Chemistry for the IB Diploma – Answers
Answers to self-assessment questions
THEME S
S1.1 Introduction to the particulate nature of matter
Page 11
1
a element
b compound
c compound
d element
e mixture of compounds
f mixture of elements with a compound
Page 13
2
a simple distillation, fractional distillation (preferable)
b filtration
c separating funnel
d chromatography
e simple distillation
f crystallisation or evaporation
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S1.2 The nuclear atom
Page 24
3
Charge on one mole of electrons = 1.602189 × 10−19 C × 6.02 × 1023 mol−1 = 9.65 × 104 C
4
a i
ii
Volume of atom = 4 × π × (10−10 m)3 / 3 = 4.2 × 10−30 m3
Volume of nucleus = 4 × π × (10−15m)3 / 3 = 4.2 × 10−45 m3
b Percentage of atom that is nucleus = (4.2 × 10−45 m3 / 4.2 × 10−30 m3) × 100 = 10−13 %
5
A hydrogen atom consists of one proton and one electron. The mass of the hydrogen atom is the sum of
the proton and electron masses (ignoring the mass defect).
mass H / mass e− = (1.672622 × 10−27 + 9.109383 × 10−31) kg / 9.109383 × 10−31 kg = 1837
6
Mass of carbon atom
= (6 × 1.672622 × 10−27 kg) + (6 × 1.674927 × 10−27 kg) + (6 × 9.109383 × 10−31 kg)
= 2.01 × 10−26 kg
7
volume = 4/3 × π × (8.41 × 10−16 m)3 = 2.49 × 10−45 m3
density = mass / volume = (1.67 × 10−27 kg / 2.49 × 10−45 m3) = 6.71 × 1017 kg m−3
Page 25
8
a
b
Protons and neutrons are present in the nucleus; electrons are found around the nucleus (in
orbit/shells/energy levels)
9
a repulsion
Particle
proton
neutron
electron
Mass relative to mass of proton
1
1
1/1836
Charge relative to charge of proton
+1
0
−1
b no force
c attraction
10 6 protons, 6 neutrons, 6 electrons and 12 nucleons
Page 31
11
Element and
nuclide symbol
Atomic Mass
Number of
number number protons
Number of
neutrons
Number of
electrons
16
8
O
8
16
8
8
8
56
28
Ni
28
56
28
28
28
41
20
Ca
20
41
20
21
20
23
11
Na
11
23
11
12
11
57
26
Fe
26
57
26
31
26
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12 a The subscript in the nuclide notation is the atomic number. Hence an atom of phosphorus contains
15 protons and, therefore, 15 electrons. However, since the ion has a net charge of −3, the ion contains
15 + 3 = 18 electrons. The difference between the atomic number and the superscript mass number is
equal to the number of neutrons: in this example, 31 − 15 = 16 neutrons.
b Since the atomic number of magnesium is 12, each atom contains 12 protons. However, since the
ion has a net charge of +2, the ion contains 12 − 2 = 10 electrons. The difference between the atomic
number and mass number is 24 − 12 = 12 neutrons.
Page 32
13 a 1p, 2n, 2e
b 1p, 2n, 0e
c 1p, 1n, 2e
d 1p, 1n, 0e
Page 34
72.1
×
100
14 Relative atomic mass of rubidium = �
15 Let %Ga-69 = x
27.9
×
100
85� + �
87� = 85.6
%Ga-71 = (100 – x), since the two isotopic percentages must sum to one hundred.
69𝑥𝑥 + 71(100 − 𝑥𝑥)
100
6970 = 69x + 71(100 – x)
69.7 =
6970 = 69x – 71x + 7100
6970 = −2x + 7100
130 = 2x
x = 65
Hence, the percentage abundance of gallium-69 is 65%.
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S1.3 Electron configurations
Page 41
16 a 5.09 × 1014 s−1 or Hz
b 3.38 × 10−19 J
c 203 kJ mol−1
Page 42
17 a radio waves
b visible
c ultraviolet
18 Ultraviolet radiation has higher frequency, smaller wavelength, greater energy and higher wavenumber.
Infrared radiation has lower frequency, greater wavelength, lower energy and lower wavenumber.
Page 44
19 c = f × λ ; 3 × 108 m s−1 = f × 600 × 10−9 m
f = 5.0 × 1014 Hz (s−1)
E = h f = 6.63 × 10−34 J s × 5.0 × 1014 Hz (s−1) = 3.315 × 10−19 J
Page 45
20 frequency, energy or wavelength
21 frequency – decreasing, energy – decreasing, wavelength – increasing, left to right
22 When sufficient energy (thermal or electrical) is supplied, electrons can be promoted (excited) to higher
energy levels in an atom. The electrons are unstable in higher levels and rapidly emit radiation as they
fall back into lower energy levels. As the energy levels are fixed, the energy lost between any higher
level and a lower level is also of a certain fixed value, so the radiation emitted will only have certain
fixed frequencies (i.e. specific colours). This means that the atomic spectrum of an element will consist
of a series of lines of different colours (on a black background). Because the energy levels of a
hydrogen atom converge, the spectral lines within a series also converge.
23 second main energy level (second shell) i.e. n = 2
24 Each element has its own characteristic line spectrum, so an element can be identified by its line
spectrum just as a criminal can be identified from a fingerprint.
Page 47
25 c = f × λ ; 3.00 × 108 m s−1 = f × 5.04 × 10−7 m
f = 5.95 × 1014 Hz
E = hf
E = 6.63 × 10−34 J s × 5.95 × 1014 s−1 = 3.95 × 10−19 J
E = 3.95 × 10−19 J × 6.02 × 1023 mol−1 = 237575 J mol−1 = 238 kJ mol−1
26 The convergence limit in the UV spectrum is due to transition from n = ∞ to n = 1 (the level occupied
by the electron in the ground state); the convergence limit in the visible spectrum is from n = ∞ to n = 2.
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Page 51
27 50 electrons
1 s-orbital, 3 p-orbitals and 5 d-orbitals
Page 54
28
Page 56
29 Ti3+
[Ar] 3d1
1s2 2s2 2p6 3s2 3p6 3d1
Cr2+
[Ar] 3d6
1s2 2s2 2p6 3s2 3p6 3d4
Cu
[Ar] 3d10 4s1
1s2 2s2 2p6 3s2 3p6 4s1 3d10
P3−
[Ne] 3s2 3p6
1s2 2s2 2p6 3s2 3p6
Cl
[Ne] 3s2 3p5
1s2 2s2 2p6 3s2 3p5
Ga
[Ar] 4s2 3d10 4p1
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p1
Mg
[Ne] 3s2
1s2 2s2 2p6 3s2
As
[Ar] 4s2 3d10 4p3
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3
Sr
[Kr] 5s2
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2
K+
[Ar]
1s2 2s2 2p6 3s2 3p6
Page 59
30 H+ has 1 proton, but the helium ion, He+ has 2 protons. The greater effective nuclear charge increases
the electrostatic force experienced by the electron in the helium ion (He+), hence it will have the greater
ionization energy. The increase in shielding (electron–electron repulsion) is less than the increase in
nuclear charge.
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S1.4 Counting particles by mass: the mole
Page 67
31 number of water molecules = 0.0100 mol × 6.02 × 1023 mol−1 = 6.02 × 1021
32 a amount of nitric(V) acid =
=
9.0 × 1023
6.02 × 1023 mol−1
= 1.5 mol
b Each molecule of nitric(V) acid contains three oxygen atoms.
Hence, 9.0 × 1023 molecules of nitric(V) acid contain 3 × 9 × 1023 = 2.7 × 1024 atoms of oxygen.
Page 72
33 a H2SO4 + 2NaOH " Na2SO4 + 2H2O
b 2Fe + 3Cl2 " 2FeCl3
c CuSO4 + 2KOH " Cu(OH)2 + K2SO4
d CuO + 2HNO3 " Cu(NO3)2 + H2O
e CuCO3 + 2HCl " CuCl2 + H2O + CO2
f Pb(NO3)2 + 2KI " PbI2 + 2KNO3
g 2Al2O3 " 4Al + 3O2
h C3H8 + 5O2 " 3CO2 + 4H2O
i Al(NO3)3 + 3NaOH " Al(OH)3 + 3NaNO3
j 3CaCl2 + 2Na3PO4 " Ca3(PO4)2 + 6NaCl
k 4Fe + 3O2 " 2Fe2O3
l Ca(OH)2 + 2HCl " CaCl2 + 2H2O
m 4HCl + MnO2 → MnCl2 + Cl2 + 2H2O
Page 76
34 amount of water molecules =
35 amount of calcium =
36 amount of water =
54 g
18.02 g mol−1
500 g
40.08 g mol−1
0.18 g
18.02 g mol−1
= 3.00 mol
= 12.48 mol
= 0.010 mol
37 mass of calcium carbonate (g) = 0.4 mol × 100.09 g mol−1 = 40.04 g
38 molar mass (g mol−1) =
1.00 g
0.00200 mol−1
= 500 g mol−1
39 one mole of propane has a mass of 44.11 g and contains 6.02 × 1023 molecules of propane.
amount of propane = 22.055 g / 44.11 g mol−1 = 0.5 mol
22.055 g of propane contains 6.02 × 1023 × 0.50 = 3.01 × 1023 molecules of propane
each molecule of propane contains 11 atoms (three carbon atoms and eight hydrogen atoms) so:
the total number of atoms is 11 × 3.01 × 1023 = 3.31 × 1024
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the total number of carbon atoms is 3 × 3.01 × 1023 = 9.03 × 1023
the total number of hydrogen atoms is 8 × 3.01 × 1023 = 2.41 × 1024
Page 78
40 amount of SO2 = (10.00 g/64.07 g mol−1) = 0.156 mol
amount of CaCO3 = 0.156 mol
mass of CaCO3 = 0.156 mol × 100.09 g mol−1 = 15.61 g
Page 83
41 mass of water driven off = 12.3 g − 6.0 g = 6.3 g
Species
MgSO4
H2O
6.0 g
6.3 g
6.0 g
120.37 g mol−1
6.3 g
18.02 g mol−1
combining masses
amount of atoms
ratio of moles
dividing through by the smallest number
0.050
:
1
:
0.350
7
The empirical formula is therefore MgSO4.7H2O.
Values of x less than 7 could be due to incomplete dehydration and values of x greater than 7 could be
due to thermal decomposition of the salt:
MgSO4(s) → MgO(s) + SO3(g).
