Acid and Base [Ch14]
1. Properties of Acid and Base
Acid
Examples
Hydrochloric acid (HCl)
Sulphuric acid (H2SO4)
Nitric acid (HNO3)
Ethanoic acid (CH3COOH)
Definition
Acid: H-containing compound which ionizes in
water to give H+ as the only positive ion
*Ionization refers to the process by which neutral
molecules split into ions
-
HCI
V
OH
im
Base
Metal oxide: e.g. Al2O3; CuO; Fe2O3
Metal hydroxide: e.g. NaOH; KOH; Fe(OH)2;
Ammonia (NH3)
Base: Compound that reacts with acid to form salt
and water only.
Alkali: soluble base; compound which dissociate/
ionize in water to form OH- as the only negative ion.
Most metal oxides and metal hydroxides are
insoluble, except NaOH, KOH and Ca(OH)2
•
•
NaOH, KOH, Ca(OH)2 and NH3 are soluble in
water. They are thus alkalis.
CuO, ZnO, Fe(OH)2 are insoluble in water. They
are bases, NOT alkalis.
e.g. 1: H2SO4 (l) ⎯water
⎯
⎯→ 2 H+ (aq) + SO4 2- (aq)
e.g. 2: CH3COOH (aq) ⇌ CH3COO- (aq) + H+(aq)
e.g. 1: NaOH (s) ⎯water
⎯
⎯→ Na+ (aq) + OH- (aq)
e.g. 2: NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq)
Role of
water
Water is essential for acid to show acidic
properties. Without water, acid cannot ionize to
gives H+ which is responsible for acidic
properties of aqueous acid.
Water is essential for alkali to show alkaline
properties. Without water, alkali cannot dissociate/
ionize to give OH- which is responsible for
alkaline properties of aqueous alkali.
Physical
properties
of aqueous
solution
Chemical
reaction
Sour in taste
Conducts electricity (mobile ions are formed upon
ionization in the presence of water)
Bitter in taste and slippery
Conducts electricity (mobile ions are formed upon
ionization/ dissociation in the presence of water)
1. acid +metal
→ salt + H2
2. acid +carbonate
→ salt + CO2 + H2O
3. acid +hydrogencarbonate → salt + CO2 + H2O
1. alkali + ammonium compound→ salt+ NH3+ H2O
(*as test for ammonium compound)
2. alkali + salt solution containing metal ion
=> insoluble metal hydroxide* + another salt
(*most metal hydroxides are insoluble in water
except group I hydroxides e.g. NaOH and KOH)
ionic equation for the precipitation of metal ion Mn+
(aq) in alkali:
Mn+ (aq) + n OH- (aq) →M(OH)n (s)
4.
acid + base → salt + water
i.
ii.
iii.
acid + metal oxide
→ salt + H2O
acid + metal hydroxide → salt + H2O
acid + ammonia
→ ammonium salt
Important!
*Salt is the compound formed when the ionizable H atom in an acid molecule is replaced by a metal ion or
other positive ion.
Exam Companion
1
2. Solubility of metal hydroxide in excess NaOH and NH3
Metal ion
Mg2+ (aq)
Al3+ (aq)
Zn2+ (aq)
Pb2+ (aq)
Fe2+ (aq)
Fe3+ (aq)
Cu2+ (aq)
Ag+ (aq)
Ionic equation to show precipitation with
NaOH (aq) / NH3 (aq)
Mg2+ (aq) + 2OH- (aq)
colourless
Al3+ (aq) + 3 OH- (aq)
colourless
Zn2+ (aq) + 2OH- (aq)
colourless
Pb2+ (aq) + 2OH- (aq)
colourless
Fe2+ (aq) + 2OH- (aq)
green
Fe3+ (aq) + 3 OH- (aq)
yellow
Cu2+ (aq) + 2OH- (aq)
blue
2 Ag+ (aq) + 2OH- (aq)
colourless
Colour of
Precipitate
Formed
White
→ Mg(OH)2 (s)
white ppt.
→ Al(OH)3 (s)
white ppt.
→ Zn(OH)2 (s)
white ppt.
