Download 3. Solution Guide to Supplementary Exercises

Topic
4
Acids and Bases
Part A Unit-based exercise
Unit 14 Acids and alkalis
Fill in the blanks
1
hydrochloric
2
Sulphuric
3
Ethanoic
4
sour
5
red; yellow
6
colourless; red
7
bases
8
dissociate; dissociation
9
hydrogen
10 mobile ions
11 basicity
12 monobasic
13 dibasic
14 a) salt; hydrogen
b) salt; water; carbon dioxide
c) salt; water; carbon dioxide
d) salt; water
e) salt; water
15 salt; water
16 alkali
17 bitter
18 ammonia
19 precipitates
20 complex
21 hygroscopic
22 deliquescent
True or false
23 F Concentrated sulphuric acid should be diluted by adding the acid to water slowly while stirring.
24 T
25 F Solid citric acid has no effect on zinc.
The aqueous solution of citric acid shows typical properties of an acid while solid citric acid does not.
26 F An acid is a hydrogen-containing substance that gives hydrogen ions as the only type of positive ions
when dissolved in water.
27 T The aqueous solution of citric acid can conduct electricity.
28 F The basicity of an acid is the maximum number of hydrogen ions produced by one acid molecule. For
example, ethanoic acid (CH3COOH) is monobasic because only the hydrogen atom in the –COOH group
can undergo dissociation.
29 T
30 F Phosphoric acid (H3PO4(aq)) is a tribasic acid.
31 T Concentrated sulphuric acid is hygroscopic.
32 T When solid citric acid dissolves in water, only a few molecules dissociate to give ions.
citric acid(s) + water
citric acid(aq)
+
citric acid(aq)
H (aq) + citrate ion(aq)
Hence an aqueous solution of citric acid contains mobile citric acid molecules.
XBUFS
DJUSJDBDJE
NPMFDVMF
DJUSBUFJPO
)
33 T
34 F Phenolphthalein is colourless in acidic solutions.
35 F Many metal hydroxides are INSOLUBLE in water.
36 T
37 F The discolorations and tight scale buildup that occur in toilet bowls are mostly calcium carbonate
deposits from hard water. Calcium carbonate reacts wtih some acids to form water soluble substances.
The solid type toilet bowl cleansers are mostly sodium hydrogensulphate while the liquid type contain
hydrochloric acid.
38 T Sodium hydroxide absorbs carbon dioxide when exposed to air. Sodium hydrogencarbonate is formed
and it undergoes dehydration to give sodium carbonate.
39 F The properties of solutions of alkalis depend on the presence of mobile hydroxide ions.
40 T Ca2+(aq) + 2OH–(aq)
Ca(OH)2(s)
white precipitate
41 F Iron(II) hydroxide does not dissolve in excess dilute aqueous ammonia.
42 F Copper(II) hydroxide does not dissolve in excess dilute sodium hydroxide solution. It dissolves in excess
dilute aqueous ammonia to give a deep blue solution.
43 T Fe3+(aq) + 3OH–(aq)
Fe(OH)3(s)
reddish brown precipitate
44 F Concentrated nitric acid tends to decompose to nitrogen dioxide gas and oxygen gas.
4HNO3(aq)
4NO2(g) + O2(g) + 2H2O(l)
45 F Concentrated sulphuric acid is NOT used to dry ammonia gas because the acid would react with
ammonia gas.
H2SO4(l) + 2NH3(g)
(NH4)2SO4(s)
Multiple choice questions
46 A
47 C Option A — Caustic soda is sodium hydroxide.
Opiton B — Drain cleansers usually contain sodium hydroxide.
Option D — Slaked lime is calcium hydroxide.
48 C Option A — Dilute acids react with carbonates to give carbon dioxide gas.
Option B — Dilute acids have a sour taste.
Option C — Dilute acids conduct electricity due to the presence of mobile ions.
Option D — Dilute acids react with reactive metals only.
49 C Option A — Acidic solutions turn litmus solution red.
Option B — Dilute solutions of alkalis have a slippery feel.
Option D — Typical acids show NO reaction with copper.
50 B Option C — Hydrochloric acid can turn methyl orange red.
Option D — Concentrated hydrochloric acid is NOT stored in brown bottles.
51 B Option A — Sulphuric acid is used to make fertilizers, but it is NOT used as fertilizers.
Option B — Concentrated sulphuric acid is hygroscopic.
Option C — Concentrated sulphuric acid should be diluted by adding the acid to water slowly while
stirring.
Option D — Dilute sulphuric acid shows NO oxidizing property.
52 D Option A — Nitric acid is NOT a drying agent.
Option B — Sulphuric acid, NOT nitric acid, is used in car batteries.
Option C — Dilute nitric acid is an oxidizing agent. It reacts with copper to give nitrogen monoxide,
NOT hydrogen.
Option D — Concentrated nitric acid is corrosive.
53 C Aqueous solution of ethanoic acid can conduct electricity.
54 C When dilute sulphuric acid reacts with calcium carbonate, insoluble calcium sulphate forms. The calcium
sulphate covers the surface of calcium carbonate and prevents further reaction.
55 A Option A — Vinegar contains ethanoic acid. It shows NO reaction with copper.
Option B — Dilute nitric acid is an oxidizing agent. It reacts with copper to give nitrogen monoxide.
Option C — Copper(II) oxide and dilute nitric acid undergo neutralization.
Option D — Copper(II) hydroxide and vinegar undergo neutralization.
56 A
57 D
58 C Solid citric acid does NOT react with magnesium because it does not contain hydrogen ions.
The aqueous solution of citric acid shows the properties of an acid while solid citric acid does not.
59 C Phosphoric acid (H3PO4) is a tribasic acid.
60 A Carbon is a reducing agent. It can reduce FeO(s) to Fe(s).
61 D Option A — Glass cleansers usually contain ammonia.
62 A Glass cleansers usually contain ammonia.
63 A Option A — Sodium hydroxide is manufactured by the electrolysis of concentrated sodium chloride
solution.
Option C — Sodium hydroxide absorbs carbon dioxide when exposed to air. Sodium hydrogencarbonate
is formed and it undergoes dehydration to give sodium carbonate.
64 A Option A — When ammonia gas dissolves in water, it reacts with water to give ammonium ions and
hydroxide ions. However, ammonia does not react with water completely. Only very few
hydroxide ions are formed.
+
–
NH3(g) + H2O(l)
NH4 (aq) + OH (aq)
Hence aqueous ammonia contains both ammonia molecules and hydroxide ions.
0)m
XBUFS
/)
/)
Option B — Metals tarnish because they react with the air to form a layer of oxide or sulphide. Acids
can be used to remove this layer. One of the oxide layers most difficultly removed is rust.
Rust removers usually contain an acid such as hydrochloric acid or phosphoric acid.
Option C — Aqueous ammonia reacts with lead(II) nitrate solution to give a white precipitate, lead(II)
hydroxide.
Option D — Aqueous ammonia gives a yellow solution with methyl orange.
65 C Dilute sodium hydroxide solution does NOT give a white precipitate with potassium chloride solution.
The mixture is a colourless solution.
66 B Aluminium hydroxide dissolves in excess dilute sodium hydroxide solution due to the formation of a
soluble complex salt.
Al(OH)3(s) + OH–(aq)
[Al(OH)4]–(aq)
aluminate ion
colourless solution
67 B
68 D
Option
Solution
Observation
A
ammonium chloride
no precipitate
B
iron(II) sulphate
green precipitate Fe(OH)2
C
magnesium chloride
white precipitate Mg(OH)2
D
nickel(II) sulphate
green precipitate Ni(OH)2
69 D The white precipitate is magnesium hydroxide.
70 D
Option
Solution
Observation
A
calcium nitrate
no precipitate
B
copper(II) sulphate
pale blue precipitate
C
lead(II) nitrate
white precipitate
D
zinc sulphate
white precipitate
71 C
72 C
Option
Hydroxide
Solubility in NaOH(aq) / NH3(aq)
A
Al(OH)3
soluble in excess NaOH(aq)
B
Cu(OH)2
soluble in excess NH3(aq)
C
Fe(OH)2
insoluble in both NaOH(aq) and NH3(aq)
D
Zn(OH)2
soluble in both NaOH(aq) and NH3(aq)
Option
Compound X
Addition of NH3(aq) to dilute solution of X
A
Fe2(SO4)3
reddish brown precipitate Fe(OH)3
B
MgCl2
white precipitate Mg(OH)2
C
NiSO4
green precipitate Ni(OH)2
D
NH4Cl
no precipitate
73 A Concentrated sodium hydroxide solution is corrosive.
74 B Concentrated nitric acid is an oxidizing agent.
75 C
76 A Option A — Concentrated hydrochloric acid and concentrated nitric acid are volatile while concentrated
sulphuric acid is NOT.
Option B — Concentrated sulphuric acid is NOT stored in brown bottles.
Option D — Concentrated sulphuric acid is NOT used to dry ammonia gas because the acid would
react with ammonia gas.
H2SO4(l) + 2NH3(g)
(NH4)2SO4(s)
77 A The aqueous solution of compound X gives a gas with dilute hydrochloric acid, hence X is probably a
carbonate, rather than a sulphate.
Solution of X does NOT give a precipitate with dilute sodium hydroxide solution. Hence X probably
contains ammonium ions, rather than iron(II) ions.
\ X is ammonium carbonate.
78 A From test 1, it can be deduced that X is a chloride.
Ag+(aq) + Cl–(aq)
From test 2, it can be deduced that X contains iron(II) ions, rather than iron(III) ions.
Fe2+(aq) + 2OH–(aq)
Fe(OH)2(s)
green precipitate
Fe3+(aq) + 3OH–(aq)
Fe(OH)3(s)
reddish brown precipitate
AgCl(s)
79 B
Option
Substances
Upon mixing the substances
A
magnesium and ethanoic acid
hydrogen gas is produced
B
magnesium hydroxide and dilute hydrochloric acid
only a salt and water are produced
C
magnesium carbonate and dilute sulphuric acid
carbon dioxide gas is produced
D
magnesium and steam
hydrogen gas is produced
\ magnesium hydroxide and dilute hydrochloric acid would NOT produce a gas when mixed.
80 C Silver nitrate solution gives a white precipitate with dilute hydrochloric acid.
Ag+(aq) + Cl–(aq)
AgCl(s)
81 D NaOH(aq) gives a precipitate Fe2+(aq) ions, but NOT K+(aq) ions.
82 C Zinc can be obtained by heating zinc oxide with carbon. Carbon dioxide gas is formed in the process.
zinc oxide + carbon
zinc + carbon dioxide
83 D The aqueous solution of citric acid shows typical properties of an acid while the citric acid crystals do
not.
When citric acid crystals (jar 3) are mixed with water (jar 5), the aqueous solution would give carbon
dioxide gas with solid hydrogencarbonate (jar 1).
HCO3–(aq) + H+(aq)
H2O(l) + CO2(g)
84 A (2) Carbonic acid is formed when carbon dioxide gas dissolves in water. Hence rainwater is slightly
acidic.
CO2(g) + H2O(l)
H2CO3(aq)
carbonic acid
85 D
86 A (1) Acids are covalent compounds when pure. For example, pure sulphuric acid and nitric acid are
colourless liquids consisting of molecules.
(2) An acid may NOT contain both hydrogen and oxygen, e.g. HCl(aq).
(3) Typical acids would NOT react with copper.
87 B (2) Sulphuric acid is used to make fertilizers, but it is NOT as fertilizers.
88 B (2) Sulphuric acid, NOT hydrochloric acid, is used in car batteries.
89 D (3) When solid citric acid dissolves in water, only a few molecules dissociate to give ions.
citric acid(s) + water
citric acid(aq)
citric acid(aq)
H+(aq) + citrate ion(aq)
Hence an aqueous solution of citric acid contains mobile citric acid molecules.
XBUFS
DJUSBUFJPO
DJUSJDBDJE
NPMFDVMF
)
90 B (2) Vinegar contains ethanoic acid. It shows NO reaction with copper.
91 D (2) Carbonic acid is formed when carbon dioxide dissolves in water.
CO2(g) + H2O(l)
(3) Sulphurous acid is formed when sulphur dioxide dissolves in water.
SO2(g) + H2O(l)
H2CO3(aq)
carbonic acid
H2SO3(aq)
sulphurous acid
92 D (1) Magnesium reacts with dilute hydrochloric acid to give hydrogen gas.
(2) and (3) Sodium carbonate and potassium hydrogencarbonate react with dilute hydrochloric acid to
give carbon dioxide gas.
93 B (2) Very dilute nitric acid acts like a typical acid. It shows NO reaction with copper.
94 D All the three metals (calcium, iron and magnesium) react with acids to give hydrogen.
95 D (1) Heating ammonium sulphate solution with dilute sodium hydroxide solution gives ammonia gas.
NH4+(aq) + OH–(aq)
(2) Iron(II) sulphate solution gives a green precipitate with dilute sodium hydroxide solution.
(3) Citric acid would undergo neutralization reaction with dilute sodium hydroxide solution.
2+
–
Fe (aq) + 2OH (aq)
NH3(g) + H2O(l)
Fe(OH)2(s)
96 D (1) When ammonia gas dissolves in water, it reacts with water to give ammonium ions and hydroxide
ions. However, ammonia does not react with water completely. Only very few hydroxide ions are
formed.
NH4+(aq) + OH–(aq)
NH3(g) + H2O(l)
Hence aqueous ammonia contains both ammonia molecules and ammonium ions.
0)m
XBUFS
/)
/)
97 B (1) Cu(OH)2 dissolves in excess dilute aqueous ammonia to give a deep blue solution.
Cu(OH)2(s) + 4NH3(aq)
2+
–
(2) Pb(OH)2 is insoluble in excess dilute aqueous ammonia.
(3) Zn(OH)2 dissolves in excess dilute aqueous ammonia to give a colourless solution.
Zn(OH)2(s) + 4NH3(aq)
[Cu(NH3)4] (aq) + 2OH (aq)
[Zn(NH3)4]2+(aq) + 2OH–(aq)
98 A (1) Heating ammonium chloride with calcium hydroxide solution liberates ammonia gas.
2NH4Cl(s) + Ca(OH)2(aq)
2NH3(g) + CaCl2(aq) + 2H2O(l)
99 A (3) Addition of dilute sodium hydroxide solution to zinc sulphate solution gives a white precipitate,
Zn(OH)2.
Zn(OH)2 dissolves in excess dilute sodium hydroxide solution to give a colourless solution.
Zn(OH)2(s) + 2OH–(aq)
[Zn(OH)4]2–(aq)
100A (2) Heating solutions of ammonium chloride and sodium hydroxide liberates ammonia.
NH4Cl(aq) + NaOH(aq)
NH3(g) + NaCl(aq) + H2O(l)
101A (1) Magnesium chloride solution gives a white precipitate with silver nitrate solution.
Ag+(aq) + Cl–(aq)
(2) Magnesium chloride solution gives a white precipitate with dilute aqueous ammonia.
(3) Copper is less reactive than magnesium. It CANNOT displace magnesium from magnesium chloride
solution.
Mg2+(aq) + 2OH–(aq)
AgCl(s)
Mg(OH)2(s)
102B (1) Zinc is more reactive than silver, but less reactive than sodium.
Zinc granules can displace silver from silver nitrate solution.
There is NO reaction between zinc granules and sodium nitrate solution.
(2) Both sodium nitrate solution and silver nitrate solution are colourless.
(3) Only silver nitrate solution gives a white precipitate with dilute hydrochloric acid.
Ag+(aq) + Cl–(aq)
AgCl(s)
103A (1) NH3(aq) and NiSO4(aq) would form a green precipitate when mixed.
Ni2+(aq) + 2OH–(aq)
(2) NaOH(aq) and FeSO4(aq) would form a green precipitate when mixed.
(3) There is no observable change when NH3(aq) and NaCl(aq) are mixed.
Fe2+(aq) + 2OH–(aq)
Ni(OH)2(s)
Fe(OH)2(s)
104D (1) Ammonia gas would react with dilute sulphuric acid.
2NH3(g) + H2SO4(aq)
(2) Carbon dioxide gas would give a white precipitate with calcium hydroxide solution.
(3) Sulphur dioxide gas would react with sodium hydroxide solution.
CO2(g) + Ca(OH)2(aq)
SO2(g) + NaOH(aq)
(NH4)2SO4(aq)
CaCO3(s) + H2O(l)
NaHSO3(aq)
105B
106C Aqueous solution of ethanoic acid can conduct electricity. Hence ethanoic acid is an electrolyte.
Ethanoic acid is a covalent compound when pure. Pure ethanoic acid consists of molecules.
107D When dilute sulphuric acid reacts with calcium carbonate, insoluble calcium sulphate forms. The calcium
sulphate covers the surface of calcium carbonate and prevents further reaction.
108B Citric acid shows its acidic properties in the presence of water.
When solid citric acid dissolves in water, only a few molecules dissociate to give ions.
citric acid(s) + water
citric acid(aq)
+
citric acid(aq)
H (aq) + citrate ion(aq)
Hence an aqueous solution of citric acid contains mobile citric acid molecules.
XBUFS
DJUSBUFJPO
DJUSJDBDJE
NPMFDVMF
)
109D
110D
111C Dilute nitric acid is an oxidizing agent. It reacts with copper to give nitrogen monoxide, NOT hydrogen.
112A Concentrated nitric acid decomposes according to the following chemical equation:
4HNO3(aq)
4NO2(g) + O2(g) + 2H2O(l)
113D Many metal hydroxides are insoluble in water.
114B When ammonia gas dissolves in water, it reacts with water to give ammonium ions and hydroxide ions.
NH4+(aq) + OH–(aq)
NH3(g) + H2O(l)
Dilute aqueous ammonia conducts electricity due to the presence of mobile ions.
115C Anhydrous calcium chloride is NOT used to dry ammonia gas because calcium chloride would react
with ammonia gas.
10 CaCl2(s) + 4NH3(g)
CaCl2•4NH3(s)
Unit 15 Molarity, pH scale and strengths of acids and alkalis
Fill in the blanks
1
molarity
2
hydrogen ions
3
neutral
4
less
5
increases
6
hydroxide ions
7
hydrogen ions; hydrogensulphate ions; sulphate ions
8
hydrogen ions; ethanoate ions; ethanoic acid molecules
9
ammonium ions; hydroxide ions; ammonia molecules
10 concentrated; strong
True or false
11 F The pH of a solution decreases as the concentration of hydrogen ions in the solution increases.
12 T
13 F The pH value of dilute aqueous ammonia is greater than 7, but less than 14.
14 T
15 F Aqueous solution of ethanoic acid can conduct electricity. Hence ethanoic acid is an electrolyte.
16 T When citric acid crystals dissolve in water, only a few molecules dissociate to give ions. Hence citric
acid is a weak acid.
17 F When we describe acids as strong and weak, we are talking about the extent of their dissocation in
water. When we talk about concentration, we are referring to the amount of an acid in a unit volume
of solution.
For example, 5 mol dm–3 nitric acid is a concentrated solution of a strong acid while 0.1 mol dm–3 nitric
acid is a dilute solution of a strong acid.
18 T
19 F 1 mol dm–3 HCl(aq) has a higher concentration of hydrogen ions than 1 mol dm–3 CH3COOH(aq) does.
Hence the pH of HCl(aq) is lower than that of CH3COOH(aq).
11
20 F NaOH(aq) + HCl(aq)
NaCl(aq) + H2O(l)
NH3(aq) + HCl(aq)
NH4Cl(aq)
According to the equations, 1 mole of NaOH / NH3 requires 1 mole of HCl for complete neutralization.
