjukkan kepekatan asid hidroklorik dan
asid etanoik.
Antara bahan berikut, yang manakah menukarkan warna kertas
litmus merah kepada biru apabila dilarutkan dalam air?
A Sulphur dioxide
Acid
Sulfur dioksida
B
Carbon dioxide
Lithium oxide
Ethanoic acid
Sodium carbonate
Which of the following statements is true about both
acids?
The table below shows the pH value of four acids which
have the same concentration.
Antara berikut, yang manakah adalah betul tentang kedua-dua
asid?
Jadual di bawah menunjukkan nilai pH empat larutan yang
mempunyai kepekatan yang sama.
A Both are strong acids
Kedua-duanya adalah asid kuat
Solution / Larutan
pH value / Nilai pH
B
Both acids are strong electrolyte
P
2
Q
7
C
The pH value of both acid are equal
R
12
S
13
Which of the following solutions has the highest
concentration of hydroxide ion?
Antara larutan berikut, yang manakah mempunyai kepekatan ion
hidroksida paling tinggi?
A P
B Q
3
A Sulphuric acid and copper(II) sulphate solution
6
A 20
B 40
7
Nitric acid and magnesium oxide
Hydrochloric acid and sodium nitrate solution
Asid hidroklorik dan larutan natrium nitrat
Asid etanoik dan larutan natrium sulfat
Antara tindak balas berikut, yang manakah tidak akan
membebaskan sebarang gas?
A Copper metal with sulphuric acid
Logam kuprum dengan asid sulfurik
B
Zinc metal with hydrochloride acid
Logam zink dengan asid hidroklorik
C
Ammonium chloride with calcium hydroxide
Ammonium klorida dengan kalsium hidroksida
D Sodium carbonate hydrochloric acid
8
C 200 cm3
D 250 cm3
Which of the following solutions have the same
concentration of hydrogen ions, H+, as in 0.1 mol dm–3
sulphuric acid, H2SO4?
Antara asid berikut, yang manakah mempunyai kepekatan ion
hidrogen, H+ yang sama dengan asid sulfurik 0.1 mol dm–3?
A 0.1 mol dm–3 hydrochloric acid
Asid hidroklorik 0.1 mol dm–3
B
0.1 mol dm–3 carbonic acid
Asid karbonik 0.1 mol dm–3
C
0.2 mol dm–3 ethanoic acid
Asid etanoik 0.2 mol dm–3
D 0.2 mol dm–3 nitric acid
Asid nitrik 0.2 mol dm–3
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What is the volume of 2.0 mol dm–3 potassium hydroxide
solution needed to prepare 500 cm3 of 0.1 mol dm–3
potassium hydroxide solution?
A 100 cm3
B 150 cm3
D Ethanoic acid and sodium sulphate solution
Which of the following reactions will not produce any
gas?
C 80
D 120
Berapakah isi padu larutan kalium hidroksida 2.0 mol dm–3
diperlukan untuk menyediakan 500 cm3 larutan kalium hidroksida
1 mol dm–3?
Asid nitrik dan magnesium oksida
4
The molarity of sodium hydroxide solution 0.5 mol dm–3.
What is the concentration of the solution in g dm–3?
[Relative atomic mass: H = 1, O =16, Na = 23]
Kemolaran larutan natrium hidroksida adalah 0.5 mol dm–3.
Apakah kepekatan larutan itu dalam g dm–3?
[Jisim atom relatif: H = 1, O =16, Na = 23]
Asid sulfurik dan larutan kuprum(II) sulfat
C
Nilai pH kedua-dua asid adalah sama
50 cm3 setiap asid memerlukan 50 cm3 larutan natrium
hidroksida 0.1 mol dm–3 untuk dineutralkan
C R
D S
Antara pasangan bahan tindak balas berikut, yang manakah akan
menghasilkan tindak balas?
Kedua-duanya adalah elektrolit yang kuat
D 50 cm3 of each acid need 50 cm3 of 0.1 mol dm–3 of
sodium hydroxide to be neutralised
Which of the following pairs of reactants would result
in a reaction?
B
0.1
Asid etanoik
Natrium karbonat
2
0.1
Asid hidroklorik
Litium oksida
D
Kepekatan / mol dm–3
Hydrochloric acid
Karbon dioksida
C
Concentration / mol dm–3
Asid
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MODULE • Chemistry Form 4
9
Which of the following sodium hydroxide solutions have
concentration of 1.0 mol dm–3?
[Relative atomic mass: H=1, O=16, Na =23]
10 The diagram below shows 25.0 cm3 of 1.0 mol dm–3 of
sulphuric acid and 50.0 cm3 of 1.0 mol dm–3 of sodium
are added hydroxide solution to form solution A.
Antara larutan natrium hidroksida berikut, yang manakah
mempunyai kepekatan 1.0 mol dm–3?
[JAR: H = 1, O = 16, Na = 23]
Rajah di bawah menunjukkan 25.0 cm3 asid sulfurik 1.0 mol dm–3 dan
50.0 cm3 larutan natrium hidroksida 1.0 mol dm–3 ditambah
bersama untuk menghasilkan larutan A.
I
5 g of sodium hydroxide dissolved in distilled water
to make 250 cm3 of solution.
50 cm3 of 1 mol dm–3 of hydroxide solution
50 cm3 larutan natrium hidroksida 1 mol dm–3
5 g natrium hidroksida dilarutkan dalam air suling
menjadikan 250 cm3 larutan.
II
20 g of sodium hydroxide dissolved in distilled water
to make 500 cm3 of solution.
20 g natrium hidroksida dilarutkan dalam air suling
menjadikan 500 cm3 larutan.
III 250 cm of 2 mol dm sodium hydroxide solution is
added to distilled water to make 1 dm3 of solution.
3
–3
250 cm3 larutan natrium hidroksida 2 mol dm ditambah air
suling menjadikan 1 dm3 larutan.
25 cm3 of 2.0 mol dm–3
sulphuric acid
25 cm3 asid sulfrik 2.0 mol dm–3
Solution A / Larutan A
–3
IV 500 cm3 of 2 mol dm–3 sodium hydroxide solution is
added to distilled water to make 1 dm3 of solution.
500 cm3 larutan natrium hidroksida 2 mol dm–3 ditambah air
suling menjadikan 1 dm3 larutan.
A I and III only
I dan III sahaja
B
II and III only
C
II and IV only
II dan III sahaja
II dan IV sahaja
Which of the following is true about the solution A?
Antara berikut, yang manakah adalah benar tentang larutan A?
A The solution has a pH value of 7
Larutan itu menpunyai nilai pH 7
B
The solution will react with any acid
C
The solution turns a red litmus paper blue
Larutan itu boleh bertindak balas dengan sebarang asid
Larutan itu menukarkan warna kertas litmus merah kepada biru
D The solution will react with zinc to produce hydrogen
gas
Larutan itu bertindak balas dengan zink untuk menghasilkan gas
hidrogen
D I, II, III and IV
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Chemistry Form 4 • MODULE
SALT
7
GARAM
PREPARATION OF SALTS / PENYEDIAAN GARAM
• THE MEANING OF SALTS / MAKSUD GARAM
– To write the meaning of salts and the formulae for all types of salt that are commonly found in this topic.
Menyatakan maksud garam dan menulis formula semua jenis garam yang biasa ditemui dalam tajuk ini.
• THE SOLUBILITY OF SALTS / KETERLARUTAN GARAM
– To determine the solubility of nitrate, sulphate, carbonate and chloride salts.
Menentukan keterlarutan semua garam nitrat, sulfat, karbonat dan klorida.
• EXPERIMENTS FOR THE PREPARATION OF SALTS BASED ON SOLUBILITY
EKSPERIMEN PENYEDIAAN GARAM BERDASARKAN KETERLARUTAN
– To determine the suitable methods for the preparation of salts based on solubility:
Menentukan kaedah yang sesuai bagi penyediaan garam berdasarkan keterlarutan:
i. Acid + metal / Asid + logam
ii. Acid + metal oxides / Asid + oksida logam
iii. Acid + alkali / Asid + alkali
iv. Acid + metal carbonate / Asid + karbonat logam
v. Double decomposition reaction / Tindak balas penguraian ganda dua
– To describe the experiments for each method of preparation and explain the rationale for each step.
Menghuraikan eksperimen bagi setiap jenis kaedah penyediaan serta menerangkan rasional setiap langkah.
CALCULATION ON QUANTITY OF REACTANTS/PRODUCTS [ QUANTITATIVE ANALYSIS ]
PENGHITUNGAN KUANTITI BAHAN/HASIL [ ANALISIS KUANTITATIF]
• CONTINUOUS VARIATIONS METHODS / KAEDAH PERUBAHAN BERTERUSAN
– To describe the methods of experiment to determine the formulae of insoluble salts.
Menghuraikan eksperimen bagi kaedah penentuan formula garam tak larut.
• SOLVING VARIOUS PROBLEMS RELATING TO QUANTITY OF REACTANTS/PRODUCTS IN SOLID, LIQUID AND GAS
FORMS
MENYELESAIKAN PELBAGAI MASALAH BERKAITAN KUANTITI BAHAN DALAM BENTUK PEPEJAL, LARUTAN DAN GAS
– Using the formula / Menggunakan formula:
i. n =
MV
1 000
ii. Mole / Mol =
Mass / Jisim
RAM/RMM/RFM / JAR/JMR/JFR
iii. The molar volume of gas at room temperature and s.t.p / Isi padu molar gas pada suhu bilik dan s.t.p
IDENTIFICATION OF IONS [ QUALITATIVE ANALYSIS ] / MENGENAL ION [ ANALISIS KUALITATIF ]
• ACTION OF HEAT ON SALTS / KESAN HABA KE ATAS GARAM
– To state the colour of the residue of lead(II) oxide, zinc oxide and copper(II) oxide.
Menyatakan warna baki bagi plumbum(II) oksida, zink oksida dan kuprum(II) oksida.
– To state the confirmatory tests for carbon dioxide and nitrogen dioxide.
Menyatakan ujian pengesahan bagi gas karbon dioksida dan nitrogen dioksida.
– To write the equations of the decomposition of carbonate and nitrate salts.
Menulis persamaan penguraian semua garam karbonat dan nitrat.
• CONFIRMATORY TEST CATIONS AND ANIONS / UJIAN PENGESAHAN KATION DAN ANION
– To state the confirmatory tests for all cations using sodium hydroxide and ammonia solution.
Menghuraikan ujian pengesahan semua kation menggunakan natrium hidroksida dan larutan ammonia.
– To state the confirmatory tests to differentiate Al3+ and Pb2+.
Menghuraikan ujian untuk membezakan Al3+ dan Pb2+.
– To state the confirmatory tests for anions of sulphate, nitrate, carbonate and chloride.
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MODULE • Chemistry Form 4
PREPARATION OF SALT / PENYEDIAAN GARAM
A salt is a compound formed when the hydrogen ion in an acid is replaced with metal ion or ammonium ion. Example:
Sodium chloride, copper(II) sulphate, potassium nitrate and ammonium sulphate.
1
Garam ialah sebatian ion yang terhasil apabila ion hidrogen daripada asid diganti oleh ion logam termasuk ion ammonium.
Contoh: natrium klorida, kuprum(II) sulfat, kalium nitrat dan ammonium sulfat.
Write the formulae of the salts in the table below by replacing hydrogen ion in sulphuric acid, hydrochloric acid, nitric
acid and carbonic acid with metal ions or ammonium ion.
2
Tuliskan formula kimia garam berikut dengan menggantikan ion hidrogen dalam asid sulfurik, asid hidroklorik, asid nitrik dan asid
karbonik dengan ion logam atau ion ammonium:
Metal ion
Ion logam
Sulphate salt (from H2SO4)
Garam sulfat (dari H2SO4 )
Chloride salt (from HCl)
Garam klorida (dari HCl)
Nitrate salt (from HNO3)
Garam nitrat (dari HNO3 )
Carbonate salt (from H2CO3)
Garam karbonat (dari H2CO3 )
Na+
Na2SO4
NaCl
NaNO3
Na2CO3
K+
K2SO4
KCl
KNO3
K2CO3
Mg2+
MgSO4
MgCl2
Mg(NO3 )2
MgCO3
Ca2+
CaSO4
CaCl2
Ca(NO3 )2
CaCO3
Al3+
Al2(SO4 )3
AlCl3
Al(NO3 )3
Al2(CO3 )3
Zn2+
ZnSO4
ZnCl2
Zn(NO3 )2
ZnCO3
Fe2+
FeSO4
FeCl2
Fe(NO3 )2
FeCO3
Pb2+
PbSO4
PbCl2
Pb(NO3 )2
PbCO3
Cu2+
CuSO4
CuCl2
Cu(NO3 )2
CuCO3
Ag+
Ag2SO4
AgCl
AgNO3
Ag2CO3
NH4+
(NH4 )2SO4
NH4Cl
NH4NO3
(NH4 )2CO3
Ba2+
BaSO4
BaCl2
Ba(NO3 )2
BaCO3
Solubility of salts in water: / Keterlarutan garam dalam air:
(a) All K+, Na+ and NH4+ salts are soluble. / Semua garam K+, Na+ dan NH4+ larut.
(b) All nitrate salts are soluble. / Semua garam nitrat larut.
(c) All carbonate salts are insoluble except K2CO3, Na2CO3 and (NH4)2CO3.
3
Semua garam karbonat tak larut kecuali K2CO3, Na2CO3 dan (NH4 )2CO3.
(d) All sulphate salts are soluble except CaSO4, PbSO4 and BaSO4.
Semua garam sulfat larut kecuali CaSO4, PbSO4 dan BaSO4.
(e) All chloride salts are soluble except PbCl2 and AgCl. / Semua garam klorida larut kecuali PbCl2 dan AgCl.
* Based on the solubility of the salts in water, shade the insoluble salts in the above table.
* Berdasarkan keterlarutan garam dalam air, lorekkan garam yang tak larut dalam jadual di atas.
Method used to prepare salt depends on the solubility of the salt.
4
Kaedah penyediaan garam bergantung pada keterlarutan garam tersebut.
Soluble salts are prepared from the reactions between an acid with a metal/ base/ metal carbonate:
Garam terlarut disediakan melalui tindak balas antara asid dengan logam/bes/karbonat logam:
i. Acid + metal / Asid + logam
salt + hydrogen / garam + hidrogen
Acid + *base
salt + water
ii. Acid + metal oxide / Asid + oksida logam
salt + water / garam + air
Asid + *bes
garam + air
iii. Acid + alkali / Asid + alkali
salt + water / garam + air
iv. Acid + metal carbonate / Asid + karbonat logam
salt + water + carbon dioxide / garam + air + karbon dioksida
* Most bases are metal oxide or metal hydroxide. / Kebanyakan bes adalah oksida logam atau hidroksida logam.
* All metal oxides and hydroxides are insoluble in water except Na2O, K2O, NaOH and KOH.
Semua oksida logam dan hidroksida logam tidak larut dalam air kecuali Na2O, K2O, NaOH dan KOH.
* Alkali is a base that soluble in water and ionises to hydroxide ion.
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Salts are prepared based on their solubility as shown in the flow chart below:
Asid + Alkali
Garam + Air
(Tindak balas Peneutralan)
– Acid + Alkali
Salt + Water
(Neutralisation Reaction)
Garam ini disediakan melalui kaedah pentitratan di antara
asid dan alkali dengan menggunakan penunjuk.
The salt is prepared by titration method of acid
and alkali using an indicator.
Method I / Kaedah I
Salts / Garam K+, Na+, NH4+
Soluble salt
Garam larut
Isi padu asid yang sama juga ditambah kepada isi padu
alkali yang sama tanpa penunjuk untuk mendapatkan
garam yang tulen dan neutral.
– The same volume of acid is then added to the
same volume of alkali without any indicator to
obtain pure and neutral salt solution.
Pentitratan dijalankan dengan menentukan isi padu asid
yang diperlukan untuk meneutralkan alkali yang isi
padunya sudah ditetapkan dengan menggunakan penunjuk.
Garam + Air + Karbon dioksida
Salt + Water + Carbon Dioxide
Keringkan baki dengan menekankan antara
kertas turas.
– Dry the residue by pressing it between
filter papers.
Bilas baki dengan air suling.
– Rinse the residue with distilled water.
Turas dengan corong turas.
– Filter using filter funnel.
Kacau dengan rod kaca.
– Stir with glass rod.
Campur dua larutan yang mengandungi kation
dan anion garam tak larut.
– Mix two solutions containing cations
and anions of insoluble salts.
Garam ini disediakan melalui kaedah
pemendakan. (Tindak balas penguraian ganda
dua).
The salt is prepared by precipitation
method. (Double decomposition
reaction)
Method III / Kaedah III
Insoluble salt
Garam tak larut
Turas dan keringkan hablur garam dengan menekan antara kertas turas.
– Cooled at room temperature / Biarkan sejuk pada suhu bilik.
– Filter and dry the salt crystals by pressing them between filter papers.
Celupkan dengan rod kaca, jika hablur terbentuk dengan serta merta, larutan adalah tepu.
– Evaporate the filtrate until it becomes a saturated solution/ Sejatkan hasil turasan hingga larutan tepu.
– Dip in a glass rod, if crystals are formed, the solution is saturated.
Turas campuran tersebut untuk mengeluarkan pepejal logam/oksida logam/karbonat
logam yang berlebihan.
– Filter the mixture to remove excess metal/metal oxide/metal carbonate
Tambah serbuk logam/oksida logam/karbonat logam ke dalam isi padu tetap asid
yang dihangatkan sehingga berlebihan.
volume of the heated acid
– Add metal/metal oxide/metal carbonate powder until excess into a fixed
Asid + Karbonat logam
Salt + Water (Neutralisation Reaction)
Garam + Air (Tindak balas Peneutralan)
– Acid + Metal carbonate
Asid + Oksida bes
Salt + Hydrogen (Displacement reaction)
Garam + Hidrogen (Tindak balas penyesaran)
– Acid + Metal oxide
Asid + Logam
– Acid + Metal
Garam ini disediakan melalui tindak balas antara asid dengan logam/oksida logam/
karbonat logam yang tak larut:
The salt is prepared by reacting acid with insoluble metal/metal oxide/
metal carbonate:
Method II / Kaedah II
Other than / Garam selain K+, Na+, NH4+
Preparation of salt / Penyediaan garam
Garam disediakan berdasarkan keterlarutannya sebagaimana yang ditunjukkan pada carta aliran di bawah:
– A titration is conducted to determine the volume
of acid needed to neutralise a fixed volume of an
alkali with the aid of an indicator.
1
PREPARATION OF SOLUBLE AND INSOLUBLE SALT / PENYEDIAAN GARAM LARUT DAN GARAM TAK LARUT
Chemistry Form 4 • MODULE
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50 – 100
yang berlebihan.