Page 87
42 amount of potassium hydroxide =
amount of phosphoric acid =
17.50
dm3
1000
20.00
dm3
1000
× 0.150 mol dm−3 = 2.63 × 10−3 mol
× 0.0656 mol dm−3 = 1.31 × 10−3 mol
The two chemicals react in a 2 to 1 molar ratio and hence the equation is:
H3PO4(aq) + 2KOH → K2HPO4(aq) + 2H2O(l)
Page 89
43 a 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l)
b amount of sodium hydroxide used in the titration =
dm3 × 1.00 mol dm−3 = 0.0250 mol
c from the equation, amount of H2SO4 = amount of NaOH ÷ 2 = 0.0125 mol in 20.00 cm3
scaling up to 1000 cm3:
× 0.0125 mol dm–3 = 0.625 mol dm–3
d scaling up from 50.0 to 1000 cm3 gives the concentration of the original H2SO4 solution:
0.625 mol dm –3 ×
= 12.5 mol dm–3
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Page 91
44 MgO(s) + 2HCl(aq) → MgCl2(aq) + H2O(l)
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
amount of NaOH = (19.70/1000) dm3 × 0.200 mol dm–3 = 3.94 × 10–3 mol
amount of HCl titrated = 3.94 × 10–3 mol
initial amount of HCl = (100.0/1000) dm3 × 2.00 mol dm–3 = 0.200 mol
amount of HCl that reacted with MgO = (0.200 mol − 3.94 × 10–3 mol) = 0.196 mol
amount of magnesium oxide that reacted =
= 0.0980 mol (1 : 2 molar ratio in equation)
molar mass MgO = 40.3 g mol −1
hence the mass MgO reacting with acid = 0.098 mol × 40.3 g mol−1 = 3.95 g
percentage purity of magnesium oxide =
3.95 g
4.08 g
× 100 = 96.8%
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S1.5 Ideal gases
Page 102
45 volume occupied = 1.35 mol × 22.7 dm3 mol−1 = 30.6 dm3
46 amount of gas =
175
� dm3
1000
22.7 dm3 mol−1
�
47 amount of carbon dioxide =
= 7.71 × 10−3 mol
= 0.126 mol
number of molecules = (0.126 × 6.02 × 1023 mol−1) = 7.56 × 1022
48 density =
49 amount of gas =
= 0.328 mol
= 63.5 g mol−1
molar mass =
50 amount of KClO3 =
0.0250 mol
amount of O2 =
= 0.037 mol
The simplest molar ratio is 2 : 3 and hence the balanced equation must be:
2KClO3(s) → 2KCl(s) + 3O2(g)
Page 105
51 Charles’s law: V1 / T1 = V2 / T2
V2 = T2 × V1 / T1
350 K × 4.50 dm3 / 300 K = 5.25 dm3
52 Boyle’s law: P1V1 = P2V2
103 kPa × 350 cm3 = 150 kPa × V2
V2 = 240 cm3
53 Gay-Lussac’s law: P1 / T1 = P2 / T2
97 000 Pa
298.15 K
=
101 325 Pa
𝑇𝑇2
T2 = 101 325 Pa ×
298.15 K
97 000 Pa
T2 = 311.4 K = (311.4 − 273) = 38.4 °C
54
𝑃𝑃2
𝑃𝑃1
𝑇𝑇
= 𝑇𝑇2 =
1
(2727+273.15) K
(27+273.15) K
P2 : P1 = 10 : 1
= 10
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Page 106
55
1.05×105 Pa × 60.0 cm3
333.15 K
56 PV = nRT
𝑛𝑛 =
𝑃𝑃𝑃𝑃
𝑅𝑅𝑅𝑅
=
=
1.00×105 Pa × 𝑉𝑉2
273.15 K
1.05 × 105 Pa × 60.0 cm3 × 273.15 K
𝑉𝑉2 =
= 51.7 cm3
1.00 × 105 Pa × 333.15 K
101325 Pa × 4.00×10−3 m3
8.31 J K–1 mol–1 × 273 K
relative formula mass =
Page 107
57 𝑀𝑀𝑟𝑟 = 𝜌𝜌
Page 108
𝑅𝑅𝑅𝑅
𝑃𝑃
= 0.1787 mol
12.64 g
0.1787 mol
= 2615 g m−3 ×
= 70.7 g mol−1
8.31 J K –1 mol–1 × 298.15 K
101325 Pa
= 63.9
58 Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
amount of H2 gas produced = 0.0147 g / 24.31 g mol−1 = 6.05 × 10−4 mol
temperature of H2 = (24.3 °C + 273.15) = 297.5 K
𝑃𝑃𝑃𝑃 101305 Pa × 13.06 × 10−6 m3
=
= 7.35 J K −1 mol−1
𝑅𝑅 =
6.05 × 10–4 mol × 297.5 K
𝑛𝑛𝑛𝑛
percentage error = 100 × (measured value – accepted value) / accepted value = 11.6%
Page 111
59
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S2.1 The ionic model
Page 115
1
a Ca
1s2 2s2 2p6 3s2 3p6 4s2
Ca2+
1s2 2s2 2p6 3s2 3p6
F
1s2 2s2 2p5
F−
1s2 2s2 2p6
Each calcium atom loses two valence electrons; each fluorine atom gains one valence electron.
b Na
1s2 2s2 2p6 3s1
Na+
1s2 2s2 2p6
O
1s2 2s2 2p4
O2−
1s2 2s2 2p6
Each sodium atom loses one valence electron; each oxygen atom gains two valence electrons.
c Al
1s2 2s2 2p6 3s2 3p1
Al3+
1s2 2s2 2p6
O
1s2 2s2 2p4
O2−
1s2 2s2 2p6
Each aluminium atom loses three valence electrons; each oxygen atom gains two valence electrons.
Page 119
2
a Fe3(PO4)2
b NH4I
c Al(NO3)3
d CaBr2
e Fe2O3
f KCl
g Na2CO3
h Fe(OH)3
i K2SO3
j MgS
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Page 120
3
Group
Element
Electron
arrangement
Electron
configuration
Number of
electrons in outer
shell
Common simple
ion
Electron
arrangement of
ion
Electron
configuration of
ion
1
Li
2
Be
13
B
14
C
15
N
16
O
17
F
18
Ne
2,1
2,2
2,3
2,4
2,5
2,6
2,7
2,8
1s2 2s1
1s2 2s2
1s2 2s2
2p1
1s2 2s2
2p2
1s2 2s2
2p3
1s2 2s2
2p4
1s2 2s2
2p5
1s2 2s2
2p6
1
2
3
4
5
6
7
8
Li+
Be2+
none
none
N3−
O2−
F−
none
2
2
2,8
2,8
2,8
1s2
1s2
1s2 2s2
2p6
1s2 2s2
2p6
1s2 2s2
2p6
Page 123
4
a chlorine–chlorine
0
covalent (non-polar)
b phosphorous–hydrogen
0
covalent (non-polar)
c carbon–chlorine
0.6
covalent (polar)
d beryllium–fluorine
2.4
ionic
e carbon–hydrogen
0.4
covalent (non-polar)
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S2.2 The covalent model
Page 138
5
6
7
Page 139
8
a There is an inverse relationship: as bond length increases, bond enthalpy decreases.
b The atoms become larger, the electron density in the valence orbital is lower, and the overlap
between larger atomic orbitals is less extensive leading to longer and weaker bonds. In larger atoms
the shared electrons forming the bond are further away from the positive charge of the nucleus.
c A value near 493 kJ mol−1. The two previous differences are I → Br = +68 and Br → Cl = +65.
Continuing the pattern, Cl → F would be expected to be +62, giving a value of 493.
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Page 143
9
PH4+ has a tetrahedral shape with angles about 109.5°.
PH3 has a tetrahedral electron pair geometry with a pyramidal shape, due to the presence of the nonbonding lone pair. The electrons in a non-bonding lone pair experience more repulsion, pushing
together the three hydrogens. The bond angle would be expected to be about 107° but as you go further
out, the hybrid orbitals tend to have more p-character and less s-character and the bond angle is actually
about 94°.
PH2− has a V shape due to an additional non-bonding lone pair. This results in a greater repulsion and a
smaller bond (about 105°).
Page 145
10 a The valence shell of the sulfur atom contains ten electrons: six from the
sulfur and one each from the four fluorine atoms. There are four bonding
pairs and one lone pair. The basic shape adopted by the electron pairs in
the molecule is trigonal bipyramidal. In this arrangement, the electron
pairs at the equatorial positions experience less repulsion compared to
axial electron pairs. Hence the lone pair occupies an equatorial position and
thus the shape of the molecule itself resembles a see-saw (sometimes also
described as ‘saw horse’). As a general rule, for a molecule where the electron
domains adopt a trigonal bipyramid structure, any lone pairs will occupy
equatorial positions.
b The valence shell of the chlorine atom contains ten electrons: seven
from the chlorine and one each from the three fluorine atoms. There are
three bonding pairs and two lone pairs. The basic shape adopted by the
electron pairs in the molecule is a trigonal bipyramid. To minimize the
repulsion between bonding pairs and lone pairs of electrons, the two lone
pairs of electrons occupy the equatorial positions. Hence, the molecule has
a T-shape.
c The valence shell of the central iodine atom contains ten electrons:
seven from the iodine atom, two from the two chlorine atoms and an
additional electron responsible for the negative charge. There are two
bonding pairs and three lone pairs. The basic shape adopted by the
molecule is a trigonal bipyramid. The three lone pairs occupy the
equatorial positions to minimize the repulsion between the bonding
pairs and the lone pairs of electrons, so the final shape is linear.
d Because one of the three Iodine atoms has a negative charge, there
are three lone pairs of electrons and two bond pairs.
The basic shape adopted by the five electron pairs in the molecule is a
trigonal bipyramid. To minimize the repulsion between bonding pairs and
lone pairs of electrons, the three lone pairs of electrons occupy the
equatorial positions. Hence, the molecule has a linear shape.
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Page 147
11
a
b
c
d
e
f
g
Molecule
C2H6
C6H6 (benzene)
CH3OH
SF4
SCl6
PCl5
BCl3
Shape (around central atom or atoms)
Tetrahedral
trigonal planar
tetrahedral and V shaped
see-saw
octahedral
trigonal bipyramidal
trigonal planar
Polarity
non-polar
non-polar
polar
polar
non-polar
non-polar
non-polar
Page 149
12 The formal charges are shown below using the rules previously outlined.
The formal charges in all three Lewis formulas sum to −1, the charge on the thiocyanate ion.
Since nitrogen (3.0) is more electronegative than carbon (2.6) or sulfur (2.6), the negative formal charge
will be located on the nitrogen atom.
In addition, the Lewis formula is selected that produces the smallest formal charges.