Hence 10 cm3 of 1 mol dm–3 NaOH(aq) and 10 cm3 of 1 mol dm–3 NH3(aq) require the same number of
moles of HCl for complete neutralization.
Multiple choice questions
21 B Number of moles of KNO3 = molarity of KNO3 solution x volume of solution
40
= 0.25 mol dm–3 x
dm3
1 000
= 1.0 x 10–2 mol
–1
22 D Molar mass of Na2CO3= [2 x 23.0 + 12.0 + 3 x 16.0) g mol
= 106.0 g mol–1
mass
molar mass
13.8 g
=
106.0 g mol–1
= 0.130 mol
Number of moles of Na2CO3=
Volume of solution =
\ the molarity of the sodium carbonate solution is 0.260 mol dm–3.
500
dm3
1 000
number of moles of Na2CO3
Molarity of Na2CO3 solution=
volume of solution
0.130 mol
=
500
dm3
1 000
(
)
= 0.260 mol dm–3
23 D Molar mass of (COOH)2•2H2O = [2 x (12.0 + 2 x 16.0 + 1.0) + 2 x (2 x 1.0 + 16.0)] g mol–1
= 126.0 g mol–1
12
mass
molar mass
3.15 g
=
126.0 g mol–1
= 0.0250 mol
Number of moles of (COOH)2•H2O present=
Molarity of ethanedioic acid solution=
number of moles of (COOH)2•2H2O
volume of solution
0.0250 mol
=
250.0
dm3
1 000
(
)
= 0.100 mol dm–3
24 B Molar mass of Ca(OH)2 = [40.1 + 2 x (16.0 + 1.0)] g mol–1
= 74.1 g mol–1
number of moles of Ca(OH)2
volume of solution
number of moles of Ca(OH)2
0.150 mol dm–3 =
250.0
dm3
1 000
Molarity of calcium hydroxide solution=
Number of moles of Ca(OH)2 = 0.150 mol dm–3 x
Mass of Ca(OH)2 = number of moles of Ca(OH)2 x molar mass of Ca(OH)2
= 0.0375 mol x 74.1 g mol–1
= 2.78 g
\ 2.78 g of calcium hydroxide are present.
(
= 0.0375 mol
)
250.0
dm3
1 000
25 D Molar mass of (NH4)2SO4 = [2 x (14.0 + 4 x 1.0) + 32.1 + 4 x 16.0)] g mol–1
= 132.1 g mol–1
number of moles of (NH4)2SO4
volume of solution
number of moles of (NH4)2SO4
1.50 dm3
Molarity of ammonium sulphate solution=
Number of moles of (NH4)2SO4 = 1.02 mol dm–3 x 1.50 dm3
= 1.53 mol
Mass of (NH4)2SO4 = number of moles of (NH4)2SO4 x molar mass of (NH4)2SO4
= 1.53 mol x 132.1 g mol–1
= 202 g
\ 202 g of ammonium sulphate are present.
1.02 mol dm–3 =
26 C Molar mass of KCl= (39.0 + 35.5) g mol–1
= 74.5 g mol–1
mass
molar mass
44.7 g
=
74.5 g mol–1
= 0.600 mol
Number of moles of KCl present =
Molarity of potassium chloride solution =
Volume of solution =
0.600 mol
2.40 mol dm–3
= 0.250 dm3
= 250 cm3
2.40 mol dm–3 =
number of moles of KCl
volume of solution
0.600 mol
volume of solution
13
27 B Number of moles of NaOH = molarity of solution x volume of solution
Option
A
3.0 mol dm–3 x
100
dm3 = 0.30 mol
1 000
B
2.5 mol dm–3 x
200
dm3 = 0.50 mol
1 000
C
1.5 mol dm–3 x
300
dm3 = 0.45 mol
1 000
D
1.0 mol dm–3 x
400
dm3 = 0.40 mol
1 000
\ 200 cm3 of 2.5 mol dm–3 NaOH(aq) contain the greatest number of moles of solute.
28 D
Number of moles of NaOH present
Option
Number of moles of MgSO4 = molarity of solution x volume of solution
A
0.40 mol dm–3 x 1.0 dm3 = 0.40 mol
B
0.30 mol dm
C
0.20 mol dm–3 x 2.0 dm3 = 0.40 mol
D
0.10 mol dm–3 x 2.5 dm3 = 0.25 mol
–3
3
x 1.5 dm = 0.45 mol
\ 2.5 dm3 of 0.10 mol dm–3 MgSO4(aq) contain the smallest number of moles of solute.
29 D Number of moles of CaCl2 in 250.0 cm3 of 0.300 mol dm–3 solution
14
= molarity of solution x volume of solution
250.0
–3
3
= 0.300 mol dm x
dm
1 000
= 0.0750 mol
Number of moles of CaCl2 in 150.0 cm of 0.180 mol dm
= molarity of solution x volume of solution
150.0
= 0.180 mol dm–3 x
dm3
1 000
= 0.0270 mol
Total number of moles of CaCl2 in the resulting solution
= (0.0750 + 0.0270) mol
= 0.102 mol
Total volume of the resulting solution = (250.0 + 150.0) cm3
= 400.0 cm3
Concentration of the resulting solution =
\ the concentration of the resulting solution is 2.55 x 10–1 mol dm–3.
3
–3
(
solution
0.102 mol
400.0
dm3
1 000
)
= 0.255 mol dm–3
30 B Number of moles of NaOH in 100.0 cm3 of 3.00 mol dm–3 solution
= molarity of solution x volume of solution
100.0
= 3.00 mol dm–3 x
dm3
1 000
= 0.300 mol
Number of moles of NaOH in 50.0 cm3 of 1.20 mol dm–3 solution
= molarity of solution x volume of solution
50.0
= 1.20 mol dm–3 x
dm3
1 000
= 0.0600 mol
Total number of moles of NaOH in the resulting solution
= (0.300 + 0.0600) mol
= 0.360 mol
Total volume of the resulting solution = (100.0 + 50.0) cm
3
= 150.0 cm
Concentration of the resulting solution =
\ the concentration of the resulting solution is 2.40 mol dm–3.
3
(
0.360 mol
150.0
dm3
1 000
)
= 2.40 mol dm–3
31 A pH of citric acid solution = –log10[H+]
= 1.85
i.e. log10[H+] = –1.85
[H+] = 10–1.85
= 0.0141 mol dm–3
\ concentration of hydrogen ions in the citric acid solution is 1.41 x 10–2 mol dm–3.
32 B pH of milk= –log10[H+]
= 6.70
i.e. log10[H+] = –6.70
[H+] = 10–6.70
= 2.00 x 10–7 mol dm–3
\ concentration of hydrogen ions in the sample of milk is 2.00 x 10–7 mol dm–3.
15
33 B Sulphuric acid dissociates completely according to the following equation:
+
2–
H2SO4(aq) –3
0.0200 mol dm 2H (aq) + SO4 (aq)
–3
? mol dm­
According to the equation, 1 mole of H2SO4 dissociates to give 2 moles of hydrogen ions.
i.e.concentration of hydrogen ions = 2 x 0.0200 mol dm–3
= 0.0400 mol dm–3
pH of acid = –log10(0.0400)
= –(–1.40)
= 1.40
\ pH of the sulphuric acid sample is 1.40.
34 A pH of sample of urine = –log10(7.94 x 10–7)
= –(–6.10)
= 6.10
\ pH of the sample of urine is 6.10.
35 C pH of sewage before treatment = 6.75
pH of sewage after treatment = 7.00
Concentration of hydrogen ions in sewage before treatment = 10
–7
–3
= 1.78 x 10 mol dm
Concentration of hydrogen ions in sewage after treatment = 10–7.00 mol dm–3
Change in concentration of hydrogen ions = (1.78 x 10–7 – 10–7) mol dm–3
= 0.78 x 10–7 mol dm–3
\ the change in the concentration of hydrogen ions in the sewage is 7.80 x 10–8 mol dm–3.
–6.75
36 B Limewater is saturated calcium hydroxide solution.
37 D Soft drinks contain carbonic acid.
38 C Option B — Glass cleanser is alkaline. Its pH is greater than 7.
Option D — The pH of sea water is about 8.
39 D Oven cleanser contains sodium hydroxide.
40 D When ammonia gas dissolves in water, it reacts with water to give ammonium ions (NH4+(aq)) and
hydroxide ions (OH–(aq)).
16
NH3(g) + water
NH3(aq)
NH3(aq) + H2O(l)
Thus, aqueous ammonia contains ammonia molecules, water molecules and a few ammonium ions and
hydroxide ions.
NH4+(aq) + OH–(aq)
41 C In general, non-metals react with oxygen to form acidic oxides and metals react with oxygen to form
basic oxides.
Option A — Calcium oxide is slightly basic. It reacts slightly with cold water to form calcium hydroxide,
which is very slightly soluble. So, only a slightly alkaline solution is formed.
Option C — Potassium oxide is strongly basic. It dissolves in water to form potassium hydroxide
solution.
K2O(s) + H2O(l)
Hence the pH of the solution of potassium oxide is higher than that of calcium oxide.
Options B and D —Oxides of carbon and sulphur are acidic. The pH values of the solutions of both
oxides are less than 7.
CaO(s) + H2O(l)
Ca(OH)2(s)
2KOH(aq)
42 D Concentration of hydrogen ions in solution X = 10–1
–3
= 0.1 mol dm
Concentration of hydrogen ions in solution Y = 10–2
= 0.01 mol dm–3
\ the concentration of hydrogen ions in solution X is ten times that in solution Y.
43 D (1) Baking soda solution is hydrogencarbonate solution. Its pH is ~9.
(2) The pH of lemon juice is ~2.
(3) The pH of milk is ~6.
44 D (1) and (2) The pH of 0.010 mol dm–3 HCl(aq) is higher than that of 0.100 mol dm–3 HCl(aq).
(3) and (4) The pH of 0.100 mol dm–3 NaOH(aq) is higher than that of 0.010 mol dm–3 NaOH(aq).
45 A Options A, C and D — Concentrated nitric acid, lemon juice and vinegar are acidic. There was no
colour change when they were added to the red solution.
46 D Option A — The concentration of hydrogen ions in lemon juice is higher than that in wine.
Option B — The concentration of hydrogen ions in lemon juice is higher than that in baking soda
solution.
Option C —
10–4 mol dm–3
concentration of hydrogen ions in wine
=
10–12 mol dm–3
concentration of hydrogen ions in drain cleanser
\ the concentration of hydrogen ions is 108 times greater in wine than in drain cleanser.
10–7 mol dm–3
concentration of hydrogen ions in distilled water
=
10–9 mol dm–3
concentration of hydrogen ions in baking soda solution
\ the concentration of hydrogen ions is 100 times greater in distilled water than in
baking soda solution.
47 B When the right amounts of an acid (pH < 7) and an alkali (pH > 7) are mixed, they react completely
to produce a salt and water only (pH = 7).
17
48 A
49 A The electrical conductivity of a solution, and hence the brightness of the bulb, is proportional to the
concentration of mobile ions.
Option
Acid
A
25 cm3 of 0.5 mol dm–3
H2SO4(aq)
B
25 cm of 0.5 mol dm
HCl(aq)
C
D
3
Remark
Concentration of ions
H2SO4(aq) is a strong acid
H2SO4(aq)
2H+(aq) + SO42–(aq)
1.5 mol dm–3
1.0 mol dm–3
3
–3
HCl(aq) is a strong acid
HCl(aq)
H+(aq) + Cl–(aq)
25 cm of 0.5 mol dm
CH3COOH(aq)
3
–3
CH3COOH(aq) is a weak acid
CH3COOH(aq)
CH3COO–(aq) + H+(aq)
< 1.0 mol dm–3
25 cm3 of 0.5 mol dm–3
NH3(aq)
NH3(aq) is a weak alkali
NH3(aq) + H2O(l)
NH4+(aq) + OH–(aq)
< 1.0 mol dm–3
\ 25 cm of 0.5 mol dm
bulb the brightest.
–3
H2SO4(aq) have the highest concentration of ions and thus will make the
50 C Options A, B and D — During the reaction of hydrochloric acid and ethanoic acid with marble chips,
marble chips would react with hydrogen ions in the acids.
Hydrochloric acid is a strong acid that completely dissociates in water.
Ethanoic acid is weak acid that only partially dissociates in water.
Therefore 1 mol dm–3 hydrochloric acid has a higher concentration of hydrogen
ions than 1 mol dm–3 ethanoic acid does. The reaction rate between marble
chips and ethanoic acid is thus lower and the reaction takes a longer time to
complete.
Option C — CaCO3(s) + 2HCl(aq)
CaCO3(s) + 2CH3COOH(aq)
According to the equations, 2 moles of HCl / CH3COOH require 1 mole of CaCO3 for
complete reaction.
100 cm3 of 1 mol dm–3 hydrochloric acid and 100 cm3 of 1 mol dm–3 ethanoic acid
require the same number of moles of CaCO3 for complete reaction.
Hence the mass of marble chips left over would be the same for both acids.
CaCl2(aq) + H2O(l) + CO2(g)
(CH3COO)2Ca(aq) + H2O(l) + CO2(g)
51 A Carbonic acid is a weak acid that only partially dissociates in water.
2H+(aq) + CO32–(aq)
H2CO3(aq)
The concentration of H2CO3(aq) is the highest in 0.10 mol dm–3 carbonic acid.
52 C Both nitric acid and hydrochloric acid are monobasic strong acids.
18
H+(aq) + NO3–(aq)
HNO3(aq)
HCl(aq)
Therefore the concentration of hydrogen ions in 1.0 mol dm–3 nitric acid and 1.0 mol dm–3 hydrochloric
acid is the same.
H+(aq) + Cl–(aq)
Mixing 100 cm3 of the nitric acid with 100 cm3 of the hydrochloric acid would NOT result in a change
in the concentration of hydrogen ions, i.e. would NOT result in a change in pH.
53 A When we describe acids as strong and weak, we are talking about the extent of their dissociation in
water. When we talk about concentration, we are referring to the amount of an acid in a unit volume
of solution.
Sulphuric acid is a strong acid. Hence 0.1 mol dm–3 sulphuric acid is a dilute solution of a strong acid.
54 B The pH of NaCl(aq) is 7. Its pH remains constant upon dilution.
55 A During the reaction between magnesium and an acid, magnesium reacts with hydrogen ions in the
acid.
From the diagrams, it can be deduced that magnesium reacts more slowly with 0.1 mol dm–3 HA(aq).
The concentration of hydrogen ions in 0.1 mol dm–3 HA(aq) is lower than that in 0.1 mol dm–3 HB(aq).
Hence acid HA is probably weaker than acid HB.
56 A (3) Washing soda is hydrated sodium carbonate (Na2CO3•10H2O). Its solution is alkaline, with a pH > 7.
57 A (2) and (3) When we describe acids as strong and weak, we are talking about the extent of their
dissocation in water. When we talk about concentration, we are referring to the amount
of an acid in a unit volume of solution.
For example, 5 mol dm–3 nitric acid is a concentrated solution of a strong acid while
0.1 mol dm–3 nitric acid is a dilute solution of a strong acid.
58 A In general, non-metals react with oxygen to form acidic oxides and metals react with oxygen to form
basic oxides.
(1) and (2) Calcium oxide and magnesium oxide are slightly basic. They react with cold water to form
hydroxides which are slightly soluble. Solutions with a pH > 7 are formed.
CaO(s) + H2O(l)
Ca(OH)2(s)
MgO(s) + H2O(l)
Mg(OH)2(s)
59 B (1) Hydrochloric acid is a strong acid while ethanoic acid is a weak acid.
(2) The pH is a measure of the concentration of hydrogen ions in a solution. Therefore acid solutions
of the same pH should have the same concentration of hydrogen ions.
(3) Hydrochloric acid almost completely dissociates in water to give hydrogen ions and chloride ions.
Ethanoic acid only partially dissociates in water, forming very few hydrogen ions.
Hence hydrochloric acid and ethanoic acid having the same pH value should have different
concentrations.
60 C (1) Adding 1 mol dm–3 hydrochloric acid to the ethanoic acid makes the ethanoic acid more acidic,
thus decreasing the pH of the ethanoic acid.
(2) and (3) Solid sodium carbonate and magnesium react with the ethanoic acid, thus both increasing
the pH of the ethanoic acid.
19
61 A (1) SO2(g) will neutralize the dilute sodium hydroxide solution, making the solution less alkaline. Thus
the pH of the solution will decrease.
(2) NaCl(s) is neutral. It will NOT lower the pH of the dilute sodium hydroxide solution.
(3) When NH3(g) dissolves in water, it reacts with water to give ammonium ions and hydroxide ions. It
will NOT lower the pH of the dilute sodium hydroxide solution.
62 B (1) The electrical conductivity of a solution is proportional to the concentration of mobile ions.
Hydrochloric acid is a strong acid while ethanoic acid is a weak acid.
Therefore 1 mol dm–3 hydrochloric acid has a higher concentration of mobile ions than ethanoic
acid does.
Hence the electrical conductivity of 1 mol dm–3 hydrochloric acid is higher than that of 1 mol dm–3
ethanoic acid.
(2) CH3COOH(aq) + NaOH(aq)
HCl(aq) + NaOH(aq)
According to the equations, 1 mole of CH3COOH / HCl requires 1 mole of NaOH for complete
neutralization.
\ 1 mol dm ethanoic acid and 1 mol dm hydrochloric acid of the same volume require the
same number of moles of NaOH for complete neutralization.
(3) For the neutralization between a strong acid and a strong alkali, the heat released is 57 kJ for 1 mole of water produced.
For neutralization in which either the acid or alkali or both are weak, the heat released is less than
57 kJ for 1 mole of water produced. The is because some energy is consumed when the weak acid
and weak alkali dissociate to give hydrogen ions and hydroxide ions before neutralization.
Ethanoic acid is a weak acid while hydrochloric acid is a strong acid.
Hence the temperature rise for the neutralization between 1 mol dm–3 ethanoic acid and NaOH(aq)
is lower than that between 1 mol dm–3 hydrochloric acid and NaOH(aq).
CH3COONa(aq) + H2O(l)
NaCl(aq) + H2O(l)
–3
–3
63 C (1) NaOH(aq) is a strong alkali while NH3(aq) is a weak alkali.
20
Hence 1 mol dm–3 NaOH(aq) is more alkaline than 1 mol dm–3 NH3(aq).
The pH of 1 mol dm
(2) The electrical conductivity of a solution is proportional to the concentration of mobile ions.
NaOH(aq) is a strong alkali while NH3(aq) is a weak alkali.
Therefore 1 mol dm–3 NaOH(aq) has a higher concentration of mobile ions than 1 mol dm–3 NH3(aq)
does.
Hence the electrical conductivity of 1 mol dm–3 NaOH(aq) is higher than that of 1 mol dm–3
NH3(aq).
–3
–3
NaOH(aq) is higher than that of 1 mol dm
NH3(aq).
(3) For the neutralization between a strong acid and a strong alkali, the heat released is 57 kJ for 1 mole of water produced.
For neutralization in which either the acid or alkali or both are weak, the heat released is less than
57 kJ for 1 mole of water produced. The is because some energy is consumed when the weak acid
and weak alkali dissociate to give hydrogen ions and hydroxide ions before neutralization.
NaOH(aq) is a strong alkali while NH3(aq) is a weak acid.
Hence the temperature rise for the neutralization between 1 mol dm–3 NaOH(aq) and HCl(aq) is
larger than that between 1 mol dm–3 NH3(aq) and HCl(aq).
64 D Carbon dioxide reacts with water to form carbonic acid.
CO2(g) + H2O(l)
H2CO3(aq)
Carbonic acid is a weak acid. It only partially dissociates in water.
H2CO3(aq)
(1) The resulting solution contains mobile ions and thus conducts electricity better than water.