3
0.5 – 2
–3
Heat
Panaskan
Acid
Asid
Salt crystals
hablur garam
dengan
Keringkan
menekan antara kertas turas.
salt crystals
• Dry the
between
pressing them
papers.
serbuk
logam/ oksida logam/ karbonat logam pada isi padu asid yang
Tambahkan
tetap sambil dihangatkan perlahan-lahan .
by
filter
Hasil turasan ialah
larutan garam
The filtrate is
salt solution
logam/
Baki adalah
logam oksida/
logam karbonat
.
.
salt solution
is poured into
evaporating dish
.
Panaskan
Heat
Sejatkan larutan sehingga
terbentuk.
are
terbentuk.
Baki adalah
hablur garam
Residue is
salt crystals
Turaskan campuran tersebut untuk mengasingkan
hablur garam
.
garam
crystals
Hablur
Salt
hablur garam
• Filter the mixture to separate
the salt crystals .
Sejukkan pada suhu bilik sehingga
crystals salts
tepu
Larutan garam
salt solutions
Saturated
larutan tepu
• Evaporate the salt solution until
saturated solution
is formed.
Larutan garam
dituangkan dalam
mangkuk penyejat
.
• The
• Cool it at room temperature until
formed.
.
The residue is
metal /metal oxide
/metal carbonate .
Turas campuran tersebut untuk
mengasingkan bahan berlebihan iaitu
logam/oksida logam/karbonat logam
larutan garam
dengan
.
• Filter the mixture to separate
metal /metal oxide
excess
/metal carbonate with the
salt solution
.
cm of
mol dm of any acid
• Measure and pour
and pour into a beaker.
50 – 100
Sukat dan tuangkan
cm3 sebarang asid berkepekatan
–3
0.5 – 2
mol dm dan tuangkan ke dalam bikar.
• Add metal/metal oxide/ metal carbonate powder into the acid and heat
gently
.
Panaskan
Heat
Logam/oksida logam/
karbonat logam
Excess of metal/
metal oxide/
metal carbonate
Tambah serbuk logam / oksida logam
/ karbonat logam kepada asid sehingga
berlebihan
.
• Add metal/metal oxide / metal
powder
carbonate
to the acid
excess
until
.
campuran
dengan
Kacau
rod kaca
menggunakan
.
• Stir the mixture with a glass rod .
Method II:/Kaedah II:
Soluble salt except K+, Na+ and NH4+ / Garam larut selain K+, Na+ dan NH4+
2 Steps to Prepare Soluble Salt/Langkah Penyediaan Garam Larut
Method I:/Kaedah I:
1 mol dm–3 sebarang
asid dititratkan kepada alkali
sehingga neutral menggunakan
penunjuk. Isi padu asid yang
digunakan dicatat.
mol dm–3 of any acid
is titrated to the alkali
until neutral by using an
indicator. The volume of
acid used is recorded.
1
Alkali
Alkali
Acid
Asid
Ulang titratan tanpa penunjuk
untuk mendapatkan larutan
garam yang
tulen dan neutral .
• Repeat the titration
without the indicator
to get pure and neutral
salt solution.
•
Sukat dan tuangkan 50 cm3
sebarang alkali berkepekatan
1 mol dm–3 ke dalam
kelalang. Tambah beberapa
titis fenolftalein.
• Measure and pour 50 cm3
of 1 mol dm–3 any
alkali into a conical
flask. Add a few drops of
phenolphthalein.
Garam larut K+, Na+ dan NH4+
Soluble salt of K+, Na+ and NH4+
MODULE • Chemistry Form 4
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Chemistry Form 4 • MODULE
3
Steps to Prepare Insoluble Salt / Penyediaan Garam Tak Larut
Insoluble salts are prepared by the precipitation method through double decomposition reactions.
Garam tak larut disediakan dengan cara pemendakan melalui tindak balas penguraian ganda dua.
(i)
In this reaction, the precipitate of insoluble salt is formed when two different solutions that contain the cation and
anion of the insoluble salt are mixed.
Dalam tindak balas ini, mendakan garam tak larut terbentuk apabila dua larutan berbeza yang mengandungi kation dan anion
garam tak terlarut dicampurkan.
(ii) The insoluble salt is obtained as a residue of a filtration.
Garam tak terlarut tersebut diperoleh daripada baki penurasan.
Method III: Preparation of Insoluble XnYm Salt by Double Decomposition Reaction
Kaedah III: Penyediaan Garam Tak Larut XnYm Melalui Tindak balas Penguraian Ganda Dua
1) Measure and pour 50 – 100 cm3
0.5 – 2
of
mol dm–3 of aqueous
solution contains X m+ cation.
2) Measure and pour 50 – 100 cm3
0.5 – 2
mol dm–3 of
of
aqueous solution contains Yn– anion
into another beaker.
Sukat dan tuangkan 50 – 100 cm3
0.5 – 2
mol dm–3
larutan berkepekatan
mengandungi kation Xm+ ke dalam bikar.
Precipitate of
salt is formed.
XnYm
Sukat dan tuangkan 50 – 100 cm3
0.5 – 2
larutan berkepekatan
mol dm–3 mengandungi anion Yn– ke dalam
bikar yang lain.
3) Mix both solutions and stir the
glass rod .
Campur dan kacaukan
rod kaca
.
XnYm
Mendakan garam
terbentuk.
The residue is
salt.
mixture
campuran
with
menggunakan
XnYm
Mendakan adalah garam
XnYm
.
4)
Filter the
with
salt.
mixture
distilled water
and
rinse
the precipitate
. The residue is
XnYm
Turas campuran dan bilas mendakan itu menggunakan air suling.
Baki ialah garam XnYm.
Salt
XnYm
5)
Press the precipitate
between
filter papers to dry it.
Tekankan mendakan antara kertas turas untuk mengeringkannya.
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MODULE • Chemistry Form 4
Complete the following table:
Lengkapkan jadual berikut:
X m+
Yn–
XnYm
Pb2+ [Pb(NO3)2]
I– [KI]
PbI2
Ba2+[ BaCl2 ]
SO4 [ Na2SO4 ]
BaSO4
Ag+ [AgNO3]
Cl– [NaCl]
AgCl
Ag+ + Cl–
Ca2+ [Ca(NO3)2]
CO32– [Na2CO3]
CaCO3
Ca2+ + CO32–
2–
Ion equation/Persamaan ion
Pb2+ + I–
Ba2+ + SO42–
PbI2
BaSO4
AgCl
CaCO3
Complete the following table by writing “S” for soluble salts and “IS” for insoluble salts. Write all the possible chemical
equations to prepare soluble salts and two chemical equations for insoluble salts.
4
Lengkapkan jadual berikut dengan menulis “L” bagi garam larut dan “TL” bagi garam tak larut. Tuliskan semua persamaan kimia dalam
penyediaan garam larut dan dua persamaan kimia bagi garam tak larut.
Salt
Garam
Zinc chloride
Zink klorida
Sodium nitrate
Natrium nitrat
Silver chloride
Argentum klorida
Copper(II) sulphate
Kuprum(II) sulfat
Lead(II) sulphate
Plumbum(II) sulfat
Aluminium nitrate
Aluminium nitrat
Lead(II) chloride
Plumbum(II) klorida
Magnesium nitrate
Magnesium nitrat
Potassium chloride
Kalium klorida
Lead(II) nitrate
Plumbum(II) nitrat
Barium sulphate
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Chemical equations
“L” / “TL”
Persamaan kimia
Zn + 2HCl
S
ZnCl2 + H2
ZnCO3 + 2HCl
ZnCl2 + CO2 + H2O
ZnCl2 + H2O
ZnO + 2HCl
S
IS
S
IS
NaOH + HNO3
NaNO3 + H2O
AgNO3 + HCl
AgCl + HNO3
AgNO3 + NaCl
AgCl + NaNO3
CuO + H2SO4
CuSO4 + H2O
CuCO3 + H2SO4
CuSO4 + CO2 + H2O
Pb(NO3)2 + H2SO4
Pb(NO3)2 + Na2SO4
2Al + 6HNO3
S
PbSO4 + 2HNO3
2Al(NO3)3 + 3H2
Al2O3 + 6HNO3
2Al(NO3)3 + 3H2O
Al2(CO3)3 + 6HNO3
IS
Pb(NO3)2 + 2HCl
Pb(NO3)2 + 2NaCl
Mg + 2HNO3
S
S
IS
KOH + HCl
PbO + 2HNO3
PbCO3 + 2HNO3
BaCl2 + H2SO4
BaCl2 + Na2SO4
2Al(NO3)3 + 3CO2 + 3H2O
PbCl2 + 2HNO3
PbCl2 + 2NaNO3
Mg(NO3)2 + H2
MgO + 2HNO3
Mg(NO3)2 + H2O
MgCO3 + 2HNO3
S
PbSO4 + 2NaNO3
Mg(NO3)2 + CO2 + H2O
KCl + H2O
Pb(NO3)2 + H2O
Pb(NO3)2 + CO2 + H2O
BaSO4 + 2HCl
BaSO4 + 2NaCl
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Barium sulfat
“S” / “IS”
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Chemistry Form 4 • MODULE
EXERCISE / LATIHAN
1
The diagram below shows the set-up of apparatus to prepare soluble salt Y.
Rajah di bawah menunjukkan susunan radas bagi menyediakan garam larut Y.
Nitric acid
Asid nitrik
25 cm3 of 1 mol dm–3 potassium hydroxide solution
+ phenolphthalein
25 cm3 larutan kalium hidroksida 1 mol dm-3 + fenolftalein
Phenolphthalein is used as an indicator in a titration between nitric acid and sodium hydroxide solution. 25 cm3 of nitric
acid completely neutralises 25 cm3 of 1 mol dm–3 potassium hydroxide solution. The experiment is repeated by reacting
25 cm3 of 1 mol dm–3 potassium hydroxide solution with 25 cm3 nitric acid without phenolphthalein. Salt Y is formed
from the reaction.
Fenolftalein digunakan sebagai penunjuk dalam pentitratan antara asid nitrik dengan larutan kalium hidroksida. 25 cm3 asid nitrik
meneutralkan 25 cm3 larutan kalium hidroksida 1 mol dm–3. Eksperimen ini diulang dengan menindakbalaskan 25 cm3 larutan kalium
hidroksida 1 mol dm–3 dengan 25 cm3 asid nitrik tanpa fenolftalein. Garam Y terbentuk daripada tindak balas ini.
(a) Name salt Y.
Nyatakan nama garam Y.
Potassium nitrate
(b) Write a balanced equation for the reaction that occurs.
Tuliskan persamaan seimbang bagi tindak balas yang berlaku.
HNO3 + KOH
KNO3 + H2O
(c) Calculate the concentration of nitric acid.
Hitungkan kepekatan asid nitrik tersebut.
1
= 0.025 mol
1 000
From the equation, 1 mol of KOH : 1 mol of HNO3
0.025 mol of KOH : 0.025 mol of HNO3
Concentration of HNO3, M
25
0.025 = M ×
1 000
M = 1 mol dm–3
Mol of NaOH = 1 ×
(d) Why is the experiment is repeated without phenolphthalein?
Mengapakah eksperimen ini diulang tanpa menggunakan fenolftalein?
To get pure and neutral salt solution Y.
(e) Describe briefly how a crystal of salt Y is obtained from the salt solution.
Huraikan secara ringkas bagaimana hablur garam Y diperoleh daripada larutan garamnya.
– The salt solution is poured into an evaporating dish.
– The solution is heated to evaporate the solution until one third its original volume// a saturated solution formed.
– The saturated solution is allowed to cool until salt crystals Y are formed.
– The crystals are filtered and dried by pressing them between filter papers.
(f)
Name two other salts that can be prepared with the same method.
Namakan dua garam lain yang boleh disediakan dengan kaedah yang sama.
Potassium/sodium/ammonium salt. Example: potassium nitrate, sodium sulphate.
(g) State the type of reaction in the preparation of the salts.
Nyatakan jenis tindak balas dalam penyediaan garam ini.
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Neutralisation
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MODULE • Chemistry Form 4
The following is the steps in the preparation of dry copper(II) sulphate crystals.
2
Berikut adalah langkah-langkah dalam penyediaan hablur garam kuprum(II) sulfat kering.
Step I:
Copper(II) oxide powder is added a little at a time with constant stirring to the heated 50 cm3 of
1 mol dm–3 sulphuric acid until some of it no longer dissolve.
Langkah I: Serbuk kuprum(II) oksida ditambahkan, sedikit demi sedikit sambil dikacau ke dalam 50 cm3 asid sulfurik
1 mol dm-3 yang dipanaskan sehingga serbuk itu tidak boleh larut lagi.
Step II:
The mixture is filtered.
Langkah II: Campuran dituras.
Step III:
The filtrate is poured into an evaporating dish and heated to evaporate the solution until one
third of its original.
Langkah III: Hasil turasan dipanaskan di dalam mangkuk penyejat sehingga isi padunya menjadi satu pertiga
daripada isi padu asal.
Step IV:
The salt solution is allowed to cool at room temperature for the crystallisation to take place.
Langkah IV: Hasil turasan itu dibiarkan sejuk ke suhu bilik sehingga penghabluran berlaku.
Step V:
The crystals formed are filtered and dried by pressing them between filter papers.
Langkah V: Hablur yang terbentuk dituraskan dan dikeringkan dengan menekan antara kertas turas.
(a) (i)
State two observations during Step I.
Nyatakan dua pemerhatian pada Langkah I.
– Black solid dissolve
– Colourless solution turns black
(ii)
Write a balance chemical equation for the reaction that occur in Step I.
Tuliskan persamaan kimia seimbang bagi tindak balas yang berlaku dalam Langkah I.
CuO + H2SO4
CuSO4 +H2O
(iii) State the type of reaction in the preparation of the salts.
Nyatakan jenis tindak balas yang berlaku dalam penyediaan garam.
Neutralisation
(b) Why is copper(II) oxide powder added until some of it no longer dissolve in Step I?
Mengapakah serbuk kuprum(II) oksida ditambah pada larutan tersebut sehingga ia tidak boleh melarut lagi dalam Langkah I?
To make sure that all sulphuric acid has reacted.
(c) What is the purpose of heating in Step III?
Apakah tujuan pemanasan dalam Langkah III?
To evaporate the water and copper(II) sulphate solution becomes saturated
(d) What is the colour of copper(II) sulphate?
Apakah warna kuprum(II) sulfat?
Blue
(e) What is the purpose of filtration in
Apakah tujuan penurasan dalam
(i) Step II? / Langkah II?
– To remove the excess copper(II) oxide.
– To obtain copper(II) sulphate solution as a filtrate
(ii)
Step V? / Langkah V?
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To obtain copper(II) sulphate crystals as a residue.
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Chemistry Form 4 • MODULE
(f)
Draw the a labelled diagram to show the set-up of apparatus used Step II and Step III.
Lukiskan gambar rajah berlabel untuk menunjukkan susunan alat radas yang digunakan dalam Langkah II dan Langkah III.
Filter paper
Excess of
copper(II) oxide
Copper(II) sulphate solution
Heat
Copper(II) sulphate
solution
(g) Can copper powder replace copper(II) oxide in the experiment? Explain your answer.
Bolehkah serbuk kuprum digunakan untuk menggantikan kuprum(II) oksida dalam eksperimen ini? Terangkan jawapan anda.
Cannot. Copper is less electropositive than hydrogen in the electrochemical series, copper cannot displace hydrogen
from the acid.
(h) Name other substance that can replace copper(II) oxide to prepare the same salt. Write a balance chemical equation
for the reaction that occur.
Namakan sebatian lain yang dapat menggantikan kuprum(II) oksida dalam penyediaan garam yang sama. Tuliskan persamaan kimia
yang seimbang bagi tindak balas yang berlaku.
Copper(II) carbonate
Substance / Garam larut :
CuCO3 + H2SO4
Balance equation / Persamaan seimbang :
3
CuSO4 + H2O + CO2
The diagram below shows the flow chart for the preparation of lead(II) nitrate and lead(II) sulphate through reaction I
and II.
Rajah di bawah menunjukkan carta aliran bagi penyediaan plumbum(II) nitrat dan plumbum(II) sulfat melalui tindak balas I dan II.
Reaction I
Tindak balas I
Lead(II) carbonate
Lead(II) nitrate
Plumbum(II) karbonat
(a) (i)
Reaction II
Tindak balas II
Plumbum(II) nitrat
Lead(II) sulphate
Plumbum(II) sulfat
What is meant by salt?
Apakah maksud garam?
Salts are ionic compounds produced when hydrogen ion from acid is replaced with metal ion including
ammonium ion.
(ii)
Based on the flow chart above, classify the above salt to soluble salt and insoluble salt.
Berdasarkan carta aliran di atas, kelaskan garam-garam tersebut kepada garam larut dan garam tak larut.
Soluble salt / Garam larut : Lead(II) nitrate
Insoluble salt / Garam tak larut : Lead(II) carbonate, Lead(II) sulphate
(b) (i)
Describe how lead(II) nitrate solution is obtained in reaction I.
Terangkan bagaimana larutan plumbum(II) nitrat diperoleh daripada tindak balas I.
Measure
–
Sukat
and pour
sebanyak
– Lead(II) carbonate
Serbuk
– Stir the
cm3 of 1 mol dm–3
cm asid
3
nitrik
nitric
mixture
Campuran
is added to the acid in the beaker until
excess
.
berlebihan
.
with a glass rod.
tersebut dikacau dengan rod kaca.
in the beaker is filtered.
dituraskan.
– The filtrate is lead(II) nitrate
larutan
solution
.
plumbum(II) nitrat.
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Hasil turasan ialah
acid in a beaker.
1 mol dm dan tuangkan ke dalam bikar.
-3
plumbum(II) karbonat ditambahkan kepada asid di dalam bikar sehingga
mixture
Campuran
– The
powder
50
50
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MODULE • Chemistry Form 4
(ii)
Write a balanced chemical equation for the reaction that occur.
Tuliskan persamaan kimia yang seimbang bagi tindak balas yang berlaku.
PbCO3 + HNO3
(c) (i)
Pb(NO3)2 + H2O + CO2
Describe how to prepare pure and dry lead(II) sulphate in reaction II.
Huraikan bagaimana cara menyediakan plumbum(II) sulfat yang tulen dan kering dalam tindak balas II.
–
50 cm3
1
mol dm–3 lead(II) nitrate solution is added to
of
sodium sulphate solution in a beaker.
50 cm3
larutan plumbum(II) nitrat
1
mol dm–3 ke dalam bikar.
– The
– The
mixture
Campuran
50 cm3
1
mol dm–3
larutan natrium sulfat
is stirred with glass rod.
Campuran
is filtered. The white precipitate of lead(II) sulphate is collected as the residue.
dituraskan. Mendakan putih plumbum(II) sulfat dikumpulkan sebagai baki.