Page 152
13 a Four bonding pairs
b coordination number = 4
c all angles are 109.5°
14 a all angles are 120°
b coordination number = 3
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c The distance between carbon atoms within a layer (0.142 nm) is less than half of the interatomic
distance between layers (0.335 nm) in graphite. This is because there is strong covalent bonding
within each layer, and very weak interactions (London forces) operating between layers.
15 The carbon–carbon bond length is intermediate between the lengths of single and double carbon-carbon
bonds, which suggests delocalization of pi electrons, which can also be described by resonance
structures.
16 There are strong covalent bonds between carbon atoms in a C60 molecule. (The bond order is between
single and double and there is some pi delocalization.)
C60 exists as a regular lattice of simple covalent molecules with instantaneous dipole–induced dipole
interactions (London (dispersion) forces) between molecules.
Page 153
17 giant structure/macromolecule/all the atoms are joined together
covalent (single or sigma bonds)
strong polar bonds/bonds difficult to break
Each silicon atom forms four single bonds and each oxygen atom forms two single bonds.
A large amount of heat energy is needed to break the bonds so it has a high melting point.
It does not burn or react with oxygen.
Page 163
18 Iodine and hexane are non-polar substances. When mixed together a solution is formed because:
strength of the attractive London forces
between iodine and hexane in the solution
>
combined strengths of the attractive London
forces in solid iodine and liquid hexane
Water is a polar solvent in which molecules form hydrogen bonds. When mixed with non-polar iodine
molecules, nearly all water molecules continue to hydrogen-bond with each other. The iodine–water
attractions are extremely weak in comparison to the combined strength of the hydrogen bonds in water
and the dispersion forces in iodine. Consequently, iodine is virtually insoluble in water.
Page 177
19 a
b
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c
19 d
e
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20 The C–O bond lengths are different in propanoic acid as one is a single covalent bond while the other is
a double covalent bond.
propanoic acid
propanoate ion
The C–O bond lengths are the same in the propanoate ion due to resonance or delocalization resulting in
equivalent C–O bonds where the negative charge is spread over the O–C–O bonds.
Page 179
21 The π bonds in penta-1,4-diene are not on alternate carbon atoms/they are next to a carbon atom with no
free p orbitals and cannot be delocalized.
The π bonds in penta-1,3-diene are on alternate carbon atoms/they are separated by a single covalent
bond and can be delocalized.
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S2.3 The metallic model
Page 186
22 a 3s
b The valence electrons are not associated with a single sodium atom, but move through the lattice
between the cations.
c The electrons are negatively charged and so move towards a positive electrode and are repelled from
a negative electrode. The net flow of electrons in a conductor is an electric current and is due to the
potential difference.
d Metallic bonding is electrostatic in nature and of comparable strength to ionic bonding.
e The sodium ions are not packed into a simple cubic lattice.
Page 188
23 a i
Most metals are silvery solids (under standard conditions). The shiny surface of alkali metals
tarnishes rapidly due to oxidation.
ii These properties are not exclusive to metals: for example, iodine and graphite have a shiny
surface (lustre).
iii Copper and gold are not silver in appearance and mercury is a liquid.
b i
Many metals have relatively high melting points.
ii This is not exclusive to metals.
iii Mercury melts at −40 °C and the alkali metals in group 1 have relatively low melting points.
c i
Most solid metals are malleable and ductile and many have high tensile strength.
ii This set of collective properties is exclusive to metals.
iii Some metals, such as manganese and zinc, are brittle.
d i
All metals are excellent electrical and thermal conductors in both the solid and liquid states.
ii These properties are almost exclusive to metals. Diamond is an excellent electrical insulator but it
is an excellent thermal conductor (because atomic vibrations, which carry the heat, are scattered
very little at room temperature).Graphite, graphene, silicon and germanium are also electrical
conductors, especially when doped (but not as good as metals).
iii All metals (including mercury) are excellent electrical conductors in the solid and liquid states.
However, titanium and mercury are about 20 times less conductive than copper and silver, which
both have a single valence s electron.
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S2.4 From models to materials
Page 198
24 a InCl3
polar covalent
b HF
polar covalent
c CsH
ionic
d AlI3
polar covalent
e S2Cl2
covalent
f KCl
ionic
g NaK
metallic
Page 202
25 An alloy is a substance with metallic properties composed of a metal combined with other metals or
non-metals.
Alloys usually have higher corrosion resistance and higher tensile strength.
26 Ferrous alloys, such as stainless steel, contain iron; non-ferrous alloys, such brass and bronze, do not
contain iron.
Page 204
27
a
b
c
d
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27 e
28 a 2,3-dimethylbut-2-ene
b 1-chloropropene
c 1,1-dichloroethene
Page 211
29
Addition polymerization
Generally involves one monomer
Polymerization does not lead to elimination
of smaller molecules
Empirical formula is the same as that of
monomer
Monomers are unsaturated and usually have
one reactive carbon–carbon double bond
Condensation polymerization
Usually involves two different monomers
Results in the formation of simple molecules such as H2O
or HCl as well as the polymer
Empirical formula is different from the constituent
monomers
Monomers are saturated and generally have reactive
functional groups at each end
Page 216
30
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Page 218
31 a
b
c
32 a benzene-1,4-diacyl dichloride (or benzene-1,4-dicarboxylic acid) and ethane-1,2-diol
b hexanediacyl dichloride (or hexane-1,6-dioic acid) and 1,6-diaminohexane
Page 219
33
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S3.1 The periodic table: classification of elements
Page 225
1
a rubidium
metal
b radon
non-metal
c
metalloid
germanium
d strontium
metal
e silicon
metalloid
f fluorine
non-metal
g copper
metal
h lead
metal
i mercury
metal
j lanthanum
metal
k boron
metalloid
Page 230
2
The outermost or valence configuration is 3s2. Hence magnesium is an s-block element. The principal
quantum number, n = 3, gives the period and the number of valence electrons (2) indicates the group.
Thus magnesium is in period 3 and group 2 of the periodic table.
3
X has the valence configuration 3s2 and hence the full electron configuration of 1s2 2s2 2p6 3s2.
4
a Gallium is an element in group 13 and period 4 of the periodic table. Hence the electronic
configuration of the valence shell is 4s2 3d10 4p1.
b Lead is an element in group 14 and period 6 of the periodic table. Hence the electronic configuration
of the valence shell is 6s2 4f14 5d10 6p2.
5
a X is magnesium: 1s2 2s2 2p6 3s2
b Y is nitrogen: 1s2 2s2 2p3
c Z is argon: 1s2 2s2 2p6 3s2 3p6
6
a X is in group 17, period 3 – chlorine
b Y is in group 17, period 4 – bromine
c Z is group 2, period 4 – calcium.
Page 235
7
a first ionization energy: H(g) → H+(g) + e−
first electron affinity:
H(g) + e− → H−(g)
b The first ionization energy is endothermic because work needs to be done to overcome the
electrostatic attraction between the electron and the proton.
The first electron affinity is exothermic because the attraction of the proton for the incoming electron
is greater than the additional electron–electron repulsion.
c 1s2
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d The two electrons in the H− ion experience electron–electron repulsion.
The electron cloud is more diffuse.
The two electrons in H− experience more shielding (screening) from the nucleus.
The electrons in H− experience a lower effective nuclear charge.
Page 237
8
a The first ionization energy is the energy required to remove one mole of electrons from one mole of
atoms in the gaseous state to form one mole of gaseous cations with a charge of +1.
b Ionization energy increases across period 2, even though electrons are removed from the same main
energy level, because the increasing nuclear charge (due to additional protons) results in a decreasing
atomic radius. This causes a stronger electrostatic attraction between the positive nucleus and each
electron.
c The beryllium atom has the electronic configuration 1s2 2s2. The boron atom has the electronic
configuration 1s2 2s2 2p1. Electrons in 2p orbitals are of higher energy and are further from the
nucleus than electrons in 2s orbitals, therefore they require less energy to remove. The electron being
lost in the boron atom also has additional shielding compared to that in the beryllium atom.
d The nitrogen atom has the electronic configuration 1s2 2s2 2p3. The oxygen atom has the electronic
configuration 1s2 2s2 2p4. For the oxygen atom, the electron is removed from a doubly occupied p
orbital – it is part of a spin pair. An electron in a doubly occupied p orbital is repelled by the other
electron and requires less energy to remove than an electron in a half-filled p orbital.
Page 239
9
a group 1, period 6, s block
b They have relatively low values of ionization energy, meaning loss of their valence electron requires
less energy than other metals.
c 2Cs(s) + 2H2O(l) → 2CsOH(aq) + H2(g)
d Bubbles of an explosive colourless gas will be observed. There will be effervescence as gas is
released at the surface.
The caesium moves rapidly around the surface and is converted into a soluble product. Heat and
sound energy are released.
e Caesium hydroxide, the strong base formed, is completely dissociated into caesium and hydroxide
ions. The high pH is due to a high concentration of hydroxide ions, OH−(aq).
f 2Cs(s) + Cl2(g) → 2CsCl(s)
4Cs(s) + O2(g) → 2Cs2O(s)
6Cs(s) + N2(g) → 2Cs3N(s)
g Caesium has two more electron shells (main energy levels) than potassium. The electrons are located
further from the nucleus than those in potassium and the additional electron shells provide shielding
that more than outweighs the increase in nuclear charge due to the additional protons, hence caesium
has a significantly lower value of ionization energy.
Page 241
10 a A group is a vertical column in the periodic table where all the atoms of the elements have the same
number of electrons in the valence shell.
A period is a horizontal row in the periodic table where all the atoms of the elements have the same
number of main energy levels (shells).
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b group 17, period 5 and p block
c They have relatively high values of electron affinity. They accept a valence electron to form a more
stable halide ion in an exothermic reaction.
d brown colour is formed from a colourless/pale yellow solution
F2(g) + 2I−(aq) → I2(aq) + 2F−(aq)
e Chlorine has two fewer electron shells (main energy levels) than iodine. The outer (valence) shell of
the chlorine is closer to the nucleus and it is less shielded than the outer shell of iodine. Since
chlorine can thus gain electrons more easily than iodine, it is more reactive. This is reflected in a
more exothermic electron affinity and higher value of electronegativity.
11 a
b 24–26 °C from the graph
The literature value is −21°C.
You should compare in terms of absolute difference and percentage difference.
c The decrease in melting points reflects the decrease in the strength of the metallic bonding as
group 1 is descended. The cations in a metal are held together by the attraction of the protons in the
cations to the delocalized valence electrons that form the sea of electrons. As the metal atoms
increase in atomic number, the cations become larger (because the reduced attraction to the nucleus
caused by the electrons being in higher shells more than balances the increased attraction due to
increased nuclear charge). This means that ions within the metal lattice are more easily pulled away
from each other by increasing heat.