(2) Carbonic acid undergoes neutralization with dilute sodium hydroxide solution.
2H+(aq) + CO32–(aq)
65 D (2) Aqueous solution of citric acid can conduct electricity. Hence citric acid is an electrolyte.
66 A
67 D Very dilute nitric acid acts as a typical acid. It reacts with magnesium.
When we describe acids as strong and weak, we are talking about the extent of their dissocation in
water. When we talk about concentration, we are referring to the amount of an acid in a unit volume
of solution.
For example, 5 mol dm–3 nitric acid is a concentrated solution of a strong acid while 0.1 mol dm–3
nitric acid is a dilute solution of a strong acid.
68 C Adding 1 mol dm–3 NaCl(aq) to 50 cm3 of 1 mol dm–3 HCl(aq) increases the total volume of the acid.
Hence the concentration of hydrogen ions in the acid decreases.
Thus the pH of the acid is affected.
69 D Ethanoic acid is a weak acid.
The basicity of ethanoic acid is 1.
)
)
0
$
$
POMZUIJTIZESPHFO
BUPNDBOVOEFSHP
EJTTPDJBUJPO
0
)
)
21
70 B Sulphuric acid is a strong acid because it almost completely dissociates in water.
Ethanoic acid is a weak acid because it only partly dissociates in water.
71 B Solid citric acid does NOT react with magnesium because it does not contain hydrogen ions.
The aqueous solution of citric acid shows the properties of an acid while solid citric acid does not.
72 B A solution can conduct electricity due to the presence of mobile ions.
When solid citric acid dissolves in water, some of the molecules dissociate to give ions.
citric acid(aq)
Hence an aqueous solution of citric acid can conduct electricity.
H+(aq) + citrate ion(aq)
73 C CH3COOH(aq) + NaOH(aq)
CH3COONa(aq) + H2O(l)
HCl(aq) + NaOH(aq)
NaCl(aq) + H2O(l)
According to the equations, 1 mole of CH 3COOH / HCl requires 1 mole of NaOH for complete
neutralization.
\ 1 mol dm–3 ethanoic acid and 1 mol dm–3 hydrochloric acid of the same volume require the same
number of moles of NaOH for complete neutralization.
When ammonia gas dissolves in water, it reacts with water to give ammonium ions and hydroxide ions.
However, ammonia does not react with water completely. Only very few hydroxide ions are formed.
NH3(g) + H2O(l)
Hence there are many mobile ammonia molecules in aqueous ammonia.
NH4+(aq) + OH–(aq)
74 C Both 1 mol dm –3 NaOH(aq) and 1 mol dm –3 NH 3 (aq) form a reddish brown precipitate with
Fe2(SO4)3(aq).
75 C CH3COOH(aq) + NaOH(aq)
CH3COONa(aq) + H2O(l)
HCl(aq) + NaOH(aq)
NaCl(aq) + H2O(l)
According to the equations, 1 mole of CH 3COOH / HCl requires 1 mole of NaOH for complete
neutralization.
Hence 10 cm3 of 1 mol dm–3 ethanoic acid and 10 cm3 of 1 mol dm–3 hydrochloric acid require the
same number of moles of sodium hydroxide for complete neutralization.
Unit 16 Salts and neutralization
Fill in the blanks
22
1
neutralization
2
exothermic
3
sodium hydrogensulphate; sodium sulphate
4
calcium; lead(II); barium
5
lead(II); silver
True or false
6
T
7
F In neutralization, salt and water are the only products. However, potassium carbonate and dilute
hydrochloric acid react to give a salt, carbon dioxide gas and water. Therefore the reaction is NOT a
neutralization reaction.
8
T
9
F Heat is absorbed during the sublimation of iodine. Hence this is NOT an exothermic process.
10 F The reaction between dilute hydrochloric acid and dilute sodium hydroxide solution produces only one
salt, sodium chloride. Sodium chloride is a normal salt.
11 T Sulphur dioxide would react with sodium hydroxide solution.
SO2(g) + 2NaOH(aq)
Na2SO3(aq) + H2O(l)
12 T A molecule of sulphuric acid can produce two hydrogen ions. If dilute sulphuric acid is allowed to react
with dilute sodium hydroxide solution, two kinds of salt can form. When one of the hydrogen ions is
replaced, the salt formed is sodium hydrogensulphate (NaHSO4). It is called an acid salt. When both
hydrogen ions are replaced, the salt formed is sodium sulphate (Na2SO4). It is called a normal salt.
acid
(sulphuric acid)
H+
+
H
acid salt
(sodium hydrogensulphate)
SO
replacing
Na+
H+
2–
4
Na+
H+
normal salt
(sodium sulphate)
SO4
SO42–
2–
Na+
replacing
Na+
Na+
(from sodium
hydroxide solution)
(from sodium
hydroxide solution)
H+
23
13 T A molecule of carbonic acid can produce two hydrogen ions. During the reaction between carbonic
acid and dilute potassium hydroxide, when one of the hydrogen ions is replaced, the acid salt
potassium hydrogen carbonate (KHCO3) is formed.
acid
(carbonic acid)
H+
+
H
acid salt
(potassium hydrogencarbonate)
H
2–
3
+
CO
replacing
K+
CO32–
+
H
K+
(from potassium
hydroxide solution)
14 F The pH of a salt solution may NOT equal 7.
Most acid salts give an acidic solution when dissolved in water. For example, sodium hydrogensulphate
(NaHSO4) solution is acidic. This is because the hydrogensulphate ion (HSO4–) can dissociate to give hydrogen
ion.
15 F Some salts are formed from the neutralization between an acid and an insoluble metal hydroxide / an
insoluble metal oxide.
For example, copper(II) oxide is an insoluble metal oxide. It reacts with dilute sulphuric acid to give copper(II)
sulphate and water.
H2SO4(aq) + CuO(s)
CuSO4(aq) + H2O(l)
16 F A few acid salts give alkaline solutions when dissolved in water.
17 T
18 F Silver iodide is yellow in colour.
19 T Lead(II) nitrate solution gives a white precipitate with dilute hydrochloric acid.
Pb2+(aq) + 2Cl–(aq)
PbCl2(s)
20 F Zinc chloride solution gives a white precipitate with sodium carbonate solution. NO gas is evolved in
the reaction.
ZnCl2(aq) + Na2CO3(aq)
ZnCO3(s) + 2NaCl(aq)
21 F When dilute sulphuric acid reacts with lead, insoluble lead(II) sulphate forms. The lead(II) sulphate covers
the surface of lead and prevents further reaction.
24
22 F Farmers neutralize acidic soil by adding quicklime. Ammonium sulphate gives an acidic solution when
dissolved in water.
23 T
24 F Sodium hydroxide is highly corrosive. It is NEVER used as an active ingredient in antacids.
25 F A lot of heat is released when concentrated hydrochloric acid reacts with concentrated aqueous
ammonia. This will cause skin burn.
Furthermore, concentrated aqueous ammonia is corrosive. The hand should be washed immediately with
plenty of water.
Multiple choice questions
26 A Iron(III) hydroxide reacts with dilute hydrochloric acid to give iron(III) chloride and water. The resulting
solution is yellow-brown in colour.
Fe(OH)3(s) + 3HCl(aq)
FeCl3(aq) + 3H2O(l)
27 D Only Zn(OH)2 dissolves in excess dilute sodium hydroxide solution due to the formation of a soluble
complex salt.
28 B Insoluble CaCO3 is formed in the reaction between solutions of Na2CO3 and Ca(NO3)2.
29 D For the neutralization between a strong acid and a strong alkali, the heat released is 57 kJ for 1 mole
of water produced.
For neutralization in which either the acid or alkali or both are weak, the heat released is less than 57
kJ for 1 mole of water produced. The is because some energy is consumed when the weak acid and
weak alkali dissociate to give hydrogen ions and hydroxide ions before neutralization.
Option
Solutions mixed
3
–3
Strength of acid and alkali
Temperature rise
—
25 cm of 1 mol dm HCl(aq) and
25 cm3 of 1 mol dm–3 KOH(aq)
neutralization between a strong
acid and a strong alkali
T
A
25 cm3 of 1 mol dm–3 CH3COOH(aq)
and 25 cm3 of 1 mol dm–3 NH3(aq)
neutralization between a weak
acid and a weak alkali
< T
B
25 cm3 of 1 mol dm–3 CH3COOH(aq)
and 25 cm3 of 1 mol dm–3 NaOH(aq)
neutralization between a weak
acid and a strong alkali
< T
C
25 cm3 of 1 mol dm–3 HNO3(aq) and
25 cm3 of 1 mol dm–3 NH3(aq)
neutralization between a strong
acid and a weak alkali
< T
D
25 cm3 of 1 mol dm–3 HNO3(aq) and
25 cm3 of 1 mol dm–3 NaOH(aq)
neutralization between a strong
acid and a strong alkali
T
\ the temperature rise for mixing 25 cm3 of 1 mol dm–3 HNO3(aq) and 25 cm3 of 1 mol dm–3
NaOH(aq) is the same as for mixing 25 cm3 of 1 mol dm–3 HCl(aq) and 25 cm3 of 1 mol dm–3
KOH(aq).
25
30 C For the neutralization between a strong acid and a strong alkali, the heat released is 57 kJ for 1 mole
of water produced.
Expt.
Acid and alkali mixed
1
100 cm3 of 1 mol dm–3
HCl(aq) and
100 cm3 of 1 mol dm–3
NaOH(aq)
100 cm3 of 2 mol dm–3
HCl(aq) and 100 cm3 of
2 mol dm–3 NaOH(aq)
2
Number of moles of HCl /
NaOH mixed
Heat
Number of moles of water formed
= molarity of solution x
released
volume of solution
–3
1 mol dm
100
x
dm3
1 000
HCl(aq) + NaOH(aq)
0.1 mol 0.1 mol
NaCl(aq) + H2O(l)
0.1 mol
5.7 kJ
100
x
dm3
1 000
HCl(aq) + NaOH(aq)
0.2 mol 0.2 mol
NaCl(aq) + H2O(l)
0.2 mol
11.4 kJ
= 0.1 mol
–3
2 mol dm
= 0.2 mol
The total volumes of the two mixtures are the same. Hence the temperature rise of the first mixture is
1
half that of the second mixture, i.e. T1 =
T2.
2
31 B For the neutralization between a strong acid and a strong alkali, the heat released is 57 kJ for 1 mole
of water produced.
Expt.
Acid and alkali mixed
1
100 cm of 1 mol dm
HCl(aq) and
3
–3
100 cm of 1 mol dm
KOH(aq)
3
–3
50 cm3 of 1 mol dm–3
HCl(aq) and 50 cm3 of
1 mol dm–3 KOH(aq)
2
Number of moles of HCl /
NaOH mixed
Heat
Number of moles of water formed
= molarity of solution x
released
volume of solution
1 mol dm–3 x
100
dm3
1 000
= 0.1 mol
–3
1 mol dm
x
50
3
dm
1 000
= 0.05 mol
HCl(aq) + KOH(aq)
0.1 mol 0.1 mol
5.7 kJ
KCl(aq) + H2O(l)
0.1 mol
HCl(aq) + KOH(aq)
0.05 mol 0.05 mol
KCl(aq) + H2O(l)
0.05 mol
2.85 kJ
The first mixture (total volume 200 cm3) is heated up by 5.7 kJ while the second mixture (total volume
100 cm3) is heated up by 2.28 kJ. Hence the two mixtures show the same temperature rise, i.e. T1 + T2.
32 D For the neutralization between a strong acid and a strong alkali, the heat released is 57 kJ for 1 mole
of water produced.
For neutralization in which either the acid or alkali or both are weak, the heat released is less than 57
kJ for 1 mole of water produced. The is because some energy is consumed when the weak acid and
weak alkali dissociate to give hydrogen ions and hydroxide ions before neutralization.
Expt.
Solutions mixed
3
26 –3
Strength of acid and alkali
Temperature rise
1
100 cm of 1 mol dm HNO3(aq) and
100 cm3 of 1 mol dm–3 NaOH(aq)
neutralization between a strong
acid and a strong alkali
T1
2
100 cm3 of 1 mol dm–3 HNO3(aq) and
100 cm3 of 1 mol dm–3 NH3(aq)
neutralization between a strong
acid and a weak alkali
T2
\ T1 > T2.
33 D For the neutralization between a strong acid and a strong alkali, the heat released is 57 kJ for 1 mole
of water produced.
For neutralization in which either the acid or alkali or both are weak, the heat released is less than 57
kJ for 1 mole of water produced. The is because some energy is consumed when the weak acid and
weak alkali dissociate to give hydrogen ions and hydroxide ions before neutralization.
Mixed
Strength of acid and alkali
Temperature rise
100 cm of 2 mol dm HNO3(aq) +
100 cm3 of 2 mol dm–3 NaOH(aq)
neutralization between a strong acid and
a strong alkali
x
100 cm3 of 2 mol dm–3 HCl(aq) +
100 cm3 of 2 mol dm–3 NH3(aq)
neutralization between a strong acid and
a weak alkali
y < x
100 cm3 of 2 mol dm–3 HCl(aq) +
100 cm3 of 2 mol dm–3 KOH(aq)
neutralization between a strong acid and
a strong alkali
z = x
3
–3
\ x = z > y
34 C Fe3+(aq) + 3OH–(aq)
Fe(OH)3(s)
reddish brown precipitate
35 A Sodium carbonate solution reacts with copper(II) sulphate solution to give a blue precipitate, copper(II)
carbonate.
36 D
Na2CO3(aq) + CuSO4(aq)
Option
CuCO3(s) + Na2SO4(aq)
A precipitate
formed
Solution mixed
Ionic equation
for reaction involved
A
barium chloride solution and silver
nitrate solution
4
Ag+(aq) + Cl–(aq)
B
barium chloride solution and dilute
sulphuric acid
4
Ba (aq) + SO4 (aq)
BaSO4(s)
C
barium chloride solution and sodium
sulphate solution
4
Ba2+(aq) + SO42–(aq)
BaSO4(s)
D
barium chloride solution and sodium
nitrate solution
8
—
2+
AgCl(s)
2–
\ Sodium nitrate solution would NOT give a precipitate with barium chloride solution.
37 A Option C — Ammonium compounds are soluble in water.
38 A
Option
Ions reacted
Precipitate formed?
A
Ba2+(aq) and SO42–(aq)
a white precipitate (BaSO4)
B
Cu2+(aq) and NO3–(aq)
no precipitate
C
Ni2+(aq) and CO32–(aq)
a green precipitate (NiCO3)
D
Na2+(aq) and OH–(aq)
no precipitate
Ionic equation involved
Ba2+(aq) + SO42–(aq)
BaSO4(s)
—
Ni2+(aq) + CO32–(aq)
NiCO3(s)
—
27
39 A K2CO3(aq) + Ba(NO3)2(aq)
reactant A
BaCO3(s) + 2HCl(aq)
white reactant B precipitate X BaCO3(s) + 2KNO3(aq)
white
precipitate X
CO2(g) + BaCl2(aq) + H2O(l)
colourless solution Y
40 B Copper(II) oxide reacts with dilute sulphuric acid to give copper(II) sulphate and water.
CuO(s) + H2SO4(aq)
CuSO4(aq) + H2O(l)
Copper(II) sulphate is soluble in water while copper is insoluble. Therefore copper can be separated
from the mixture by filtration.
41 C Option A — AgNO3(aq) forms a white precipitate (Ag2CO3) with CO32–(aq) ions.
2Ag+(aq) + CO32–(aq)
AgNO3(aq) also forms a white precipitate (Ag2SO4) in concentrated solution of SO4 (aq)
ions.
Option B — Ba(NO3)2(aq) forms a white precipitate with both SO42–(aq) ions and CO32–(aq) ions.
Ba2+(aq) + SO42–(aq)
Ba (aq) + CO3 (aq)
Option C — Fe(NO3)3(aq) would give a precipitate with CO32–(aq) ions, but NOT with SO42–(aq) ions.
Ag2CO3(s)
2–
2+
2–
BaSO4(s)
BaCO3(s)
Hence Fe(NO3)3(aq) can be used to separate the anions SO42–(aq) and CO32–(aq).
42 D Option A — Calcium carbonate gives gas bubbles with dilute sulphuric acid. However, it is insoluble in
water.
Option B — Lead(II) sulphate does not give gas bubbles with dilute sulphuric acid. It is also insoluble
in water.
Option C — Magnesium chloride does not give gas bubbles with dilute sulphuric acid.
Option D — Potassium carbonate is soluble in water and gives gas bubbles (carbon dioxide) with dilute
sulphuric acid.
43 D Option A — Calcium sulphate is insoluble in water.
Option B — Copper(II) oxide is not white in colour. It is also insoluble in water.
Option C — Iron(II) chloride is not white in colour.
Option D — Neutralization occurs when potassium hydroxide is mixed with dilute sulphuric acid. Heat
is released.
44 B Copper(II) oxide is black in colour. When copper(II) oxide is mixed with dilute sulphuric acid, a blue
solution results.
28
CuO(s) + H2SO4(aq)
CuSO4(aq) + H2O(l)
45 C Lead(II) carbonate gives a colourless gas (carbon dioxide) with dilute nitric acid.
PbCO3(s) + 2HNO3(aq)
Pb(NO3)2(aq) + H2O(l) + CO2(g)
The resulting solution contains lead(II) ions. The solution gives a white precipitate (lead(II) chloride) with
dilute hydrochloric acid.
Pb2+(aq) + 2Cl–(aq)
PbCl2(s)
46 C Zinc reacts with dilute H2SO4(aq) (reactant X) to give a solution containing Zn2+ ions.
Zn(s) + H2SO4(aq)
ZnSO4(aq) + H2(g)
The solution gives a precipitate (ZnCO3(s)) with K2CO3(aq) (reactant Y).
Zn2+(aq) + CO32–(aq)
ZnCO3(s)
47 A Calcium reacts with water to give calcium hydroxide and hydrogen
Ca(s) + 2H2O(l)
Ca(OH)2(s) + H2(g)
Calcium hydroxide is slightly soluble in water. Hence the clear solution contains calcium ions. It gives a
white precipitate with sodium carbonate solution.
Ca2+(aq) + CO32–(aq)
CaCO3(s)
48 B From Test 1, it can be deduced that solid X is a sulphate. A sulphate gives a white precipitate (barium
sulphate) with barium nitrate solution.
Ba2+(aq) + SO42–
From Test 2, it can be deduced that solid X contains aluminium ions rather than calcium ions.
Aluminium ions give a white precipitate (zinc hydroxide) with dilute sodium hydroxide solution. The
precipitate dissolves in excess alkali due to the formation of a complex salt.
Al3+(aq) + 3OH–(aq)
Al(OH)3(s) + OH–(aq)
BaSO4(s)
Al(OH)3(s)
[Al(OH)4]–(aq)
49 C Lead(II) sulphate is an insoluble salt. It can be prepared by adding lead(II) nitrate to dilute sulphuric
acid.
Pb2+(aq) + SO42–(aq)
PbSO4(s)
50 C Options A, B and D — Oxides or hydroxides of copper, iron and zinc are insoluble bases. Sulphates of
copper, iron and zinc are NOT prepared by an acid-alkali titration method.
Option C — Sodium sulphate can be prepared by reacting dilute sulphuric acid with dilute sodium
hydroxide solution via an acid-alkali titration method.
/VNCFSPGJPOT
51 D
FRVJWBMFODFQPJOU
"NPVOUPGBDJE
BEEFE
29
As sulphuric acid was added, it removed both barium ions (by precipitation) and hydroxide ions (by
neutralization).
Ba(OH)2(aq) + H2SO4(aq)
At the equivalence point, all the barium ions and hydroxide ions had been removed. Hence the number
of ions in the mixture falls until the equivalence point.