→ Pb(OH)2 (s)
white ppt.
→ Fe(OH)2 (s)
dirty green ppt.
→ Fe(OH)3 (s)
reddish brown ppt.
→ Cu(OH)2 (s)
blue ppt.
→ Ag2O (s) + H2O (l)
brown ppt.
* Ag+ (aq) ions do not form AgOH(s) with OH
−
White
White
White
Dirty green
Reddish
brown
Blue
Brown
Action of excess
NaOH(aq)
Action of excess
NH3(aq)
insoluble
insoluble
ppt. redissolves to form
insoluble
colourless solution
ppt. redissolves to form ppt. redissolves to form
colourless solution
colourless solution
ppt. redissolves to form
insoluble
colourless solution
insoluble
insoluble
insoluble
insoluble
ppt. redissolves to form
deep blue solution
ppt. redissolves to form
colourless solution
insoluble
insoluble
(aq) because AgOH does not exist; Ag2O is formed instead.
Addition of NaOH (aq) / NH3 (aq) is often used in qualitative analysis to identify certain metal ions.
Concentration of Solution [Ch15]
Concentration of solution =
amount of solute
Note:
volume of solution
1 dm = 10 cm
(1 dm)3 = (10 cm)3
1 dm3 = 1000 cm3
⚫
The concentration of a solution is usually expressed in: (a) g cm-3 (b) g dm-3 (c) mol dm-3 (or M)
⚫
Molarity of a solution refers to its concentration expressed as number of moles of solute per dm3 of the
solution. The unit of molarity is mol dm-3 or M.
Molarity of a solution (mol dm-3 or M ) =
no. of moles of solute (mol)
volume of solution (dm3)
By rearrangement,
No. of moles
of solute (mol)
=
=
Molarity of
solution (mol dm-3)

Volume of
solution (dm3)
M  V
Example 1
15.0 g of NaOH is dissolved in 250 cm3 of water.
Calculate the molarity of the NaOH solution.
[relative atomic masses: H – 1.0; O – 16.0; Na – 23.0]
Example 2
Find the mass of sodium carbonate (in g) in 50 cm3 of
3.5 M Na2CO3 (aq).
[relative atomic masses: C – 12.0; O – 16.0; Na – 23.0]
Strategy: Express concentration in mol dm-3
Concentration
15 g
= 250 cm3
=
15
mol
23+16+1
250
dm3
1000
= 1.5 mol dm-3
Exam Companion
Carry out unit
conversion, 1 at a time
No. of moles of Na2CO3
50
=(1000) (3.5)
= 0.175
Mass of Na2CO3
= 0.175 x (23x2+12+16x3)
= 18.55 g
No. of moles of solute = MV
Volume must be in dm3
2
Example 3
A concentrated sulphuric acid solution contains 98% by mass of sulphuric acid. Density of this concentrated
sulphuric acid is 1.83 g cm-3.Calculate the molarity of the sulphuric acid solution.
[relative atomic masses: H – 1.0; O – 16.0; S – 32.1]
Strategy:
Mass of 1 dm3 of solution
1.83 g
=
x 1000 dm3
1 cm3
= 1830 g
A solution is made of
solute (H2SO4) and solvent (water).
The solution contains 98% by mass of
H2SO4 means 100 g of the solution has
98 g of H2SO4 in it.
mass of 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧
Density of solution =
volume of solution
Density = 1.83 g cm-3 means 1 cm3 of solution has a
mass of 1.83 g (which is made up of water and
H2SO4)
Mass of H2SO4 in 1 dm3 of solution
= 1830 x 98%
= 1793.4 g
Molarity of H2SO4 (aq)
1793.4
=2+32.1+16 (4)
= 18.2 M
Molarity of solution
no.of moles of 𝐬𝐨𝐥𝐮𝐭𝐞
= 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (𝑖𝑛 𝑑𝑚3 )
Dilution
No. of moles of solute before dilution = No. of moles of solute after dilution
M1V1 (before dilution) = M2V2 (after dilution)
Example 1
Example 2
Calculate the molarity of the diluted acid solution when
600 cm3 of water is added to 400 cm3 of 8M of hydrochloric
acid solution.