The same packing arrangement (type of lattice) is assumed.
Page 243
12 Cl2O7(l) + H2O(l) → 2HClO4(aq)
2Cl2O7(l) → 2Cl2(g) + 7O2(g)
Page 249
13 a +5
b +2
c +3
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d +5
e +5
f +4
g +5
h +3
Page 251
14 a
The arrows can be half arrows or full arrows but must point in opposite directions if there are two in
a box.
b Co2+ (accept +2, 2+, cobalt(II), II)
1s2 2s2 2p6 3s2 3p6 3d7
c Cobalt is a d-block element because it is located in group 9 where a d sublevel is being filled across
the period.
Cobalt is a transition metal because it forms two cations, both of which have an incomplete d
sublevel; Co3+ has 4 unpaired d electrons and Co2+ has 3 unpaired d electrons.
Page 254
15 a The electrons in the 3d and 4s sublevels are close in energy; in iron, there are eight electrons in the
sea of delocalized valence electrons. Magnesium has only two delocalized valence electrons forming
the metallic bonding with dipositive cations. The bonding in magnesium requires less energy to
overcome than the stronger bonding in iron, so iron has a higher melting point (>1500 °C).
b Iron has valence electrons in the 3d and 4s sublevels. These are close in energy so different numbers
of valence electrons are lost to form ions of different oxidation states. Hence iron can form two
stable cations, Fe2+ and Fe3+.
A third electron removed from a magnesium atom would come from a 2p orbital in a shell that is
closer to the nucleus than that occupied by the valence electrons. This requires a significant increase
in energy. Hence magnesium forms only one cation, Mg2+.
Page 256
16 a +2
b water and ammonia molecules
c coordination bonds
d 6
e ligand replacement/exchange
no change in oxidation state of nickel (remains at +2)
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S3.2 Functional groups: classification of compounds
Page 265
17
Page 267
18
19 a HO(CH2)3CH(CH3)CH(CH3)2
b HOCH(CN)2
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20
21
Page 271
22 phenyl, hydroxy, amido (N-substituted amide), ester, alkoxy, carbonyl
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Page 276
23 a
b The boiling point increases with the number of carbon atoms due to the increase in the number of
electrons available for polarization and hence stronger dispersion forces.
The percentage increase in the molar mass decreases with the addition of each –CH2– unit and hence
the percentage increase in boiling point becomes progressively smaller.
c The molar mass suggests a formula of C14H30.
From the graph, this compound would have a boiling point of about 254 °C.
d They decompose before boiling, or undergo sublimation.
e They are unable to form hydrogen bonds; the interaction between alkane molecules is stronger than
the interaction between alkane and water molecules. It is not energetically or thermodynamically
favourable to replace strong hydrogen bonds between water molecules with weaker interactions
between water and alkane molecules.
Page 294
24 Infrared absorption of 1300 cm−1 occurs in alcohols, esters and ethers.
Infrared absorption of 1740 cm−1 occurs in ketones, aldehydes, carboxylic acids and esters.
The only molecule that shows both lines is an ester. There are five isomers of this ester. It cannot be a
>C=O in a carboxylic acid because there are no bands due to the presence of –OH.
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Page 300
25
Page 301
26
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Chemistry for the IB Diploma – Answers
THEME R
R1.1 Measuring enthalpy changes
Page 313
1
a
b
Page 315
2
a 38 000 J
b 2900 J
c 30 000 J
d 56 000 J
e 6800 J
3
Q = 800 g × 4.18 J g−1 K−1 × 70 K = 234 kJ
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4
240 000
700 ×4.18
= ∆𝑇𝑇 = 82 K
100°C – 82°C = 18°C
Page 319
5
−52 kJ mol−1
6
−13 100 kJ mol−1
Page 322
7
a 3.25 × 10−3 mol
b 6290 J
c −1940 kJ mol−1
Page 324
8
a
FeSO4(aq) + Mg(s) → Fe(s) + MgSO4(aq)
b
moles FeSO4 = 0.500 mol dm-3 × 0.0500 dm3 = 0.0250 mol
moles Mg =
0.800 𝑔𝑔
24.31 𝑔𝑔 𝑚𝑚𝑚𝑚𝑚𝑚−1
= 0.0329 mol
FeSO4 is limiting
Q=
c
− 50.0 𝑔𝑔 × 4.18 𝐽𝐽 𝑔𝑔−1 𝐾𝐾−1 × 20.6 𝐾𝐾
0.0250 𝑚𝑚𝑚𝑚𝑚𝑚
×
1 𝑘𝑘𝑘𝑘
1000 𝐽𝐽
= 172 kJ mol−1
The density of the iron(II) sulfate solution is 1.00 g cm−3.
The specific heat capacity of the iron(II) sulfate solution is identical to pure water (4.18 J g−1 K−1).
The heat capacity of the polystyrene cup and solid are small, so can be ignored.
No energy is lost to evaporation.
Energy is not transferred to the air surrounding the mixture.
Page 329
9 a
b
Q = (20.0 g + 25.0 g) × 4.18 J g−1 °C−1 × 11.8 °C
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= 2.22 kJ
n(NaOH) = 2.00 mol dm−3 × 0.0200 dm3 = 0.0400 mol
c
ΔH = −
2.22
0.0400
= −55.5 kJ mol−1
experimental error =
|−55.5+ 57.9|
|−57.9|
× 100 = 4.1%
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R1.2 Energy cycles in reactions
Page 338
10 a −99 kJ mol−1
b +78 kJ mol−1
c +156 kJ mol−1
11 I–Cl = 232 kJ mol−1
Page 344
12 78.2 kJ mol−1
13 −395 kJ mol−1
14 −25 kJ mol−1
15 −2222 kJ mol−1
Page 346
1
2
16 a ∆𝐻𝐻c ⦵ of H2(g) = H2(g) + O2(g) → H2O(l)
1
2
b ∆𝐻𝐻c ⦵ of C2H6(g) = C2H6(g) + 3 O2(g) → 2CO2(g) + 3H2O(l)
1
2
c ∆𝐻𝐻c ⦵ of C8H18(l) = C8H18(l) + 12 O2(g) → 8CO2(g) + 9H2O(l)
d ∆𝐻𝐻c ⦵ of C6H12O6(s) = C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)
1
2
3
2
e ∆𝐻𝐻f ⦵ of NH3 = N2(g) + H2(g) → NH3(g)
1
8
f ∆𝐻𝐻f ⦵ of SO2 = S8(s) + O2(g) → SO2(g)
1
2
g ∆𝐻𝐻f ⦵ of Na2O = 2Na(s) + O2(g) → Na2O(s)
1
2
3
2
h ∆𝐻𝐻f ⦵ of CHCl3 = C(s) + H2(g) + Cl2(g)→ CHCl3(l)
Page 350
17 a −166 kJ mol−1
b −124 kJ mol−1
c −554 kJ mol−1
d −424 kJ mol−1
e −127 kJ mol−1
18 a −312 kJ mol−1
b −206 kJ mol−1
c −635 kJ mol−1
d −44 kJ mol−1
e −493 kJ mol−1
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19 a ∆𝐻𝐻f ⦵ of NH4Cl(s) = −314 kJ mol−1
b ∆𝐻𝐻f ⦵ of MgCl2(s) = −641 kJ mol−1
c ∆𝐻𝐻f ⦵ of (CH3CO)2O(l) = −1794 kJ mol−1
Page 355
20 a Fe(g) → Fe+(g) + e−
b Zn(s) → Zn(g)
c MgO(s) → Mg2+(g) + O2−(g)
d Cl(g) + e− → Cl−(g)
e
1
Br2(l)
2
→ Br(g)
f Ca+(g) → Ca2+(g) + e−
g O−(g) + e− → O2−(g)
h Li2O(s) → 2Li+(g) + O2−(g)
21 a ∆𝐻𝐻f ⦵(CaCl2)
b ∆𝐻𝐻L ⦵(CaCl2)
c 2∆𝐻𝐻at ⦵(O) / ∆𝐻𝐻BE ⦵(O=O)
d ∆𝐻𝐻IE1 ⦵(Zn)
e ∆𝐻𝐻IE1 ⦵(Zn) + ∆𝐻𝐻IE2 ⦵(Zn)
f ∆𝐻𝐻at ⦵(Zn) + ∆𝐻𝐻IE1 ⦵(Zn) + ∆𝐻𝐻IE2 ⦵(Zn)
g ∆𝐻𝐻at ⦵(F) + ∆𝐻𝐻EA1 ⦵ (F)
h ∆𝐻𝐻EA1 ⦵(O) + ∆𝐻𝐻EA2 ⦵ (O)
Page 359
22 ∆𝐻𝐻L ⦵ (Na2 O) = (2 × 109) + (2 × 496) + 248 −141 +753 − −414
∆𝐻𝐻L ⦵ (Na2 O) = 2482 kJ mol−1
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23 −524 + 2420 = 150 + 736 + 1450 + 30.9 + ∆𝐻𝐻BE ⦵ (Br–Br) + (2 × −342)
∆𝐻𝐻BE ⦵ (Br–Br) = 213 kJ mol−1 (3 s.f.)
24 Magnesium chloride would have the higher lattice enthalpy because a magnesium ion is more highly
charged and smaller than a sodium ion.
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R1.3 Energy from fuels
Page 364
25
a C3H8 + 5O2 → 3CO2 + 4H2O
b 2C3H6 + 9O2 → 6CO2 + 6H2O
c C6H12O6 + 6O2 → 6CO2 + 6H2O
d 2C3H6S + 11O2 → 6CO2 + 6H2O + 2SO2
or
C3H6S +
e C6H5OH + 7O2 → 6CO2 + 3H2O
11
O2
2
→ 3CO2 + 3H2O + SO2
26 a carbon monoxide
b Incomplete combustion occurs due to an insufficient supply of oxygen to the fuel. This may be
because of insufficient oxygen in the atmosphere, or because the fuel has a high demand for oxygen
and it is not able to mix well with the oxygen because it is not volatile.
c Carbon monoxide, a poisonous gas, is produced, and solid particulates (of unburnt fuel or carbon)
which cause lung disease are produced.
d Soot will be produced, and a flame will appear orange.