After the equivalence point, the number of ions in the mixture rises as excess sulphuric acid was
added.
BaSO4(s) + 2H2O(l)
52 C Option A — Ammonium sulphate gives an acidic solution when dissolved in water. Farmers may add it
to soil which has become too alkaline.
Option C — Slaked lime (calcium hydroxide) can neutralize the acidic waste.
53 A
54 A
55 B (2) This is NOT a neutralizaton reaction.
In neutralization, salt and water are the only products.
56 A (1) Heat is released in neutralization reactions.
(2) The product is sodium nitrate, a normal salt.
(3) Sodium nitrate is an ionic compound.
57 C (1) When hydrogen chloride gas dissolves in water, almost all the hydrogen chloride molecules
dissociate to give hydrogen ions and chloride ions. The hydrochloric acid formed contains manly
hydrogen ions (H+(aq)), chloride ions (Cl–(aq)) and water molecules. Hydrochloric acid is a strong
acid.
HCl(aq)
H+(aq) + Cl–(aq)
(2) A solution of hydrogen chloride in water can conduct electricity due to the presence of mobile
ions.
(3) A solution of hydrogen chloride and sodium hydroxide solution undergo a neutralization reaction
when mixed. The reaction is exothermic.
58 D Iron reacts with dilute sulphuric acid to give a rion(II) sulphate solution and hydrogen gas.
Fe(s) + H2SO4(aq)
FeSO4(aq) + H2(g)
59 A (1) Heat would be released when concentrated sulphuric acid was mixed with water.
30
(2) Dilute sulphuric acid and dilute aqueous ammonia underwent a neutralization reaction when mixed.
Heat would be released.
(3) A white precipitate (barium sulphate) appeared upon mixing dilute sulphuric acid and barium
chloride solution.
Ba2+(aq) + SO42–(aq)
BaSO4(s)
60 C (1) There is NO reaction between copper and dilute ethanoic acid.
61 A Heat is absorbed during the evaporation of ethanol. Hence this is NOT an exothermic process.
62 B For the neutralization between a strong acid and a strong alkali, the heat released is 57 kJ for 1 mole
of water produced.
Expt.
For neutralization in which either the acid or alkali or both are weak, the heat released is less than 57
kJ for 1 mole of water produced. The is because some energy is consumed when the weak acid and
weak alkali dissociate to give hydrogen ions and hydroxide ions before neutralization.
Acid and alkali Number of moles of acid /
mixed
alkali mixed
= molarity of solution x
volume of solution
Number of moles of
water formed
Strength of acid Heat
and alkali
released
Temperature rise
20 cm3 of 1 mol
dm–3 HCl(aq) and
20 cm3 of 1 mol
dm–3 NaOH(aq)
Number of moles of HCl /
H+(aq) + OH–(aq)
NaOH
0.02 mol 0.02 mol 20
= 1 mol dm–3 x
dm3
H2O(l)
1 000
= 0.02 mol
0.02 mol
neutralization
between a
strong acid and
a strong alkali
1.14 kJ
1
20 cm3 of 1 mol
dm–3 HCl(aq) and
20 cm3 of 1 mol
dm–3 NH3(aq)
Number of moles of HCl /
H+(aq) + OH–(aq)
NH3
0.02 mol 0.02 mol 20
= 1 mol dm–3 x
dm3
H2O(l)
1 000
= 0.02 mol
0.02 mol
neutralization
between a
strong acid and
a weak alkali
<1.14 kJ This mixture (total
volume 40 cm3)
is heated up by
<1.14 kJ.
Hence the
temperature rise
is <T.
2
40 cm of 1 mol
–3
dm HCl(aq) and
3
40 cm of 1 mol
–3
dm NaOH(aq)
Number of moles of HCl /
H (aq) + OH (aq)
NaOH
0.04 mol 0.04 mol 40
–3
3
= 1 mol dm x
dm
H2O(l)
1 000
= 0.04 mol
0.04 mol
neutralization
between a
strong acid and
a strong alkali
2.28 kJ
This mixture (total
3
volume 80 cm ) is
heated up by 2.28
kJ.
Hence the
temperature rise
is T.
3
10 cm3 of 2 mol
dm–3 HCl(aq) and
10 cm3 of 2 mol
dm–3 NaOH(aq)
Number of moles of HCl /
H+(aq) + OH–(aq)
NaOH
0.02 mol 0.02 mol 10
–3
3
= 2 mol dm x
dm
H2O(l)
1 000
= 0.02 mol
0.02 mol
neutralization
between a
strong acid and
a strong alkali
1.14 kJ
This mixture (total
volume 20 cm3) is
heated up by 1.14
kJ.
Hence the
temperature rise
is 2T.
3
+
–
This mixture (total
volume 40 cm3) is
heated up by 1.14
kJ.
Assume that the
temperature rise
is T.
63 D (1) A solution containing iron(III) ions is yellow-brown in colour.
31
64 D
The white precipitate formed
with lead(II) nitrate solution
Reagent
Ionic equation involved
(1)
Dilute sulphuric acid
lead(II) sulphate
Pb2+(aq) + SO42–(aq)
(2)
Potassium chloride solution
lead(II) chloride
Pb (aq) + 2Cl (aq)
(3)
Sodium carbonate solution
lead(II) carbonate
Pb2+(aq) + CO32–(aq)
65 D
Precipitate formed with
silver nitrate solution
Solution
(1)
KBr(aq)
silver bromide
(2)
Na2CO3(aq)
silver carbonate
(3)
HCl(aq)
silver chloride
66 D
2+
PbCl2(s)
PbCO3(s)
Ionic equation involved
Ag+(aq) + Br–(aq)
2Ag+(aq) + CO32–(aq)
+
–
Ag (aq) + Cl (aq)
White precipitate formed
with dilute sulphuric acid
Substance
–
PbSO4(s)
AgBr(s)
Ag2CO3(s)
AgCl(s)
Ionic equation involved
(1)
saturated limewater
calcium sulphate
Ca2+(aq) + SO42–(aq)
CaSO4(s)
(2)
barium chloride solution
barium sulphate
Ba2+(aq) + SO42–(aq)
BaSO4(s)
(3)
lead(II) nitrate solution
lead(II) sulphate
Pb2+(aq) + SO42–(aq)
PbSO4(s)
67 C (1) Ammonium iodide is an ionic compound. It contains ammonium ions and iodide ions.
(2) Ammonium iodide solution gives a precipitate (silver iodide) with silver nitrate solution.
Ag+(aq) + I–(aq)
68 D
AgI(s)
Solutions mixed
(1)
Precipitate formed
Pb(NO3)2(aq) and K2SO4(aq)
PbSO4
Ionic equation involved
2+
2–
2+
–
Pb (aq) + SO4 (aq)
PbSO4(s)
(2)
NH3(aq) and Fe(NO3)2(aq)
Fe(OH)2
Fe (aq) + 2OH (aq)
Fe(OH)2(s)
(3)
(NH4)2CO3(aq) and CaCl2(aq)
CaCO3
Ca2+(aq) + CO32–(aq)
CaCO3(s)
69 A (1) Potassium iodide solution conducts electricity due to the presence of mobile ions (potassium ions
and iodide ions).
32
(2) Potassium iodide solution gives a yellow precipitate (silver iodide) with silver nitrate solution.
(3) Potassium iodide solution is colourless.
Ag+(aq) + I–(aq)
AgI(s)
70 B (1) Iron is more reactive than silver, but less reactive than zinc.
Iron can displace silver from silver nitrate solution.
There is NO reaction between iron and zinc nitrate solution.
(3) Potassium chloride solution gives a white precipitate (silver chloride) with silver nitrate solution.
Ag+(aq) + Cl–(aq)
There is NO observable change when potassium chloride solution and zinc nitrate solution are
mixed.
AgCI(s)
71 A The student can obtain calcium chloride crystals from the reaction mixture using the steps below:
• Remove the excess calcium carbonate by filtration (using filter paper and filter funnel).
• Heat the filtrate to evaporate about half of the water (using Bunsen burner and evaporating basin).
Set the concentrated solution aside to cool and crystallize.
Filter (using filter paper and filter funnel) the crystals from the remaining solution.
72 B (1) Adding excess aqueous ammonia will leave an alkaline solution after reacting with the acid.
(2) Adding excess calcium carbonate will not leave an alkaline solution after reacting with the acid
because calcium carbonate is insoluble in water.
(3) Sodium chloride does not react with the acid.
The following equation represents the reaction between dilute aqueous ammonia and dilute
hydrochloric acid:
NH3(aq) + HCl(aq)
Ammonium chloride is produced in the reaction.
NH4Cl(aq)
73 A (3) Ammonium chloride is soluble in water. NO precipitate is formed in the reaction between dilute
aqueous ammonia and dilute hydrochloric acid.
74 C (1) CuO is a base, NOT a salt.
75 D Zinc sulphate can be prepared by reacting dilute sulphuric acid with either zinc, zinc oxide or zinc
carbonate.
Zn(s) + H2SO4(aq)
ZnO(s) + H2SO4(aq)
ZnCO3(s) + H2SO4(aq)
ZnSO4(aq) + H2(g)
ZnSO4(aq) + H2O(l)
ZnSO4(aq) + H2O(l) + CO2(g)
76 B (2) The chemical formula of calcium hydrogenphosphate is CaHPO4, NOT CaH2PO4.
77 B (2) Ammonium ethanoate is a normal salt.
It is formed from ethanoic acid, a monobasic acid. A monobasic acid can form normal salts only.
(3) Aqueous solution of ammonium ethanoate can conduct electricity due to the presence of mobile
ions.
33
78 A (1) Antacids contain bases. They will react with acidic substances, e.g. vinegar.
79 A
80 C The number of moles of Na+ ions in 10 cm3 of 1 mol dm–3 sodium hydroxide solution
10
= 1 mol dm–3 x
dm3
1 000
= 0.01 mol
When 10 cm3 of 1 mol dm–3 hydrochloric acid are added to 10 cm3 of 1 mol dm–3 sodium hydroxide
solution,
total volume of the solution mixture = (10 + 10) cm3
= 20 cm3
Concentration of Na+ ions in the solution mixture =
\ the concentration of Na+ ions would change.
(
0.01 mol
20
3
dm
1 000
)
= 0.5 mol dm–3
81 B Sodium chloride is a salt formed from hydrochloric acid.
Hydrochloric acid is a monobasic acid. It can form normal salts only.
82 D Some salts are insoluble in water.
Some salts are formed from the neutralization between an acid and an insoluble metal hydroxide an
insoluble metal oxide.
For example, copper(II) oxide is an insoluble metal oxide. It reacts with dilute sulphuric acid to give
copper(II) sulphate and water.
H2SO4(aq) + CuO(s)
CuSO4(aq) + H2O(l)
83 D Only the hydrogen atom in the –COOH group of the ethanoic acid can undergo dissociation. Ethanoic
acid is a monobasic acid. It cannot form acid salt.
84 C There is NO reaction between copper and dilute hydrochloric acid.
85 B Ammonium chloride is a salt formed from the reaction between dilute aqueous ammonia and dilute
hydrochloric acid.
Ammonium chloride can conduct electricity in aqueous solution due to the presence of mobile ions.
86 C Nickel(II) carbonate is insoluble in water.
87 C Sodium hydroxide is corrosive and thus NEVER used in antacids.
88 C A lot of heat is released when vinegar reacts with concentrated sodium hydroxide solution. This will
cause skin burn.
34
The student should wash the affected area with plenty of water.
Unit 17 Concentration of solutions and volumetric analysis
Fill in the blanks
1
a) electronic balance
b) volumetric flask
c) pipette
d) burette
2
standard
3
pipette; burette
4
equivalence
5
methyl orange
6
Phenolphthalein
True or false
7
F To dilute 100 cm3 of 1.0 mol dm­–3 hydrochloric acid to 0.10 mol dm–3, add water until the total
volume of the solution is 1 000 cm3.
8
F The volumetric flask should be washed only with distilled water before use.
9
F A volumetric flask is used to prepare a solution of accurately known volume, e.g. 250.0 cm3.
Washing the volumetric flask with the acid before use would increase the number of moles of solute it
holds, making the concentration of the sulphuric acid obtained higher than calculated.
10 T
11 F Potassium hydroxide absorbs moisture from the air and cannot be weighed accurately. It is unsuitable
for preparing a standard solution.
12 F A solution with an accurately known concentration is a standard solution.
13 T If we wash the pipette or burette with distilled water only, water droplets remaining on the inside of
the glassware will dilute the solution that the glassware is going to contain.
14 F The conical flask is to hold a specific volume of a solution (usually 25.0 cm3), i.e. a specific amount
of the solute. It should NOT be washed with the solution it is to contain before use because the
additional amount of solute remaining in the flask will affect the titration results.
15 F The last drop of the solution in the pipette should NOT be blown out.
35
16 T The following diagram shows the titration curve for the titration of 20.0 cm3 of 0.1 mol dm–3 HCl(aq)
with 0.1 mol dm–3 NaOH(aq).
QIFOPMQIUIBMFJO
DIBOHFTDPMPVS
XJUIJOUIJTQ)SBOHF
Q)
FRVJWBMFODF
QPJOU
7PMVNFPG/B0)BR
BEEFEDN Phenolphthalein changes colour within the pH range of the vertical part of the titration curve.
Hence phenolphthalein is a suitable indicator for the titration.
17 F The following diagram shows the titration curve for the titration of 0.1 mol dm–3 CH3COOH(aq) with 0.1 mol dm–3 NaOH(aq).
Q)
FRVJWBMFODF
QPJOU
7PMVNFPG/B0)BR
BEEFEDN
Methyl orange does NOT change colour within the pH range of the vertical part of the titration curve.
Hence methyl orange is NOT a suitable indicator for the titration.
18 F During the titration of an acid and an alkali, the pH of the solution mixture may NOT be 7.00.
For example, during the titration of a weak acid and a strong alkali, the pH of the solution mixture is
greater than 7.00.
19 F During a titration the end point and the equivalence point usually occur at slightly different times.
However, we have to assume that the end point is the equivalence point and recognize this assumption
as a source of error.
36
20 T
Number of moles of acid used
= molarity of solution x volume of solution
Number of ions moles of NaOH
required for complete neutralization
Number of moles of CH3COOH
20
3
= 1 mol dm–3 x
dm
1 000
= 0.02 mol
CH3COOH(aq) + NaOH(aq)
CH3COONa(aq) + H2O(l)
0.02 mol ? mol
\ 20 cm3 of 1 mol dm–3 CH3COOH(aq) require 0.02 mole
of NaOH for complete neutralization.
Number of moles of H2SO4
10
3
= 1 mol dm–3 x
dm
1 000
= 0.01 mol
H2SO4(aq) + 2NaOH(aq)
Na2SO4(aq) + 2H2O(l)
0.01 mol ? mol
\ 10 cm3 of 1 mol dm–3 H2SO4(aq) require 0.02 mole of
NaOH for complete neutralization.
Multiple choice questions
21 C Molar mass of (COOH)2•2H2O = [2 x (12.0 + 2 x 16.0 + 1.0) + 2 x (2 x 1.0 + 16.0)] g mol–1
–1
= 126.0 g mol
mass
molar mass
5.04 g
=
126.0 g mol–1
= 0.0400 mol
Number of moles of (COOH)2•2H2O used=
Concentration of acid solution=
=
= 0.160 mol dm–3
number of moles of (COOH)2•2H2O
volume of solution
0.0400 mol
250.0
dm3
1 000
(
)
22 C Consider 1 000 cm3 (i.e. 1.00 dm3) of the sample.
Mass of 1 000 cm3 of the sample = 1.41 g cm–3 x 1 000 cm3
= 1 410 g
Mass of HNO3 in 1 000 cm3 of sample
= mass of 1 000 cm3 of sample x percentage by mass of HNO3 in sample
= 1 410 g x 61.5%
= 867 g
Molar mass of HNO3 = (1.0 + 14.0 + 3 x 16.0) g mol–1
= 63.0 g mol–1
Number of moles of HNO3 in 1.00 dm3 of sample =
mass
molar mass
867 g
=
63.0 g mol–1
= 13.8 mol
37
number of moles of HNO3
volume of solution
13.8 mol
=
1.00 dm3
= 13.8 mol dm–3
Molarity of nitric acid =
\ the concentration of the nitric acid sample is 13.8 mol dm–3.
23 A Mass of HNO3 in 60.0 m3 acid = 68 000 000 g x 35.5%
= 24 100 000 g
Molar mass of HCl = (1.0 + 35.5) g mol–1
= 36.5 g mol–1
Number of moles of HCl =
Molarity of hydrochloric acid =
\ the concentration of the hydrochloric acid is 11.0 mol dm–3.
mass
molar mass
24 100 000 g
=
36.5 g mol–1
= 660 000 mol
number of moles of HCl
volume of solution
660 000 mol
=
60 000 dm3
= 11.0 mol dm–3
24 B Mass of H2SO4 in 58.0 m3 acid = 55 000 000 g x 94.0%
= 51 700 000 g
38
Molar mass of H2SO4 = (2 x 1.0 + 32.1 + 4 x 16.0) g mol–1
= 98.1 g mol–1
Number of moles of H2SO4=
Molarity of the sulphuric acid =
\ the molarity of the sulphuric acid is 9.08 mol dm–3.
mass
molar mass
51 700 000 g
=
98.1 g mol–1
= 527 000 mol
number of moles of H2SO4
volume of solution
527 000 mol
=
58 000 dm3
= 9.08 mol dm–3
25 D One mole of K3PO4 contains 3 moles of K+ ions and 1 mole of PO43– ions.
Total number of mole of ions = 4 x 0.100 mol
= 0.400 mol
Total ion concentration in solution=
=
= 0.800 mol dm–3
\ the total ion concentration in the solution is 0.800 mol dm–3.
number of moles of ions
volume of solution
0.400 mol
500.0
dm3
1 000
(
)
26 D Number of moles of Na2CO3 present = molarity of solution x volume of solution
= 2.00 mol dm–3 x 2.00 dm3
= 4.00 mol
One mole of Na2CO3 contains 2 moles of Na+ ions and 1 mole of CO32– ions.
Total number of ions present = 3 x 4.00 mol x 6.02 x 1023 mol–1
= 7.22 x 1024
\ the total number of ions present is 7.22 x 10–4.
27 A pH of acid = –log10[H+]
= 2.70
i.e. log10[H+] = –2.70
[H+] = 10–2.70
= 1.995 x 10–3 mol dm–3
Sulphuric acid dissociates completely according to the following equation:
H2SO4(aq) Concentration of sulphate ions =
\ tht concentration of sulphate ions in the acid is 9.98 x 10–4 mol dm–3.
2H+(aq) + SO42–(aq)
1.995 x 10–3 mol dm–3
concentration of hydrogen ions
2
1.995 x 10–3
=
mol dm–3
2
= 9.98 x 10–4 mol dm–3
39
28 A
Substance
= 0.010 mol
Fe2(SO4)3
= 0.0025 mol
= 0.0040 mol
0.080 mol dm–3 x
MgSO4
0.050 mol dm–3 x
0.080 mol dm–3 x
CaCl2
= 0.0080 mol
3
–3
\ 100 cm of 0.10 mol dm­
number of ions.