M1V1 = M2V2
400
600+400
8 (1000) = M2 ( 1000 )
V2 refers to the total volume after
M2 = 3.2 M
mixing.
Find the volume of water added to 50.0 cm3 of 1.6 M
nitric acid, in order to prepare 0.8M nitric acid.
M1V1 = M2V2
50
50+V
1.6 (
) = 0.8 (
)
1000
1000
3
V = 50 cm
V refers to volume of water
added
Example 3
A mixture is prepared by mixing 50.0 cm3 0.1 M Fe2(SO4 )3 (aq) and 50.0 cm3 0.2 M FeSO4 (aq). Calculate the
molarity of the SO42- in the resulting mixture.
Molarity of the SO42=
=
no.of moles of SO42− (mol)
volume of solution (dm3 )
(0.05x 0.1 x 3)+(0.05x0.2)
(0.05+0.05)
1 mole of Fe2(SO4)3 contains 3
moles of SO42-
1 mole of FeSO4 contains 1
mole of SO42-
= 0.25 M
➢
➢
➢
➢
Hazard of acids and alkalis : corrosive
Safety precautions: wear safety goggles, rubber gloves and laboratory gown
A large amount of heat is released when strong acids, like H2SO4 and HCl, are mixed with water.
In the dilution of acids, the small amount of concentrated acid should be added into large amount of water
with constant stirring.
Exam Companion
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pH and indicators [Ch 16]
Acidic sol.
Alkaline sol
Neutral sol
conc. of H + >
conc. of H + <
conc. of H + =
Methyl orange
Phenolphthalein
pH = - log [H+ (aq)]
[H+ (aq)] = 10 − pH
conc. of OH −
conc. of OH −
conc. of OH −
In acidic medium
Red
Colourless
pH at 25oC
<7
>7
=7
In alkaline medium
Yellow
Pink
where [H+(aq)] is concentration of H+ (aq) in mol dm-3 (or M)
minus pH
Example 1
Calculate the pH of 0.3 M HCl (aq)
Solution:
pH = - log [H+ (aq)]
= - log (0.3)
= 0.52
Example 2
Calculate [H (aq)] of a solution of pH = 4.3
Solution
[H+ (aq)] = 10− pH = 10- 4.3 = 5.01 x 10-5 M
+
Strength of acids/ alkalis [Ch 17]
➢ Acids and alkalis can be classified as strong or weak based on their extent of ionization.
➢ Acids and alkalis which completely ionize are strong.
➢ Acids and alkalis which partially/ slightly ionize are weak.
Strong acid
HCl, HNO3, H2SO4
Weak acid
CH3COOH and other carboxylic acid
Strong alkali
NaOH, KOH, Ca(OH)2
Weak alkali
NH3
Ionization in water:
CH3COOH (aq) ⇌ CH3COO- (aq) + H+ (aq)
⇌ means the reaction is incomplete
Ionization in water:
NH3 (aq) + H2O (l) ⇌ NH3 (aq) + H2O (l)
pH of solution of acid depends on the following factors:
➢ basicity of acid (maximum no. of H+ ions formed from 1 acid molecule upon ionization)
➢ concentration of acid
➢ strength of acid
To compare the strength of acids, we can compare the following provided that the acids used are of the same
basicity and the solutions used are of the same molarity.
➢ Electrical conductivity [which depends on concentration of mobile ions]
➢ pH [which depends on the concentration of H+(aq)]
➢ Rate of reaction with metal [depends on concentration of H+(aq)]
Strength of alkalis can also be studied similarly by comparing electrical conductivity and pH.
Exam Companion
4
Neutralization [Ch 18]
➢ Acid react with base to form salt and water ONLY. This reaction is known as neutralization.
➢ Neutralization is the combination of hydrogen ion and hydroxide ion (or oxide ion) to form water
molecules.
➢ Neutralization reaction is exothermic.