– The precipitate is rinsed with
Mendakan tersebut dibilas dengan
– The precipitate is
(ii)
mol dm–3 ditambahkan kepada
of
tersebut dikacau dengan rod kaca.
mixture
Mendakan tersebut
1
50 cm3
pressed
ditekan
distilled water
air suling
.
.
between sheets of
antara
kertas turas
filter papers
to dry it.
.
Write an ionic equation the reaction that occur.
Tuliskan persamaan ion bagi tindak balas yang berlaku.
Pb2+ + SO42–
PbSO4
(iii) Name the type of reaction that occur in reaction II.
Namakan jenis tindak balas yang berlaku dalam tindak balas II.
Double decomposition reaction
(iv) What is the step taken to make sure that pure lead(II) sulphate in reaction II is pure?
Apakah langkah yang diambil untuk memastikan plumbum(II) sulfat dalam tindak balas II tulen?
The precipitate is rinsed with distilled water.
(d) (i)
Can lead(II) sulphate be prepared by adding excess of lead(II) nitrate to calcium(II) sulphate followed by
filtration. Explain your answer.
Bolehkah plumbum(II) sulfat disediakan dengan menambahkan plumbum(II) nitrat berlebihan kepada kalsium(II) sulfat dan
diikuti dengan penurasan? Terangkan jawapan anda.
– Cannot.
– Calcium sulphate is insoluble salt, it cannot form a solution and there are no free moving ions.
– Double decomposition reaction cannot occur.
(ii)
Can lead(II) sulphate be prepared by adding excess of lead(II) oxide to sulphuric acid. Explain your answer.
Bolehkah plumbum(II) sulfat disediakan dengan menambahkan plumbum(II) oksida berlebihan kepada asid sulfurik? Terangkan
jawapan anda.
– Cannot.
– Lead(II) sulphate and lead(II) oxide are insoluble, both cannot be separated by filtration.
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– The insoluble lead(II) sulphate will prevent lead(II) oxide to undergo further reaction with sulphuric acid.
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Chemistry Form 4 • MODULE
4
The diagram below shows the flow chart for the preparation of zinc carbonate and zinc sulphate through reactions I
and II.
Rajah di bawah menunjukkan carta aliran bagi penyediaan garam zink karbonat dan zink sulfat melalui tindak balas I dan tindak balas II.
Zinc nitrate
Reaction I
Tindak balas I
Zinc carbonate
Zink nitrat
Reaction II
Tindak balas II
Zinc sulphate
Zink karbonat
Zink sulfat
(a) Based on the flow chart above, classify the above salt to soluble salt and insoluble salt.
Berdasarkan carta aliran di atas, kelaskan garam di atas kepada garam larut dan garam tak larut.
Soluble salt / Garam larut : Zinc nitrate, zinc sulphate
Insoluble salt / Garam tak larut : Zinc carbonate
(b) (i)
State the reactant for the preparation of zinc carbonate from zinc nitrate in reaction I.
Nyatakan bahan tindak balas untuk penyediaan zink karbonat dalam tindak balas I.
Sodium carbonate solution / potassium carbonate solution / ammonium carbonate solution
(ii)
State the type of reaction the occurs in reaction I.
Nyatakan jenis tindak balas yang berlaku dalam tindak balas I.
Double decomposition
(iii) Describe the preparation zinc carbonate from zinc nitrate in the laboratory through reaction I.
Huraikan penyediaan zink karbonat dari zink nitrat melalui tindak balas I.
– 50 cm3 of 1 mol dm–3 zinc nitrate solution is added to 50 cm3 of 1 mol dm–3 sodium carbonate solution in a
beaker.
– The mixture is stirred with a glass rod and a white solid, ZnCO3 is formed.
– The mixture is filtered and the residue is rinsed with distilled water.
– The white precipitate is dried by pressing it between filter papers.
(iv) Write the chemical equation for the reaction in (b)(iii).
Tuliskan persamaan kimia untuk tindak balas yang berlaku dalam (b)(iii).
Zn(NO3)2 + Na2CO3
(c) (i)
ZnCO3 + 2NaNO3
State the reactant for the preparation of zinc sulphate from zinc carbonate in reaction II.
Nyatakan bahan tindak balas bagi penyediaan zink sulfat dari zink karbonat dalam tindak balas II.
Sulphuric acid
(ii)
Describe laboratory experiment to prepare zinc sulphate from zinc carbonate through reaction II.
Huraikan eksperimen dalam makmal untuk menyediakan zink sulfat dari zink karbonat melalui tindak balas II.
– 50 cm3 of 1 mol dm–3 of sulphuric is measured and poured into acid in a beaker.
– The white precipitate from reaction I/ zinc carbonate powder is added to the acid until in excess.
– The mixture is stirred with a glass rod.
– The excess white precipitate is filter out.
– The filtrate is poured into an evaporating dish.
– The salt solution is gently heated until saturated.
– The hot saturated salt solution is allowed to cool for crystals to form.
– The crystals formed are filtered and dried by pressing it between sheets of filter papers.
(iii) Write the chemical equation for the reaction in (c)(ii).
Tuliskan persamaan kimia untuk tindak balas yang berlaku dalam (c)(ii).
ZnSO4 + H2O + CO2
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ZnCO3 + H2SO4
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MODULE • Chemistry Form 4
Constructing Ionic Equation for the Formation of Insoluble Salt
Membina Persamaan Ion bagi Pembentukan Garam Tak Larut
The ionic equation for the formation of insoluble salt can be constructed if the number of moles of anion and cation to
form 1 mol of insoluble salt are known.
1
Persamaan kimia untuk pembentukan garam tak terlarut dapat ditulis jika bilangan mol anion dan kation untuk membentuk 1 mol garam tak
larut diketahui.
The number mol of cation and anion which combined to form 1 mol of insoluble salt is determined experimentally by a
continuous method:
2
Bilangan mol kation dan anion yang bergabung untuk membentuk 1 mol garam tak terlarut dapat ditentukan secara eksperimen menggunakan
kaedah perubahan berterusan:
(a) A fixed volume of a solution A contains cations, Xm+ of the insoluble salt reacts with increasing volume of another
solution B contains the anions, Yn– of the insoluble salt.
Isi padu tetap larutan A mengandungi kation, X m+ daripada garam tak terlarut bertindak balas dengan isi padu yang meningkat larutan
B yang mengandungi anion, Y n– daripada garam tak terlarut.
(b) The volume of solution B needed to completely react with fixed volume of solution A is determined.
Isi padu larutan B yang diperlukan untuk bertindak balas dengan isi padu larutan A yang ditetapkan ditentukan.
(c) The number of mol of Xm+ react with Yn– is calculated based on the result of the experiment.
Bilangan mol X m+ yang bertindak balas dengan Y n– dihitung berdasarkan keputusan eksperimen.
(d) The simplest ration of mol of Xm+: mol of Yn– is calculated.
Nisbah di antara bilangan mol X m+: bilangan mol Y n– dihitung.
(e) Use the ratio to construct ionic equation.
Gunakan nisbah tersebut untuk membina persamaan ion.
Example: / Contoh:
5.0 cm³ of 0.5 mol dm–3 copper(II) sulphate solution is poured to 8 test tubes with the same size. Different volume of
1.0 mol dm–3 potassium hydroxide solution are added to each test tube. The test tubes are stoppered and shaken well.
The test tubes are left for 30 minutes. The height of precipitate formed in each test tube is measured.
The graph below is obtained when the height of precipitate is plotted against the volume of potassium hydroxide
solution.
3
5.0 cm³ larutan kuprum(II) sulfat 0.5 mol dm–3 dituang ke dalam setiap 8 tabung uji yang mempunyai saiz yang sama. Larutan kalium
hidroksida 1.0 mol dm–3 yang berlainan isi padu ditambah kepada setiap tabung uji. Tabung uji tersebut digoncangkan dan dibiarkan selama
30 minit. Tinggi mendakan yang terbentuk dalam setiap tabung uji diukur.
Graf di bawah diperoleh apabila ketinggian mendakan diplot melawan isi padu larutan kalium hidroksida.
Height of precipitate (cm) / Tinggi mendakan (cm)
5
0
(a) (i)
Volume of potassium hydroxide /cm3
1
2
3
4
5
6
7
8
9
Isi padu kalium hidroksida /cm3
Name the precipitate formed.
Nyatakan nama mendakan yang terbentuk.
Copper(II) hydroxide
(ii)
What is the colour of the precipitate?
Apakah warna mendakan?
Blue
(b) Based on the above graph, what is the volume of potassium hydroxide solution needed to completely react with
copper(II) sulphate solution?
Berdasarkan graf di atas, apakah isi padu larutan kalium hidroksida yang diperlukan untuk bertindak balas dengan larutan kuprum(II)
sulfat secara lengkap?
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Chemistry Form 4 • MODULE
(c) (i)
Calculate the number of moles of copper(II) ions in 5.0 cm³ of 0.5 mol dm–3 copper(II) sulphate solution.
Hitung bilangan mol ion kuprum(II) dalam 5.0 cm³ larutan kuprum(II) sulfat 0.5 mol dm–3.
CuSO4
Cu2+ + SO42–
5 × 0.5
= 0.0025 mol
Mol of CuSO4 =
1 000
From the equation,
1 mol CuSO4 : 1 mol Cu2+
0.0025 mol CuSO4 : 0.0025 mol Cu2+
(ii)
Calculate the number of mol of hydroxide ions needed to react with 5.0 cm³ of 0.5 mol dm–3 copper(II) sulphate
solution.
Hitung bilangan mol ion hidroksida yang diperlukan untuk bertindak balas dengan 5.0 cm³ larutan kuprum(II) sulfat
0.5 mol dm–3.
KOH
K+ + OH–
Mol of KOH = 5 × 1.0 = 0.005 mol
1 000
From the equation,
1 mol KOH : 1 mol OH–
0.005 mol KOH : 0.005 mol OH–
(iii) How many moles of hydroxide ions react with one mole of copper(II) ions to form a precipitate?
Berapakah bilangan mol ion hidroksida yang bertindak balas dengan satu mol ion kuprum(II) untuk membentuk mendakan?
0.0025 mol Cu2+ : 0.005 mol OH–
1 mol Cu2+
: 2 mol of OH–
(d) Calculate the number of mol of hydroxide ions needed to react with 5.0 cm³ of 0.5 mol dm–3 copper (II) sulphate
solution.
Tuliskan persamaan ion bagi pembentukan mendakan.
Cu2+ + 2OH–
Cu(OH)2
Solving Numerical Problems Involving the Salt Preparation
Penghitungan Pelbagai Masalah Melibatkan Penyediaan Garam
Mass in gram
Jisim dalam gram
÷ (RAM/RMM/RFM) g mol–1
÷ (JAR/JMR/JFR) g mol–1
Solution concentration in mol dm–3 (M)
and volume in cm3 (V)
Kepekatan larutan dalam mol dm–3 (M)
dan isi padu dalam cm3 (V)
MV
n = 1000
× (RAM/RMM/RFM) g mol–1
× (JAR/JMR/JFR) g mol–1
× 24 dm3 mol–1/22.4 dm3 mol–1
Number of mol (n)
Volume of gas in dm3
Bilangan mol (n)
Isi padu gas dalam dm3
÷ 24 dm mol /22.4 dm mol
3
–1
3
–1
Gas occupies the volume of 24 dm3 at room temperature and 22.4 dm3 at s.t.p (standard temperature and pressure).
1 mol sebarang gas menempati isipadu 24 dm3 pada suhu bilik dan 22.4 dm3 pada s.t.p (suhu dan tekanan piawai).
Calculation steps: / Langkah-langkah pengiraan:
S1 Write a balanced equation.
L1
Tuliskan persamaan seimbang.
S2 Write the information from the question above the equation.
L3
Tuliskan maklumat daripada soalan di atas persamaan tersebut.
S3 Write the information from the chemical equation below the equation (the number of moles of reactants/products).
L3
Tuliskan maklumat daripada persamaan kimia di bawah persamaan tersebut (bilangan mol bagi bahan/hasil tindak balas).
S4 Change the information in S2 into moles by using the method shown in the chart below.
L4
Tukar maklumat dalam L2 menjadi mol dengan menggunakan kaedah yang ditunjukkan dalam carta di atas.
S5 Use the relationship between number of moles of substance involved in S3 to find the answer.
L5
Gunakan perhubungan bilangan mol bahan terlibat dalam L3 untuk mendapatkan jawapan.
S6 Change the information to the unit required using the chart below.
Tukar maklumat tersebut kepada unit yang dikehendaki mengikut carta di atas.
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MODULE • Chemistry Form 4
EXERCISE / LATIHAN
50 cm3 of 2 mol dm–3 sulphuric acid is added to an excess of copper(II) oxide powder. Calculate the mass of copper(II)
sulphate formed in the reaction. [Relative atomic mass: H = 1, O = 16, Cu = 64, S = 32]
1
50 cm3 asid sulfurik 2 mol dm–3 ditambah kepada serbuk kuprum(II) oksida berlebihan. Hitungkan jisim kuprum(II) sulfat yang terbentuk
dalam tindak balas itu. [Jisim atom relatif: H = 1, O = 16, Cu = 64, S = 32]
M = 2 mol dm–3
V = 50 cm3
CuO(aq) + H2SO4(aq)
?g
CuSO4(ak) + 2H2O(l)
2 × 50
Number of moles of sulpuric acid =
= 0.1 mol
1 000
From the equation, 1 mol CuO : 1 mol CuSO4
0.1 mol CuO : 0.1 mol CuSO4
Mass of CuSO4 = 0.1 mol × [64 + 32 + (16 × 4)] g mol–1 = 16 g
27.66 g of lead(II) iodide is precipitated when 2.0 mol dm–3 of aqueous lead(II) nitrate solution is added to an excess
of aqueous potassium iodide solution. Calculate the volume of aqueous lead(II) nitrate solution used. [Relative atomic
mass: I = 127, Pb = 207]
2
27.66 g plumbum(II) iodida termendak apabila 2.0 mol dm–3 larutan plumbum(II) nitrat akueus ditambahkan kepada larutan kalium iodida
akueus berlebihan. Hitungkan isi padu plumbum(II) nitrat yang digunakan. [Jisim atom relatif: I = 127, Pb = 207]
M = 2 mol dm–3
V = ? cm3
Pb(NO3)2(aq) + 2KI(aq)
Mol of PbI2 =
25 g
PbI2(s) + 2KNO3(aq)
27.66
= 0.06 mol
(207 + 2 × 127)
From the equation,
1 mol PbI2 : 1 mol Pb(NO3)2
0.06 mol PbI2 : 0.06 mol Pb(NO3)2
Volume of Pb(NO3)2 =
n mol
0.06 mol
=
= 0.03 dm3 = 30 cm3
M mol dm–3
2 mol dm–3
Zinc oxide powder is added to 100 cm3 of 2 mol dm–3 nitric acid to form zinc nitrate. Calculate
3
Serbuk zink oksida ditambahkan kepada 100 cm3 asid nitrik 2 mol dm–3 untuk membentuk zink nitrat. Hitungkan
(i)
the mass of zinc oxide that has reacted.
jisim zink oksida yang bertindak balas.
(ii) the mass of zinc nitrate produced. [Relative atomic mass: H = 1, O = 16, Cl = 35.5, Zn = 65]
jisim zink nitrat yang terhasil. [Jisim atom relatif: H = 1, O = 16, Cl = 35.5, Zn = 65]
(i) 2HNO3(aq) + ZnO(s)
Zn(NO3)2(aq) + H2O(l)
100 × 2
= 0.2 mol
1 000
From the equation, 2 mol of HNO3 : 1 mol of ZnO
0.2 mol of HNO3 : 0.1 mol of ZnO
Mass of ZnO = 0.1 × [65 + 16] = 8.1 g
Number of moles of HNO3 =
2 mol of HNO3 : 1 mol of Zn(NO3)2
0.2 mol of HNO3 : 0.1 mol of Zn(NO3)2
Mass of Zn(NO3)2 = 0.1 mol × [65 +[14 + (16 × 3)] × 2] g mol–1 = 0.1 × 189 = 18.9 g
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(ii) From the equation,
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Chemistry Form 4 • MODULE
4
200 cm3 of 1 mol dm–3 barium chloride solution reacts 100 cm3 of 1 mol dm–3 silver nitrate solution. Calculate the mass
of precipitate produced. [Relative atomic mass Ag = 108, Cl = 35.5]
200 cm3 larutan barium klorida 1 mol dm–3 bertindak balas dengan 100 cm3 larutan argentum nitrat 1 mol dm–3. Hitungkan jisim mendakan
yang terbentuk. [Jisim atom relatif: Ag = 108, Cl = 35.5]
M = 0.1 mol dm–3
V = 100 cm3
M = 0.2 mol dm–3
V = 100 cm3, ? g
2AgNO3
2AgCl + Ba(NO3)2
1 × 200
Mol of barium chloride =
= 0.2 mol (excess)
1 000
1 × 100
= 0.1 mol
Mol of silver nitrate =
1 000
From the equation,
1 mol of BaCl2 : 2 mol of AgNO3 : 2 mol of AgCl
0.2 mol of BaCl2 (lebih) : 0.1 mol of AgNO3 : 0.1 mol of AgCl
Mass of AgCl = 0.1 mol × [108 + 35.5] g mol–1 = 14.35 g
BaCl2
+
Qualitative Analysis of Salts / Analisis Kualitatif Garam
1
Qualitative analysis of a salt is a chemical technique to identify the ions present in a salt.
Analisis kualitatif garam ialah suatu teknik dalam kimia yang digunakan untu mengenal pasti ion-ion yang hadir dalam garam.
2
The qualitative analysis consists of the following steps:
Analisis kualitatif terdiri daripada langkah-langkah berikut:
(a) Observe the physical properties on salt.
Perhatikan sifat-sifat fizik garam.
(b) The action of heat on salts.
Kesan haba ke atas garam.
(c) Prepare aqueous solution of salts and conduct confirmatory test for cation and anion present.
Sediakan larutan akueus garam dan menjalankan ujian pengesahan untuk kation dan anion yang hadir.
Physical Properties of Salt
Sifat-Sifat Fizik Garam
1
Physical properties such as colour and solubility indicate the possibility of the presence of certain cations, anions or metal
oxide.
Sifat-sifat fizikal seperti warna dan keterlarutan menunjukkan kemungkinan kehadiran kation, anion atau oksida logam tertentu.