Number of ions in one
formula unit of substance
Number of moles
of ions present
2
2 x 0.010 mol = 0.020 mol
50
3
dm
1 000
5
5 x 0.0025 mol = 0.013 mol
50
3
dm
1 000
3
3 x 0.0040 mol = 0.012 mol
100
3
dm
1 000
2
2 x 0.0080 mol = 0.016 mol
100
3
dm
1 000
0.10 mol dm–3 x
NaCl
Number of moles
of substance present
NaCl(aq) contain the greatest number of moles of ions, i.e. the greatest
29 C Number of moles of ZnCl2 in ZnCl2(aq) = molarity of solution x volume of solution
150.0
= 2.00 mol dm–3 x
dm3
1 000
= 0.300 mol
40
One mole of ZnCl2 contain 2 moles of Cl– ions.
\ number of moles of Cl (aq) ions in ZnCl2(aq) = 2 x 0.300 mol
= 0.600 mol
Number of moles of NaCl in NaCl(aq) = molarity of solution x volume of solution
50.0
= 1.20 mol dm–3 x
dm3
1 000
= 0.0600 mol
One mole of NaCl contains 1 mole of Cl– ions.
\ number of moles of Cl–(aq) ions in NaCl(aq) = 0.0600 mol
Total number of moles of Cl–(aq) ions = (0.600 + 0.0600) mol
= 0.660 mol
Total volume of solution X = (150.0 + 50.0) cm3
= 200.0 cm3
Concentration of Cl–(aq) ions in solution X=
\ the concentration of Cl–(aq) ions in solution X is 3.30 mol dm–3.
–
(
0.660 mol
200.0
dm3
1 000
)
= 3.30 mol dm–3
30 C One mole of sodium sulphate (Na2SO4) contains 2 moles of Na+ ions and 1 mole of SO42– ions.
number of moles of Na+ ions
2
2.00 x 10–2
=
mol
2
= 1.00 x 10–2 mol
Number of moles of Na2SO4 in solution=
Molarity of the sodium sulphate solution=
\ the molarity of the sodium sulphate solution is 5.00 x 10–1 mol dm–3.
number of moles of Na2SO4
volume of solution
1.00 x 10–2 mol
=
20.0
dm3
1 000
(
)
= 0.500 mol dm–3
31 D Number of moles of Na2CO3 in 2.50 mol dm–3 solution= molarity of solution x volume of solution
200.0
= 2.50 mol dm–3 x
dm3
1 000
= 0.500 mol
One mole of Na2CO3 contain 2 moles of Na+ ions.
\ number of moles of Na+(aq) ions in the solution= 2 x 0.500 mol
= 1.00 mol
Number of moles of Na2CO3 in 1.00 mol dm–3= molarity of solution x volume of solution
50.0
= 1.00 mol dm–3 x
dm3
1 000
= 0.0500 mol
Number of moles of Na+(aq) ions in the solution= 2 x 0.0500 mol
= 0.100 mol
Total number of moles of Na+(aq) ions = (1.00 + 0.100) mol
= 1.10 mol
Total volume of solution = (200.0 + 50.0) cm3
= 250.0 cm3
Concentration of Na+(aq) ions in the resulting solution=
\ the concentration of Na+(aq) ions in the resulting solution is 4.40 mol dm–3.
(
1.10 mol
250.0
dm3
1 000
)
= 4.40 mol dm–3
41
32 C One mole of magnesium chloride (MgCl2) contains 1 mole of Mg2+ ions and 2 moles of Cl– ions.
number of moles of Cl– ions
2
2.0 x 10–2
=
mol
2
= 1.0 x 10–2 mol
Number of moles of MgCl2 in solution =
Molarity of the magnesium chloride solution=
=
= 0.20 mol dm–3
\ the molarity of the solution is 2.0 x 10–1 mol dm–3.
number of moles of MgCl2
volume of solution
1.0 x 10–2 mol
50.0
dm3
1 000
(
)
33 C Number of moles of Fe2(SO4)3 in Fe2(SO4)3(aq) = molarity of solution x volume of solution
50
= 0.20 mol dm–3 x
dm3
1 000
= 0.010 mol
42
One mole of Fe2(SO4)3 contain 3 moles of SO42– ions.
\ number of moles of SO42– ions in Fe2(SO4)3(aq)= 3 x 0.010 mol
= 0.030 mol
Number of moles of K2SO4 in K2SO4(aq)= molarity of solution x volume of solution
200
= 0.25 mol dm–3 x
dm3
1 000
= 0.050 mol
One mole of K2SO4 contains 1 mole of SO42– ions.
\ number of moles of SO42– ions in K2SO4(aq) = 0.050 mol
Total number of moles of SO42– ions in solution X= (0.030 + 0.050) mol
= 0.080 mol
Total volume of solution X = (50 + 200) cm3
= 250 cm3
Concentration of SO42–(aq) ions in solution X=
\ the concentration of SO42–(aq) ions in solution X is 0.32 mol dm–3.
(
0.080 mol
250
dm3
1 000
)
= 0.32 mol dm–3
34 B (MV) before dilution = (MV) after dilution, where M = molarity, V = volume
5.0 x
V
250
= 0.40 x
1 000
1 000
3
V = 20 cm
35 A Volume of diluted solution = 3.75 dm3 + 250.0 cm3
= 4.00 dm3
(MV) before dilution = (MV) after dilution, where M = molarity, V = volume
3.20 x
\ the molarity of the diluted solution is 0.200 mol dm–3.
250.0
= M x 4.00
1 000
M = 0.200
36 B (MV) before dilution = (MV) after dilution, where M = molarity, V = volume
120
V
= 2 x
1 000
1 000
V = 600
10 x
Volume of the final solution = 600 cm3
\ volume of water added = (600 – 120) cm3
= 480 cm3
37 B (MV) before dilution = (MV) after dilution, where M = molarity, V = volume
100
V
= 0.10 x
1 000
1 000
V = 500 cm3
0.50 x
Volume of the final solution = 500 cm3
\ volume of water added = (500 – 100) cm3
= 400 cm3
38 B Volume of dilute solution of acid X = (40 + 10) cm3
= 50 cm3
(MV) before dilution = (MV) after dilution, where M = molarity, V = volume
1.0 x
\ the concentration of the dilute solution of acid X is 0.80 mol dm–3.
40
50
= M x
1 000
1 000
M = 0.80 mol dm–3
43
39 B Volume of diluted solution of acid Y= (25 + 25) cm3
= 50 cm3
(MV) before dilution = (MV) after dilution, where M = molarity, V = volume
1.6 x
\ the concentration of the dilute solution of acid Y is 0.80 mol dm–3.
Hence the concentration of dilute solution of acid X equals that of acid Y.
During the reaction between magnesium and an acid, magnesium reacts with hydrogen ions in the
acid.
From the diagrams, it can be deduced that magnesium reacts more slowly with dilute solution of acid X.
The concentration of hydrogen ions in dilute solution of acid X is lower than that in dilute solution of acid Y.
As the concentration of dilute solution of acid X equals that of acid Y, it can be concluded that acid X
is weaker than acid Y.
25
50
= M x
1 000
1 000
M = 0.80
40 C Volume of solution X = (20 + 80) cm3
= 100 cm3
(MV) before dilution = (MV) after dilution, where M = molarity, V = volume
0.50 x
\ concentration of acid A in solution X is 0.10 mol dm–3.
20
100
= M x
1 000
1 000
M = 0.10
41 A Volume of solution Y = (60 + 40) cm3
= 100 cm3
44
(MV) before dilution = (MV) after dilution, where M = molarity, V = volume
0.25 x
\ concentration of acid B in solution Y is 0.15 mol dm–3.
Option A — During the reaction between magnesium and an acid, magnesium reacts with hydrogen
ions in the acid.
From the diagrams, it can be deduced that magnesium reacts more quickly with solution X.
The concentration of hydrogen ions in solution X is higher than that in solution Y.
Although the concentration of acid A in solution X is lower than the concentration of
acid B in solution Y, it can be concluded that acid A is stronger than acid B.
Option D — As the concentration of acid A in solution X is lower than the concentration of acid B
in solution Y, the number of moles of acid A in solution X is lower than the number of
moles of acid B in solution Y.
60
100
= M x
1 000
1 000
M = 0.15
When both solutions of monobasic acids react with magnesium, solution X would give
less hydrogen than solution Y.
42 A 2NH3(aq) + H2SO4(aq) 1.60 mol dm–3
? mol dm–3
25.0 cm3
32.0 cm3
(Na4)2SO4(aq)
3
Number of moles of NH3 in 1.60 cm solution = molarity of solution x volume of solution
25.0
= 1.60 mol dm–3 x
dm3
1 000
= 0.0400 mol
According to the equation, 2 moles of NH3 require 1 mole of H2SO4 for complete neutralization.
Number of moles of H2SO4 in 32.0 cm3 solution =
Concentration of sulphuric acid =
=
= 0.625 mol dm–3
Molar mass of H2SO4 = (2 x 1.0 + 32.1 + 4 x 16.0) g mol–1
= 98.1 g mol–1
Concentration of sulphuric acid= 98.1 g mol–1 x 0.625 mol dm–3
= 61.3 g dm–3
\ the concentration of the sulphuric acid is 61.3 g dm–3.
0.0400
mol
2
= 0.0200 mol
number of moles of H2SO4
volume of solution
0.0200 mol
32.0
dm3
1 000
(
)
43 A We can represent the dibasic acid solution by H2X(aq).
H2X(aq) + 2NaOH(aq) 0.0600 mol dm–3 0.150 mol dm–3
25.0 cm3
? cm3
Na2X(aq) + 2H2O(l)
Number of moles of H2X in 25.0 cm solution = molarity of solution x volume of solution
25.0
= 0.0600 mol dm–3 x
dm3
1 000
= 0.00150 mol
According to the equation, 1 mole of H2X requires 2 moles of NaOH for complete neutralization.
i.e.number of moles of NaOH= 2 x number of moles of H2X
= 2 x 0.00150 mol
= 0.00300 mol
Volume of sodium hydroxide solution required for complete neutralization
number of moles of NaOH
=
molarity of solution
0.00300 mol
=
0.150 mol dm–3
= 0.0200 dm3
= 20.0 cm3
3
45
44 D K2O(s) + 2HCl(aq) 6.12 g
0.50 mol dm–3
? cm3
–1
Molar mass of K2O= (2 x 39.1 + 16.0) g mol
–1
= 94.2 g mol
Number of moles of K2O=
According to the equation, 1 mole of K2O requires 2 moles of HCl for complete neutralization.
i.e.number of moles of HCl= 2 x 0.0650 mol
= 0.130 mol
Volume of hydrochloric acid required for complete neutralization=
45 B 2Na(s) + 2H2O(l)
0.182 g
46
2KCl(aq) + H2O(l)
mass
molar mass
6.12 g
=
94.2 g mol–1
= 0.0650 mol
number of moles of HCl
molarity of solution
0.130 mol
=
0.50 mol dm–3
= 0.26 dm3
= 260 cm3
2NaOH(aq) + H2(g)......(1)
NaOH(aq) + HCl(aq)
0.240 mol dm–3
? cm3
NaCl(aq) + H2O(l)......(2)
Number of moles of Na=
Number of moles of NaOH formed from Na= number of moles of Na
= 0.00791 mol
According to equation (2), 1 mole of NaOH requires 1 mole of HCl for complete neutralization.
i.e.number of moles of HCl = 0.00791 mol
Volume of hydrochloric acid required for complete neutralization=
mass
molar mass
0.182 g
=
23.0 g mol–1
= 0.00791 mol
number of moles of HCl
molarity of solution
0.00791 mol
=
0.240 mol dm–3
= 0.0330 dm3
= 33.0 cm3
46 B CaCO3(s) + 2HCl(aq) 7.50 g
2.00 mol dm–3
50.0 cm3
CaCl2(aq) + H2O(l) + CO2(g)
–1
Molar mass of CaCO3 = (40.1 + 12.0 + 3 x 16.0) g mol
–1
= 100.1 g mol
Number of moles of CaCO3 =
Number of moles of HCl in 50.0 cm3 solution= molarity of solution x volume of solution
50.0
= 2.00 mol dm–3 x
dm3
1 000
= 0.100 mol
According to the equation, 1 mole of CaCO3 reacts with 2 moles of HCl to produce 1 mole of CO2.
During the reaction, 0.100 mole of HCl reacts with 0.0500 mole of CaCO3. Therefore CaCO3 is in
excess. The amount of HCl limits the amount of CO2 evolved.
Number of moles of CO2 evolved=
\ 0.0500 mole of carbon dioxide is evolved.
mass
molar mass
7.50 g
=
100.1 g mol–1
= 0.0749 mol
0.100
mol
2
= 0.0500 mol
47 A Zn(s) + 2HCl(aq) 1.80 g 0.250 mol dm–3
100.0 cm3
ZnCl2(aq) + H2(g)
mass
molar mass
1.80 g
=
65.4 g mol–1
= 0.0275 mol
Number of moles of Zn=
Number of moles of HCl in 100.0 cm3 solution= molarity of solution x volume of solution
100.0
= 0.250 mol dm–3 x
dm3
1 000
= 0.0250 mol
According to the equation, 1 mole of Zn reacts with 2 moles of HCl to produce 1 mole of H2. During
the reaction, 0.0250 mole of HCl reacted with 0.0125 mole of Zn. Therefore Zn was in excess. The
amount of HCl limited the amount of H2 formed.
Number of moles of H2 formed=
Mass of H2 formed = number of moles of H2 x molar mass of H2
= 0.0125 mol x 2.0 g mol–1
= 0.0250 g
\ 0.0250 g of hydrogen was formed.
0.0250
mol
2
= 0.0125 mol
47
48 C CuO(s) + 2HNO3(aq) 4.00 g
1.40 mol dm–3
30.0 cm3
Cu(NO3)2(aq) + H2O(l)
–1
Molar mass of CuO = (63.5 + 16.0) g mol
–1
= 79.5 g mol
Number of moles of CuO =
Number of moles of HNO3 in 30.0 cm3 solution = molarity of solution x volume of solution
30.0
= 1.40 mol dm–3 x
dm3
1 000
= 0.0420 mol
According to the equation, 1 mole of CuO requires 2 moles of HNO3 for complete reaction. During the
reaction, 0.0420 mole of HNO3 reacts with 0.0210 mole of CuO. Therefore CuO is in excess.
Number of moles of CuO left unreacted= (0.0503 – 0.0210) mol
= 0.0293 mol
Mass of CuO left unreacted= number of moles x molar mass
= 0.0293 mol x 79.5 g mol–1
= 2.33 g
\ 2.33 g of copper(II) oxide are left unreacted.
mass
molar mass
4.00 g
=
79.5 g mol–1
= 0.0503 mol
49 B H2SO4(aq) + 2NaOH(aq)
Na2SO4(aq) + 2H2O(l)
According to the equation, 1 mole of H2SO4 requires 2 moles of NaOH for complete neutralization.
Option
Solutions mixed
A
100 cm3 of 1 mol dm–3 H2SO4(aq) and
100 cm3 of 1 mol dm–3 NaOH(aq)
Number of moles of acid /
alkali
= molarity of solution x
volume of solution
number of moles of H2SO4
100
= 1 mol dm–3 x
dm3
1 000
= 0.1 mol
Complete
neutralization?
not enough NaOH(aq)
to neutralize H2SO4(aq)
completely
number of moles of NaOH
100
= 1 mol dm–3 x
dm3
1 000
= 0.1 mol
B
100 cm3 of 1 mol dm–3 H2SO4(aq) and
100 cm3 of 2 mol dm–3 NaOH(aq)
number of moles of H2SO4
100
= 1 mol dm–3 x
dm3
1 000
= 0.1 mol
number of moles of NaOH
100
= 2 mol dm–3 x
dm3
1 000
= 0.2 mol
48
complete neutralization
occurs and a neutral
solution is obtained.
C
3
–3
100 cm of 2 mol dm H2SO4(aq) and
3
–3
50 cm of 2 mol dm NaOH(aq)
number of moles of H2SO4
100
–3
3
= 2 mol dm x
dm
1 000
= 0.2 mol
not enough NaOH(aq)
to neutralize H2SO4(aq)
completely
number of moles of NaOH
50
–3
3
= 2 mol dm x
dm
1 000
= 0.1 mol
D
3
–3
200 cm of 2 mol dm H2SO4(aq) and
3
–3
100 cm of 1 mol dm NaOH(aq)
number of moles of H2SO4
200
–3
3
= 2 mol dm x
dm
1 000
= 0.4 mol
not enough NaOH(aq)
to neutralize H2SO4(aq)
completely
number of moles of NaOH
100
–3
3
= 1 mol dm x
dm
1 000
= 0.1 mol
50 D Solid sodium hydroxide is deliquescent.
51 C
52 B Phenolphthalein is red in an alkali and colourless in an acid.
53 D HCl(aq) + NaOH(aq) 0
.200 mol dm–3 ? mol dm–3
27.5 cm3
25.0 cm3
NaCl(aq) + H2O(l)
Number of moles of HCl in 27.5 cm3 solution= molarity of solution x volume of solution
27.5
= 0.200 mol dm–3 x
dm3
1 000
= 0.00550 mol
According to the equation, 1 mole of HCl requires 1 mole of NaOH for complete neutralization.
i.e.number of moles of NaOH = 0.00550 mol
Molarity of sodium hydroxide solution=
=
= 0.220 mol dm–3
number of moles of NaOH
volume of solution
0.00550 mol
25.0
dm3
1 000
(
)
54 D Methyl orange is red in acid and yellow in an alkali.
55 B Burette reading = (29.6 – 0.8) cm3
= 28.8 cm3
49
56 B H2SO4(aq) + 2NaOH(aq) 10.0 cm3
0.320 mol dm–3
28.8 cm3
Na2SO4(aq) + 2H2O(l)
250.0 cm3
(used) 25.0 cm3
Number of moles of NaOH in 28.8 cm3 solution = molarity of solution x volume of solution
28.8
= 0.320 mol dm–3 x
dm3
1 000
= 0.00922 mol
According to the equation, 1 mole of H2SO4 requires 2 moles of NaOH for complete neutralization.
i.e. number of moles of H2SO4 in 25.0 cm3 diluted cleanser=
Number of moles of H2SO4 in 250.0 cm3 diluted cleanser= 0.00461 mol x
Number of moles of H2SO4 in 10.0 cm3 undiluted cleanser = 0.0461 mol
Molarity of H2SO4 in undiluted cleanser=
=
= 4.61 mol dm–3
\ the concentration of sulphuric acid in the undiluted toilet cleanser is 4.61 mol dm–3.
0.00922
mol
2
= 0.00461 mol
250.0 cm3
25.0 cm3
= 0.0461 mol
number of moles of H2SO4
volume of solution
0.0461 mol
10.0
dm3
1 000
(
)
57 A Phenolphthalein is colourless in an acid and red / pink in an alkali.
50
58 D H3PO4(aq) + 2NaOH(aq) 25.0 cm3
0.0200 mol dm–3
22.0 cm3
Na2HPO4(aq) + 2H2O(l)
250.0 cm3
(used) 25.0 cm3
Number of moles of NaOH in 22.0 cm3 solution = molarity of solution x volume of solution
22.0
= 0.0200 mol dm–3 x
dm3
1 000
= 4.40 x 10–4 mol
According to the equation, 1 mole of H3PO4 reacts with 2 moles of NaOH.
i.e. number of moles of H3PO4 in 25.0 cm3 diluted acid=
Number of moles of H3PO4 in 250.0 cm3 diluted acid= 2.20 x 10–4 x
Number of moles of H3PO4 in 25.0 cm3 original acid = 2.20 x 10–3 mol
Molarity of original phosphoric acid=
\ the concentration of the original phosphoric acid was 8.80 x 10–2 mol dm–3.