Page 370
27
12 000 kJ mol−1
299 g mol−1
Page 372
28 a
= 40.1 kJ g −1
Mass of natural gas =
15 000 kWh
15.4 kWh kg−1
Mass of CO2 = 974 kg × 0.75 ×
b Mass of heavy fuel oil =
44.01 g mol−1
12.01 g mol−1
15 000 kWh
11.6 kWh kg−1
Mass of CO2 = 1293 kg × 0.85 ×
Page 376
= 974 kg
= 2680 kg
= 1293 kg
44.01 g mol−1
12.01 g mol−1
= 4030 kg
29 a Solar radiation, largely in the visible region of the electromagnetic spectrum, reaches and warms the
Earth’s surface. The Earth radiates energy back into the atmosphere in the IR region. Some
wavelengths of the IR radiation are absorbed by greenhouse gases and converted into vibrational and
translational energy resulting in heating of the lower atmosphere. In addition, some of the absorbed
IR radiation is re-radiated back to the surface of the Earth by greenhouse gases such as carbon
dioxide. The result is a warming of the surface of the Earth.
b Carbon dioxide absorbs IR radiation. This energy causes the bonds in carbon dioxide to change
length (stretch) or the bond angles to change (bend). This stretching or bending must be
accompanied by a change in polarity or dipole moment of the molecule.
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R1.4 Entropy and spontaneity (HL only)
Page 392
30
a +20 J K−1 mol−1
b −149 J K−1 mol−1
c −169 J K−1 mol−1
31 +446 J K−1
32 = +44.1 +244.1 J K-1 mol-1
Page 394
33 A, D and E
34
a little change
b decrease
c decrease (more order is being gained / decrease in number of particles)
d increase
e increase
f decrease
g decrease
h increase
i
decrease
j increase
Page 398
35 a ΔS(total) = −433 J K−1 mol−1
b No, not spontaneous at 298 K because there is a decrease in total entropy.
c ∆𝐺𝐺 ⦵, 129 kJ mol−1
Page 402
36 a The reaction is significantly exothermic in nature and at low temperatures the enthalpy change
contribution (ΔH) will dominate over the entropy change contribution (TΔS), thus making ΔG
negative.
b 592 K
Page 405
37
a 3.34 × 10−30
b 163 kJ mol−1 (ln 1 = 0)
c Equilibrium
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Chemistry for the IB Diploma – Answers
R2.1 How much? The amount of chemical change
Page 407
1
NO(g)
20 molecules
2000 molecules
1.204 × 1024 molecules
0.2 mol
60.02 g
O2(g)
10 molecules
1000 molecules
6.020 × 1023 molecules
0.1 mol
32.00 g
NO2(g)
20 molecules
20000 molecules
1.204 × 1024 molecules
0.2 mol
92.02 g
Page 408
2
a
2LiOH + H2SO4 → Li2SO4 + 2H2O
b 2Na + MgF2 → 2NaF + Mg
c Cu + 2AuNO3 → 2Au + Cu(NO3)2
d C12H22O11 + 12O2 → 12CO2 + 11H2O
e MgCO3 + 2HBr → MgBr2 + H2O + CO2
f Zn + Pb(NO3)2 → Pb + Zn(NO3)2
g 2AlBr3 + 3Cl2 → 2AlCl3 + 3Br2
h 2Na3PO4 + 3CaCl2 → 6NaCl + Ca3(PO4)2
i
3Ca(OH)2 + 2H3PO4 → Ca3(PO4)2 + 6H2O
j Mn + H2SO4 → MnSO4 + H2
3
a
2Hg(SCN)2(s) → 2HgS(s) + CS2(l) + C3N4(s)
b CS2(l) + 3O2(g) → CO2(g) + 2SO2(g)
c 2C3N4(s) → 3(CN)2(g) + N2(g)
Page 411
4 4Al(s) + 3O2(g) → 2Al2O3(s)
Oxygen and aluminium oxide are in a 3:2 ratio by amount
Hence, amount of aluminium oxide formed = (4.00 mol × 2/3) = 2.67 mol
Page 412
5 2KClO3(s) → 2KCl(s) + 3O2(g)
The potassium chlorate(V) and oxygen amounts are in a 2:3 ratio by moles.
Amount of KClO3 =
49.00 g
122.55 g mol−1
= 0.3998 mol
3
2
Amount of O2 = 0.3998 mol × = 0.599 8 mol
Mass of O2 = 0.599 8 mol × 32.00 g mol−1 = 19.19 g
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Page 413
6
Amount of sucrose = (17.117 g/342.34 g mol−1) = 0.050 00 mol
1 mole of sucrose, C12H22O11, reacts with 12 moles of oxygen, O2, to form 12 moles of carbon dioxide,
CO2, so, with 0.050 00 mol of sucrose we will obtain 0.050 00 ×12 = 0.6000 moles of carbon dioxide,
CO2.
Volume of CO2 = (0.6000 mol × 22.7 dm−3 mol) = 13.62 dm3.
Page 414
7
a N2(g)
+
3H2(g) → 2NH3(g)
b 1 mol
3 mol
2 mol
1 vol
3 vol
2 vol
12.5 dm3
37.5 dm3 25.0 dm3
Page 417
8
Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g), or
Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g) (1:1 ratio by moles)
Amount of H2SO4(aq) = (100.00/1000) dm3 × 2.0 mol dm−3 = 0.200 mol
Amount of ZnSO4(aq) = 0.200 mol
Mass of ZnSO4(s) = 0.200 mol × 161.45 g mol−1 = 32.29 g
Page 420
9
Amount of magnesium atoms = (4.862 g/24.31 g mol−1) = 0.2000 mol
Amount of sulfur atoms = (3.207 g/32.07 g mol−1) = 0.1000 mol
The coefficients in the equation indicate that 1 mole of magnesium atoms reacts with 1 mole of sulfur
atoms to form 1 mole of magnesium sulfide (formula units). The amounts indicate that sulfur is the
limiting reactant and magnesium is present in excess.
Amount of magnesium unreacted = 0.2000 mol − 0.1000 mol = 0.1000 mol
Mass of magnesium unreacted = 0.1000 mol × 24.31 g mol−1 = 2.431 g
Page 421
10
a C2H5Cl + 3O2 → 2CO2 + 2H2O + HCl
b Oxygen is the limiting reagent, since 3 mol × 3 = 9 mol of O2 are needed to react with 3 mol of
C2H5Cl, and only 3 mol of O2 present.
2
3
c Amount of carbon dioxide = 3 × = 2.0 mol
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11 Given 90% in each of the three steps:
1 mol of Na2SO4(s) yields 0.9 mol of Na2S(s)
0.9 mol of Na2S(s) yields (2 × 0.81 × 0.9) = 0.81 mol of Na2CO3(s)
0.81 mol of Na2CO3(s) yields 1.458 mol of NaHCO3(s)
mole ratio of Na2SO4(s): NaHCO3(s) = 1:1.458
100 000
×
142.05
mass of NaHCO3 = (
Page 422
1.458) × 84.01 = 86 228 g = 86.2 kg
12 n(Fe2O3) = (5 × 107 g/159.70 g mol−1) = 313 100 mol (limiting reactant)
Hence n(Fe) = (2 × 313 100 mol) = 626 200 mol
n(Fe) = (626 200 mol × 55.85 g mol−1) = 34 970 000 g =34.97 tonnes
Percentage yield of iron = (30.00 t/34.97 t) × 100 = 85.79%
Page 427
13 C(s)
12.01
+
2H2O(g) →
CO2(g) +
2H2(g)
[2 × 18.02]
44.01
[2 × 2.02]
Recall that the Ar or Mr in grams is 1 mole, so:
total molar mass of products = (44.01 g mol−1 + 4.04 gmol−1) = 48.05 g mol−1
(note that this is the same as the reactants: (12.01 g mol−1 + 36.04 g mol−1) = 48.05 g mol−1)
molar mass of hydrogen = 4.04 g mol−1
atom economy = (4.04 g mol−1/48.05 g mol−1) × 100 = 8.41%
14 TiCl4(g) + 2Mg(l) → Ti(s) + 2MgCl2(s)
Mass of all starting materials in stoichiometric equation (via molar masses – mass in grams of one mole
of each substance)
= M(Ti) + 2M(MgCl2) = (47.87 g mol−1 + 190.42 g mol−1) = 238.29 g mol−1 = 238.29 g
Mass of desired product (titanium) = 47.87 g mol−1 = 47.87 g
Therefore: atom economy = (47.87 g/238.29 g) × 100% = 20.1%
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Chemistry for the IB Diploma – Answers
R2.2 How fast? The rate of chemical change
Page 432
15
Collision 1 is the only successful type of collision because the orientation of the molecule is correct
both in terms of the H–Cl molecule interacting with the pi electron cloud of the carbon–carbon double
bond and the polarization of the attacking molecule. The hydrogen atom is slightly positive (δ+) and
interacts with the negative pi cloud.
Collisions 2–4 are unsuccessful because either the orientation of the H–Cl molecule is incorrect (the H
does not point towards the alkene) or the collision does not occur with the reacting part of ethene (the
collision needs to be with the carbon–carbon double bond.
Page 437
16
a See diagram, paler curve
b Shown on diagram by darker curve
c Labelled on diagram as Ea
17 a Energy difference between reactants and products = −92 kJ mol−1.
Energy difference between transition state peak and products = 335 kJ mol−1.
Ea = (335 – 92) kJ mol−1 = +243 kJ mol−1
b The activation energy will be lowered to a value less than +243 kJ mol−1 (due to the presence of an
alternative pathway).
18
a A graph similar to Figure R2.14
b Activation energy is the minimum combined kinetic energy the colliding molecules must have in
order to react.
c See Figure R2.14
d See Figure R2.19; new line at a lower energy value than the first Ea
e Area under the graph between the two values for activation energy shaded; see Figure R2.19
f See Figure R2.15
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19
a At higher pressure the same number of molecules are packed closer together and will collide more
frequently (the concentration of the gases is effectively increased) and so there is more chance of a
reaction.
b At higher concentration there are more particles in a given volume and so the collision frequency is
increased and so there is more chance of reaction.
c When the solid is finely divided there is a greater surface area exposed (for the same mass) and so
there is more chance of reaction.
d There are two factors here arising from the molecules having greater kinetic energy: they are moving
faster, on average, and so collide more frequently; also, when they do collide, the combined kinetic
energy is more likely to equal or exceed the activation energy and so there is more chance of the
collision producing a reaction. The second factor is the more significant.