➢ Applications of neutralization:
e.g. treatment of soil/ industrial waste/ indigestion; cleaning action of toothpaste
reluhenf
Solubilities of Some Common Compounds in Water
Soluble
Insoluble
1. All hydrogencarbonates*
2. All sodium salts
3. All potassium salts
4. All ammonium salts
SPANISH
5. All nitrates
6. All sulphates, except
SH
7. All halides (e.g.Cl, Br, I), except
8. Sodium carbonate
Potassium carbonate
Ammonium carbonate
9. Sodium hydroxide
Potassium hydroxide
Calcium hydroxide (slightly soluble)
10. Sodium oxide
Potassium oxide
Calcium oxide
AgX, BaSO4, CaSO4, PbSO4, PbX2
ABCPP
XSSSX
All other common
•
carbonates
•
hydroxides
•
oxides
Alkalis: soluble bases
SH
[Note: when dissolved in water, they react with water
immediately to form soluble metal hydroxide]
*For hydrogencarbonates, only those of group 1 exist in solid form, the others are too unstable and can only exist
in solution forms
Common insoluble compounds
1. ABCPP XSSSX
2. CHOO (carbonates, hydroxides and oxides) except SPA (sodium,
potassium and ammonium) [Note: CaO and Ca(OH)2 are soluble]
[Note:inCaO
and
Ca(OH)
soluble]
2 arein
Precipitation reaction: reaction
which
two
substances
aqueous form react to form an insoluble product.
precipitation of insoluble ionic compounds: solution containing the cation + solution containing the anion
e.g. Fe(NO3)2 (aq) + 2NaOH (aq) → Fe(OH)2 (s) + NaNO3 (aq)
e.g. KCl (aq) + AgNO3 (aq) → AgCl (s) + KNO3 (aq)
e.g. HCl (aq) + AgNO3 (aq) → AgCl (s) + HNO3 (aq)
Exam Companion
5
8.
The following shows a flow chart that can help us to decide how to prepare a particular salt.
Their corresponding
metals, bases and
carbonates are all
soluble.
For salts except those of
Na+, K+, NH4+
For salts of Na+, K+, NH4+
Acid +
 excess metal/
 excess insoluble/
base/
 excess insoluble
carbonate
Acid +
Mixing two solutions
 alkali/
(one with the cation;
 carbonate
the other with the
anion) to get a
(via Titration)
precipitate
i.e. precipitation*
(Remove insoluble reactant)
(Remove soluble impurities)
IMPORTANT!
[Reminders: students are expected to suggest reagents needed to prepare a salt and describe how to purify the
salt prepared from the reaction mixture.]
Exam Companion
6
Volumetric Analysis [Ch19]
Definition: Standard solution refers to solution of accurately known concentration
important
Standard solution can be prepared by
(i) dissolving a known amount of solute in known volume of water which usually involves the following
stages:
1. weigh the required mass of solute accurately using electronic balance
2. dissolve the solute in water in beaker
3. transfer the solution into a volumetric flask with the help of filter funnel and glass rod
4. after rinsing, transfer all washings into the volumetric flask
5. fill the volumetric flask up to the graduation mark with distilled water
6. stopper the flask, invert it and swirl
(ii) diluting a concentrated solution of known molarity which usually involves the following stages:
1. transfer the required volume of the solution using pipette (or burette) into a volumetric flask
2. make up the solution to the graduation mark by adding distilled water
3. stopper the flask, invert it and swirl
graduation mark
pipette
Stock solution of
known concentration
[Reminders: students are expected to state what should be used to rinse the apparatus with]
Exam Companion
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Titration
aA+bB ⎯
⎯→ c C + d D
no. of
particles of
A
Mass of A
molar mass
no. of
particles of
B
Avogadro's no.
no. of moles
of A
molarity
Mass of B
Avogadro's no.
molar mass
no. of moles
of B
mole ratio
molarity
no. of particles
no. of moles =
Avogadro' s number
Volume of
Solution A
Volume of
Solution B
mass (g)
no.of moles =
molar mass (g mol -1 )
molarity (M) =
no. of moles of solute (mol)
volume of solution (dm3 )
Strategy:
1. Write a balanced equation.
2. Write down the given data underneath the corresponding reactants/products.
3. Identify the ‘given’ and the ‘target’
4. Work out the steps to be taken to calculate the unknown from known quantities.
Usual steps involved:
(i) find no. of moles of the “given“ (e.g. A)
(ii) use the mole ratio to find the no. of moles of the “target” (e.g. B)
(iii) calculate what the question is asking for.