4.40 x 10–4
mol
2
= 2.20 x 10–4 mol
= 2.20 x 10–3
250.0
mol
25.0
mol
number of moles of H3PO4
volume of solution
2.20 x 10–3 mol
=
25.0
dm3
1 000
(
)
= 0.088 mol dm–3
59 C NaOH(aq) + HA(aq) 0.100 mol dm–3 ? mol dm–3
22.5 cm3
30.0 cm3
NaA(aq) + H2O(l)
Number of moles of NaOH in 22.5 cm3 solution = molarity of solution x volume of solution
22.5
= 0.100 mol dm–3 x
dm3
1 000
= 0.00255 mol
According to the equation, 1 mole of HA requires 1 mole of NaOH for complete neutralization.
i.e.number of moles of HA = 0.00225 mol
Molarity of HA(aq) =
=
= 0.0750 mol dm–3
\ the concentration of HA(aq) is 7.50 x 10–2 mol dm–3.
number of moles of HA
volume of solution
0.00225 mol
30.0
dm3
1 000
(
)
60 B Only phenolphthalein changes colour within the pH range of the vertical part of the titration curve.
61 A Wash a pipette and a burette first with distilled water and then with the solution they are going to
contain.
Wash a conical flask with distilled water only.
51
62 A A conical flask should be washed with distilled water only.
Washing the conical flask with the tablet solution increased the number of moles of solute it held. This
would increase the average burette reading (i.e. the volume of hydrochloric acid used in the titration).
63 C There was problem with Step 3.
The conical flask should NOT be rinsed with alkali.
Rinsing the conical flask with the alkali increased the number of moles of solute it held. This would
increase the volume of acid used to neutralize the alkali, making the calculated concentration of the
alkali too high.
64 A Na2CO3(aq) + 2HCl(aq)
3.57 g 1.20 mol dm–3
48.0 cm3
2NaCl(aq) + CO2(g) + H2O(l)
Molar Mass of Na2CO3•H2O = (2 x 23.0 + 12.0 + 3 x 16.0 + 18.0n) g mol–1
Number of moles of Na2CO3•H2O=
Number of moles of HCl in 48.0 cm3 solution = molarity of solution x volume of solution
48.0
= 1.20 mol dm–3 x
dm3
1 000
= 0.0576 mol
According to the equation, 1 mole of Na2CO3 requires 2 moles of HCl for complete reaction.
i.e.number of moles of Na2CO3•H2O=
Number of moles of Na2CO3•H2O =
mass
molar mass
3.57 g
=
(106.0 + 18.0n) g mol–1
0.0576
mol
2
= 0.0288 mol
3.57
(106.0 + 18.0n)
mol = 0.0288 mol
n = 1
65 C Let n be the basicity of the acid.
52
We can represent the acid solution by HnX(aq).
HnX(aq) + nNaOH(aq) 0.125 mol dm–3 0.30 mol dm–3
20.0 cm3
25.0 cm3
Number of moles of NaOH in 25.0 cm3 solution = molarity of solution x volume of solution
25.0
= 0.30 mol dm–3 x
dm3
1 000
= 0.0075 mol
NanX(aq) + nH2O(l)
Number of moles of HnX in 20.0 cm3 solution = molarity of solution x volume of solution
20.0
= 0.125 mol dm–3 x
dm3
1 000
= 0.0025 mol
According to the equation, 1 mole of HnX requires n moles of NaOH for complete neutralization.
1
Number of moles of HnX
=
n
Number of moles of NaOH
0.0075 mol
=
0.0025 mol
n = 3
\ the basicity of the acid is 3.
66 D We can represent the dibasic acid by H2X.
H2X(aq) + 2NaOH(aq)
2.62 g 0.220 mol dm–3
18.9 cm3
Na2X(aq) + 2H2O(l)
250.0 cm3
(used) 25.0 cm3
Let m g mol–1 be the molar mass of H2X.
Number of moles of H2X in 2.62 g solid =
Number of moles of H2X in 25.0 cm3 solution =
Number of moles of NaOH in 18.9 cm solution= molarity of solution x volume of solution
18.9
= 0.220 mol dm–3 x
dm3
1 000
= 0.00416 mol
According to the equation, 1 mole of H2X requires 2 moles of NaOH for complete neutralization.
i.e. number of moles of H2X =
Number of moles of H2X =
\ the molar mass of the acid is 126 g mol–1.
mass
molar mass
2.62 g
=
m g mol­–1
1
2.62
x
mol
10
m
3
0.00416
mol
2
= 0.00208 mol
1
2.62
x
mol = 0.00208 mol
10
m
m = 126
53
67 B Number of moles of OH– in 25.0 cm3 of 1.00 mol dm–3 NaOH solution
= molarity of solution x volume of solution
25.0
= 1.00 mol dm–3 x
dm3
1 000
= 0.0250 mol
68 C
69 A NaOH(aq) + HCl(aq)
? mol 0.550 mol dm–3
28.0 cm3
NaCl(aq) + H2O(l)
Number of moles of HCl in 28.0 cm3 solution = molarity of solution x volume of solution
28.0
= 0.550 mol dm–3 x
dm3
1 000
= 0.0154 mol
According to the equation, 1 mole of NaOH requires 1 mole of HCl for complete neutralization.
i.e.number of moles of OH– in the filtrate = 0.0154 mol
\ 0.0154 mole of hydroxide ions is present in the solution obtained in Stage II.
70 C Number of moles of OH– used for precipitation= (0.0250 – 0.0154) mol
= 0.00960 mol
Ni2+(aq) + 2OH–(aq)
0.00960 mol
According to the equation, 1 mole of Ni2+ ions react with 2 moles of OH– ions to give 1 mole of
Ni(OH)2.
0.00960
\ number of moles of Ni2+ ion =
mol
2
= 0.00480 mol
Molarity of nickel(II) sulphate solution=
\ the molarity of the nickel(II) sulphate solution is 0.192 mol dm–3.
Ni(OH)2(s)
number of moles of Ni2+ ions
volume of solution
0.00480 mol
=
25.0
dm3
1 000
= 0.192 mol dm–3
(
)
71 B Methyl orange is yellow in alkaline solution and red in acidic solutions.
72
54
D
Na2CO3(aq) + H2SO4(aq)
3.20 g 1.08 mol dm–3
(with
24.8 cm3
impurity)
Na2SO4(aq) + CO2(g) + H2O(l)
Number of moles of H2SO4 in 24.8 cm3 solution = molarity of solution x volume of solution
24.8
= 1.08 mol dm–3 x
dm3
1 000
= 0.0268 mol
According to the equation, 1 mole of Na2CO3 requires 1 mole of H2SO4 for complete reaction.
i.e. number of moles of Na2CO3 in the sample = 0.0268 mol
Molar mass of Na2CO3 = (2 x 23.0 + 12.0 + 3 x 16.0) g mol–1
= 106.0 g mol–1
Mass of Na2CO3 in the sample = number of moles of Na2CO3 x molar mass of Na2CO3
= 0.0268 mol x 106.0 g mol–1
= 2.84 g
2.84 g
\ percentage purity of Na2CO3 in the sample=
x 100%
3.20 g
= 88.8%
\ the percentage purity of sodium carbonate in the sample is 88.8%.
73 C Fe(s) + H2SO4(aq)
1.60 mol dm–3
100 cm3
FeSO4(aq) + H2(g)
? g
Number of moles of H2SO4 in 100 cm3 solution = molarity of solution x volume of solution
100
= 1.60 mol dm–3 x
dm3
1 000
= 0.160 mol
According to the equation, 1 mole of H2SO4 reacts with 1 mole of Fe to give 1 mole of FeSO4.
i.e. number of moles of FeSO4 = 0.160 mol
Molar mass of FeSO4•7H2O= [55.8 + 32.1 + 4 x 16.0 + 7 x (2 x 1.0 + 16.0)] g mol–1
= 277.9 g mol–1
Mass of FeSO4•7H2O obtained = number of moles of FeSO4•7H2O x molar mass of FeSO4•7H2O
= 0.160 mol x 277.9 g mol–1
= 44.5 g
\ 44.5 g of crystals are obtained.
74 C Ca(OH)2(aq) + 2HCl(aq) 0.0200 mol dm–3 0.400 mol dm–3
? cm3
25.0 cm3
CaCl2(aq) + 2H2O(l)
Number of moles of HCl in 25.0 cm3 solution= molarity of solution x volume of solution
25.0
= 0.400 mol dm–3 x
dm3
1 000
= 0.0100 mol
According to the equation, 1 mole of Ca(OH)2 requires 2 moles of HCl for complete neutralization,
giving 1 mol of CaCl2.
55
0.0100
mol
2
= 0.00500 mol
i.e. number of moles of Ca(OH)2 =
Volume of Ca(OH)2 solution =
Number of moles of CaCl2 = 0.00500 mol
Volume of the resulting solution = (25.0 + 250) cm3
= 275 cm3
Concentration of CaCl2 in the resulting solution=
=
= 0.0182 mol dm–3
\ the concentration of calcium chloride in the resulting solution is 1.82 x 10–2 mol dm–3.
number of moles of Ca(OH)2
molarity of solution
0.00500 mol
=
0.0200 mol dm–3
= 0.250 dm–3
= 250 cm–3
75 A Mg(s) + 2HCl(aq) 2.00 g 0.800 mol dm–3
200.0 cm3
number of moles of CaCl2
volume of solution
0.00500 mol
275
dm3
1 000
(
)
MgCl2(aq) + H2(g)
mass
molar mass
2.00 g
=
24.3 g mol–1
= 0.0823 mol
Number of moles of Mg =
Number of moles of HCl in 200.0 cm3 solution= molarity of solution x volume of solution
200.0
= 0.800 mol dm–3 x
dm3
1 000
= 0.160 mol
According to the equation, 1 mole of Mg reacts with 2 moles of HCl to produce 1 mole of MgCl2.
During the reaction, 0.160 mole of HCl reacted with 0.0800 mole of Mg. Therefore Mg was in excess.
The amount of HCl limited the amount of MgCl2 formed.
Number of moles of MgCl2=
Concentration of MgCl2 in the resulting solution =
=
= 0.400 mol dm–3
56 \ the concentration of magnesium chloride in the resulting solution is 0.400 mol dm–3.
0.160
mol
2
= 0.0800 mol
number of moles of MgCl2
volume of solution
0.0800 mol
200.0
dm3
1 000
(
)
76 D Copper(II) sulphate solution and sodium carbonate solution react to give a green precipitate (copperIII)
carbonate).
2+
2–
Cu (aq) + CO3 (aq)
CuCO3(s)
77 C K2CO3(aq) + MgCl(aq)
MgCO3(s) + 2KCl(aq)
According to the equation, 1 mole of K2CO3 reacts with 1 mole of MgCl2 to produce 1 mole of MgCO3, i.e.
equal volumes of 1 mol dm–3 K2CO3(aq) and 1 mol dm–3 MgCl2(aq) would react.
Option
Volume of K2CO3(aq) /
MgCl2(aq) reacted
Solutions mixed
3
–3
3
–3
10 cm of 1 mol dm
B
15 cm3 of 1 mol dm–3 K2CO3(aq) + 25 cm3 of 1 mol dm–3 MgCl2(aq)
15 cm3
C
20 cm3 of 1 mol dm–3 K2CO3(aq) + 20 cm3 of 1 mol dm–3 MgCl2(aq)
20 cm3
D
30 cm of 1 mol dm
3
–3
K2CO3(aq) + 30 cm of 1 mol dm
3
–3
K2CO3(aq) + 10 cm of 1 mol dm
MgCl2(aq)
3
A
MgCl2(aq)
10 cm
3
10 cm
The greatest volumes of K2CO3(aq) and MgCl2(aq) react in option C. Thus the greatest amount of
precipitate would be produced.
78 D
Beaker
1
Number of moles of
mass
CaCO3 =
molar mass
number of moles of CaCO3
4 g
–1
100.1 g mol
= 0.04 mol
=
2
Number of moles of acid
= molarity of solution x
volume of solution
Reaction between CaCO3 and acid
number of moles of HCl
100
3
= 1 mol dm–3 x
dm
1 000
= 0.1 mol
CaCO3(s) + 2HCl(aq)
0.04 mol 0.1 mol
CaCl2(aq) + H2O(l) + CO2(g)
number of moles of CH3COOH
100
–3
3
= 1 mol dm x
dm
1 000
= 0.1 mol
CaCO3(s) + 2CHCOOH(aq)
0.04 mol 0.1 mol
(CH3COO)2Ca(aq)
+ H2O(l) + CO2(g)
According to the equations, 1 mole of CaCO3 reacts with 2 moles of HCl / CH3COOH to produce 1 mole of CO2. During the reaction, 0.04 mole of CaCO3 reacts with 0.08 mole of HCl / CH3COOH.
Therefore HCl / CH3COOH is in excess. The amount of CaCO3 limits the amount of CO2 produced.
Number of moles of CO2 produced = 0.04 mol
Option A — No calcium carbonate remains in both beakers.
Options B and C — Hydrochloric acid and ethanoic acid in both beakers are in excess. They do not
react completely.
Option D — The same number of moles of gas, i.e. the same amount of gas, is produced by both
reacting mixtures.
57
79 C
Number of moles of Mg
mass
=
molar mass
Beaker
1
Number of moles of acid
= molarity of solution x
volume of solution
number of moles of HCl
100
–3
3
= 1 mol dm x
dm
1 000
= 0.1 mol
Mg(s) + 2HCl(aq)
0.062 mol 0.1 mol
MgCl2(aq) + H2(g)
According to the equation, 1 mole
of Mg reacts with 2 moles of HCl
to produce 1 mole of H2. During the
reaction, 0.1 mole of HCl reacted
with 0.05 mole of Mg. Therefore
Mg was in excess. The amount
of HCl limited the amount of H2
produced.
Number of moles of H2 produced
= 0.05 mol
number of moles of H2SO4
100
3
= 1 mol dm–3 x
dm
1 000
= 0.1 mol
Mg(s) + H2SO4(aq)
0.062 mol 0.1 mol
MgSO4(aq) + H2(g)
According to the equation, 1 mole
of Mg reacts with 1 mole of H2SO4
to produce 1 mole of H2. During
the reaction, 0.062 mole of Mg
reacted with 0.062 mole of H2SO4.
Therefore H2SO4 was in excess. The
amount of Mg limited the amount
of H2 produced.
Number of moles of H2 produced
= 0.062 mol
number of moles of Mg
1.5 g
=
–1
24.3 g mol
= 0.062 mol
2
Reaction between Mg and acid
Option A — The magnesium in Beaker 1 was in excess.
Option B — The sulphuric acid in Beaker 2 was in excess.
Option C — A greater number of moles of hydrogen, i.e. a greater amount of hydrogen, was
produced by the reacting mixture in Beaker 2.
Option D — Some magnesium remained in Beaker 1, but none remiained in Beaker 2.
80 D Option A — HCl(aq) is a monobasic acid while H2SO4(aq) is a dibasic acid.
58
H+(aq) + Cl–(aq)
HCl(aq)
H2SO4(aq)
1 mol dm–3 H2SO4(aq) has a higher concentration of hydrogen ions than 1 mol dm–3
HCl(aq) does.
Hence 1 mol dm–3 H2SO4(aq) has a pH value lower than 1 mol dm–3 HCl(aq) does.
Option B — Methyl orange is red in acids.
Option C — During the reaction of zinc and an acid, zinc reacts with hydrogen ions in the acid.
2H+(aq) + SO42–(aq)
As 1 mol dm–3 H2SO4(aq) has a higher concentration of hydrogen ions than 1 mol dm–3
HCl(aq) does, the reaction rate between zinc and 1 mol dm–3 H2SO4(aq) is higher than
that between zinc and 1 mol dm–3 H2SO4(aq).
Option D — Number of moles of acid = molarity of
Reaction between acid and NaOH
solution x volume of solution
number of moles of HCl
20
–3
3
dm
= 1 mol dm x
1 000
= 0.02 mol
HCl(aq) + NaOH(aq)
NaCl(aq) + H2O(l)
0.02 mol
According to the equation, 1 mole of HCl requires
1 mole of NaOH for complete neutralization
\ number of moles of NaOH required = 0.02 mol
number of moles of H2SO4
10
3
= 1 mol dm–3 x
dm
1 000
= 0.01 mol
H2SO4(aq) + 2NaOH(aq)
Na2SO4(aq) + H2O(l)
0.01 mol
According to the equation, 1 mole of H2SO4
requires 2 moles of NaOH for complete
neutralization
\ number of moles of NaOH required
= 2 x 0.01 mol
= 0.02 mol
3
–3
3
–3
\ 20 cm of 1 mol dm HCl(aq) and 10 cm of 1 mol dm
number of moles of NaOH for complete neutralization.
H2SO4(aq) require the same
81 D
82 B The following diagram shows the titration curve for the titration of 20.0 cm 3 of 0.1 mol dm –3
CH3COOH(aq) with 0.1 mol dm–3 NaOH(aq).
Q)
FRVJWBMFODF
QPJOU
7PMVNFPG/B0)BR
BEEFEDN
During the titration of CH3COOH(aq) (a weak acid) with NaOH(aq) (a strong alkali), the pH at the
equivalence point is greater than 7.
83 C Options A and D — These two curves are INCORRECT as the initial pH of the 0.1 mol dm–3 weak
alkali is greater 7.
Option B — This curve shows the pH change as a strong acid is added to a strong alkali. There is a
marked change in pH at the equivalence point.
59
Option C — The following diagram shows the titration curve for the titration of 20.0 cm 3 of
0.1 mol dm–3 NH3(aq) with 0.1 mol dm–3 HCl(aq).
Q)
FRVJWBMFODF
QPJOU
7PMVNFPG)$MBR
BEEFEDN
During the titration of a weak alkali with a strong acid, the pH at the equivalence point
is less than 7.
84 D The following diagram shows the titration curve for the titration of 20.0 cm3 of 0.1 mol dm–3 KOH(aq)
with 0.1 mol dm–3 CH3COOH(aq).
Q)
QIFOPMQIUIBMFJO
DIBOHFTDPMPVS
XJUIJOUIJTQ)SBOHF
FRVJWBMFODF
QPJOU
7PMVNFPG$)$00)BR
BEEFEDN
During the titration of a strong alkali with a weak acid, the pH at the equivalence point is greater than 7.
Phenolphthalein changes colour within the pH range of the vertical part of the titration curve.
Hence phenolphthalein is a suitable indicator for the titration.
85 A The pH of solution X is less than 7. Thus solution X should be an acid.
60
As the pH of the mixture levels off at 7 with excess solid X, it can be concluded that solid X is
insoluble in water. Otherwise the mixture will be alkaline.
86 D
Indicator
Colour in solution with a pH of 5
Methyl orange
yellow
Bromothylmol blue
yellow
Phenolphthalein
colourless
\ a yellow colour would be observed.
87 B The following graph shows the relationship between the temperature of the mixture and the volume of
sodium hydroxide solution added.
5FNQFSBUVSFž$
FRVJWBMFODFQPJOUPGUJUSBUJPO
UFNQFSBUVSFJTUIFIJHIFTU
:
;
9
FRVJWBMFODF
QPJOU
7PMVNFPG/B0)BEEFEDN
When sodium hydroxide solution is added, it reacts with the hydrochloric acid. Heat is released.
Therefore the temperature of the solution mixture rises from X to Y. Neutralization is completed at
Y. No more heat is produced. The excess alkali added also cools the solution mixture. Therefore the
temperature of the solution mixture falls from Y to Z.
88 D As sulphuric acid was added, it removed both the barium ions (by precipitation) and hydroxide ions (by
neutralization).
Ba(OH)2(aq) + H2SO4(aq)
BaSO4(s) + 2H2O(l)
At the equivalence point, all the barium ions and hydroxide ions had been removed. Hence the
electrical conductivity of the mixture fell to almost zero.
After the equivalence point, the number of ions in the mixture rises as excess sulphuric acid was
added. Hence the electrical conductivity of the mixture increased.