Page 442
20
Moles of CO2 produced =
Final concentration of CO2 =
21
1 𝑚𝑚𝑚𝑚𝑚𝑚
4 𝑑𝑑𝑑𝑑3
0.250 𝑚𝑚𝑚𝑚𝑚𝑚 𝑑𝑑𝑑𝑑−3 – 0 𝑚𝑚𝑚𝑚𝑚𝑚 𝑑𝑑𝑑𝑑−3
Rate =
44.02 𝑔𝑔
44.02 𝑔𝑔 𝑚𝑚𝑚𝑚𝑚𝑚 −1
15.00 𝑠𝑠−0 𝑠𝑠
= 1.000 mol
= 0.25 mol dm-3
= 0.0167 mol dm-3 s-1
0.06 mol dm−3 of iodine produced in 30 s
Rate =
Page 445
0.06 mol dm−3
30 s
= 2 × 10−3 mol dm−3 s−1
22 a
b
Hydrogen peroxide concentration / mol dm−3
Rate / mol dm−3 min−1
0.16
0.12
0.076 0.049
0.08
0.04
0.033
0.019
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Chemistry for the IB Diploma – Answers
c
d
Measuring rates from a concentration–time graph involves drawing a graph and measuring the
gradients to the curve at a number of points, at least five. This can be an inaccurate process at any
point, but especially towards the end of the reaction when it is slowing down and the change in rate
is relatively small.
Pages 454–455
23 a A diagram similar to Figure R2.37 but without the small tube on the string and with a strip of
magnesium ribbon in the acid.
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
b
Initial rate = 1.25 cm3 s−1
c Reaction is fastest at the start as the concentration of the acid is greatest at this point. The graph
curves as the rate decreases because the acid (and magnesium) is being used up and the collision
frequency decreases. Eventually the curve flattens to a plateau as the magnesium is all used up and
there is no further reaction.
d Total volume of gas produced in 150 s = 75.6 cm3
Average rate of reaction = 75.6 cm3/150 s = 0.504 cm3 s−1
e The magnesium has all reacted and no more gas is produced. The acid (15 × 10−3 mol) is in excess of
the magnesium (4.17 × 10−3 mol).
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Chemistry for the IB Diploma – Answers
f, g, h
i
A diagram similar to Figure R2.15; the first experiment is at the higher temperature. Draw in the
areas representing the activation energy.
At the higher temperature more particles will have energy greater than or equal to the activation
energy and therefore more collisions will result in product formation. Hence the rate of reaction
will increase.
24 a Average temperatures: A 24 °C; B 32 °C; C 39 °C; D 49 °C; E 57 °C
b
c The rate is fastest in experiment E.
d The rate is fastest at the highest temperature. The particles are moving fastest and therefore collide
more frequently. More importantly, a greater proportion of collisions will result in a reaction as more
collisions involve particles whose total kinetic energy is equal to or greater than the activation
energy for the reaction.
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Chemistry for the IB Diploma – Answers
e These factors, including the dimensions of the flask, must be kept constant so that the experiments
are comparable, with only one variable changed; the cross must be viewed through the same depth of
solution each time.
f Extrapolate your curve and estimate a value for the time taken at 70 °C.
It should be around 12 s.
h
Use an ice-water bath to achieve a temperature between 0 °C and 5 °C; equilibrate the solutions to
this temperature before mixing them to react.
Page 458
25
Overall order = 2; units of k are mol−1 dm3 s−1
26
Rate = k[propene] [bromine] = 30(0.040)2
= 30(1.6 × 10−3) = 0.048 mol−1 dm3 s−1
27
Rate = k[HI]2
2.50 × 10−4 = k × (2.00)2
k = 6.25 × 10−5 mol−1 dm3 s−1
Number of molecules used up in 1 dm3 in 1 s
= (2.50 × 10−4 mol dm-3 s−1) × (6.02 × 1023 mol−1) = 1.51 × 1020
Page 463
28
Comparing experiments 1 and 3: when [NO] is kept constant while [O2] doubles, the rate also doubles.
Therefore, the reaction is first order with respect to [O2].
Comparing experiments 3 and 4: when [O2] and [NO] are doubled, the rate increases by 8 times.
Doubling [O] doubles the rate, so doubling [NO] quadrupled the rate. Therefore, the reaction is second
order with respect to [NO].
Rate = k[NO]2[O2]
Page 464
29 a 5 minutes = 300 seconds
Average rate of reaction = (0.27/300) = 9 × 10−4 mol dm−3 s−1
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Chemistry for the IB Diploma – Answers
b
Initial rate = 0.056 mol dm−3 min−1 or 9.33 × 10-4 mol dm-3 s−1
c
Draw tangents to find the gradient for the other specified concentrations.
This gives values of:
1.20 mol dm−3 ÷ 25 min = 0.048 mol dm−3 min−1 or 0.008 mol dm−3 s−1
(1.40 – 0.16) mol dm−3 ÷ (34 − 1) min = 0.038 mol dm−3 min−1 or 0.0063 mol dm−3 s−1
(1.40 – 0.40) mol dm−3 ÷ (44 − 2) min = 0.024 mol dm−3 min−1 or 0.0004 mol dm−3 s−1
d The concentrations of cyclopropane are 1.50, 1.20, 0.90, and 0.60 mol dm−3 respectively.
e
The graph is a straight line of gradient k.
Gradient 𝑘𝑘 =
0.064−0.008
1.70−0.10
= 0.035 min−1 or 5.83 × 10-4 s-1
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Chemistry for the IB Diploma – Answers
Page 473
30 a 2XY2→ X2 + 2Y2
b Rate = k [XY2]2
c Units of k = mol−1 dm3 s−1
d
31 a NO2Cl → NO2 + Cl
NO2Cl + Cl → NO2 + Cl2
slow
fast
b O3 + NO2 → NO3 + O2
slow
NO3 + NO2 → N2O5
fast
c I2 ⇌ 2I
2I + H2 → 2HI
fast
slow
or H2 + I2 → 2HI
d 2NO ⇌ N2O2
fast
N2O2 + H2 → N2O + H2O
slow
N2O + H2 → N2 + H2O
fast
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Chemistry for the IB Diploma – Answers
Page 479
32 a Arrhenius plot
b From the graph, gradient = −6098 K
gradient = −
𝐸𝐸a
𝑅𝑅
−Ea = −6098 K × 8.31 J mol−1 K−1 = 50.7 kJ mol−1
c y-intercept = −8.28
Intercept c = ln A = -8.28
A ≈ 2.54×10-4 mol−1 dm3 s−1. The units are that of k in a second order rate equation.
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Chemistry for the IB Diploma – Answers
R2.3 How far? The extent of chemical change
Page 488
33 a
b
c
d
e
34
[HF]2
[H2 ][F2 ]
[SO3 ]2
[O2 ][SO2 ]2
[Ag+ ][NH3 ]2
[[(Ag(NH3 )2 ]+ ]
[P4 ][F2 ]10
[PF5 ]4
[NO2 ]2
[N2 ][O2 ]
2O3(g) ⇌ 3O2(g)
Page 489
35
K1 = ([CO(g)] × [Cl2(g)])/[COCl2(g)]
K2 = [COCl2(g)]/([CO(g)] × [Cl2(g)])
K1 =
1
𝐾𝐾2
(or K2 =
Page 492
36 a i
1
);
𝐾𝐾1
reciprocals of each other
Amount of ethyl ethanoate = 0.500 mol (3 s.f.)
Amount of water = 2.00 mol (3 s.f.)
ii
Amount of sodium hydroxide used in titration = 2.95 × 10−2 mol
Amount of ethanoic acid in 25 cm3 of the equilibrium mixture = 2.95 × 10−2 mol
Amount of ethanoic acid in 250 cm3 of the equilibrium mixture = 0.295 mol
iii
If V = volume of the reaction mixture:
At start / moles
At equilibrium / moles
At equilibrium*/ mol dm−3
iv
ethyl
ethanoate
water
ethanoic acid
ethanol
0.50
2.0
–
–
0.205
1.705
0.295
0.295
0.205 / V
1.705 / V
0.295 / V
0.295 / V
If K = [CH3COOH] × [C2H5OH]/[CH3COOC2H5] × [H2O]
Since / V appears in each term, and there are the same number of terms on the top and bottom,
it cancels out.
0.2952
K = 0.205 × 1.705 = 0.249
b The protons or hydrogen ions (H+) will still be available to titrate with the alkali and interfere with
the result.
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c Add a defined and accurate volume of catalyst of known concentration using a micro-syringe or
automated micropipette. An alternative would be to do a blank titration at the start of the reaction.
d Phenolphthalein would be a suitable indicator.
Page 505
37
Reaction
A
B
C
Effect of increased temperature on
equilibrium position
Shift to the right – more ozone; K
increased
Shift to the left – less SO3; K
decreased
Shift to the right – more CO; K
increased
Effect of increased pressure on equilibrium
position
Fewer molecules on the right; therefore, shift
to right – more O3
Fewer molecules on the right; therefore, shift
to right – more SO3
No change as there are the same number of
molecules on both sides
38 a The equilibrium will shift to the left, favouring the reactants, to restore equilibrium concentrations
that fit the same value of K.
b The equilibrium position will shift to the left, to replace ethanol and restore the equilibrium at the
constant value of K.
39 a If the concentration of Fe2+ ions is increased then the equilibrium position will shift to the right to
use the added ions and restore equilibrium concentrations that fit the same K.
b The addition of water reduces the concentration of all the ions in solution equally; there will be no
change in the equilibrium position.
40 a Increasing pressure will shift the equilibrium to the left as the reactants occupy less volume (there
are fewer moles of gas on the reactants side of the equation).
b There will be no change in the equilibrium position as there are equal numbers of moles of gas on
the two sides of the equation.
c Increasing the pressure will favour the reactants side of the equation. The equilibrium position will
shift to the left as there are fewer moles of gas on this side of the equation; the reactants occupy less
volume than the products.
41
Decreasing the pressure will favour the products, shifting the equilibrium to the right, as this results in
an expansion (the products will occupy a greater volume).
42 a Increasing the temperature always favours the endothermic process (that absorbs the heat), so the
equilibrium will shift to the left (favouring the reactants) as the reverse reaction is endothermic.
b The forward reaction is endothermic, so this will be favoured by increasing the temperature; the
equilibrium position will shift to the right.