Pepejal
Larutan akueus
Aqueous
Salts/ Cation/Metal oxide
White
Putih
Colourless
Tanpa warna
K+, Na+, Ca2+, Mg2+, Al3+, Zn2+, Pb2+, NH4+
Green
Hijau
Insoluble
Tak larut
CuCO3
Light green
Hijau muda
Light Green
Hijau muda
Fe2+, contoh: FeSO4, FeCl2, Fe(NO3)2
Blue
Biru
Blue
Biru
CuSO4, Cu(NO3 )2 dan CuCl2
Brown
Perang
Brown
Perang
Fe3+
Black
Hitam
Insoluble
Tak larut
CuO
Yellow when hot, white when cold
Kuning apabila panas, putih apabila sejuk
Insoluble
Tak larut
ZnO
Brown when hot, yellow when cold
Perang apabila panas, kuning apabila sejuk
Insoluble
Tak larut
PbO
Garam/Kation/Oksida logam
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MODULE • Chemistry Form 4
Action of Heat on Salt / Kesan Haba ke atas Garam
Some salts decompose when they are heated:
1
Beberapa jenis garam terurai apabila dipanaskan:
2
Salt
metal oxide
Garam
oksida logam
gas
+
gas
Common Gas Identification: / Pengesahan Gas yang biasa:
Gas
Gas
Observation/ Test
Pemerhatian/Ujian
Nitrogen
dioxide, NO2
Nitrogen
dioksida, NO2
Oxygen,O2
Oksigen,O2
Carbon
dioxide, CO2
Karbon
dioksida, CO2
Inference
Inferens
– Brown gas.
Wasap perang.
– Place a moist blue litmus paper at the mouth of the boiling
tube, blue litmus paper turns red.
Letakkan kertas litmus biru lembap pada mulut tabung didih,
kertas litmus biru bertukar menjadi merah.
– Nitrogen dioxide gas is produced by
heating nitrate salt.
Nitrogen dioksida terhasil apabila garam
nitrat dipanaskan.
– Nitrate ion, NO3– present.
Ion nitrat, NO3– hadir.
– Colourless gas.
Gas tanpa warna.
– Put a glowing wooden splinter near to the mouth of a boiling
tube, the glowing wooden splinter is relighted.
Dekatkan kayu uji berbara ke mulut tabung didih, kayu uji
berbara menyala.
– Oxygen gas is produced by heating
nitrate or chlorate(V) salt.
Gas oksigen terhasil apabila garam nitrat
atau klorat(V) dipanaskan.
– Nitrate ion, NO3– present or ClO3– ion
present.
Ion nitrat, NO3– atau ion ClO3– hadir.
– Colourless gas.
Gas tanpa warna.
– Pass the gas through lime water, lime water turns chalky.
Lalukan gas pada air kapur, air kapur menjadi keruh.
– Draw the set-up of apparatus to conduct the test:
Lukiskan susunan radas untuk menjalankan ujian:
– Produced by heating carbonate salt.
Terhasil apabila garam karbonat
dipanaskan.
– Carbonate ion, CO3– present.
Ion karbonat, CO3– hadir.
Calcium carbonate
Heat
Lime water
Ammonia, NH3
Ammonia, NH3
– Colourless gas with pungent smell.
Gas tanpa warna dengan bau yang sengit.
– Place a moist red litmus paper at the mouth of the boiling.
tube, red litmus paper turns blue.
Letakkan kertas litmus merah lembap pada mulut tabung didih,
kertas litmus merah bertukar menjadi biru.
– Produced by heating ammonium salt
with alkali.
Terhasil apabila garam ammonium
dipanaskan dengan alkali.
– Ammonium ion NH4+ present.
Ion ammonium NH4+ hadir.
Action of heat on nitrate and carbonate salts.
3
Kesan haba ke atas garam nitrat dan garam karbonat.
Cation
Kation
Nitrate (NO3–) / Nitrat ( NO3–)
Decompose to oxygen gas and metal nitrite
when heated
Terurai kepada gas oksigen dan logam nitrit apabila
dipanaskan
K+
m
2NaNO3
White solid
Pepejal putih
2KNO2
+ O2
White solid
Pepejal putih
2NaNO2
+ O2
White solid
Pepejal putih
Does not decompose when heated
Tidak diuraikan apabila dipanaskan
–
–
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Na+
2KNO3
White solid
Pepejal putih
Carbonate (CO32–) / Karbonat (CO32–)
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Chemistry Form 4 • MODULE
Decompose to oxygen gas, nitrogen dioxide
gas and metal oxide when heated
Terurai kepada gas oksigen, gas nitrogen dioksida dan
oksida logam apabila dipanaskan
Ca2+
Mg2+
Al3+
2Ca(NO3)2
White solid
Pepejal putih
2CaO
2Mg(NO3)2
White solid
Pepejal putih
2MgO
2Al2O3
Zn2+
Pb2+
Cu2+
4
MgCO3
2PbO
+
4NO2 + O2
Perang bila panas,
kuning apabila sejuk
2CuO
Black solid
Pepejal hitam
+
4NO2 + O2
+
6CO2
Turn lime water chalky
Air kapur menjadi keruh
ZnO
+
CO2
Turn lime water chalky
Air kapur menjadi keruh
Kuning apabila panas,
putih apabila sejuk
PbO
+
CO2
Brown when hot
Turn lime water chalky
Yellow when cold
Air kapur menjadi keruh
Perang apabila panas,
kuning apabila sejuk
CuO
Black solid
CuCO3
Green solid
Pepejal hijau
Brown fume
Wasap perang
CO2
Turn lime water chalky
Air kapur menjadi keruh
Yellow when hot
white when cold
PbCO3
White solid
Pepejal
Putih
Brown fume
Wasap perang
+
2Al2O3
White solid
Pepejal
Putih
Kuning apabila panas,
putih apabila sejuk
CO2
Turn lime water chalky
Air kapur menjadi keruh
White solid
Pepejal putih
ZnCO3
Brown gas
Gas perang
Brown when hot
yellow when cold
MgO
White solid
Pepejal putih
4NO2 + O2
+
White solid
Pepejal putih
2Al2 (CO3)3
Brown fume
Wasap perang
+
CaO
White solid
Pepejal putih
White solid
Pepejal putih
+ 12NO2 + O2
2ZnO
2Cu(NO3)2
Blue solid
Pepejal biru
White solid
Pepejal putih
Yellow when hot
white when cold
2Pb(NO3)2
White solid
Pepejal
Putih
CaCO3
Brown fume
Wasap perang
Brown fume
Wasap perang
White pepejal
Pepejal putih
2Zn(NO3)2
White solid
Pepejal
putih
4NO2 + O2
+ 4NO2 + O2
White pepejal
Pepejal putih
4Al(NO3 )3
White solid
Pepejal putih
+
White pepejal
Pepejal putih
Decompose to carbon dioxide gas and metal oxide
when heated
Terurai kepada gas karbon dioksida dan oksida logam apabila
dipanaskan
+
Pepejal hitam
CO2
Turn lime water chalky
Air kapur menjadi keruh
Sulphate salts are more stable, they are not easily decompose when heated.
Garam sulfat lebih stabil kerana ia tidak terurai dengan mudah apabila dipanaskan.
5
Chloride salts do not decompose except NH4Cl: NH4Cl(s)
Garam klorida tidak terurai kecuali NH4Cl: NH4Cl(p)
6
NH3(g) + HCl(g)
NH3(g) + HCl(g)
Complete the following table:
Lengkapkan jadual berikut:
Observation
Pemerhatian
A white salt is heated.
Inference/conclusion
Inferens/kesimpulan
Gas
Garam berwarna putih dipanaskan.
– Brown gas is released, the gas turns moist blue
litmus paper red.
Gas perang dibebaskan, menukar kertas litmus biru lembap
kepada merah.
– Residue is yellow when hot and white when cold.
Garam berwarna hijau dipanaskan.
– Colourless gas released, the gas turns lime water
chalky.
Gelembung gas dibebaskan, ia menukar air kapur menjadi
keruh.
– Residue is black
zinc
zinc nitrate
zink nitrat
Garam putih ialah
Carbon dioxide
Gas
Baki ialah
kuprum(II)
– The green salt is
Garam hijau ialah
hadir.
.
Carbonate
dibebaskan. Ion
copper(II)
– The residue is
zink
.
gas released.
karbon dioksida
hadir.
ion present.
oksida. Ion
– The white salt is
–
Zinc
oxide.
zink
Baki ialah
nitrat
dibebaskan. Ion
oxide.
Copper(II)
copper(II) carbonate
hadir.
ion present.
kuprum(II)
oksida. Ion
kuprum(II) karbonat
ion present
karbonat
hadir.
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Baki berwarna hitam.
gas released. Nitrate ion present.
nitrogen dioksida
– The residue is
Baki berwarna kuning apabila panas dan putih apabila sejuk
A green salt is heated.
Nitrogen dioxide
–
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MODULE • Chemistry Form 4
A white salt is heated.
–
Garam berwarna putih dipanaskan.
Carbon dioxide
Gas
– Colourless gas released, the gas turns lime water
chalky.
Gelembung gas dibebaskan, ia menukar air kapur menjadi
keruh.
Baki ialah
plumbum(II)
Baki berwarna perang apabila panas dan kuning apabila sejuk.
zinc
– The residue is
– Colourless gas released, the gas turns lime water
chalky.
oksida. Ion
zink
Baki ialah
Garam putih ialah
– Residue is yellow when hot and white when cold.
.
zinc
oxide.
ion present.
zink
zinc carbonate
zink karbonat
hadir.
.
oksida. Ion
– The white salt is
Gelembung gas dibebaskan, ia menukar air kapur menjadi
keruh.
hadir.
ion present.
plumbum(II)
plumbum(II) karbonat
Garam putih ialah
A white salt is heated.
Garam berwarna putih dipanaskan.
Lead(II)
lead(II) carbonate
– The white salt is
– Residue is brown when hot and yellow when cold.
oxide.
ion present
karbonat
dibebaskan. Ion
lead(II)
– The residue is
Carbonate
gas released.
karbon dioksida
hadir.
.
.
Baki berwarna kuning apabila panas dan putih apabila sejuk.
A blue salt is heated.
– Nitrogen dioxide gas released. Nitrate ion present.
Garam berwarna biru dipanaskan.
Gas nitrogen dioksida dibebaskan. Ion nitrat hadir.
– Brown gas is released, the gas turns moist blue
litmus paper red.
– The residue is copper(II) oxide. Copper(II) ion present.
Gas perang terbebas menukar warna kertas limus biru menjadi
merah.
– Residue is black.
Baki ialah kuprum(II) oksida. Ion kuprum(II) hadir.
– The blue salt is
Garam biru ialah
Baki berwarna hitam.
A white salt is heated.
copper(II) nitrate
.
kuprum(II) nitrat
.
– Nitrogen dioxide gas released. Nitrate ion present.
Garam berwarna putih dipanaskan.
Gas nitrogen dioksida dibebaskan. Ion nitrat hadir.
– Brown gas is released, the gas turns moist blue
litmus paper red.
– The residue is lead(II) oxide. Lead(II) ion present.
Gas perang terbebas menukar warna kertas limus biru menjadi
merah.
– Residue is brown when hot and yellow when cold.
Baki ialah plumbum(II) oksida. Ion plumbum(II) hadir.
– The blue salt is
Garam putih ialah
Baki berwarna perang apabila panas dan kuning apabila sejuk.
A white salt is heated.
–
Garam berwarna putih dipanaskan.
– Colourles gas released, the gas turns lime water
chalky.
Gelembung gas dibebaskan, ia menukar air kapur menjadi
keruh.
lead(II) nitrate
plumbum(II) nitrat
Carbon dioxide gas released.
karbon dioksida
Gas
.
.
Carbonate
dibebaskan. Ion
ion present.
karbonat
hadir.
– The possible residue are ZnO/PbO/MgO/Al2O3
Baki yang mungkin adalah CaOl/MgO/Al2O3.
– Residue is white
Baki berwarna putih.
lead(II) nitrate ,
From the above table, action of heat on heat on salt can be used to identify
zinc nitrate
zinc
carbonate
copper(II)
nitrate
copper(II)
carbonate .
,
,
and
–
Daripada jadual di atas, kesan haba ke atas garam boleh digunakan untuk mengenal garam
zink nitrat
–
,
zink karbonat
,
kuprum(II) nitrat
dan
plumbum(II) nitrat
kuprum(II) karbonat
lead(II) carbonate ,
, plumbum(II) karbonat ,
.
Confirmatory test for other cations and anions is carried out by Confirmatory Tests for Anions and Cations
Ujian pengesahan untuk kation dan anion lain boleh dijalankan dengan menggunakan Ujian Pengesahan Anion dan Kation.
Confirmatory Tests for Cations
Ujian Pengesahan bagi Kation
Chemical tests is conducted for confirmation of cations in aqueous form.
1
Ujian-ujian kimia dijalankan bagi pengesahan kation dalam bentuk akueus.
Confirmatory test is carried out by adding a small amount of sodium hydroxide solution / ammonia solution followed
by excess sodium hydroxide / ammonia solution to the solution contains the cation.
2
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Ujian pengesahan dijalankan dengan menambah sedikit larutan natrium hidroksida / larutan ammonia diikuti dengan larutan natrium
hidroksida / larutan ammonia berlebihan kepada larutan yang mengandungi kation.
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Chemistry Form 4 • MODULE
Sodium hydroxide solution
Larutan natrium hidroksida
Cations
Kation
small amount
sedikit
K+
Na+
Ca2+
No change
No change
No change
Tiada perubahan
Tiada perubahan
Tiada perubahan
No change
No change
No change
No change
Tiada perubahan
Tiada perubahan
Tiada perubahan
Tiada perubahan
White precipitate
Insoluble in excess
No change
No change
Mendakan putih
Tak larut dalam berlebihan
Tiada perubahan
Tiada perubahan
Insoluble in excess
White precipitate
Soluble in excess
White precipitate
White precipitate
Soluble in excess
White precipitate
Soluble in excess
Mendakan putih
Larut dalam berlebihan
Mendakan putih
Larut dalam berlebihan
White precipitate
Soluble in excess
White precipitate
Soluble in excess
Mendakan putih
Larut dalam berlebihan
Mendakan putih
Larut dalam berlebihan
Green precipitate
Insoluble in excess
Green precipitate
Soluble in excess
Mendakan hijau
Tak larut dalam berlebihan
Mendakan hijau
Larut dalam berlebihan
Brown precipitate
Insoluble in excess
Brown precipitate
Soluble in excess
Mendakan perang
Tak larut dalam berlebihan
Mendakan perang
Larut dalam berlebihan
Blue precipitate
Insoluble in excess
Blue precipitate
Soluble in excess
Mendakan biru
Tak larut dalam berlebihan
Mendakan biru
Larut dalam berlebihan
Zn2+
White precipitate
Mendakan putih
Mendakan putih
Fe2+
Fe3+
Cu2+
NH4+
excess
berlebihan
No change
White precipitate
Pb2+
small amount
sedikit
excess
berlebihan
Tiada perubahan
Mg2+
Al3+
Ammonia solution
Larutan ammonia
Tak larut dalam berlebihan
Mendakan putih
Larut dalam berlebihan
Insoluble in excess
Tak larut dalam berlebihan
Soluble in excess
Mendakan putih
Larut dalam berlebihan
No change
No change
No change
No change
Tiada perubahan
Tiada perubahan
Tiada perubahan
Tiada perubahan
(a) Reaction with small amount until excess of sodium hydroxide solution: (refer to the above table)
Tindak balas dengan larutan natrium hidroksida sedikit demi sedikit sehingga berlebihan: (rujuk jadual di atas)
Pungent smell, moist red litmus paper turn to blue
Bau sengit, menukarkan kertas litmus
merah lembap kepada biru
Solution
contains:
Larutan
mengandungi:
Heat
Add a little sodium
hydroxide solution
K+, Ca2+, Mg2+,
Al , Zn , Pb ,
3+
2+
2+
Fe2+, Fe3+, Cu2+,
NH4+
NH4+
Tambahkan sedikit
larutan natrium
hidroksida
K , NH4
+
Panaskan
+
No precipitate
Tiada mendakan
No changes
K+
Tiada perubahan
Precipitate formed
Mendakan terbentuk
Cu2+ (blue),
Fe2+ (green),
Fe3+ (brown)
Coloured precipitate
Mendakan berwarna
Add excess sodium
hydroxide solution
White precipitate
Mendakan putih
Pb2+, Al3+,
Zn2+, Ca2+,
Mg2+
Soluble
Larut
Tambahkan larutan
natrium hidroksida
berlebihan
Insoluble
Zn2+, Al3+, Pb2+
Ca2+, Mg2+
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MODULE • Chemistry Form 4
(b) Reaction with small amount until excess of ammonia solution:
Tindak balas dengan larutan ammonia sedikit demi sedikit sehingga berlebihan:
Add a little
solution of
ammonia
Solution
contains:
Larutan
mengandungi:
Tambah
sedikit larutan
ammonia
No precipitate
Cu2+ (blue),
Fe2+ (green),
Fe3+ (brown)
Tiada mendakan
K+, Na+, Ca2+,
Mg2+, Al3+, Zn2+,
Pb2+, Fe2+, Fe3+,
Cu2+
Add excess
aqueous
ammonia
Ca2+, K+, Na+
Precipitate
formed
Mendakan
terbentuk
Tambahkan
larutan ammonia
berlebihan
Soluble
Larut
Insoluble
Coloured precipitate
Add excess
aqueous
ammonia
Mendakan berwarna
White precipitate
Mendakan putih
Pb , Al ,
Zn2+, Mg2+
2+
3+
Tambahkan
larutan ammonia
berlebihan
Cu2+
Fe2+, Fe3+
Tak larut
Soluble
Larut
Insoluble
Zn2+
Mg2+, Al3+,
Pb3+
Tak larut
(c) Conclusion of the confirmatory test for colourless/white cations:
Kesimpulan ujian pengesahan bagi kation tanpa warna/putih:
(i)
Zn2+: White precipitqte, soluble in excess of sodium hydroxide and ammonia solution
(ii)
Mg2+: White precipitate, insoluble in excess of sodium hydroxide and ammonia solution
(iii) Al3+: White precipitate, soluble in excess of sodium hydroxide and insoluble in excess ammonia solution
(iv) Ca2+: White precipitate insoluble in excess of sodium hydroxide and no precipitate with ammonia solution
(v)
NH4+: No precipitate with sodium hydroxide solution and pungent smell released when heated
(d) Conclusion of the confirmatory test for coloured cations.
Kesimpulan untuk ujian pengesahan bagi kation berwarna.
(i)
Cu2+: Blue precipitate insoluble in excess of sodium hydroxide solution and soluble in excess ammonia solution
(ii)
Fe2+: Green precipitate, insoluble in excess of sodium hydroxide and ammonia solution
(iii) Fe3+: Brown precipitate, insoluble in excess of sodium hydroxide and ammonia solution
(e) All cations can be identified with confirmatory test using sodium hydroxide solution and ammonia solution except
Al3+ and Pb2+.
Semua kation boleh dikenal pasti dengan ujian pengesahan menggunakan larutan natrium hidroksida dan larutan ammonia kecuali
Al3+ dan Pb2+.