This method works
in most titration
problems.
!!! Be particularly careful about questions
which involve dilution/ extraction of
part of the solution for titration.
Example 1
50.0 cm3 of a sample of sulphuric acid were diluted to 250.0 cm3.
25.0 cm3 of the diluted sulphuric acid required 50.0 cm3 0.160 M sodium hydroxide solution for complete
neutralization.
Calculate the molarity of sulphuric acid in the original sample.
0.160 M NaOH (aq)
Stage 1
Stage 2
dilution
Extract part of
the solution
( 1/10) for
titration
50.00 cm3
H2SO4
250.00 cm3
diluted H2SO4
no. of moles of solute the same;
concentration becomes 1/10 of the original
Stage 1:
H2SO4 (aq) → H2SO4 (aq)
50.00 cm3
250.00 cm3
?M
diluted
Solute particle
25.00 cm3
diluted H2SO4
no. of moles of solute becomes 1/10 of the original;
same concentration;
Stage
2: part of the solution ( 1/ 10) for titration
Extract
H2SO4 (aq) + 2 NaOH (aq) →Na2SO4(aq)+ 2 H2O (l)
25.00 cm3
19.50 cm3
diluted
0.160 M
No. of moles of NaOH = 0.16 x 0.0195 = 3.12 x 10-3
No. of moles of H2SO4 in 25.0 cm3 of the diluted acid = 3.12 x 10-3 x
1
2
= 1.56 x 10-3
No. of moles of H2SO4 in 250.0 cm3 of the diluted acid = 1.56 x 10-3 x 10 = 0.0156
Molarity of H2SO4 in the original acid = 0.015725 / 0.05 = 0.312 M
Exam Companion
8
Example 2
2.98 g of an impure sample of ethanedioic acid, of chemical formula H2C2O4• 2H2O, are dissolved in water. The
solution required 20.0 cm3 of 2.00 M potassium hydroxide solution for complete neutralization. Assuming that the
impurity in the sample does not react with potassium hydroxide, calculate the percentage purity of
H2C2O4• 2H2O in the sample. (Relative atomic mass: C=12.0, H= 1.0, O=16.0)
H2C2O4• 2H2O (s) → H2C2O4 (aq)
2.98 g
H2C2O4 + 2 KOH → K2C2O4 + 2 H2O
20.0 cm3
2.00 M
% purity of H2C2O4• 2H2O
=
mass of H2 C2 O4 • 2H2 O
mass of the whole sample
x 100%
No. of mole of KOH = 0.05 (2) = 0.04
No. of moles of H2C2O4 = 0.04 / 2 = 0.02
No. of mole of H2C2O4• 2H2O = 0.02 (since 1 mole of H2C2O4• 2H2O forms 1 mole of H2C2O4)
Mass of H2C2O4• 2H2O = 0.02 x [2+12 x 2 + 16 x4 + 2 x (2+16)] = 2.52 g
% purity of H2C2O4• 2H2O =(2.52 / 2.98) x 100% = 84.6 %
Example 3
X is an acid. 25.0 cm3 of 0.20 M solution X requires 30.0 cm3 of 0.50 M sodium hydroxide solution for complete
neutralisation. Determine the basicity of acid.
Let the basicity of acid be n.
acid X + n NaOH => salt + water
Strategy: find the mole ratio of acid
and alkali
no. of moles of acid X = 0.025 x 0.20 = 0.005
no. of moles of NaOH = 0.03 x 0.50 = 0.015
Acid X and NaOH react in the ratio of 1: n
1
0.005
= 0.015
n
n=3
Back Titraiton
The technique of back titration is used when the mass of a solid sample to be analysed is insoluble in water.
Analogy:
Consider the following scenario and find out how much a lollipop cost. Ans: ____________________ $40
Mary had cash $100
She bought 2 lollipops of
unknown price
She then used all the remaining cash
to buy two ice-cream cones, which
cost $10 each.