89 C Sodium hydroxide solution and sulphuric acid react according to the following equation:
2NaOH(aq) + H2SO4(aq)
Na2SO4(aq) + 2H2O(l)
According to the equation, 2 moles of NaOH require 1 mole of H2SO4 for complete neutralization, i.e.
2V cm3 of 1.00 mol dm–3 NaOH(aq) would react with V cm3 of 1.00 mol dm–3 H2SO4(aq).
61
Option
Solutions mixed
A
10.0 cm of NaOH(aq) + 20.0 cm of H2SO4(aq)
10.0 cm of NaOH(aq) + 5.0 cm of H2SO4(aq)
B
15.0 cm3 of NaOH(aq) + 15.0 cm3 of H2SO4(aq)
15.0 cm3 of NaOH(aq) + 7.5 cm3 of H2SO4(aq)
C
20.0 cm3 of NaOH(aq) + 10.0 cm3 of H2SO4(aq)
20.0 cm3 of NaOH(aq) + 10.0 cm3 of H2SO4(aq)
D
25.0 cm of NaOH(aq) + 5.0 cm of H2SO4(aq)
3
Volumes of NaOH(aq) and H2SO4(aq) reacted
3
3
3
3
3
3
3
10.0 cm of NaOH(aq) + 5.0 cm of H2SO4(aq)
The greatest volumes of NaOH(aq) and H2SO4(aq) react in Option C. The greatest amount of heat is
released and thus the temperature rise is the greatest.
90 D Lead(II) nitrate solution and potassium sulphate solution react to form a precipitate (lead(II) sulphate)
according to the following equation:
Pb(NO3)2(aq) + K2SO4(aq)
PbSO4(s) + 2KNO3(aq)
The height of precipitate levels off when 6.0 cm3 of lead(II) nitrate solution are added to the potassium
sulphate solution. Hence it can be concluded that 6.0 cm3 of lead(II) nitrate solution are required to
react completely with the potassium sulphate solution.
Pb(NO3)2(aq) + K2SO4(aq) ? mol dm–3
1.2 mol dm–3
6.0 cm3
10.0 cm3
Number of moles of K2SO4 in 10.0 cm3 solution = molarity of solution x volume of solution
10.0
= 1.2 mol dm–3 x
dm3
1 000
= 0.012 mol
According to the equation, 1 mole of K2SO4 requires 1 mole of Pb(NO3)2 for complete reaction.
i.e. number of moles of Pb(NO3)2 in 6.0 cm3 solution = 0.012 mol
Molarity of lead(II) nitrate solution=
\ the concentration of lead(II) nitrate solution is 2.0 mol dm–3.
PbSO4(s) + 2KNO3(aq)
number of moles of Pb(NO3)2
volume of solution
0.012 mol
=
6.0
dm3
1 000
(
)
= 2.0 mol dm–3
91 A (1) The precipitate lead(II) sulphate is white in colour.
(3) A burette should be used to measure the 1.0 cm3 portions of the lead(II) nitrate solution.
92 C (1) Number of moles of Fe2(SO4)3 = molarity of solution x volume of solution
250.0
= 0.100 mol dm–3 x
dm3
1 000
= 0.0250 mol
62
(2) and (3) One formula unit of Fe2(SO4)3 contains 2 Fe3+ ions and 3 SO42– ions.
\ number of moles of Fe
\ number of moles of SO42– ions= 3 x 0.0250 mol
= 0.0750 mol
3+
ions = 2 x 0.0250 mol
= 0.0500 mol
93 C (1) Conical flask is NOT required for dilution process.
94 A (1) Wash a pipette first with distilled water and then with the solution they are going to contain.
(2) and (3) Wash a conical flask and a volumetric flask with distilled water only.
95 A (1) 1 mole of tribasic acid (H3A) requires 3 moles of NaOH for complete neutralization.
H3A(aq) + 3NaOH(aq)
Na3A(aq) + 3H2O(l)
(2) The number of hydrogen atoms 1 molecule of an acid contains may be different from the basicity
of the acid.
For example, a molecule of ethanoic acid (CH3COOH) contains four hydrogen atoms but only the
hydrogen atom in the –COOH group can undergo dissociation. Therefore it is a monobasic acid.
)
)
0
$
$
POMZUIJTIZESPHFO
BUPNDBOVOEFSHP
EJTTPDJBUJPO
0
)
)
96 D The following diagram shows the titration curve for the titration of dilute sodium hydroxide solution
with dilute hydrochloric acid.
Q)
FRVJWBMFODF
QPJOU
NFUIZMSFEDIBOHFTDPMPVSXJUIJOUIJTQ)SBOHF
CSPNPDSFTPMHSFFODIBOHFTDPMPVSXJUIJOUIJTQ)SBOHF
CSPNPUIZMNPMCMVFDIBOHFTDPMPVSXJUIJOUIJTQ)SBOHF
7PMVNFPG)$MBR
BEEFEDN
All the indicators change colour within the pH range of the vertical part of the titration curve. Hence
all the indicators are suitable for the titration.
63
97 A
Beaker
Number of moles of H2SO4 = molarity of solution x volume of solution
x
100
3
dm = 0.1 mol
1 000
–3
x
50
3
dm = 0.1 mol
1 000
1 mol dm
B
2 mol dm
(1) Equal numbers of moles of H2SO4 are present in both beakers. Equal masses of zinc react in both
beakers. Hence equal masses of zinc remain in both beakers.
(2) Equal amounts of zinc and H 2SO 4(aq) react in both beakers. Hence equal masses of gas are
produced in both cases.
(3) Equal masses of zinc sulphate are produced in both cases.
As the volumes of solution in the two beakers are different, zinc sulphate solutions of different
concentrations are produced.
98 A
Reaction
Number of moles of Mg
mass
=
molar mass
1
1 g
–1
24.3 g mol
2
64
–3
A
= 0.04 mol
Number of moles of acid
= molarity of solution x
volume of solution
Reaction between Mg and acid
number of moles of HCl
50
–3
3
= 2 mol dm x
dm
1 000
= 0.1 mol
Mg(s) + 2HCl(aq)
0.04 mol 0.1 mol
MgCl2(aq) + H2(g)
number of moles of CH3COOH
50
–3
3
= 2 mol dm x
dm
1 000
= 0.1 mol
Mg(s) + 2CH3COOH(aq)
0.04 mol 0.1 mol
(CH3COO)2Mg(aq) + H2(g)
According to the equations, 1 mole of Mg reacts with 2 moles of HCl / CH3COOH to produce 1 mole
of H2. During the reaction, 0.04 mole of Mg reacted with 0.08 mole of HCl / CH3COOH. Therefore HCl
/ CH3COOH was in excess. The amount of Mg limited the amount of H2 produced.
Number of moles of H2 produced = 0.04 mol
(1) Magnesium reacted completely in both cases. Hence magnesium disappeared in both reactions at
the end.
(2) Equal amounts of hydrogen were produced in both reactions.
(3) HCl(aq) is a strong acid while CH3COOH(aq) is a weak acid.
2 mol dm–3 HCl(aq) had a higher concentration of hydrogen ions than 2 mol dm–3 CH3COOH(aq).
Hence HCl(aq) reacted more quickly with magnesium than CH3COOH(aq) did. The two reactions
took different times to complete.
99 A
Beaker
A
Number of moles of acid
= molarity of solution x
volume of solution
Number of moles of alkali
= molarity of solution x
volume of solution
number of moles of HCl
50
–3
3
= 1.2 mol dm x
dm
1 000
= 0.060 mol
number of moles of NaOH
60
3
= 1.0 mol dm–3 x
dm
1 000
= 0.060 mol
number of moles of CH3COOH
60
–3
3
= 1.0 mol dm x
dm
1 000
= 0.060 mol
Reaction between acid and
alkali
HCl(aq) + NaOH(aq)
0.060 mol 0.060 mol
NaCl(aq) + H2O(l)
CH3COOH(aq) + NaOH(aq)
0.060 mol 0.060 mol
CH3COONa(aq)
+ H2O(l)
According to the equations, 1 mole of HCl / CH 3COOH requires 1 mole of NaOH for complete
neutralization.
(1) During the reactions, 0.060 mole of HCl / CH3COOH reacts with 0.060 mole of NaOH.
(2) For the neutralization between a strong acid and a strong alkali, the heat released is 57 kJ for
1 mole of water produced.
For neutralization in which either the acid or alkali or both are weak, the heat released is less than
57 kJ for 1 mole of water produced.
HCl(aq) is a strong acid and CH3COOH(aq) is a weak acid.
The heat released in the neutralization between HCl(aq) and NaOH(aq) is higher than that in the
neutralization between CH3COOH(aq) and NaOH(aq).
Futhermore, the total volume of the reaction mixture in Beaker A is less than that of the reaction
mixture in Beaker B.
Hence the temperature rise for the reaction mixture in Beaker A is higher than that for the reaction
mixture Beaker B.
(3) Equal numbers of Na+(aq) ions are present in both reaction mixtures.
However, the total volumes of the reaction mixtures are different.
Hence concentrations of Na+(aq) ions in the reaction mixtures are different.
Hence both acids are completely neutralized.
100D The molarity of a solution is the number of moles of solute dissolved in 1 dm3 of the solution.
101B Ethanedioic acid crystals are suitable for preparing standard acid solutions because it has the following
characteristics:
• it is obtainable in a very pure form;
• it has a known chemical formula;
• it dissolves in water completely at room temperature;
• it is stable and does not absorb moisture from the air; and
• it has a high molar mass to minimize weighing errors.
65
102C Solid sodium hydroxide is NOT suitable for preparing a standard solution because it absorbs moisture
from the air and cannot be weighed accurately.
103A Wash a burette first with distilled water and then with the acid it is going to contain.
104D Rinsing the conical flask with the solution it is about to contain would increase the number of moles
of solute it holds.
It is not necessary to dry the conical flask as any water remaining in the flask will NOT change the
number of moles of solute it holds.
105D H2SO4(aq) + 2NaOH(aq)
Na2SO4(aq) + 2H2O(l)
100 cm3 of 1 mol dm–3 sulphuric acid require 200 cm3 of 1 mol dm–3 sodium hydroxide solution for
complete neutalization.
Sulphuric acid is a dibasic acid. The number of hydrogen ions in 100 cm3 of 1 mol dm–3 sulphuric acid
is twice as that of hydroxide ions in 100 cm3 of 1 mol dm–3 sodium hydroxide solution.
3
106B The following diagram shows the titration curve for the titration of 20.0 cm of 0.1 mol dm
–3
with 0.1 mol dm HCl(aq).
–3
NH3(aq)
Q)
NFUIZMPSBOHF
DIBOHFTDPMPVS
XJUIJOUIJTQ)SBOHF
FRVJWBMFODF
QPJOU
7PMVNFPG)$MBR
BEEFEDN
107C
66
Methyl orange is suitable indicator for the titration because the indicator changes colour within the pH
range of the vertical part of the titration curve.
Number of moles of acid
= molarity of solution x volume of solution
Reaction between acid and alkali
number of moles of H2SO4
10
3
= 1 mol dm–3 x
dm
1 000
= 0.01 mol
H2SO4(aq) + 2NaOH(aq)
Na2SO4(aq) + H2O(l)
0.01 mol
number of moles of NaOH required for neutralization
= 2 x 0.01 mol
= 0.02 mol
number of moles of CH3COOH
20
–3
3
= 1 mol dm x
dm
1 000
= 0.02 mol
CH3COOH(aq) + NaOH(aq)
CH3COONa(aq) + H2O(l)
0.02 mol
number of moles of NaOH required for neutralization
= 0.02 mol
\ the acids require the same number of moles of NaOH for complete neutralization.
Unit 18 Rate of reaction
Fill in the blanks
1
concentration; time
2
colorimeter
3
a) surface area
b) concentration
c) temperature
4
catalyst; chemical
True or false
5
T In the oxidation of oxalate ions by permanganate ions, the intensity of the purple colour of
permanganate ions decreases as the reaction proceeds.
When we shine light upon the reaction mixture, the absorbance of the reaction mixture is directly
proportional to the colour intensity of the reaction mixture and the concentration of the permanganate
ions in the reaction mixture.
Hence the progress of the reaction can be followed by a colorimeter.
6
T During the decomposition of hydrogen peroxide solution, oxygen gas is formed.
2H2O2(aq)
If the reaction vessel is a closed system, the pressure inside the vessel will increase. We can follow the
progress of the reaction by measuring the pressure inside the vessel with a pressure sensor connected to
a data-logger interface and a computer.
7
F
8
T The rate of a reaction decreases with time. At the start, there are plenty of reactant particles per unit
volume. As the reactant particles are consumed gradually, there are fewer particles per unit volume, i.e.
the concentration of reactants falls. So, the reaction slows down.
9
F The volume of a liquid would NOT affect the rate of a reaction.
2H2O(l) + O2(g)
10 F The surface area of marble chips is smaller than that of powdered marble of the same mass.
The reaction rate between dilute hydrochloric acid and marble chips is slower than that between the acid
and powdered marble.
11 F During the reaction between magnesium and an acid, magnesium would react with hydrogen ions in
the acid.
Hydrochloric acid is a strong acid while ethanoic acid is a weak acid. Hence 2 mol dm–3 hydrochloric acid
has a higher concentration of hydrogen ions than 2 mol dm–3 ethanoic acid does.
Thus the reaction between magnesium and 2 mol dm–3 hydrochloric acid is faster than that between
67
magnesium and 2 mol dm–3 ethanoic acid.
12 T
13 T
14 F The physical state of a catalyst may be the same as those of the reactants.
For example, iron acts as a catalyst in the reaction between nitrogen gas and hydrogen gas to produce
ammonia gas.
15 T Manganese(IV) oxide can catalyze the decomposition of hydrogen peroxide.
Multiple choice questions
16 D Option D — A limestone status damage caused by acid rain may take years.
17 D Option D — The reaction between two solutions has the greatest reaction rate.
0.105 mol dm–3
20 s
–3
–3 –1
= 5.25 x 10 mol dm s
18 C Average reaction rate =
19 B
$PODFOUSBUJPONPMENm
m
NPMENm
m
T
5JNFT
(0.165 – 0.09) mol dm–3
Instantaneous rate of reaction at 30 s =
(60 – 0) s
= 1.25 x 10–3 mol dm–3 s–1
0.30
mol
2
= 0.15 mol
20 A Number of moles of Fe2O3 produced in 10.0 seconds =
68
Rate of consumption of Fe2O3 =
\ the rate of consumption of Fe2O3 is 1.5 x 10–2 mol s–1.
0.15 mol
10.0 s
= 0.015 mol s–1
1
2
d[N2O5(g)]
dt
1 d[NO2(g)]
=
4
dt
d[O2(g)]
=
dt
21 B Rate = –
Instantaneous rate of decomposition of N2O5(g) = –
d[N2O5(g)]
dt
2 d[NO2(g)]
=
4
dt
2
=
(3.8 x 10–5 mol dm–3 s–1)
4
= 1.9 x 10–5 mol dm–3 s–1
d[O2(g)]
dt
1 d[NO2(g)]
=
4
dt
1
=
(3.8 x 10–5 mol dm–3 s–1)
4
= 9.50 x 10–6 mol dm–3 s–1
22 D Instantaneous rate of decomposition of O2(g) =
23 A
24 C The sealed flask was a closed system. Therefore the mass of the flask plus its contents remained the
same.
25 A The rate of a reaction decreass with time. At the start, there are plenty of reactant particles per unit
volume. As the reactant particles are consumed gradually, there are fewer particles per unit volume, i.e.
the concentration of reactants fall. So, the reaction slows down.
26 B In the oxidation of oxalate ions by permanganate ions, the intensity of the purple colour of
permanganate ions decreases as the reaction proceeds.
When we shine light upon the reaction mixture, the absorbance of the reaction mixture is directly
proportional to the colour intensity of the reaction mixture and the concentration of the permanganate
ions in the reaction mixture.
Hence the progress of the reaction can be followed by a colorimeter.
27 C Consider a reaction in which a gas is produced. If the reaction vessel is a closed system, the pressure
inside the vessel will increase. We can follow the progress of the reaction by measuring the pressure
inside the vessel with a pressure sensor connected to a data-logger interface and a computer.
Option C — NO gas is produced in the reaction between copper(II) oxide and dilute nitric acid. Hence
the progress of the reaction CANNOT be followed by using a data-logger with a pressure
sensor.
69
28 B Option A — The antacid tablets in Experiments I and II have the same surface area. So, they CANNOT
be used for comparison.
Option B — The only difference between Experiments I and IV is the surface area of the antacid
tablets.
Option C — The antacid tablets in Experiments II and III have the same surface area. So, they CANNOT
be used for comparison.
Option D — The concentration and temperature of the acids in Experiments III and IV are different. So,
they CANNOT be used for comparison.
29 C Option A — The concentration and temperature of the acids in Experiments I and II are different. So,
they CANNOT be used for comparison.
Option C — The only difference between Experiments II and III is the concentration of the acids.
Option D — The surface area of the tablets, concentration and temperature of the acids in Experiments
III and IV are different. So, they CANNOT be used for comparison.
30 B The hydrochloric acid was in excess in each case. Hence the amount of calcium carbonate limited the
amount of carbon dioxide produced, i.e. the total loss in mass of contents of the reaction flask was
the same in both cases.
Powdered marble had a greater surface area than marble lumps of the same mass.
The rate of the reaction increased when the surface area of marble was increased.
31 D The initial rate of reaction would decrease when hydrochloric acid of a lower concentration is used.
32 D During the reaction between marble and an acid, marble would react with hydrogen ions in the acid.
Hydrochloric acid is a strong acid while ethanoic acid is a weak acid.
Therefore 1 mol dm–3 hydrochloric acid has a higher concentration of hydrogen ions than 1 mol dm–3
ethanoic acid.
Hence the initial rate of reaction would decrease when ethanoic acid is used.
33 A The initial rate of the reaction would increase with an increase in temperature.
34 C The total volume of hydrogen formed increased as the reaction proceeded.
After some time, no more hydrogen was produced. So, its volume no longer changed. The reaction
was over.
35 B Option A — All the magnesium disappeared after reaction. Magnesium was the limiting reactant. The
amount of magnesium limited the amount of hydrogen formed.
70
Hence increasing the volume of hydrochloric acid used would NOT affect the final volume
of hydrogen formed.
Option B — Using a greater volume of acid of the same concentration had NO effect on the rate of
reaction.
36 D Options A and B — During the reaction between sodium carbonate and an acid, sodium carbonate
would react with hydrogen ions in the acid.
Sulphuric acid is a dibasic acid while nitric acid is a monobasic acid. Hence
sulphuric acid has a higher concentration of hydrogen ions.
Thus the reaction rate between sodium carbonate and 1 mol dm–3 sulphuric acid
is higher than that between sodium carbonate and 1 mol dm–3 nitric acid.
Options C and D — The rate of reaction at 40 °C is higher than that at 20 °C.
37 A
Number of moles of Zn
mass
=
molar mass
Reaction
mixture
Number of moles of H2SO4
= molarity of solution x
volume of solution
–3
2 mol dm
1
5 g
–1
65.4 g mol
2
= 0.076 mol
x
100
3
dm
1 000
Zn(s) + H2SO4(aq)
0.076 mol 0.2 mol
ZnSO4(aq) + H2(g)
x
100
3
dm
1 000
Zn(s) + H2SO4(aq)
0.076 mol 0.1 mol
ZnSO4(aq) + H2(g)
= 0.2 mol
–3
1 mol dm
= 0.1 mol
Reaction between Zn and H2SO4
According to the equation, 1 mole of Zn reacts with 1 mole of H2SO4 to produce 1 mole of H2. During
the reaction, 0.076 mole of Zn reacted with 0.076 mole of H2SO4. Therefore the amount of Zn limited
the amount of H2 produced (0.076 mole in both cases).