(f)
To differentiate between Al3+ and Pb2+:
Untuk membezakan Al3+ dengan Pb2+:
– Al3+ and Pb2+ are differentiated by double decomposition reaction. An aqueous solution containing SO42–/ Cl–/ I–
anion is used to detect the presence of Al3+ and Pb2+.
Al3+ dan Pb2+ boleh dibezakan dengan menggunakan tindak balas pernguraian ganda dua. Larutan akueus yang mengandungi
anion SO42–/ Cl– / I– digunakan untuk mengesan kehadiran Al3+ dan Pb2+.
– Precipitate is formed when solution containing SO42–/ Cl–/ I– added to Pb2+.
Mendakan terbentuk apabila larutan mengandungi SO42–/ Cl–/ I– ditambah kepada Pb2+.
– No precipitate when solution containing SO42–/ Cl– / I– added to Al3+.
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Tiada mendakan terbentuk apabila larutan mengandungi SO42–/ Cl–/ I– ditambah keepada Al3+.
d.
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Chemistry Form 4 • MODULE
(g) Write the ionic equations for the formation of precipitates:
Tuliskan persamaan ion bagi pembentukan mendakan:
Al3+ and Pb2+
Al3+ dan Pb2+
Add sodium
sulphate solution
Add potassium
iodide solution
Tambahkan larutan
natrium sulfat
No changes
Add sodium
chloride solution
White precipitate
Tiada perubahan
Mendakan putih
Pb2+ + SO42–
No changes
Tiada perubahan
Tambahkan larutan
natrium klorida
Pb2+
Al3+
Tambahkan larutan
kalium iodida
Yellow precipitate
Mendakan kuning
Pb2+
Al3+
Pb2+ + 2I–
PbSO4
No changes
PbI2
White precipitate
Tiada perubahan
Mendakan putih
Al3+
Pb2+
Pb2+ + 2Cl–
PbCl2
Confirmatory tests for Anions
Ujian Pengesahan untuk Anion
Anion/Anion
Procedure/Prosedur
Remark/Catatan
– 2 cm3 of dilute hydrochloric acid / nitric acid /sulphuric acid
is added to 2 cm3 of carbonate salt.
2 cm3 asid nitrik/asid sulfurik cair ditambah kepada 2 cm3 garam
karbonat.
– The gas given off is passed through lime water:
Draw a labelled diagram to conduct the test:
Ion karbonat, CO32–
Lukiskan gambar rajah berlabel untuk menjalankan ujian:
Acid
Carbonate salt
Gas tersebut ialah karbon dioksida.
Ionic equation: / Persamaan ion:
H2O + CO2
CO32– + 2H+
Lime water
– 2 cm3 of dilute nitric acid is added to 2 cm3 solution of
chloride ions followed by 2 cm3 of silver nitrate solution.
2 cm3 asid nitrik cair ditambah kepada 2 cm3 larutan ion klorida diikuti
dengan 2 cm3 larutan argentum nitrat.
Chloride ion, Cl–
Pembuakan berlaku dan air kapur menjadi keruh.
Inference: / Inferens:
The gas is carbon dioxide.
Gas yang terbebas dilalukan air kapur.
Carbonate ion,
CO32–
Observation: / Pemerhatian:
Effervescence occurs and lime water turns
chalky.
Ion klorida, Cl–
Observation: / Pemerhatian:
A white precipitate is formed.
Mendakan putih terbentuk.
Inference: / Inferens:
The precipitate is silver chloride
Mendakan ialah argentum klorida.
Ionic equation: / Persamaan ion:
AgCl
Ag+ + Cl–
– 2 cm3 of dilute hydrochloric / nitric acid is added to 2 cm3
of sulphate solution followed by 2 cm3 of barium chloride
solution / barium nitrate solution.
Sulphate ion, SO4
Ion sulfat SO4
2–
2–
2 cm3 asid sulfurik asid/asid nitrik cair ditambah kepada 2 cm3 larutan
sulfat diikuti dengan 2 cm3 larutan barium klorida/larutan barium
nitrat.
Observation: / Pemerhatian:
A white precipitate is formed.
Mendakan putih terbentuk.
Inference: / Inferens:
The precipitate is barium sulphate
Mendakan tersebut ialah barium sulfat.
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Ionic equation: / Persamaan ion:
BaSO4
Ba2+ + SO42–
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MODULE • Chemistry Form 4
– 2 cm3 of dilute sulphuric acid is added to 2 cm3 solution of
nitrate ions followed by 2 cm3 of iron(II) sulphate solution.
2 cm3 larutan ion nitrat ditambah kepada 2 cm3 asid sulfurik cair
diikuti dengan 2 cm3 larutan ferum(II) sulfat.
Nitrate ion, NO3
–
Ion nitrat, NO3–
– The mixture is shaken.
Campuran digoncang.
– The test tube is slanted and held with a test tube holder.
Observation: / Pemerhatian:
A brown ring is formed between two layers.
Cincin perang terbentuk di antara dua lapisan.
Inference: / Inferens:
Nitrate ion present.
Ion nitrat hadir.
Tabung uji dicondongkan dan diapit dengan pemegang tabung uji.
– A few drops of concentrated H2SO4 acid is dropped along the
wall of the test tube and is held upright.
Beberapa titis H2SO4 pekat dititiskan melalui dinding tabung uji dan
ditegakkan.
EXERCISE / LATIHAN
(a) Substance A is white in colour. When A is strongly heated, a brown gas, B and gas C are released. These gases lighted
a glowing wooden splinter. Residue D which is yellow in colour when hot and white when cold is formed.
1
Bahan A berwarna putih. Apabila A dipanaskan dengan kuat, gas berwarna perang B dan gas C dibebaskan. Gas C menyalakan kayu
uji berbara. Baki D yang berwarna kuning apabila sejuk dan putih apabila sejuk terbentuk.
(i)
Name substances A, B, C and D.
Namakan bahan A, B, C dan D.
A:
(ii)
Zinc nitrate
B:
Nitrogen dioxide
C:
Oxygen
D:
Zinc oxide
Write the chemical equation when substance A is heated.
Tuliskan persamaan kimia apabila bahan A dipanaskan.
2Zn(NO3)2
2ZnO + 4NO2 + O2
(b) Write the chemical equation when substance E is heated.
Larutan tanpa warna E memberi keputusan berikut apabila dijalankan beberapa siri ujian:
S1 – Add sodium hydroxide solution, a white precipitate is formed. The precipitate is soluble in excess sodium
hydroxide solution.
L1 – Apabila ditambah dengan larutan natrium hidroksida, mendakan putih terbentuk. Mendakan ini larut apabila ditambah natrium
hidroksida berlebihan.
S2 – Add ammonia solution, a white precipitate is formed. The precipitate is insoluble in excess ammonia solution.
L2 – Apabila ditambah larutan ammonia, mendakan putih terbentuk dan mendakan ini tidak larut dalam larutan ammonia
berlebihan.
S3 – Add potassium iodide solution, a yellow precipitate F, is formed.
L3 – Apabila ditambah dengan larutan kalium iodida, mendakan kuning F terbentuk.
(i)
What are the possible cations present in substance E as a result of S1 test?
Apakah kation-kation yang mungkin hadir dalam bahan E hasil ujian L1?
Pb2+, Al3+ and Zn2+
(ii)
What are the possible cations present in solution E as a result from S1 and S2 tests?
Apakah kation yang mungkin hadir dalam larutan E hasil ujian L1 dan L2?
Pb2+ and Al3+
(iii) What is the ion present in E after S3 test has been done? Write an ionic equation for the formation of substance
F.
Apakah ion yang disahkan hadir dalam E setelah dilakukan ujian L3? Tulis persamaan ion bagi pembentukan bahan F.
2+
Ion present /Ion hadir : Pb
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Ionic equation/Persamaan ion : Pb2+ + 2I–
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Chemistry Form 4 • MODULE
2
The diagram below shows the flow chart for Test I and Test II on colourless solution P.
Rajah di bawah menunjukkan carta aliran bagi ujian I dan ujian II ke atas larutan tanpa warna P.
Gas Q with a pungent smell is
released and turns moist red
litmus paper blue.
Test I
Ujian I
Gas Q berbau sengit terbebas dan
menukarkan warna kertas litmus
merah lembap kepada biru.
Test II
Colourless
solution P
Larutan tanpa
warna P
Effervescence occurs
and gas S is released
Ujian II
Add dilute
hydrochloric acid
Pembuakan berlaku dan
membebaskan gas S.
Tambah asid
hidroklorik cair
(a) Identify gas Q and state its chemical properties.
Kenal pasti gas Q dan nyatakan sifat kimia yang ditunjukkan oleh gas Q.
Ammonia, alkaline gas
(b) State the reagent used in test I and state how the test is carried out.
Nyatakan bahan uji yang digunakan dalam ujian I serta huraikan bagaimana ujian dilakukan.
Add sodium hydroxide solution, heat it.
(c) (i)
Name gas S and write the ionic equation that occurred in Test II:
Namakan gas S dan tuliskan persamaan ion bagi tindak balas yang berlaku dalam ujian II:
Gas S/Gas S : Carbon dioxide
+
2–
Ionic equation/Persamaan ion: CO3 + 2H
(ii)
H2O + CO2
Explain how you confirmed gas S.
Terangkan bagaimana anda mengesahkan gas S.
Pass the gas through lime water, lime water turns chalky.
(iii) Name salt P based on the results of tests I and II.
Namakan garam P berdasarkan keputusan ujian I dan II.
Ammonium carbonate
3
The table below shows the colour of five solutions labelled A, B, C, D and E added with small amount until excess of
ammonia solution and sodium hydroxide solution.
Jadual di bawah menunjukkan warna lima larutan berlabel A, B, C, D dan E yang ditambah dengan larutan natrium hidroksida dan larutan
ammonia sedikit demi sedikit sehingga berlebihan.
Solution
Larutan
A
B
C
Colour
With sodium hydroxide solution
Warna
Dengan larutan natrium hidroksida
With ammonia solution
Dengan larutan ammonia
Blue
Blue precipitate insoluble in excess
Blue precipitate soluble in excess
Biru
Mendakan biru tidak larut dalam berlebihan
Mendakan biru larut dalam berlebihan
Colourless
White precipitate soluble in excess
White precipitate soluble in excess
Tanpa warna
Mendakan putih larut dalam berlebihan
Mendakan putih larut dalam berlebihan
Light green
Green precipitate
Dirty green precipitate
Hijau muda
Mendakan hijau kotor
Mendakan hijau kotor
D
Colourless
White precipitate soluble in excess
White precipitate insoluble in excess
Tanpa warna
Mendakan putih larut dalam berlebihan
Mendakan putih tidak larut dalam berlebihan
E
Colourless
White precipitate insoluble in excess
White precipitate insoluble in excess
Tanpa warna
Mendakan putih tidak larut dalam berlebihan
Mendakan putih tidak larut dalam berlebihan
(a) What are the cations present in
Apakah kation yang terdapat dalam
2+
B: Zn
2+
C: Fe
2+
E: Mg
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2+
A: Cu
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MODULE • Chemistry Form 4
(b) State another test to identify C.
Nyatakan satu lagi ujian bagi mengenali C.
Add potassium hexacyanoferrate(II) solution, light blue precipitate formed
(c) What are the possible cations present in solution D?
Apakah kation-kation yang mungkin terdapat dalam larutan D?
Al3+, Pb2+
(d) Describe briefly a test that can differentiate the cations present in solution D.
Terangkan secara ringkas satu ujian yang boleh digunakan untuk membezakan kation-kation yang hadir dalam larutan D.
– Add a few drops of potassium iodide / sodium chloride / sodium sulpahte solution into 1 cm3 of solution D.
– Yellow/white precipitate formed, lead(II) ion / Pb2+ present
– No precipitate, aluminium ion / Al3+ present.
You are given lead(II) carbonate, zinc(II) carbonate and copper(II) carbonate.
Without using any reagents, describe how you can differentiate the three substances in the laboratory.
4
Anda diberi plumbum(II) karbonat, zink(II) karbonat dan kuprum(II) karbonat. Tanpa menggunakan sebarang bahan uji, terangkan
bagaimana anda membezakan ketiga-tiga bahan tersebut di dalam makmal.
•
Heat
strongly
kuat
Panaskan dengan
bakinya:
boiling tube
one spatula of each salt in a
and observe the residue:
tabung didih
satu spatula setiap jenis garam dalam
dan perhatikan baki-
– If the residue is yellow when hot and white when cold, then zinc oxide is formed. The salt is zinc carbonate .
Jika baki berwarna kuning apabila panas dan putih apabila sejuk, maka
zink karbonat
.
adalah
– If the residue is black, then
copper(II) oxide
Jika baki berwarna hitam, maka
zink oksida
is formed. The salt is
kuprum(II) oksida
terbentuk. Garam tersebut
copper(II) carbonate .
kuprum(II) karbonat
terbentuk. Garam tersebut adalah
– If the residue is brown when hot and yellow when cold, then
lead(II) carbonate
.
Jika baki berwarna perang apabila panas dan kuning apabila sejuk, maka
plumbum(II) karbonat
.
tersebut adalah
lead(II) oxide
.
formed. The salt is
plumbum(II) oksida
terbentuk. Garam
The diagram below shows the flow chart of changes that took place beginning from solid M. Solid M is a zinc salt. When
solid M is heated strongly, it decomposes into solid Q which is yellow when hot and white when cold.
5
Rajah di bawah menunjukkan carta aliran bagi perubahan yang berlaku bermula daripada pepejal M. Pepejal M adalah suatu garam bagi
zink. Apabila pepejal M dipanaskan dengan kuat, ia terurai kepada suatu pepejal Q yang berwarna kuning apabila panas dan putih apabila
sejuk.
Reaction I
Tindak balas I
Solid M
Pepejal M
Reaction II
Tindak balas II
Add dilute nitric acid/Tambah asid nitrik cair
Panaskan
Heat
Solid Q + carbon dioxide gas
Pepejal Q + gas karbon dioksida
Solution S
Larutan S
+
Carbon dioxide gas
Gas karbon dioksida
+
Water
Air
Reaction III + Magnesium
Tindak balas III + Magnesium
Zinc metal + Magnesium nitrate solution / Logam zink + Larutan magnesium nitrat
(a) (i)
Berikan satu ujian kimia bagi gas karbon dioksida.
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Passed the gas through lime water, lime water turns chalky
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Chemistry Form 4 • MODULE
(ii)
Draw a diagram of the apparatus set-up to carry out reaction I.
Lukiskan gambar rajah susunan radas untuk menjalankan tindak balas I.
Solid M
Heat
Lime water
(b) Name solids M and Q.
Nyatakan nama pepejal M dan Q.
M : Zinc carbonate
Q: Zinc oxide
(c) State the observations made when excess ammonia solution is added to solution S.
Nyatakan pemerhatian yang dibuat apabila larutan ammonia berlebihan ditambahkan kepada larutan S.
White precipitate, soluble in excess of ammonia solution
(d) (i)
Write the chemical equation for reaction II.
Tuliskan persamaan kimia bagi tindak balas II.
ZnCO3 + 2HNO3
(ii)
Zn(NO3)2 + H2O + CO2
For reaction II, calculate the volume of carbon dioxide gas released at room condition if 12.5 g solid M
decomposes completely. [Relative atomic mass: C =12, O =16, Zn = 65, 1 mole of gas occupies 24 dm3 at
room condition]
Bagi tindak balas II, hitungkan isi padu gas karbon dioksida yang dibebaskan pada keadaan bilik, jika 12.5 g pepejal M terurai
dengan lengkap. [Jisim atom relatif: C = 12, O = 16, Zn = 65, 1 mol gas menempati 24 dm3 pada suhu bilik]
12.5
= 0.1 mol
125
From the equation, 1 mol M : 1 mol CO2
0.1 mol M : 0.1 mol CO2
Volume of CO2 = 0.1 mol × 24 dm3 mol–1 = 2.4 dm3
Mol of solid M =
(e) Name reaction III.
Namakan tindak balas III.
Displacement reaction
(f)
Describe a chemical test to determine the presence of anion in the magnesium nitrate solution.
Huraikan ujian kimia untuk menentukan kehadiran anion dalam larutan magnesium nitrat.
– About
Masukkan
2
cm3 of magnesium nitrate solution is poured into a test tube.
2
cm3 larutan magnesium nitrat ke dalam tabung uji.
sulphuric acid
– 2 cm3 of dilute
2 cm
asid sulfurik
3
digoncang
Campuran
– A
brown ring
Gelang perang
– Anion present is
.
and held with a test tube holder.
dan dipegang dengan pemegang tabung uji.
concentrated
sulphuric acid is dropped along the wall of the test tube and is held upright.
pekat
Beberapa titis asid sulfurik
ferum(II) sulfat
solution.
.
dicondongkan
– A few drops of
iron(II) sulphate
.
slanted
– The test tube is
Tabung uji
cair ditambah kepada larutan diikuti dengan larutan
shaken
– The mixture is
is added to the solution followed by 2 cm3 of
dititiskan melalui dinding tabung uji dan ditegakkan.
is formed between two layers.
terbentuk antara dua lapisan.
nitrate
ion.
nitrat
.
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Anion yang hadir adalah ion
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MODULE • Chemistry Form 4
The diagram below shows list of chemical substances.
6
Rajah di bawah menunjukkan senarai bahan-bahan kimia.
Hydrochloric acid, 1.0 mol dm–3
Barium chloride solution, 1.0 mol dm–3
Larutan asid hidroklorik, 1.0 mol dm
Larutan barium klorida, 1.0 mol dm–3
Iron(II) sulphate solution, 1.0 mol dm–3
Solid copper(II) oxide
Solid calcium carbonate
Larutan ferum(II) sulfat, 1.0 mol dm–3
Pepejal kuprum(II) oksida
Pepejal kalsium karbonat
–3
(a) (i)
Choose two solutions that can be used to prepare insoluble salts.
Pilih dua larutan yang digunakan untuk menyediakan garam tak terlarutkan.
Barium chloride and iron(II) sulpahate
(ii)
What is the type of reaction for the preparation of the salt in (a)(i)?
Apakah jenis tindak balas bagi penyediaan garam di (a)(i)?
Double decomposition reaction
(iii) Write the ionic equation for the production of the salt in (a)(i).
Tulis persamaan ion bagi penghasilan garam di (a)(i).
Ba2+ + SO42–
BaSO4
(iv) Describe how to collect the pure salt produced.
Huraikan bagaimana anda mendapatkan pepejal garam tulen yang terhasil.
Filter the mixture and rinse with distilled water
(b) State the observations when sodium hydroxide solution is added in small amount until in excess into iron(II)
sulphate solution./ Nyatakan pemerhatian anda apabila larutan natrium hidroksida ditambah sedikit sehingga berlebihan kepada
larutan ferum(II) sulfat.
Green precipitate formed, insoluble in excess of sodium hydroxide solution
(c) (i)
Choose two chemical substances that can react to produce carbon dioxide gas.