In shop 1
In shop 2
Exam Companion
9
Back titration usually involves the following stages:
Acid A remained is then
titrated with B (known amount)
acid A +
solid X
(known excess amount) (unknown amount)
(Reaction 1)
(Reaction 2 – Titration)
To be determined
known excess amount of A
Reaction 1
Amount of A reacted with
X (of unknown amount) in
Reaction 1
Reaction 2
Amount of A reacted with
B (of known amount) in
Reaction 2
Strategy:
1. Find no. of moles of unreacted A after reaction 1. (based on titration results).
2. Find no. of moles of A reacted with X. (original amount of A - remaining amount of A)
3. Find no. of moles of X. (based on reaction 1)
[Example 1] – determine the number of moles of CaCO3 in a limestone sample
➢ A limestone sample was fully dissolved in 0.04 moles of HCl. [Reaction 1]
➢ After the reaction, the acid solution remained was then titrated with 1.0 M NaOH. It was found that
10.0 cm3 NaOH was required to reach end point. [Reaction 2]
➢ Determine the number of moles of CaCO3 in the limestone sample.
Reaction 1
Reaction 2
10.0 cm3
1.0 M NaOH
0.04 moles of
HCl
(in excess)
Unreacted HCl
limestone sample
0.04 moles of HCl
Reaction 1
Reacted with
limestone
Reaction 2
Reacted with
NaOH
Strategy:
1. Write balanced equations for the reactions involved.
Reaction 1:
CaCO3 + 2HCl → CaCl2 + CO2 + H2O
Reaction 2: (Titration)
HCl + NaOH → NaCl + H2O
2. Calculate
(i)
no. of moles of HCl originally present before reaction 1
= 0.04
(ii) no. of moles of HCl remained after reaction 1 based on the titration result
= (1) (10/1000) = 0.01
(iii) no. of moles of HCl used up during reaction 1
= 0.04 – 0.01 = 0.03
(iv) no. of moles of CaCO3 present in the solid by considering coefficient ratio of reaction 1
= 0.03 / 2 = 0.015
Exam Companion
10
[Example 2] – determine the mass of insoluble CaCO3 by back titration
50.0 cm3 of 2.0 M HCl(aq) was added to 5.50 g of an impure sample of CaCO3. After the reaction, the
remaining acid in the reaction mixture was then titrated with 1.0 M NaOH(aq). It was found that 22.0 cm3
of NaOH(aq) was required to reach the end point. Determine the mass of CaCO3 in the sample.
[Molar mass of CaCO3 = 100.1 g]
back-titrate with standard NaOH(aq)
excess original no. of moles of HCl
22.0 cm3
1.0M NaOH (aq)
NaOH
remaining HCl
HCl reacted
C
remaining
HCl
determine
C
amount of CaCO3
Strategy:
1. Write down 2 balanced equations for
the reactions involved.
2. Find no. of moles
1M, 22 cm3
of remaining HCl
(back titration,
Reaction 2)
NaOH
Reaction 1:
Reaction 2:
CaCO3 + 2HCl → CaCl2 + CO2 + H2O
HCl + NaOH → NaCl + H2O
(titration)
remaining
Refer to reaction 2, in titration
no. of moles of remaining HCl
= no. of moles of NaOH
= 1 x 22 x 10-3
remaining
HCl
C
3. Find no. of moles of HCl reacted
with CaCO3 (by subtracting no. of
moles of remaining HCl from the
No. of moles of HCl reacted with CaCO3
= 2 x 50 x 10-3 (original) - 1 x 22 x 10-3 (remaining)
= 0.078
original no. of moles of HCl)
4. Find no. of moles of CaCO3 (which
can be determined from no. of moles
Refer to reaction 1,
No. of moles of CaCO3 = ½ x 0.078
of HCl reacted, Reaction 1)
5. Find the required quantity of CaCO3
Mass of CaCO3 = ½ x 0.078 x 100.1 = 3.904 g
(e.g. mass)
Exam Companion
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