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Chemistry for the IB Diploma – Answers
Page 510
43
N2(g) + 3H2(g) ⇌ 2NH3(g)
Initial amount / mol
1
2
0
Change in amount / mol
−0.6
−1.8
+1.2
Equilibrium amount / mol
0.4
0.2
1.2
0.2
0.1
0.6
O2
Equilibrium concentration / mol dm−3
[NH3 ]2
(0.6)2
𝐾𝐾 =
=
= 1800
[N2 ][H2 ]3 0.2 × (0.1)3
Page 514
44
2SO2
Initial amount / mol
2.00
1.40
⇌ 2SO3
Equilibrium amount / mol
1.70
1.25
0.30
1.70/3.00 =
0.567
1.25/3.00 =
0.416
0.30/3.00
= 0.100
Equilibrium concentration / mol dm−3
45
𝐾𝐾 =
+
0.00
[SO3 ]2
(0.100)2
0.01
=
=
= 0.075
2
2
[SO2 ] [O2 ] (0.567) (0.416) 0.311 × 0.416
SO3
Initial amount / mol
Equilibrium concentration / mol
dm−3
𝐾𝐾 = 6.78 =
+
NO
0.030
0.030
(0.030 – x)
(0.030 – x)
⇌ NO2 + SO2
0
0
x
x
[NO2 ][SO2 ]
𝑥𝑥 2
=
(0.030 − 𝑥𝑥)2
[SO3 ][NO]
Taking square roots of both sides:
√6.78 =
𝑥𝑥
0.030 − 𝑥𝑥
2.60(0.030 – x) = x
0.078 − 2.60x = x
Therefore 3.60x = 0.078
x = 0.022
[SO3] = [NO] = 8 × 10−3 mol dm−3 and [NO2] = [SO2] = 0.022 mol dm−3
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52
Chemistry for the IB Diploma – Answers
46
H2(g)
Initial amount / mol
+
CO2(g)
2.00
2.00
Equilibrium amount / mol
(2.00 − x)
(2.00 − x)
Equilibrium concentration
/ mol dm−3
[H2 O][CO]
𝐾𝐾 = 2.10 =
[H2 ][CO2 ]
(2.00 – 𝑥𝑥)
10
(2.00 – 𝑥𝑥)
10
⇌
H2O(g) +
CO(g)
1.00
1.00
(1.00 + x)
(1.00 + x)
(1.00 + 𝑥𝑥) (1.00 + 𝑥𝑥)
10
10
Note that the volumes will cancel out.
Therefore:
2.10 =
(1.00 + 𝑥𝑥)2
(2.00 − 𝑥𝑥)2
Taking square roots of both sides:
√2.10 =
1.00 + 𝑥𝑥
2.00 − 𝑥𝑥
1.45(2.00 − x) = (1.00 + x)
1.90 = 2.45x
x = 0.78
[H2O] = [CO] =
[H2] = [CO2] =
Page 515
1.78
10
1.22
10
= 0.178 mol dm−3
= 0.122 mol dm−3
47 The equilibrium constant, K, is a mathematical relationship that shows how the concentrations of the
products vary with the concentrations of the reactants.
If the value of K is greater than 1, the products in the reaction are favoured. If the value of K is less than
1, the reactants in the reaction are favoured.
48 K = 1.08 ×105
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53
Chemistry for the IB Diploma – Answers
R3.1 Proton transfer reactions
Page 524
1
a
Na2CO3
Fe(NO3)2
BaCl2
K2SO4
NaOH
NH4OH or NH3
AgNO3
FeCl3
Pb(NO3)2
CuSO4
✗
✓
✓
✓
✓
✓
✗
✗
✗
✗
✗
✗
✗
✓
✗
✗
✗
✓
✓
b Fe2+(aq) + 2OH−(aq) → Fe(OH)2(s)
✗
✓
✓
✓
✓
✓
2Ag+(aq) + CO32−(aq) → Ag2CO3(s)
Ag+(aq) + Cl−(aq) → AgCl(s)
Ag+(aq) + OH−(aq) → AgOH(s) (or 2AgOH(s) → Ag2O(s) + H2O(l))
2Fe3+(aq) + 3CO32−(aq) → Fe2(CO3)3(s)
Fe3+(aq) + 3OH−(aq) → Fe(OH)3(s)
Pb2+(aq) + CO32−(aq) → PbCO3(s)
Pb2+(aq) + 2Cl−(aq) → PbCl2(s)
Pb2+(aq) + SO42−(aq) → PbSO4(s)
Pb2+(aq) + 2OH−(aq) → Pb(OH)2(s)
Cu2+(aq) + CO32−(aq) → CuCO3(s)
Ba2+(aq) + SO42−(aq) → BaSO4(s)
Cu2+(aq) + 2OH−(aq) → Cu(OH)2(s)
Page 526
2
a H2SO4(aq) + CuCO3(s) → CuSO4(aq) + H2O(l) + CO2(g)
b 2HBr(aq) + Ca(HCO3)2(s or aq) → CaBr2(aq) + 2H2O(l) + 2CO2(g)
c 2H3PO4(aq) + 3Na2CO3(s or aq) → 2Na3PO4(aq) + 3H2O(l) + 3CO2(g)
d 2CH3COOH(aq) + Ca(s) → Ca(CH3COO)2(aq) + H2(g)
e 2(NH4)3PO4(aq) + 3Ba(OH)2(s or aq) → 6NH3(g) + 3H2O(l) + Ba3(PO4)2(aq)
f C2H5NH2(g or aq) + HCl(g or aq) → C2H5NH3Cl(s or aq)
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Chemistry for the IB Diploma – Answers
Page 528
3
a Ca(CH3COO)2 calcium hydroxide/calcium oxide and ethanoic acid
b KNO3 potassium hydroxide/carbonate and nitric acid
c Na2SO3 sodium hydroxide and sulfurous acid
d FeSO4 iron/carbonate/hydroxide and sulfuric acid
e K2CO3 potassium hydroxide and carbonic acid (carbon dioxide bubbled into the alkali)
f BaCl2 barium hydroxide/carbonate/oxide and hydrochloric acid
Page 530
4
a base H2O; acid [Al(H2O)6]3+
b base H2O; acid HCO3–
c base H2NCONH2; acid H2O
d base HSO4–; acid H3O+
e acid NH4+; base OH–
Page 536
5
a H2O is the acid; OH− is the conjugate base; CO32− is the base and HCO3− is the conjugate acid.
b H2SO4 is the acid and SO42− is the conjugate base; H2O is the base and H3O+ is the conjugate acid.
Page 538
6
a I–
b NO2–
c HSO4–
d SO42–
e S2–
7
a HBr
b H2S
c HCO3–
d H2SO4
e H3O+
f N2H5+
Page 539
8
a H2O + H2O
acid
b NH3 +
base
base
H3O+
acid
⇌
⇌
c HCO3− + OH− ⇌
acid
base
OH− +
H3O+
conjugate base conjugate acid
NH4+ +
H2O
conjugate acid conjugate base
CO32− +
H2O
conjugate base conjugate acid
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55
Chemistry for the IB Diploma – Answers
d NH3 + HCl
base
9
acid
⇌
NH4+ +
Cl−
conjugate acid conjugate base
a NH4+ + OH− ⇌ NH3 + H2O
b H3O+ + SO42− ⇌ H2O + HSO4−
c H2O + H− ⇌ OH− + H2
d HCO3− + OH− ⇌ CO32−+ H2O
Page 541
10 a alkaline
b acidic
c acidic
d neutral
11 The one with the pH value of 14
12 The one with the pH value of –1
Page 542
13 pH= –log10 (0.01) = 2
14 pH= –log10 (0.100) = 1.000
15 pH= –log10 (3.00 × 10–7) = 6.523
16 pH = –log10 (1.00) = 0.00
17 pH = –log10 (0.02) = 1.7
18 Amount of hydrogen chloride = (3.646 g/36.46 g mol–1) = 0.1000 mol
0.1000 mol H+(aq) in 250 cm3 solution, hence concentration is 0.40 mol dm–3 and pH is 0.40.
19 amount of HCl(aq) = 2.500 × 10–3 mol
amount of NaOH = 1.500 × 10–3 mol
amount of excess HCl(aq)/H+(aq) = 1.000 × 10–3 mol in 40.00 cm3
concentration = (1000/40.00) × 1.000 × 10–3 mol = 0.02500 mol dm–3
pH = –log10 (0.02500) = 1.602
Page 543
20 pOH= –log10 (0.100) = 1
21 [OH–(aq)] = 10−pOH; [OH–(aq)] = 10–2; [OH–(aq)] = 0.01 mol dm–3
Page 544
22 a pH probe and meter; indicator paper or solution
b pH probe and meter (if calibrated correctly)
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Chemistry for the IB Diploma – Answers
Page 545
23 [OH−(aq)] = 0.01 mol dm–3
[H+(aq)] = 1.00 × 10–14 ÷ 0.01 mol dm–3 = 1 × 10–12 mol dm–3
pH = 12
24 [OH−(aq)] = 2.50 × 10–3 mol dm–3
[H+(aq)] = 1.00 × 10–14 ÷ 2.50 × 10–3 mol dm–3 = 4.00 × 10–12 mol dm–3
pH = –log10 (4.00 × 10–12) = 11.4
25 [OH−(aq)] = 0.0200 mol dm–3
[H+(aq)] = 1.00 × 10–14 ÷ 0.0200 mol dm–3 = 5.00 × 10–13 mol dm–3
pH = –log10 (5.00 × 10–13) = 12.3
26 [OH−(aq)] = 0.010 × 2 = 0.020 mol dm–3
[H+(aq)] = 1.00 × 10–14 ÷ 0.020 mol dm–3 = 5.0 × 10–13 mol dm–3
pH = –log10 (5.0 × 10–13) = 12.3
Page 547
27 pH + pOH = 14; pH + 1 = 14; pH = (14 – 1) = 13
28 pH + pOH = 14; 2 + pOH = 14; pOH = (14 – 2) = 12
29 pOH = –log10 (0.100 × 2) = 0.699; pH = 14.00 – 0.699 = 13.30
Page 551
30 a to ensure that the ions from the acid and alkali were mixed together and the solution was
homogeneous
b Solution X is acidic; pH is near 2 and less than 7.
c Solution Y is alkaline; pH is near 12 and more than 7.
d It increases from close to 2 and then when Y added approaches 10.0 cm3 there is a rapid increase to
close to 11 followed by a much slower increase to 12.0.
e The graph would be flipped around so it starts at 12 and finishes near 2 with the vertical region in the
middle again.