Pilih dua bahan yang boleh bertindak balas untuk menghasilkan gas karbon dioksida.
Calcium carbonate and hydrochloric acid
(ii)
Write a balanced chemical equation for the reaction in (c)(i).
Tulis persamaan kimia seimbang bagi tindak balas di (c)(i).
CaCO3 + 2HCl
CaCl2 + H2O + CO2
You are given zinc chloride crystals. Describe how you would conduct a chemical test in the laboratory to identify the ions
presence ions in zinc chloride crystals./ Anda diberi hablur zink klorida. Huraikan bagaimana anda boleh menjalankan ujian kimia di
7
dalam makmal untuk mengenal pasti ion-ion yang hadir dalam hablur zink klorida.
Dissolve
–
1
spatula zinc chloride crystals in 10 cm3 of
2
distilled
water.
1 spatula hablur zink klorida di dalam 10 cm3
air suling
.
2
Larutan
solution
is poured in three test tubes./
tersebut dituang ke dalam tiga tabung uji.
Larutkan
– The
sodium hydroxide
solution are added to zinc chloride
– Add a few drops
sodium
hydroxide
precipitate soluble in excess of
solution.
natrium hidroksida
Tambahkan beberapa titik larutan
ke dalam
Mendakan putih
natrium
hidroksida
larut dalam larutan
larutan
solution
until excess. A white
zink klorida sehingga
berlebihan
.
berlebihan.
ammonia
solution
solution are added to another zinc chloride
until excess. A white
– Add a few drops
excess
ammonia
zinc
ions
precipitate soluble in
of
solution. Ions present are
.
ammonia
Tambahkan beberapa titik larutan
Mendakan putih
larut dalam larutan
ke dalam
ammonia
larutan
zink klorida yang lain sehingga
ion zink
berlebihan. Ion yang hadir adalah
nitric acid
is added to 2 cm3 solution of chloride ions followed by 2 cm3 of
– About 2 cm3 of dilute
solution. White precipitate formed. Ions present are chloride ions.
m
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silver nitrate
argentum nitrat
.
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a
asid nitrik
2 cm3
cair ditambahkan kepada 2 cm3 larutan ion klorida diikuti dengan 2 cm3 larutan
Mendakan putih terbentuk. Ion yang hadir adalah ion klorida.
berlebihan
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Chemistry Form 4 • MODULE
8
The diagram below shows the formation of zinc nitrate and the changes to other compounds.
Rajah berikut menunjukkan pembentukan zink nitrat dan perubahannya kepada sebatian lain.
Heat
+ Substance X
Zinc oxide
+ Bahan X
Zink oksida
Panaskan
Zinc nitrate
Zink nitrat
Brown gas
Gas perang
+ Potassium carbonate solution/ + Larutan kalium karbonat
Precipitate Z
+
Potassium nitrate
Mendakan Z
(a) (i)
Kalium nitrat
Zinc oxide reacts with substance X to form zinc nitrate. State the name of substance X.
Zink oksida bertindak balas dengan bahan X untuk membentuk zink nitrat. Namakan sebatian X.
Nitric acid
(ii)
Write the chemical equation for the reaction in (a)(i).
Tuliskan persamaan kimia untuk tindak balas dalam (a)(i).
ZnO + HNO3 → Zn(NO3)2 + H2O
(b) (i)
State the name of the brown gas formed.
Namakan gas perang yang terbentuk.
Nitrogen dioxide
(ii)
Write the chemical equation for the reaction in (b)(i).
Tuliskan persamaan kimia untuk tindak balas dalam (b)(i).
2Zn(NO3)2 → 2ZnO + 2NO2 + O2
(c) When potassium carbonate solution added to zinc nitrate solution, precipitate Z and potassium nitrate formed.
Apabila larutan kalium karbonat ditambah kepada larutan zink nitrat, mendakan Z dan kalium nitrat terbentuk.
(i)
State the type of reaction occurs.
Namakan jenis tindak balas yang berlaku.
Precipitation
(ii)
Write the ionic equation for the formation of compound Z.
Tulis persamaan ion untuk pembentukan sebatian Z.
Zn2+ + CO32– → ZnCO3
(iii) State how the precipitate Z separated from potassium nitrate.
Nyatakan bagaimana mendakan Z diasingkan daripada kalium nitrat.
Filtration
(d) Excess of zinc nitrate solution is added to 100 cm3 of 1 mol dm–3 potassium carbonate. Calculate the mass of zinc
carbonate formed. [Relative atomic mass: Zn = 65, C = 12, O = 16]
Larutan zink nitrat berlebihan ditambah kepada 100 cm3 larutan kalium karbonat 1 mol dm–3. Hitungkan jisim zink karbonat yang
terbentuk. [Jisim atom relatif: Zn = 65, C = 12, O = 16]
Zn(NO3)2 + K2CO3 → ZnCO3 + 2KNO3
100
Mol of K2CO3 = 1×
= 0.1 mol
1 000
From the equation,
1 mol K2CO3 : 1 mol ZnCO3
0.1 mol K2CO3 : 0.1 mol ZnCO3
Mass of ZnCO3 = 0.1 mol × 125 g mol–1 = 12.5 g
(e) Sodium hydroxide solution is added until excess to zinc nitrate solution. State the observation that can be made.
Larutan natrium hidroksida ditambah sedikit demi sedikit hingga berlebihan kepada larutan zink nitrat. Nyatakan pemerhatian yang
dapat dibuat.
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White precipitate soluble in excess of sodium hydroxide solution.
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MODULE • Chemistry Form 4
Objective Questions / Soalan Objektif
Which of the following is a salt?
1
5
Antara berikut, yang manakah adalah garam?
A Lead(II) oxide
Plumbum(II) oksida
B Calcium hydroxide
Kalsium hidroksida
C Barium sulphate
Barium sulfat
D Tetrachloromethane
Tetraklorometana
Antara tindak balas berikut, yang manakah akan menghasilkan
kuprum(II) klorida?
Which of the following salts can be prepared by double
decomposition reaction?
Antara garam berikut, yang manakah boleh disediakan dengan
kaedah pemendakan?
A Copper(II) chloride
Kuprum(II) klorida
B Lead(II) nitrate
Plumbum(II) nitrat
C Barium sulphate
Barium sulfat
D Zinc sulphate
Zink sulfat
Which pair of substances represented by the following
formulae react to produce salt?
4
Antara pasangan bahan tindak balas berikut, yang manakah
dapat bertindak balas menghasilkan garam?
I
II
III
IV
HNO3(aq) + NaOH(aq)
HCl(aq) + NaCl(aq)
H2SO4(aq) + MgSO4(aq)
H2CO3(aq) + KOH(aq)
A I and II only
I dan II sahaja
B
I and IV only
C
I, II and IV only
I dan IV sahaja
I, II dan IV sahaja
D
I, II, III and IV
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Copper(II) oxide and hydrochloric acid
Kuprum(II) oksida dan asid hidroklorik
Kuprum(II) karbonat dan asid hidroklorik
IV Copper(II) sulphate and sodium chloride
Kuprum(II) sulfat dan natrium klorida
A
I and II only
B
II and III only
C
III and IV only
D
I, II, III and IV
I dan II sahaja
II dan III sahaja
III dan IV sahaja
I, II, III dan IV
6
If 0.2 mole of calcium carbonate is heated until no further
change, what is the mass of calcium oxide produced?
[Relative atomic mass of C=12, O=16, Ca=40]
Jika 0.2 mol kalsium karbonat dipanaskan sehingga tiada
perubahan, berapakah jisim kalsium oksida, CaO yang terhasil?
[Jisim atom relatif: C = 12, O = 16, Ca = 40]
A 5.6 g
B 11.2 g
C 16.8 g
D 22.4 g
7
The diagram below shows observations when white
solid X heated strongly.
Rajah di bawah menunjukkan pemerhatian apabila pepejal X
dipanaskan dengan kuat.
White solid X / Pepejal putih X
Heat strongly/Panaskan dengan kuat
– Brown gas is released/ Gas perang terbebas
– Residue is a solid which is yellow when hot and white when
cold/ Baki perang apabila panas dan kuning apabila sejuk.
Which of the following substance is X?
Antara berikut, yang manakah adalah bahan X?
A Zinc nitrate
Zink nitrat
B Zinc carbonate
Zink karbonat
C Lead(II) nitrate
Plumbum(II) nitrat
D Lead(II) carbonate
Plumbum(II) karbonat
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I, II, III dan IV
Copper and hydrochloric acid
III Copper(II) carbonate and hydrochloric acid
Antara garam berikut, yang manakah larut dalam air?
A Iron(II) sulphate
Ferum(II) sulfat
B Silver chloride
Argentum klorida
C Calcium carbonate
Kalsium karbonat
D Lead(II) bromide
Plumbum(II) bromida
3
I
Kuprum dan asid hidroklorik
Which of the following salts is soluble in water?
2
Which of the following reactions will produce copper(II)
chloride?
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Chemistry Form 4 • MODULE
8
The diagram below shows a series of tests carried out on
solution Y.
Rajah di bawah menunjukkan satu siri ujian kimia ke atas larutan
Y.
Solution
Larutan
Y
Sodium hydroxide
solution
Green
precipitate
Larutan natrium
hidroksida
Mendakan
hijau
10 The diagram below shows the reaction between 20 cm3
of 0.5 moldm–3 of sodium chloride solution is and to
20 cm3 of 1.0 moldm–3 silver to produce silver chloride
precipitate and solution X.
Rajah di bawah menunjukkan tindak balas antara 20 cm3 larutan
natrium klorida 0.5 mol dm–3 dengan 20 cm3 larutan argentum
nitrat 1.0 mol dm–3 untuk menghasilkan mendakan argentum
klorida dan larutan X.
20 cm3 of 1.0 moldm–3 silver nitrate solution
Dilute nitric acid followed by silver nitrate solution
20 cm3 argentum nitrat 1.0 mol dm–3
Asid nitrik cair diikuti dengan larutan argentum nitrat
White precipitate/Mendakan putih
Which of the following is solution Y?
Antara berikut, yang manakah adalah bahan Y?
A Iron(II) chloride
C Copper(II) chloride
Ferum(II) klorida
Kuprum(II) klorida
B Iron(II) sulphate
D Copper(II) carbonate
Ferum(II) sulfat
Kuprum(II) karbonat
9
The diagram below shows two bottles of aqueous
solutions.
Rajah di bawah menunjukkan dua botol mengandungi larutan
garam aluminium nitrat dan larutan plumbum(II) nitrat.
20 cm3 of 0.5 moldm–3
of sodium chloride
solution
Solution X
Larutan X
20 cm3 larutan natrium
klorida 0.5 mol dm–3
Silver chloride precipitate
Mendakan argentum klorida
Which of the following ions are present in the solution
X?
Antara ion berikut, yang manakah yang hadir dalam larutan X?
Aluminium
nitrate
solution
Larutan
aluminium
nitrat
Lead(II)
nitrate
solution
Larutan
plumbum(II)
nitrat
Which of the following substances can be used to
differentiate between and aluminium nitrate solution
and lead(II) nitrate solution?
Antara bahan berikut, yang manakah dapat digunakan untuk membezakan
larutan aluminium nitrat dan larutan plumbum(II) nitrat?
A
Sodium hydroxide solution
B
Ammonia solution
C
Potassium chloride solution
D
Barium nitrate solution
I
II
III
IV
Na+
Ag+
NO3–
Cl–
A I and III only
I dan III sahaja
B
II and III only
C
I, II and III only
D
I, II, and IV only
II dan III sahaja
I, II dan III sahaja
I, II dan IV sahaja
Larutan natrium hidroksida
Larutan ammonia
Larutan kalium klorida
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MODULE • Chemistry Form 4
8
MANUFACTURED SUBSTANCES IN INDUSTRY
BAHAN KIMIA DALAM INDUSTRI
• SULPHURIC ACID/ASID SULFURIK
– Write an equation for Contact process and Haber process, stating the temperature, pressure and catalyst required.
Menulis persamaan untuk Proses Sentuh dan Proses Haber, menyatakan suhu, tekanan dan mangkin yang diperlukan.
• AMMONIA/AMMONIA
– List the uses of sulphuric acid and ammonia.
Menyenaraikan kegunaan asid sulfurik dan ammonia.
– Explain how sulphur dioxide causes environmental pollution.
Menerangkan bagaimana sulfur dioksida menyebabkan pencemaran alam.
•
–
–
–
–
ALLOY/ALOI
State the meaning of an alloy. / Menyatakan maksud aloi.
Draw the arrangement of atoms in metals and alloys. / Melukis susunan atom di dalam aloi dan logam.
Explain why an alloy is stronger than its pure metal. / Menerangkan mengapa aloi lebih kuat daripada logam tulennya.
Design an experiment to investigate the hardness of a material and its alloy.
Mereka bentuk eksperimen untuk mengkaji kekerasan aloi dan logam tulennya.
– List the examples of alloys, compositions and properties of alloys. / Menyenaraikan contoh aloi, komposisi dan sifat aloi.
– Relate properties of alloys to their uses. / Mengaitkan sifat aloi dengan kegunaannya.
•
–
–
–
–
POLYMERS/POLIMER
Sate the meaning of polymers. / Menyatakan maksud polimer.
List naturally occurring polymers and synthetic polymers. / Menyenaraikan polimer semula jadi dan polimer sintetik.
State the uses of synthetic polymers. / Menyatakan kegunaan polimer sintetik.
Explain the effect of environmental pollution caused by the disposal of synthetic polymers.
Menghuraikan kesan pembuangan polimer sintetik ke atas pencemaran alam sekitar.
– Ways to reduce pollution caused by synthetic polymers. / Cara-cara mengurangkan pencemaran yang disebabkan polimer sintetik.
•
–
–
–
GLASS AND CERAMICS/KACA DAN SERAMIK
List uses of glass and ceramics. / Menyenaraikan kegunaan kaca dan seramik.
List types of glass and their properties. / Menyenaraikan jenis-jenis kaca dan kegunaannya.
State properties of ceramics. / Menyenaraikan sifat-sifat seramik.
• COMPOSITE MATERIALS/BAHAN KOMPOSIT
– State the meaning of composite materials. / Menyatakan maksud bahan komposit.
– List examples of composite materials and their components and uses.
Menyenaraikan contoh-contoh bahan komposit dan komponen dan kegunaannya.
– Compare and contrast properties of composite materials with those of their original component
Membanding dan membezakan sifat bahan komposit dengan bahan asalnya.
– Design an experiment to produce composite materials.
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Chemistry Form 4 • MODULE
Sulphuric Acid / Asid Sulfurik
1
Sulfuric acid is manufactured through the Contact Process. This process consists of three stages.
Asid sulfurik dihasilkan melalui Proses Sentuh. Proses ini terdiri daripada tiga peringkat.
Sulphur
Sulfur
Sulphur dioxide
SO2
Sulphur trioxide
SO3
Oxygen
Sulfur dioksida SO2
Sulfur trioksida SO3
Oleum
H 2 S2 O7
Asid sulfurik
H2SO4
Oleum
H2 S2 O7
Oksigen
Stage I/Peringkat I
Sulphuric acid
H2SO4
Stage II/Peringkat II
Stage III/Peringkat III
Concentrated sulphuric acid
Asid sulfurik pekat
Waste gas
Gas terbuang
Molten sulphur
Sulfur lebur
SO3
Dry air
Udara kering
Burner
Pembakar
SO2 + O2
H2S2O7 (Oleum)
Catalytic converter
Bekas mangkin
H2S2O7 (Oleum)
Water/Air
H2SO4
Stage I/Peringkat I
2
Stage II/Peringkat II
Stage III/Peringkat III
Based on the above diagram, explain each stage and state the conditions required. Include all the balanced chemical
equations involve in each stage.
Berdasarkan rajah di atas, terangkan setiap peringkat serta keadaan yang diperlukan. Sertakan semua persamaan kimia yang seimbang yang
terlibat dalam setiap peringkat.
Stage
Explanation/Equation
Peringkat
Stage I: / Peringkat I:
sulphur dioxide
Production of
Penghasilan
sulfur dioksida
Stage II: / Peringkat II:
sulphur trioxide
Production of
Penghasilan
sulfur trioksida
Penerangan/Persamaan kimia
– Molten sulphur is burnt in dry air to produce sulphur dioxide.
Sulfur lebur dibakar dalam udara kering untuk menghasilkan sulfur dioksida.
Balanced equation: / Persamaan seimbang:
S + O2
SO2
– In a converter, sulphur dioxide and excess oxygen are passed through
vanadium(V) oxide .
Di dalam bekas mangkin, sulfur dioksida dan oksigen dialirkan melalui
Balanced equation: / Persamaan seimbang:
2SO2 + O2
vanadium(V) oksida
.
2SO3
– Optimum conditions for maximum amount of product are:
Keadaan optimum untuk penghasilan sulfur trioksida yang maksimum adalah:
Temperature / Suhu:
450 – 500 °C
2 – 3 atm
Catalyst / Mangkin:
vanadium(V) oxide, V2O5
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Pressure / Tekanan:
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MODULE • Chemistry Form 4
Stage III: / Peringkat III:
sulphuric acid
Production of
Penghasilan
Sulphur trioxide
–
Sulfur trioksida
asid sulfurik
is dissolved in concentrated sulphuric acid to form oleum.
Balanced equation: / Persamaan seimbang:
SO3 + H2SO4
Oleum
–
Oleum
.
H2S2O7
is diluted in water to produce concentrated sulphuric acid.
asid sulfurik pekat
dilarutkan dalam air untuk menghasilkan
Balanced equation: / Persamaan seimbang:
H2O + H2S2O7
*
oleum
dilarutkan dalam asid sulfurik pekat untuk menghasilkan
.
2H2SO4
Note that directly dissolving SO3 in water is impractical due to the highly exothermic nature of the reaction. Acidic vapour or mists are
formed instead of a liquid.
Melarutkan sulfur dioksida dalam air secara terus tidak dapat dilakukan kerana pembebasan haba yang sangat banyak. Ini kerana tindak balas tersebut
adalah eksotermik. Asid yang terhasil adalah dalam bentuk wap air dan bukannya cecair.
State five main uses of sulphuric acid.
3
Nyatakan lima kegunaan utama asid sulfurik.
(i)
To manufacture detergents
(iv)
As electrolyte in car batteries
(ii)
To manufacture fertilizers
(v)
To manufacture synthtetic fibers
(iii) To manufacture paints
Sulphur dioxide and environmental pollution:
4
Sulfur dioksida dan pencemaran alam:
(a) Major sources of sulphur dioxide in the air is combustion of fuel in power station or factories.
Punca utama kehadiran sulfur dioksida di udara adalah pembakaran bahan bakar di stesen janakuasa dan kilang.