f 10.0 cm3
Page 552
31 pH = –log10 [H+(aq)]
[H+(aq)] = −antilog (3.10) or 10–3.10 = 7.94 × 10–4 mol dm–3
𝐾𝐾a =
[H + (aq)] × [C6 H5 COO− (aq)]
[C6 H5 COOH(aq)]
but since [H+(aq)] = [C6H5COO–(aq)]
𝐾𝐾a =
𝐾𝐾a =
[H + (aq)]2
[C6 H5 COOH(aq)]
2
�7.94×10−4 �
0.01
= 6.3 × 10–5 mol dm–3
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Chemistry for the IB Diploma – Answers
32 𝐾𝐾a =
[H+ (aq)] × [CH3 COO− (aq)]
[CH3 COOH(aq)]
but since [H+(aq)] = [CH3COO−(aq)];
𝐾𝐾a =
[H + (aq)]2
[CH3 COOH(aq)]
then rearranging:
[H+(aq)] = �[CH3 COOH(aq)] × 𝐾𝐾a
[H+(aq)] = √0.1 × 1.8 × 10−5 = 1.34 × 10–3 mol dm–3
and then pH = –log10 [H+(aq)] = –log10 (1.34 × 10–3 mol dm–3) = 2.87
Page 554
33 NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH–(aq)
Kb = [OH–(aq)] × [NH4+(aq)] / [NH3(aq)]
= [OH–]2/ [NH3(aq)]
1.8 × 10–5 = [OH–]2/0.50
pH = 11.48
Page 556
34 a CH3COONa(aq) alkaline
b NH4Br(aq) acidic
c KCN(aq) alkaline
d CH3CH2CH2CH2COOK(aq) alkaline
35 a Na2SO4(aq) pH ≈ 7
b NH4CH3COO(aq) pH ≈ 7
c Na2S(aq) pH > 7
d KNO2(aq) pH > 7
Page 565
36 pH = pKa + log10 ([salt]/[acid])
5.6 = –log10 (7.20 × 10–4) + log10 ([NaF]/0.0042)
[NaF] = 1.204 mol dm–3
mass of NaF = 1.204 × (250/1000) × (22.99 + 19.00) = 12.64 g
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Chemistry for the IB Diploma – Answers
R3.2 Electron transfer reactions
Page 572
37 The conversion of Al(OH)3 to Al involves hydrogen loss, so could be oxidation, but the reaction also
involves oxygen loss, so could also be reduction.
Page 581
38 Two arsenic atoms oxidized from +3 to +5 per mole of As2O3 (loss of 4 electrons); therefore, if 5 moles
oxidized, total 20 electrons lost / change in oxidation number = 20; therefore, 4 moles MnO4− reduced,
total 20 electrons gained / change in oxidation number = 20, each Mn(VII) gains 5 electrons / change in
oxidation number is 5; hence Mn(II) / Mn2+
39 Amount of Cr2O72− = (24.00/1000) dm3 × 0.100 mol dm−3 = 0.0024 mol
Amount of Fe2+ in 20.00 cm3 = 6 × 0.0024 mol = 0.0144 mol
Amount of Fe2+ in 500 cm3 solution = (500/20) × 0.0144 mol = 0.36 mol
Amount of Fe2+ = amount of FeSO4.xH2O
Molar mass = mass / amount
Molar mass = 101.2 g/0.36 mol = 281.1 g mol−1
Molar mass of FeSO4 = 151.91 g mol−1
Molar mass of water molecules = 281.1 g mol−1 – 151.91 g mol−1 = 129.2 g mol−1
Number of water molecules = 129.2 g mol−1/18.02 g mol−1 = 7; x = 7
40 Amount of S2O32− = (26.20/1000) dm3 × 0.500 mol dm−3 = 0.0131 mol
Amount of I2 reacting with S2O32− = 1/2 × 0.0131 mol = 0.00655 mol
Amount of MnO4−/KMnO4 = 2/5 × 0.00655 mol = 0.00262 in 25.00 cm3
Concentration of KMnO4 = (1000/25.00) × 0.00262 mol = 0.105 mol dm−3
41 n(S2O32−) reacted = 23.60 × 10−3 × 0.02 = 4.72 × 10−4 mol
n(I2) reacted = ½ × 4.72 × 10−4 = 2.36 × 10−4 mol
initial n(I2) = 40.00 × 10−3 × 0.01 = 4.00 × 10−4 mol
n(I2) reacted with SO2 = 4.00 × 10−4 – 2.36 × 10−4 = 1.64 × 10−4 mol
n(SO2) in wine = n(I2) reacted = 1.64 × 10−4 mol
[SO2] in wine = 1.64 × 10−4/50.00 × 10−3 = 3.28 × 10−3 mol dm−3
Page 590
42 a
b
c
d
e
f
g
h
i
j
2I− → I2 + 2e−
Fe2+ → Fe3+ + e−
Fe → Fe2+ + 2e−
2Cl− → Cl2 + 2e−
SO42− + 4H+ + 2e− → SO2 + 2H2O
MnO4− + 8H+ + 5e− → Mn2+ + 4H2O
SO42− + 10H+ + 8e− → H2S + 4H2O
H2O2 → O2 + 2H+ + 2e−
2Cr3+ + 7H2O → Cr2O72− + 14H+ + 6e−
C2O42− → 2CO2 + 2e−
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Chemistry for the IB Diploma – Answers
k Hg22+ + 2e− → 2Hg
l 2IO3− + 12H+ + 10e− → I2 + 6H2O
43 a
b
c
d
e
f
2MnO4− + 16H+ + 10Cl− → 2Mn2+ + 5Cl2 + 8H2O
2MnO4− + 16H+ + 5C2O42− → 2Mn2+ + 10CO2 + 8H2O
SO42− + 10H+ + 8I− → H2S + 4I2 + 4H2O
2MnO4− + 6H+ + 5H2O2 → 2Mn2+ + 5O2 + 8H2O
2IO3− + 12H+ + 10I− → 6I2 + 6H2O
Cr2O72− + 14H+ + 6Fe2+ → 2Cr3+ + 6Fe3+ + 7H2O
44
In the reaction hydrazine, N2 H4, is the reducing agent as it is oxidized to nitrogen, N2 .
Similarly, the bromate(V) ion, BrO3− , is the oxidizing agent at it is reduced to Br − .
The corresponding half-equations are:
Oxidation half-reaction:
Reduction half-reaction:
Oxidation
3N2 H4 �⎯⎯⎯⎯⎯⎯� 3N2 + 12H + + 12e−
Reduction
2BrO3 − + 12H + + 12e− �⎯⎯⎯⎯⎯⎯� 2Br − + 6H2 O
45 a H2O2 + I2 →+ 2H+ + 2I− + O2
b 4Cl2 + S2O32− + 5H2O → 8Cl− + 2SO42− + 10H+
Page 593
46 a lithium and bromine
b caesium and nitrogen
c iron and bromine
Page 607
47 X = propan-1-ol
Y = 2-methylpropan-2-ol
Z = 2,2-dimethylpropan-1-ol
48 a CH3CH2OHCH2CH2CH3
b CH3CH2(CH3)CH2OH
c CH3CH2OH
d (C6H5)CH2CHOHCH2OH
Page 608
49 a
b
c
d
e
2
2
2
4
2
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Chemistry for the IB Diploma – Answers
Page 616
50 a Fe(s) + Ni2+(aq) → Fe2+(aq) + Ni(s); E⦵cell = (0.45 V) + (−0.26 V) = +0.19 V
b 2MnO4–(aq) + 16H+(aq) + 10I─(aq) → 5I2(s) + 2Mn2+(aq) + 8H2O(l);
E⦵cell = (−0.54 V) + (1.51 V) = +0.97 V
c 2Cr3+(aq) + 7H2O(l) + 3F2(g) → 6F−(g) + Cr2O72─(aq) + 14H+(aq); E⦵cell = (2.87 V) + (−1.36 V) =
+1.51 V
d 2Ag+(aq) + Cu(s) → Cu2+(aq) + 2Ag(s); E⦵cell = (0.80 V) + (−0.34 V) = +0.46 V
⦵
⦵
51 Given that 𝐸𝐸Ag
+ ∣Ag (= +0.80 V) > 𝐸𝐸Pb2+ ∣Pb (= −0.13 V)
the Ag ∣ Ag + electrode will be the cathode (right-hand electrode) and Pb ∣ Pb2+ will act as the anode
(left-hand electrode). Thus, the voltaic cell obtained by these two electrodes can be represented as:
Pb ∣ Pb2+ ∥ Ag + ∣ Ag
anode
cathode
Page 619
52 a
E⦵cell = +0.43 V
∆G⦵ = −nFE⦵cell l = (−4 × 96 500 C mol−1 × 0.43 V)
= −165 980 J mol−1 = −166 kJ mol−1
b The reaction is spontaneous (in the forward direction) under standard conditions.
c The E⦵ value for this related reaction will not change because standard electrode potentials are an
intensive property, this means they are independent of amount. The standard electrode potentials of
the two half equations are not halved even though the stoichiometry in this reaction is halved. The
∆G⦵ value will be halved as the amount of electrons transferred in the reaction is halved.
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Chemistry for the IB Diploma – Answers
R3.3 Electron sharing reactions
Page 626
53
54
Page 629
55
Page 634
56 Initiation:
Cl2 → 2Cl•
Propagation:
C6H12 + Cl• → C6H11• + HCl
C6H11• + Cl2 → C6H11Cl + Cl•
Termination:
2Cl• → Cl2
Cl• + C6H11• → C6H11Cl
2C6H11• → C12H22
Page 636
57 a
b
c
4; chloromethane, dichloromethane, trichloromethane, tetrachloromethane
3; chloromethylbenzene, dichloromethylbenzene, trichloromethylbenzene
9; chloroethane, 1,1-dichloroethane, 1,2-dichloroethane, 1,1,1-trichloroethane, 1,1,2-trichloroethane,
1,1,1,2-tetrachloroethane, 1,1,2,2-tetrachloroethane, pentachloroethane, hexachloroethane
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Chemistry for the IB Diploma – Answers
R3.4 Electron pair sharing reactions
Page 641
58 a (CH3)3C+ = electrophile Br− = nucleophile
b R3CLi = nucleophile
RCHO = electrophile
c RCHO = electrophile cyclic enolate ion = nucleophile
59 Nucleophilic sites circled in red:
60 Electrophilic sites circled in red:
Page 643
61 a
b
c
d
e
radical substitution
nucleophilic substitution
nucleophilic substitution
nucleophilic substitution
radical substitution
Page 649
62
Propanone is a Lewis acid in this reaction.
Page 650
63
Lewis acid
Lewis base
a
NH4
OH−
b
HCO3−
H2O
c
H2O
H2NCONH2
d
H3O
HSO4−
e
H2O
H−
+
+
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Chemistry for the IB Diploma – Answers
64 a
b
c
d
e
Page 651
65 a
b
c
d
e
f
g
h
+2
+2
+2
+3
+2
+3
+2
+3
Page 656
66 a SN2
b SN1
c SN2
67 a
b
nucleophilic substitution (SN2)
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