(b) Sulphur dioxide dissolve in rainwater to form sulphurous acid which will cause acid rain, balanced equation:
Sulfur dioksida larut dalam air hujan untuk membentuk asid sulfurus yang menghasilkan hujan asid, persamaan seimbang:
SO2 + H2O
H2SO3
Oxidation of sulphurous acid in the air will produce sulphuric acid which will also cause acid rain.
Pengoksidaan asid sulfurus di udara akan menghasilkan asid sulfurik yang juga merupakan penyebab kepada hujan asid.
(c) Effect of acid rain:
Kesan hujan asid:
corrodes
building, monuments and statues made from marble (calcium carbonate) because
– Acid rain
calcium carbonate react with acid to produce salt, water and carbon dioxide, balanced equation:
mengkakis
Hujan asid
bangunan, monumen dan tugu yang diperbuat daripada marmar (kalsium karbonat) kerana
kalsium karbonat bertindak balas dengan asid menghasilkan garam, air dan karbon dioksida, persamaan seimbang:
CaCO3 + H2SO4
CaSO4 + H2O + CO2
corrodes
structures of the buildings or bridges which are made from
– Acid rain
iron rusts faster with the presence of sulphuric acid.
metal
. The
mengkakis
Hujan asid
struktur bangunan-bangunan dan jambatan-jambatan yang diperbuat daripada logam. Besi
berkarat lebih cepat dengan kehadiran asid sulfurik.
– Acid rain
Hujan asid
– Acid rain
Hujan asid
increases
the acidity of lakes and river that causes aquatic organism to die.
meningkatkan
increases
keasidan tasik-tasik dan sungai-sungai yang menyebabkan kematian hidupan akuatik.
the acidity of soil. Acidic soil is not suitable for the growth of plants.
meningkatkan
keasidan tanah. Tanah yang berasid tidak sesuai untuk pertumbuhan tanam-tanaman.
(d) Ways to reduce production of sulphur dioxide and effect of acid rain:
Cara-cara mengurangkan penghasilan sulfur dioksida dan kesan-kesan hujan asid:
– Gas released from power station and factories are sprayed with powdered limestone ( calcium carbonate ).
Gas yang dilepaskan dari stesen janakuasa dan kilang boleh disembur dengan serbuk batu kapur (
– Add lime (
calcium oxide
m
).
) and limestone ( calcium carbonate ) to the lake or river.
kalsium oksida
) dan batu kapur (
kalsium karbonat
) ke tasik atau sungai.
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Chemistry Form 4 • MODULE
Ammonia / Ammonia
1
In industry, ammonia is manufactured through the Haber Process:
Dalam industri, ammonia dihasilkan melalui Proses Haber.
N2 + 3H2
Balanced equation of reaction / Persamaan seimbang tindak balas:
Ferum
Catalyst / Mangkin :
Temperature / Suhu :
Pressure / Tekanan :
2
2NH3
400 – 500°C
200 atm
Ammonia is used in the manufacture of:
Ammonia digunakan dalam pembuatan:
(a) Synthetic fertilizer such as ammonium sulphate, ammonium nitrate, ammonium phosphate and urea
Baja sintetik seperti ammonium sulfat, ammonium nitrat, ammonium fosfat dan urea.
(b) Nitric acid in Ostwald Process.
Asid nitrik dalam Proses Ostwald.
(c) Synthetic fiber and nylon.
Gentian kaca sintetik dan nilon.
(d) Liquid form of ammonia is used as cooling agent in refrigerators.
Cecair ammonia digunakan sebagai penyejuk dalam peti sejuk.
(e) Prevent coagulation of latex.
Mencegah penggumpalan lateks.
3
Ammonia is a colourless gas with pungent smell and very soluble in water.
Ammonia adalah gas yang tidak berwarna dengan bau yang sengit dan sangat larut di dalam air.
4
Chemical properties of ammonia:
Sifat-sifat kimia ammonia:
Property
Sifat
Dissolve in water to form weak
alkali
Larut di dalam air membentuk alkali
lemah
Effect on moist red litmus paper
Kesan ke atas kertas litmus merah
Neutralise any acid to form
ammonium salt
Meneutralkan asid untuk membentuk
garam ammonium
Chemical equation / Observation
Persamaan kimia / Pemerhatian
NH3(g) + H2O(ce)
NH4+(ak) + OH –(ak)
The presence of hydroxide ions causes aqueous solution of ammonia to become alkaline.
Kehadiran ion hidroksida menyebabkan larutan ammonia akueus menjadi alkali.
Turn moist red litmus paper to blue
Ammonia reacts with sulphuric acid to form ammonium sulphate salt.
Ammonia bertindak balas dengan asid sulfurik untuk membentuk garam ammonium sulfat.
Balanced equation: / Persamaan seimbang:
2NH3 + H2SO4
(NH4)2SO4
Alloy / Aloi
1
Complete the following table:
Lengkapkan jadual di bawah:
Questions
Soalan
1 What is the meaning of alloy?
Apakah maksud aloi?
Facts / Elaboration / Drawing
Fakta / Penerangan / Lukisan
mixture
elements
of two or more
with a certain
Alloy is a
fixed/specific
composition. The major component in the mixture is a metal.
tetap
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campuran
unsur
dua atau lebih
dengan komposisi yang
Aloi ialah
logam
Komponen utama dalam campuran tersebut ialah
.
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MODULE • Chemistry Form 4
2 Relate the arrangement of atoms
in pure metals to their ductile and
malleable properties.
Nyatakan hubungan antara susunan atom
dalam logam tulen dengan sifat mulur dan
mudah ditempa.
Force/Daya
Pure metals/Logam tulen
atoms
.
Pure metal is made up of one type of
atom
Logam tulen terbentuk daripada satu jenis
layers
Atoms in pure metals are all the same
saiz
Atom-atom dalam logam tulen mempunyai
The same
size
.
.
yang sama.
atoms are orderly arranged in layers.
saiz
Atom-atom yang mempunyai
yang sama ini tersusun dalam
lapisan
force
is applied to the pure metal, layers of atoms
When
easily over one another.
daya
Apabila
sama lain.
3 Draw the arrangement of atoms in
Lukiskan susunan atom dalam
dikenakan ke atas logam tulen, lapisan atom
.
slide
menggelongsor
di antara satu
(b) Steel / Keluli
(a) Bronze / Gangsa
(a) Bronze (90% copper and 10% tin)
Gangsa (90% kuprum dan 10% timah)
Carbon
(b) Steel (99% iron and 1% of carbon)
Keluli (99% besi dan 1% karbon)
[Relative atomic mass: Cu = 64,
Sn = 119, Fe = 56; C = 12]
[Jisim atom relatif: Cu = 64,
Sn = 119, Fe = 56, C = 12]
4 Explain why an alloy is stronger
than its pure metal in terms of the
arrangement of atoms in metals and
alloys.
Terangkan mengapa aloi lebih kuat daripada
logam tulen dari segi susunan atom dalam
logam dan aloi.
Iron
Copper
Tin
Atoms of other element added to the pure metal to make an alloy are
These atoms
disrupts
Nyatakan tiga sebab mengapa logam tulen
dialoikan sebelum digunakan.
in size.
the orderly arrangement of atoms in pure metal.
mengganggu
Atom-atom ini
susunan atom yang teratur dalam logam tulen.
force
is applied to an alloy, the presence of added other atoms
When
prevent
sliding
layers of atoms from
.
daya
dikenakan ke atas aloi, kehadiran atom-atom asing ini
Apabila
menggelongsor
atom-atom ini daripada
.
5 State three reason why pure metals are
alloyed before used.
different
Atom-atom unsur lain yang ditambah dalam logam tulen membentuk aloi yang terdiri daripada atom-atom
berlainan
saiz.
yang
strength
(a) To increase the
kekuatan
Meningkatkan
dan
(b) To increase the resistance to
Mencegah
kakisan
(c) To improve the
Membaiki
kekerasan
corrosion
lapisan
of pure metals.
logam tulen.
of a pure metals.
logam tulen.
appearance
rupa
hardness
and
menghalang
of a pure metal.
logam tulen.
Experiment to compare the hardness of brass and pure copper.
5
Eksperimen untuk membandingkan kekerasan loyang dengan kuprum tulen.
(a) Hypothesis: / Hipotesis:
Brass is harder than copper
(b) Manipulated variable: / Pemboleh ubah dimanipulasi:
Copper and brass block
(c) Responding variable: / Pemboleh ubah bergerak balas:
Hardness of the copper and brass block
(d) Fixed variable: / Pemboleh ubah dimalarkan:
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Chemistry Form 4 • MODULE
(e) Apparatus: / Alat radas:
Retort stand and clamp, 1 kg weight, string, metre ruler.
Materials: / Bahan-bahan:
Steel ball, copper block, brass block
(f)
Procedure: / Prosedur:
1. A steel ball bearing is tapped onto a copper block.
Set-up of the apparatus: / Susunan alat radas:
Satu bola keluli dilekatkan di atas sebuah bongkah kuprum.
2. A 1 kg weight is hung at a height of 50 cm above the copper
block as shown in the diagram.
Sebiji pemberat 1 kg digantung setinggi 50 cm di atas bongkah kuprum
seperti yang ditunjukkan.
Retort stand
3. Drop the 1 kg weight on the steel ball.
Pemberat 1 kg dijatuhkan ke atas bebola keluli.
String
4. Measure the diameter of the dent formed on the copper block
with a ruler.
1 kg weight
Diameter lekuk yang terbentuk di atas bongkah kuprum diukur dengan
pembaris.
5. Repeat the experiment three times on the other part of the
copper block.
Steel ball
Eksperimen diulang tiga kali, pada ruang berbeza pada bongkah kuprum
yang sama.
Cellophane tape
6. Steps 1 to 5 are repeated using a brass block to replace the
copper block.
Copper block
Langkah 1 hingga 5 diulang dengan menggunakan bongkah loyang,
menggantikan bongkah kuprum.
(g) Results: / Keputusan:
Experiment
Average diameter/cm
Eksperimen
1
2
3
Diameter of dent on copper block/cm
a
b
c
a+b+c
=x
3
Diameter of dent on brass block/cm
d
e
f
d+e+f
=y
3
Diameter purata / cm
(h) Discussion: / Perbincangan:
The average diameter of dent on copper, x is larger than the average diameter of dent on brass, y.
(i)
Conclusion: / Kesimpulan:
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MODULE • Chemistry Form 4
Flow chart shows the composition, properties and uses of some alloys.
Carta aliran di bawah menunjukkan komposisi, sifat-sifat dan kegunaan aloi-aloi.
ALLOY / ALOI
Major component / Komponen utama
COPPER / KUPRUM
IRON / FERUM
Type of alloy/Jenis aloi
Type of alloy/Jenis aloi
BRONZE/GANGSA
(90% Cu, 10% Sn)
– Hard and strong,
does not corrode,
(shiny surface)
–
BRASS/LOYANG
(70% Cu, 30% Zn)
– Hard and strong.
STEEL/KELULI
(99% Fe, 1% C)
– Hard and strong.
Keras dan kuat.
Uses: / Kegunaan:
–
Keras dan kuat. Tidak
berkarat (permukaan
bersinar)
Uses: / Kegunaan:
–
Musical instrument
and Kitchenware
KELULI TAHAN KARAT
(74% Fe, 8% C, 18% Cr)
– Shiny, strong and does
Keras dan kuat.
Uses: / Kegunaan:
not rust
Construction of
building and bridge
and railway tracks
Alat muzik dan perkakas
dapur
Building statue or
monuments, medal,
swords and artistic
materials
STAINLESS STEEL
–
Bersinar, kuat dan tidak
berkarat.
Uses: / Kegunaan:
Making cutlery and
surgical instrument
Pembinaan bangunan
dan jambatan serta
landasan keretapi.
Membuat sudu, garpu dan
alat-alat pembedahan.
Pembuatan tugu atau
monumen pingat, pedang
dan bahan hiasan
ALUMINIUM / ALUMINIUM
Type of alloy
Jenis aloi
CUPRONICKEL
KUPRONIKEL
(75% Cu, 25% Ni)
– Shiny, hard and does
–
TIN / TIMAH
Type of alloy
Jenis aloi
DURALUMIN
DURALUMIN
not corrode
(93% Al, 3% Cu & 1% Mn)
– Light and strong
Making coins
– Uses: / Kegunaan:
Building body of aeroplane
and bullet train.
Bersinar, keras dan tidak
berkarat.
Uses: / Kegunaan:
Membuat duit syiling
Ringan dan kuat.
Membuat rangka kapal terbang dan
keretapi laju.
PEWTER / PEWTER
(96% Sn, 3% Cu,
1% Sb)
– Luster, shiny and
strong
–
Berkilau, bersinar dan kuat.
Uses: / Kegunaan:
Making souvenirs.
Membuat cenderamata.
SYNTHETIC POLYMERS / POLIMER SINTETIK
Polymer is a long chain molecules made up of a
monomer.
1
large
Polimer ialah molekul berantai panjang yang terbentuk daripada gabungan
monomer.
Monomer is small identical
2
Monomer adalah unit kecil yang
repeating
berulang
number of small repeating
banyak
unit kecil yang
identical
sama
unit of
dipanggil
units in the polymer.
dalam polimer.
Polymers can be naturally occurring or synthetic.
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Polimer boleh didapati secara semula jadi atau sintetik.
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Chemistry Form 4 • MODULE
4
Example of naturally occurring polymers and their monomers are:
Contoh polimer semula jadi dan monomernya:
5
Synthetic Polymer / Polimer
Monomer / Monomer
Protein / Protein
Amino Acid / Asid amino
Starch / Kanji
Glucose / Glukosa
Rubber / Getah
Isoprene / Isoprena
Synthetic polymers are made polymers. The monomers are usually obtained from petroleum after refining and cracking
process.
Polimer sintetik adalah polimer buatan. Monomer biasanya adalah daripada petroleum yang telah mengalami penyulingan dan peretakan.
6
Example of synthetic polymers, their monomers and uses:
Contoh polimer sintetik, monomernya dan kegunaannya:
Synthetic polymer
Monomer
Polimer sintetik
Example of uses
Monomer
Contoh kegunaan
Ethene, C2H4
Polythene
Politena
Polypropene
Polipropena
Plastic bags, shopping bags, plastic containers and plastic toys
Etena, C2H4
Beg plastik, beg membeli belah, bekas plastik dan permainan plastik
Propene, C3H6
Plastic bottles, plastic tables and chairs, car batteries casing and ropes
Botol plastik, meja dan kerusi plastik, bekas bateri kereta dan tali
Propena, C3H6
Waterproof materials such as rain clothes, bags, shoes, artificial leather.
Polyvinylchloride (PVC)
Polivinil klorida (PVC)
Bahan kalis air seperti baju hujan, beg, kasut dan kulit tiruan.
Chloroethene, C2H3Cl
Insulation for electric wiring.
Bahan penebat pendawaian wayar elektrik.
Kloroetena, C2H3Cl
Making water pipes because it does not rust.
Paip air sebab ia tidak berkarat.
Styrene, C2H3C6H5
Polystyrene
Polistirena
Packaging materials, disposable cups and plates
Stirena, C2H3C6H5
Bahan pembungkus, cawan dan pinggan pakai buang.
Perspex
Methylmetacrylate
Safety glass, car lamps and lens
Perspeks
Metil metakrilat
Kaca keselamatan, lampu kereta dan kanta
Hexane-1, 6-diol
Heksana-1, 6-diol
Terylene (polyester)
Clothing, sails, sleeping bags, ropes and fishing net
Benzene-1, 4-dicarboxylic
acid
Terilena (poliester)
Pakaian, kain layar, tali dan jala
Benzena-1, 4-dikarboksilik asid
7
joining
Polymerisation is the process of
Pempolimeran ialah proses
penggabungan
together the large number of monomers to form a polymer.
monomer-monomer untuk membentuk polimer.
Example: / Contoh:
(a) Polymerisation of ethene:
Pempolimeran etena:
H
H
n C = C
H
H
H
– C – C –
H
H
n, n is large number up to a few thousands
Polythene
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MODULE • Chemistry Form 4
(b) Polymerisation of propene:
(c) Polymerisation of chloroethene:
Pempolimeran propena:
H
Pempolimeran kloroetena:
CH3
H
n C = C
H
CH3
H
– C – C –
H
H
Propene / Propena
H
H
H
n C = C
H
n
– C – C –
Cl
H
Chloroethene / Kloroetena
Polypropene
H
Cl
n
Polyvinylchloride
Complete the following table related to issues of the use of polymers in everyday life.
8
Lengkapkan jadual di bawah berkaitan isu penggunaan polimer sintetik dalam kehidupan seharian.
Advantages of
synthetic polymers
Kebaikan polimer sintetik
(a) Very stable and do not
corrode
.
Sangat stabil dan tidak
berkarat
.
chemical
(b) Inert to
reaction.
Reducing pollution of
synthetic polymers
Environmental pollution from synthetic polymers
Pencemaran alam sekitar dari penggunaan polimer sintetik
Pengurangan pencemaran dari
polimer sintetik
(a) Disposal of synthetic polymers such as plastic bottles and
blockage
containers cause
of drainage systems and
river thus causing flash floods .
(a) Reduce,
reuse
recycle
and
the synthetic
polymers.
Mengurangkan, mengitar semula
dan mengguna semula polimer
sintetik.
Pembuangan polimer sintetik seperti botol plastik dan bekas
tersekat
yang
menyebabkan sistem saliran dan sungai
banjir kilat
mengakibatkan
.
biodegradable
Lengai terhadap tindak balas (b) Open burning of polymers will release acidic and poisonous (b) Using
polimer.
kimia
gas that will cause air pollution:
.
Menggunakan polimer
Pembakaran polimer sintetik secara terbuka membebaskan gas
strong
.
(c) Light and
berasid dan beracun yang menyebabkan pencemaran udara:
terbiodegradasi
.
kuat
.
Ringan dan
– Burning most of the synthetic polymers will produce:
Pembakaran kebanyakan polimer sintetik menghasilkan:
(c) On-going research to produce
(d) Cheap.
Murah.
shaped
(e) Easily
and coloured.
dibentuk
Mudah
dan diwarnakan.
(i) carbon dioxide gas which cause green house effect .
karbon dioksida yang menyebabkan
(ii) carbon monoxide which is
karbon monoksida yang
kesan rumah hijau
.
poisonous .
beracun
.
– Burning of PVC will release hydrogen chloride gas which
acid rain .
will cause
Pembakaran PVC membebaskan gas hidrogen klorida yang
hujan asid
.
menyebabkan
– Burning of synthetic polymers contains carbon and
nitrogen such as nylon will produce highly poisonous
hydrogen cynide .
gas such as
Pembakaran polimer sintetik mengandungi karbon dan
nitrogen seperti nilon membebaskan gas sangat beracun seperti
hidrogen sianida .
cheap biodegradable polymers.
Penyelidikan berterusan
untuk menghasilkan polimer
terbiodegradasi yang murah.
(d) Disintegrate plastics by
pyrolysis : Plastic can
be disintegrated by heating at
temperature between
400 – 800°C without oxygen.
Penguraian plastik secara
pirolisis : Plastik boleh diuraikan
dengan pemanasan pada suhu
antara 400 – 800 °C tanpa oksigen.
(c) Plastic containers that are left in open area collect rain
water will become breeding ground for mosquito
which will cause diseases such as dengue fever.
Bekas plastik yang ditinggalkan di tempat terbuka menakung air
nyamuk
yang menyebabkan
hujan menjadi tempat pembiakan
penyebaran penyakit seperti demam denggi.
Glass / Kaca
1 Name the element which forms
the major component of glass.
Silicon dioxide
, SiO2 which exist naturally in
Namakan unsur yang membentuk
komponen utama kaca.
Silikon dioksida
, SiO2 yang boleh didapati secara semula jadi di dalam
2 List the property of glass.
Senaraikan sifat-sifat kaca.
sand
.
pasir
.
Properties: / Sifat-sifat:
Transparent, hard but brittle, non-porous, heat insulator, electric insulator, resistant to
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chemical, easy to clean, can withstand compression
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Chemistry Form 4 • MODULE
Complete the table below.
Lengkapkan jadual di bawah.
Types of
glass
Jenis kaca
Soda-lime
glass
Kaca soda
kapur
Composition
Komposisi
Silicon dioxide,
sodium carbonate
or calcium calcium
carbonate
Special properties
Sifat istimewa
Borosilicate
glass
Kaca
borosilikat
Silikon dioksida, boron
dioksida, natrium
oksida, aluminium
oksida
durability
kimia
Tahan kakisan bahan
High
–
tinggi
Pekali pengembangan haba
haba
Tidak tahan
.
chemical
– Good
Low
kimia
mirrors, glass containers
.
.
Making cookware and laboratory
thermal expansion.
Pekali pengembangan haba
heat
haba
rendah
.
glassware such as boiling tube and
when heated to
– Resistant to
high temperature.
Tahan
tinggi.
Making flat glass, electrical bulbs,
durability
Tahan kakisan bahan
–
.
termal expansion but does not
heat
.
withstand
Silikon dioksida,
natrium karbonat,
kalsium karbonat
Silicion dioxide,
boron dioxide,
sodium oxide,
aluminum oxide
chemical
– Good
Uses
Kegunaan
beakers.
apabila dipanaskan pada suhu
– Optically transparent.
Lut sinar.
chemical
– Good
Fused glass
Kaca silika
terlakur
durability
kimia
Tahan kakisan bahan
Silicon dioxide
– Low thermal expansion
Silikon dioksida
Pekali pengembangan haba
.
Laboratory glassware, lenses,
rendah
telescope mirrors, optical fibres.
.
high
temperature
– Can be heated to
and resistance to thermal shock.
tinggi
Boleh dipanaskan pada suhu yang
tahan terhadap pertukaran suhu yang cepat.
Lead glass
Kaca plumbum
Silicon dioxide,
sodium oxide,
lead(II) oxide
Silikon dioksida,
natrium oksida,
plumbum(II) oksida
refractive
– High
Indeks
biasan
Glittering
–
Kelihatan
index and
dan
density
ketumpatan
appearance.
berkilat
,
.
yang tinggi
Tableware, crystal glass ware and
decorative glassware.
.
Ceramics / Seramik
1
Name the elements found in ceramic.
Namakan unsur-unsur yang terkandung dalam seramik.
Aluminium, silicon, oxygen and hydrogen
2
Ceramics are made from clay. Name the main component of clay.
Seramik dibuat daripada tanah liat. Namakan komponen utama tanah liat.
which is rich in
Kaolin
yang mengandungi
hydrated aluminium silicate
aluminium silikat terhidrat
, Al2O32SiO2.2H2O.
, Al2O32SiO2.2H2O.
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MODULE • Chemistry Form 4
Complete the following table for the properties and uses of ceramic.
3
Lengkapkan jadual berikut untuk menunjukkan sifat-sifat dan kegunaan seramik.
Property/Sifat
Uses/Kegunaan
Hard and strong.
Building materials such as
Keras dan kuat.
Bahan binaan seperti
Chemically inert and non-corrosive.
cement
simen
tiles
,
jubin
,
– Kitchenware such as cooking pots and plates.
periuk
Perkakas dapur seperti
pinggan
dan
Tidak reaktif secara kimia dan tidak mudah
menghakis.
– Decorative items such as vases and pottery.
Have high melting point and good
insulator of heat, remain stable under
high temperature.
Insulation such as
parts.
lining
Penebat haba seperti
enjin
bahagian
melapik
Mempunyai takat lebur yang tinggi dan
penebat haba yang baik serta stabil dalam
suhu yang tinggi.
Penebat elektrik yang baik.
.
Barang hiasan seperti pasu dan lain-lain.
of furnace, wall of
nuclear reactor
dinding
dinding relau,
and
engine
bagi reaktor nuklear dan
.
electric plugs
Electric insulator in electrical items such as
electric cables
.
Good insulator electric.
, bricks, roof and toilet bowl.
, batu-bata, atap dan tandas.
Penebat elektrik bagi alat-alat elektrik seperti
kabel elektrik
.
plug elektrik
,
oven
and
ketuhar
,
dan
Medical
dental
and
apparatus such as orthopedic joint replacement,
dental restoration and bone implants.
Non compressible.
perubatan
Alat-alat
palsu dan pemindahan tulang.
Tidak boleh dimampatkan.
pergigian
dan
seperti penukaran sendi ortopedik, gigi
Composite Materials / Bahan Komposit
(a) Composite materials are structural materials that are formed by combining two or more different substances such as
metal
alloys
ceramic
glass
polymer
,
,
,
and
.
1
Bahan-bahan komposit adalah bahan yang diperbuat daripada gabungan dua atau lebih bahan berbeza seperti
aloi
seramik
kaca
polimer
,
,
dan
.
superior
(b) Composite materials have properties that are
Bahan-bahan komposit mempunyai sifat-sifat yang
logam
,
than those of the original components.
lebih baik
berbanding dengan komponen-komponen asal.
Complete the table below:
2
Lengkapkan jadual di bawah:
Types of
composite
materials
Components
Special properties
Komponen
Sifat istimewa
Example of uses
Contoh kegunaan
Jenis bahan komposit
Superconductors
Super konduktor
Reinforced
concrete
Konkrit yang
diperkukuhkan
Copper(II) oxide, barium
carbonate and Yttrium oxide
heated to form a type of
ceramic known as perovoskyte
Conduct electricity with
no resistance
when it is
cooled at low temperature.
Kuprum(II) oksida, barium
karbonat dan natrium oksida
dipanaskan membentuk sejenis
seramik dipanggil perovoskit
amat rendah.
Concrete ( cement , sand
and pebbles) reinforced with
steel
and polymer
fibers
m
Very
strong
moulded
and can be
into any shape.
devices(MRI), generators,
transformers, computer parts and
bullet train
Construction of building, bridges
and oil platforms
kuat
dan boleh
Sangat
dibentuk
menjadi pelbagai bentuk.
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Konkrit ( simen , pasir
dan batu kerikil) diperkukuhkan
dengan keluli dan polimer
gentian.
Boleh mengalirkan arus elektrik
tanpa rintangan
pada suhu yang
Used in medical magnetic-imaging
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Chemistry Form 4 • MODULE
High tensile strength,
low
density, easily moulded in
Plastic reinforced with
glass fiber .
Fibre glass
Plastik yang
diperkukuhkan
dengan kaca
Plastik yang diperkukuhkan dengan
gentian kaca .
thin
Kaca fotokromik
boats, helmets
layers.
regangan
tinggi, ketumpatan
Daya
rendah
, mudah dibentuk menjadi
lapisan
Photochromic
glass
Making water storage tanks,
nipis
.
Darken
Photochromic substance like
silver chloride embedded
in glass/transparent polymers
when exposed to bright
clear
when
light and becomes
exposed to dim light.
Bahan fotokromik seperti
argentum klorida digabungkan
gelap
apabila dikenakan
Menjadi
cerah
cahaya cerah dan menjadi
dalam cahaya malap.
dengan kaca atau polimer lut sinar.
Making optical lens, car wind
shield light intensity meters
EXERCISE / LATIHAN
1
The diagram below shows the reaction involve in the production of fertilizer Z in industry.
Rajah berikut menunjukkan tindak balas yang terlibat dalam pembuatan baja Z dalam industri.
Ammonia
Process X
Proses X
Ammonia
Process Y
Sulphuric acid
Compound Z
Sebatian Z
Asid sulfurik
Proses Y
(a) (i)
Reaction P
Tindak balas P
Name Process X and Process Y.
Namakan Proses X dan Proses Y.
Haber process
Process X / Proses X:
(ii)
Process Y / Proses Y:
Contact process
Complete the following table related to process X and Y.
Lengkapkan jadual berikut yang berkaitan dengan proses X dan Y.
Process
Proses
Catalyst
Mangkin
Process X
Iron
Proses X
Besi
Process Y
Vandaium(V) oxide
Proses Y
Vanadium(V) oksida
Temperature/°C
Suhu/°C
Pressure/
atm
Tekanan / atm
Balanced equation for the reaction
that Involve a catalyst
Persamaan kimia tindak balas yang melibatkan
mangkin
400 – 500
200
N2 + 3H2
2NH3
450 – 500
2–3
2SO2 + O2
2SO3
(b) Ammonia react with sulphuric acid through reaction P to produce compound Z.
Ammonia bertindak balas dengan asid sulfurik melalui tindak balas P menghasilkan sebatian Z.
(i)
Write a balance equation for reaction P.
Tuliskan persamaan seimbang bagi tindak balas P.
NH3 + H2SO4
(ii)
(NH4 )2SO4
What is the type of reaction that takes place?
Apakah jenis tindak balas yang berlaku?
Neutralisation
(iii) State one important use of compound Z.
Nyatakan satu kegunaan penting sebatian Z.
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MODULE • Chemistry Form 4
(iv) Calculate the percentage by mass of nitrogen in compound Z.
[Relative atomic mass: N = 14, S = 32, O = 16, H = 1]
Hitungkan peratusan jisim nitrogen dalam sebatian Z.
[Jisim atom relatif: N = 14, S = 32, O = 16, H = 1]
%N =
2 × 14
× 100% = 21.2%
2(14 + 4 × 1) + 32 + 4 × 16
The table shows the examples and component of four types of manufactured substances in industry.
2
Jadual berikut menunjukkan contoh-contoh dan komponen bagi empat jenis bahan buatan dalam industri.
Type of manufactured
substances
Example
Jenis bahan buatan
P
Component
Contoh
Komponen
Reinforced concrete
Konkrit yang diperkukuhkan
Q
Bronze / Gangsa
Polymer / Polimer
R
Glass / Kaca
S
Cement, sand, small pebbles and steel
Simen, pasir, batu kecil dan keluli
Copper and tin / Kuprum dan stanum
Chloroethene / Kloroetena
Silicon dioxide, sodium carbonate, calcium carbonate
Silikon dioksida, natrium karbonat, kalsium karbonat
(a) State the name of P, Q, R and S.
Namakan P, Q, R dan S.
P: Composite materials
Q: Alloy
R: Polyvinyl chloride
S: Soda-lime glass
(b) (i)
State two uses of reinforced concrete.
Nyatakan dua kegunaan konkrit yang diperkukuhkan.
To make framework of buildings and bridges.
(ii)
What is the advantage of using reinforced concrete compared to concrete?
Apakah kelebihan konkrit yang diperkukuhkan berbanding dengan konkrit?
Reinforced concrete can withstand higher pressure/support heavier loads/ stronger/ higher tensile strength
than concrete.
(c) (i)
Draw the arrangement of particles in
Lukis susunan atom dalam
Pure copper / Kuprum tulen
Bronze / Gangsa
Copper
Copper
(ii)
Tin
Bronze is harder than pure copper. Explain.
Gangsa lebih keras daripada kuprum. Terangkan.
– Atoms of pure copper metal are the of same size, they arranged orderly in layers.
– Layers of atoms are easily slide over each other when external force is applied on them.
– The size of tin atoms which are bigger than copper in bronze disrupt the orderly arrangement of copper
atoms.
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– Layers of metal atoms are prevented from sliding each other when external force is applied.
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Chemistry Form 4 • MODULE
(d) The diagram shows the structure of R. / Rajah berikut merupakan struktur bagi R.
(i)
H
H
C
– C
H
C1
n
Draw the structural formula for monomer R. / Lukiskan formula struktur bagi monomer R.
H
H
C = C
H
(ii)
C1
State one use of polymer R.
Nyatakan satu kegunaan polimer R.
Pipe / wire cables / bags / footwear
(iii) State two ways how R causes environmental pollution.
Nyatakan dua cara R menyebabkan pencemaran alam.
– R is non biodegradable, it can cause blockage of drainage system and flash flood.
– Burning of R produces hydrogen chloride gas which is poisonous and acidic.
(e) (i)
Explain why glass containers are more suitable for storing acid in the laboratory.
Terangkan mengapa bekas kaca lebih sesuai digunakan untuk menyimpan asid di dalam makmal.
Glass is chemically inert/ glass is non-reactive
(ii)
Soda-lime glass cannot withstand high temperature. State the name of another type of glass that is more heat
resistant.
Kaca soda kapur tidak tahan suhu yang tinggi. Namakan jenis kaca lain yang lebih tahan haba.
Borosilicate glass
Objective Questions / Soalan Objektif
1
Which of the following are the uses of sulphuric acid?
Antara berikut, yang manakah adalah kegunaan asid sulfurik?
I
Detergent
II
Cat
Fertiliser
IV Synthetic fiber
Baja
A
B
2
Gentian sintetik
C
I and II only
I dan II sahaja
D
III and IV only
III dan IV sahaja
I, II dan IV sahaja
I, II, III and IV
I, II, III dan IV
SO2
II
SO3
III
H2S2O7
IV
Fe
2NH3
Which of the following is the function of iron, Fe in the
process?
Antara berikut, yang manakah adalah fungsi besi, Fe dalam proses itu?
Rajah di bawah menunjukkan peringkat I, II, III dan IV dalam Proses
Sentuh.
I
N2 + 3H2
I, II and IV only
The diagram below shows the stages I, II, III and IV in the
Contact Process.
S
The equation below shows chemical equation to produce
ammonia in Haber Process.
Persamaan tindak balas berikut menunjukkan persamaan kimia untuk
menghasilkan ammonia dalam Proses Haber.
III Paint
Detergen
3
H2SO4
Which of the following stages requires the use of a catalyst?
A
B
C
D
To lower the pressure required for the process.
Merendahkan tekanan yang diperlukan untuk proses itu.
To lower the temperature required for the process.
Merendahkan suhu yang diperlukan untuk proses itu.
To increase the rate of production of ammonia.
Untuk meningkatkan kadar pengeluaran ammonia.
To increase the percentage of production of ammonia.
Untuk meningkatkan peratus penghasilan ammonia.
Antara peringkat berikut, yang manakah memerlukan mangkin?
I
II
C
D
III
IV
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MODULE • Chemistry Form 4
4
The diagram below shows the arrangement of atoms in alloy
X.
8
Rajah di bawah menunjukkan susunan atom dalam aloi X.
Which of the following are the characteristics of synthetic
polymers that causes environmental pollution?
Antara berikut, yang manakah adalah ciri-ciri polimer sintetik yang
menyebabkan pencemaran alam sekitar?
I
Copper/Kuprum
Polymers are non biodegradable
Polimer adalah tidak terbiodegradasi
II
Zinc/Zink
Polymers dissolve in water and increase pH of water
Polimer larut dalam air dan meningkatkan pH air
III
IV
Apakah aloi X?
A
B
5
C
Brass
Loyang
D
Bronze
Gangsa
A
Kupronikel
Duralumin
B
Duralumin
C
D
Aloi Y digunakan untuk membuat badan kapal terbang.
Antara berikut, yang manakah adalah aloi Y dan komponen utamanya?
Alloy Y
Major component
Duralumin
Magnesium
Duralumin
Magnesium
Duralumin
Aluminium
Duralumin
Aluminium
Bronze
Copper
Gangsa
Kuprum
C
Cupronickel
Copper
Kupronikel
Kuprum
D
Aloi Y
B
C
D
6
Komponen utama
Which type of glass is suitable for making beakers and test
tubes that can be used for heating?
Kaca yang manakah adalah sesuai untuk membuat bikar dan tabung uji
yang boleh digunakan untuk pemanasan?
A
B
7
Lead glass
Kaca plumbum
Soda-lime glass
Kaca soda kapur
C
D
Disposal of polymers promote excessive growth of algae
Pembuangan polimer meningkatkan pertumbuhan alga berlebihan
Cupronickel
An alloy Y is used to make a body of an aeroplane.
Which of the following is alloy Y and its major component?
A
Burning of polymers release toxic gas
Pembakaran polimer membebaskan gas beracun
What is alloy X?
9
A
B
What is glass X?
Hard and strong
Keras dan kuat
Good insulator electric
Penebat elektrik yang baik
Remain stable under high temperature
Kekal stabil pada suhu tinggi
Chemically inert and non corrosive
Lengai terhadap bahan kimia dan tidak terkakis
Maklumat berikut adalah berkaitan dengan bahan Z yang digunakan
dalam keretapi laju.
Conducts electricity with no resistance at low
temperature.
Photochromic glass
Apabila kaca X dipanaskan dengan kuat dan seterusnya
dimasukkan ke dalam air sejuk, kaca itu tidak pecah.
II, III and IV only
II, III dan IV sahaja
10 The following information is about substance Z which is used
in bullet train.
Mengkonduksi elektrik tanpa rintangan pada suhu rendah.
Kaca fotokromik
When the glass X is heated to a high
temperature and plunged into cold water, the
glass does not crack.
I, III and IV only
I, III dan IV sahaja
Seramik digunakan untuk membuat dinding reaktor nuklear. Antara
berikut, yang manakah adalah ciri seramik untuk penggunaan itu?
Borosilicate glass
Maklumat di bawah menunjukkan sifat kaca X.
II and III only
II dan III sahaja
Ceramic is used to make wall of reactor nuclear. Which of the
following is the characteristic of ceramic for the usage?
Kaca borosilikat
The information below shows the property of a glass X.
I and III only
I dan III sahaja
What is substance Z?
Apakah bahan Z?
A
B
C
D
Fiber glass
Duralumin
Superconductors
Super konduktor
Polyvinylchloride
Polivinil klorida
Fibre glass
Plastik yang diperkukuhkan dengan kaca
Apakah kaca X?
A
m
Soda-lime glass
Kaca soda kapur
C
D
Fused glass
Kaca silika terlakur
Borosilicate glass
Kaca borosilikat
Publica
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a
B
Lead crystal glass
Kaca plumbum
d.
Bh
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