Option A — The hydrogen produced would escape. Thus the mass of each reaction mixture would
decrease.
Options B, C and D — The rate of reaction between zinc and 2 mol dm–3 sulphuric acid was higher
than that between zinc and 1 mol dm–3 sulphuric acid.
As the same amount of gas was produced in both reactions the total mass lost of both
reaction mixtures would be the same.
Hence the reaction between zinc and 2 mol dm–3 sulphuric acid took less time
to complete.
38 C As calcium carbonate disappeared in both experiments after reaction, it could be deduced that calcium
carbonate was the limiting reactant, i.e. its amount limited the amount of carbon dioxide formed.
Curve Y showed a smaller loss in mass, and thus the mass of calcium carbonate used was smaller (i.e.
Option C).
Furthermore, the tangent to curve Y was less steep than that to curve X, i.e. the rate of the reaction
represented by curve Y was lower.
This agreed with the change stated in Option C. The rate of the reaction would decrease when a lump
of calcium carbonate was used instead of calcium carbonate powder.
39 C Curve II showed a greater volume of oxygen formed. Thus the number of moles of hydrogen peroxide
decomposed was greater, i.e. adding some hydrogen peroxide solution (Option C).
Furthermore, the tangent to curve II was less steep than that to curve I, i.e. the rate of decomposition
represented by curve II was lower.
This agreed with the change stated in Option C. Adding some 0.1 mol dm–3 hydrogen peroxide
solution to the original 1 mol dm–3 hydrogen peroxide solution would dilute the original hydrogen
71
peroxide solution. The rate of decomposition would decrease.
40 D During the reaction between magnesium and dilute hydrochloric acid, hydrogen was formed.
If the reaction vessel was a closed system, the pressure inside the vessel would increase. We could
follow the progress of the reaction by measuring the pressure inside the vessel with a pressure sensor
connected to a data-logger interface and a computer.
As magnesium ribbons disappeared in both experiments after reaction, it could be deduced that
magnesium was the limiting reactant, i.e. its amount limited the amount of hydrogen formed.
The tangent to curve Y was steeper than that to curve X, i.e. the rate of reaction represented by curve
Y was higher.
Option A — Lowering the temperature would decrease the rate of the reaction.
Option B — Using a greater volume of hydrochloric acid had NO effect on the rate of the reaction.
Option C — Using more magnesium ribbon would increase the amount of hydrogen formed, i.e.
increase the pressure change.
Option D — Using more concentrated hydrochloric acid would increase the rate of the reaction.
41 C The second experiment took a longer time to complete, i.e. the rate of reaction of the second
experiment is lower.
Option A — Using a catalyst would increase the rate of the reaction.
Option B — Using powdered Mg would increase the rate of the reaction.
Option C — HCl(aq) is a strong acid while CH3COOH(aq) is a weak acid.
Therefore 1 mol dm–3 HCl(aq) has a higher concentration of hydrogen ions than 1 mol
dm–3 ethanoic acid does.
During the reaction between Mg and an acid, Mg would react with hydrogen ions in the
acid.
Hence the rate of the reaction between Mg and CH3COOH(aq) (i.e. Option C) would be
lower than that between Mg and HCl(aq).
Option D — Increasing the temperature would increase the rate of the reaction.
42 D
Number of moles of CaCO3
mass
Reaction
=
molar mass
Number of moles of HCl
= molarity of solution x
volume of solution
–3
1 mol dm
1
8 g
–1 = 0.08 mol
100.1 g mol
100
3
dm
1 000
= 0.1 mol
–3
2
x
2 mol dm
100
3
x
dm
1 000
= 0.2 mol
72
Reaction between CaCO3 and HCl
CaCO3(s) + 2HCl(aq)
0.08 mol 0.1 mol
CaCl2(aq) + H2O(l)
+ CO2(g)
CaCO3(s) + 2HCl(aq)
0.08 mol 0.2 mol
CaCl2(aq) + H2O(l)
+ CO2(g)
According to the equation, 1 mole of Mg reacts with 2 moles of HCl to produce 1 mole of CaCl2.
In Reaction 1, 0.1 mole of HCl reacted with 0.05 mole of CaCO3 to give 0.05 mole of CaCl2.
Thus CaCO3 was in excess, i.e. Option A was INCORRECT.
In Reaction 2, 0.08 mole of CaCO3 reacted with 0.16 mole of HCl to give 0.08 mole of CaCl2.
Thus HCl was in excess, i.e. Option B was INCORRECT.
Option C — Different amounts of CaCl 2 , and hence calcium chloride solution of different
concentrations, were produced in the two reactions.
Option D — The initial rate of the reaction between CaCO3 and a less concentrated HCl(aq) (Reaction 1)
is smaller than that between CaCO3 and a more concentrated HCl(aq) (Reaction 2).
43 B HCl(aq) and CH3COOH(aq) react with calcium carbonate according to the following equations:
Beaker A CaCO3(s) + 2HCl(aq)
1 mol dm–3 x
Beaker B CaCO3(s) + 2CH3COOH(aq)
1 mol dm–3 x
Option A — During the reaction between CaCO3 and an acid, CaCO3 reacts with hydrogen ions in the
acid.
HCl(aq) is a strong acid while CH3COOH(aq) is a weak acid.
1 mol dm –3 HCl(aq) has a higher concentration of hydrogen ions than 1 mol dm –3
CH3COOH(aq) does.
Hence the initial rate of reaction in Beaker A is higher than that in Beaker B.
Option B — As equal numbers of moles of HCl and CH3COOH are used, the same amount of marble
chips would be consumed in both cases.
Options C and D — As excess marble chips are used, HCl(aq) and CH3COOH(aq) are the limiting
reagents. Their amounts limit the amount of carbon dioxide gas produced.
CaCl2(aq) + H2O(l) + CO2(g)
100
dm3 = 0.1 mol
1 000
(CH3COO)2Ca(aq) + H2O(l) + CO2(g)
100
dm3 = 0.1 mol
1 000
Hence equal masses of marble chips remain in the two beakers after reaction.
0.1 mole of HCl and 0.1 mole of CH3COOH would produce the same amount of
gas.
44 B Options A and B —HCl is a monobasic and H2SO4 is a dibasic acid.
Therefore 2 mol dm–3 HCl(aq) has a lower concentration of hydrogen ions than 2
mol dm–3 sulphuric acid does.
During the reaction between Zn and an acid, Zn would react with hydrogen ions
in the acid.
Hence the rate of the reaction between Zn and HCl(aq) is lower than that
between Zn and H2SO4(aq).
73
Options C and D —
Number of moles of Zn
mass
=
molar mass
Reaction
1
10 g
–1
65.4 g mol
= 0.15 mol
2
Number of moles of acid
= molarity of solution x
volume of solution
Reaction between Zn and acid
number of moles of HCl
100
–3
3
= 2 mol dm x
dm
1 000
= 0.2 mol
Zn(s) + 2HCl(aq)
0.15 mol 0.2 mol
ZnCl2(aq) + H2(g)
number of moles of H2SO4
100
3
= 2 mol dm–3 x
dm
1 000
= 0.2 mol
Zn(s) + H2SO4(aq)
0.15 mol 0.2 mol
ZnSO4(aq) + H2(g)
According to the first equation, 1 mole of Zn reacts with 2 moles of HCl.
In Reaction 1, 0.2 mole of HCl reacted with 0.1 mole of Zn.
Thus Zn was in excess, i.e. Option C was INCORRECT.
According to the second equation, 1 mole of Zn reacts with 1 mole of H2SO4.
In Reaction 2, 0.15 mole of Zn reacted with 0.15 mole of H2SO4.
Thus sulphuric acid was in excess, i.e. Option D was INCORRECT.
45 B The tangent to the curve for experiment I at the start was less steep than that for experiment 2, i.e.
the initial rate of reaction of experiment I was lower than that of experiment 2.
Thus Options A and C are INCORRECT.
The curve for experiment 2 went flat easiler than that for experiment I, i.e. the rate of reaction for
experiment 2 fell to zero easiler. Hence the graph shown in Option B is correct.
46 B HCl(aq) and CH3COOH(aq) react with calcium carbonate according to the following equations:
74
Experiment 1 CaCO3(s) + 2HCl(aq)
Experiment 2 CaCO3(s) + 2CH3COOH(aq)
As excess CaCO3 is used, HCl(aq) and CH3COOH(aq) are the limiting reagents. Their amounts limit the
amount of carbon dioxide gas produced.
0.11 mole of HCl and 0.11 mole of CH3COOH would produce the same amount of gas (i.e. Options A
and C are INCORRECT).
During the reaction between CaCO3 and an acid, CaCO3 reacts with hydrogen ions in the acid.
HCl(aq) is a strong acid while CH3COOH(aq) is a weak acid.
1.2 mol dm–3 HCl(aq) has a higher concentration of hydrogen ions than 0.9 mol dm–3 CH3COOH(aq)
does.
Hence the initial rate of reaction in Experiment 1 is higher than that in Experiment 2.
\ the curves in Option B show the results of the experiments.
1.2 mol dm–3 x
0.9 mol dm–3 x
CaCl2(aq) + H2O(l) + CO2(g)
90
dm3 = 0.11 mol
1 000
(CH3COO)2Ca(aq) + H2O(l) + CO2(g)
120
dm3 = 0.11 mol
1 000
47 D
Experiment
Number of moles of Zn
mass
=
molar mass
1
5 g
–1
65.4 g mol
2
= 0.076 mol
Number of moles of acid
= molarity of solution x
volume of solution
Reaction between Zn and acid
number of moles of HCl
100
–3
3
= 1 mol dm x
dm
1 000
= 0.1 mol
Zn(s) + 2HCl(aq)
0.076 mol 0.1 mol
ZnCl2(aq) + H2(g)
number of moles of H2SO4
100
3
= 1 mol dm–3 x
dm
1 000
= 0.1 mol
Zn(s) + H2SO4(aq)
0.076 mol 0.1 mol
ZnSO4(aq) + H2(g)
According to the first equation, 1 mole of Zn reacts with 2 moles of HCl to produce 1 mole of H2.
In Experiment 1, 0.1 mole of HCl reacted with 0.05 mole of Zn to produce 0.05 mole of H2.
According to the second equation, 1 mole of Zn reacts with 1 mole of H2SO4.
In Experiment 2, 0.076 mole of Zn reacted with 0.076 mole of H2SO4 to produce 0.076 mole of H2.
Thus the volume of gas produced in Experiment 1 was smaller than that produced in Experiment 2 (i.e.
Options A and B are INCORRECT).
During the reaction between Zn and an acid, Zn reacts with hydrogen ions in the acid.
HCl(aq) is a monobasic acid while H2SO4(aq) is a dibasic acid.
1 mol dm–3 HCl(aq) has a lower concentration of hydrogen ions than 1 mol dm–3 H2SO4(aq) does.
Hence the initial rate of reaction in Experiment 1 is lower than that in Experiment 2.
\ the curves in Option D show the results of the experiments.
48 D The concentration of the brown colour iodine in the reaction mixture decreased as the reaction
proceeded. The reaction mixture became lighter in colour gradually. Thus the reaction mixture absorbed
less and less light and so the absorbance went down.
The concentration of I2 in Sample 1 was the same as that in Sample 2.
Hence the initial absorbance of Sample 1 was the same as that of Sample 2 (i.e. Options B and C are
INCORRECT).
The concentration of propanone in Sample 1 was lower than that in Sample 2.
The rate of reaction for Sample 1 was lower and thus the tangent to its curve would be less steep.
\ the curves in Option D show the results.
49 B
50 D Option D — Manganese(IV) oxide can act as a catalyst in the decomposition of hydrogen peroxide.
51 C (1) Following the progress of a reaction using physical property may require the use of expensive
instructments, e.g. colorimeter.
52 D
75
53 A During the reaction between Mg and an acid, Mg reacts with hydrogen ions in the acid.
(1) HCl(aq) is a monobasic acid while H2SO4(aq) is a dibasic acid.
2 mol dm–3 H2SO4(aq) has a higher concentration of hydrogen ions than 2 mol dm–3 HCl(aq) does.
Hence the initial rate of reaction would increase when using 2 mol dm–3 sulphuric acid instead of 2
mol dm–3 hydrochloric acid.
(2) HCl(aq) is a strong acid while CH3COOH(aq) is a weak acid.
1 mol dm–3 HCl(aq) has a higher concentration of hydrogen ions than 1 mol dm–3 CH3COOH(aq)
does.
Hence the initial rate of reaction would decrease when using 2 mol dm–3 ethanoic acid instead of 2
mol dm–3 hydrochloric acid.
(3) Using a greater volume of 1 mol dm–3 hydrochloric acid would NOT affect the initial rate of the
reaction.
54 B Zinc reacts with H2SO4(aq) according to the following equation:
Zn(s) + H2SO4(aq)
ZnSO4(aq) + H2(g)
Beaker A excess 1 mol dm–3 x
Beaker B excess 2 mol dm–3
As zinc granules remain in both beaker after reaction, it can be deduced that zinc is in excess in both
cases.
(1) The initial rate of reaction is higher in Beaker B as sulphuric acid of a higher concentration is used.
(2) The hydrogen produced would escape. Thus the mass of each reaction mixture would decrease.
As zinc is in excess, the amount of acid limits the amount of hydrogen produced.
Each reaction mixture contains 0.1 mole of H2SO4. Hence the same amount of hydrogen would
form. Thus the total losses in mass of the two reaction mixtures after reaction are the same.
(3) Each reaction mixture contains 0.1 mole of H2SO4. Hence the same amount of ZnSO4 would be
produced in each case. However, the total volumes of the reaction mixtures are different.
100
dm3 = 0.1 mol
1 000
50
x
dm3 = 0.1 mol
1 000
Thus zinc sulphate solutions of different concentrations are produced in both beakers.
55 B
Reaction
Number of moles of Mg
mass
=
molar mass
Number of moles of HCl
= molarity of solution x
volume of solution
–3
2 mol dm
1
1 g
–1
24.3 g mol
2
76
x
50
3
dm
1 000
= 0.1 mol
= 0.04 mol
Number of moles of H2SO4
50
3
4 mol dm–3 x
dm
1 000
= 0.2 mol
Reaction between Mg and acid
Mg(s) + 2HCl(aq)
0.04 mol 0.1 mol
MgCl2(aq) + H2(g)
Mg(s) + 2HCl(aq)
0.04 mol 0.2 mol
MgCl2(aq) + H2(g)
According to the equation, 1 mole of Mg reacts with 2 moles of HCl to produce 1 mole of H2.
In Reaction 1, 0.04 mole of Mg reacted with 0.08 mole of HCl to give 0.04 mole of H2.
Thus HCl was in excess.
In Reaction 2, 0.04 mole of Mg reacted wtih 0.08 mol of HCl to give 0.04 mole of H2.
Thus HCl(aq) was in excess.
(1) Magnesium was the limiting reactant in both reactions. Thus magnesium disappeared in both cases
after reaction.
(2) The same amount of hydrogen (0.04 mole) was produced in both reactions.
(3) The initial rate of Reaction 2 was greater as a more concentrated hydrochloric acid was used.
56 A
57 B Carbon dioxide gas is produced in the reaction between sodium carbonate and dilute hydrochloric acid.
As the carbon dioxide gas escapes, the reaction mixture gets lighter as the reaction proceeds.
58 B In the reaction between iodine and propanone, the intensity of the brown colour of iodine decreases
as the reaction proceeds.
I2(aq) + CH3COCH3(aq)
brown
colourless CH3COCH2I(aq) + HI(aq)
Hence the progress of the reaction can be followed by using a colorimeter.
colourless
59 A The rate of a reaction decreases with time. At the start, there are plenty of reactant particles per unit
volume. As the reactant particles are consumed gradually, there are fewer particles per unit volume, i.e.
the concentration of reactants fall. So, the reaction slows down.
60 D During the reaction between calcium carbonate and an acid, calcium carbonate reacts with hydrogen
ions in the acid.
Nitric acid is a monobasic acid while sulphuric acid is a dibasic acid.
1 mol dm–3 nitric acid has a lower concentration of hydrogen ions than 1 mol dm–3 sulphuric acid
does.
Hence the rate of reaction between 1 mol dm–3 nitric acid and calcium carbonate is smaller than that
between 1 mol dm–3 sulphuric acid and calcium carbonate.
61 A
62 D In most cases, the rate of a reaction increases when the temperature is increased.
63 C A catalyst CANNOT change the amount of product formed in a reaction.
77
Part B Topic-based exercise
Multiple choice questions
1
B Option A — Phenolphthalein is colourless in acids.
Option B — Citric acid is a weak acid. It only partially dissociates in water.
citric acid(aq)
Hence an aqueous solution of citric acid contains both citric acid molecules and hydrogen
ions.
Option C — An aqueous solution of citric acid reacts with sodium hydrogencarbonate to give carbon
dioxide gas.
Option D — There is NO reaction between an aqueous solution of citric acid and copper.
2
A Option A — Dilute sodium hydroxide solution turns methyl orange yellow.
3
D Zn (aq) + 2OH (aq)
4
D
5
6
78
Option C — Sodium hydroxide is a strong alkali.
2+
B
C
+
H (aq) + citrate ion(aq)
–
Zn(OH)2(s)
Option
Solution
Addition of NH3(aq) to solution
A
calcium chloride solution
no precipitate
B
chromium(III) sulphate
green precipitate
C
iron(III) sulphate
reddish brown precipitate
D
potassium chloride
no precipitate
Option
Solution
Addition of Na2CO3(aq) to solution
A
aluminium sulphate solution
white precipitate
B
ammonium nitrate solution
no precipitate
C
lead(II) nitrate solution
white precipitate
D
magnesium sulphate solution
white precipitate
Option
Solution
Addition of NH3(aq) to solution
A
CuSO4(aq)
a pale blue precipitate
B
Na2CO3(aq)
no precipitate
C
Pb(NO3)2(aq)
a white precipitate, insoluble in excess alkali
D
ZnCl2(aq)
a white precipitate, soluble in excess alkali
7
C
Option
Solution
Reaction with Ba(NO3)2(aq)
Reaction with NaOH(aq)
A
calcium chloride
no precipitate
white precipitate Ca(OH)2
B
iron(II) chloride
no precipitate
green precipitate Fe(OH)2
C
magnesium sulphate
white precipitate BaSO4
white precipitate Mg(OH)2
D
potassium sulphate
white precipitate BaSO4
no precipitate
8
C Options A and C — Carbonates give a gas (carbon dioxide) with dilute hydrochloric acid.
Upon the addition of dilute aqueous ammonia, a solution containing zinc ions
gives a white precipitate which is soluble in excess alkali.
\ X is zinc carbonate.
Options B and D — Sulphates do NOT give a gas with dilute hydrochloric acid.
9
C A solution containing iron(II) ion gives a green precipitate (Fe(OH)2) with dilute sodium hydroxide
solution.
The green iron(II) hydroxide turns brown on prolonged standing in air due to the formation of iron(III)
hydroxide.
4Fe(OH)2(s) + 2H2O(l) + O2(g)
4Fe(OH)3(s)
10 D Barium chloride solution reacts with sodium carbonate soluton to give a white precipitate, barium
carbonate.
Ba2+(aq) + CO32–(aq)
Barium carbonate is soluble in dilute nitric acid because it can react with the acid.
BaCO3(s) + 2H+(aq)
BaCO3(s)
Ba2+(aq) + H2O(l) + CO2(g)
11 C Option C — Concentrated sulphuric acid is a hygroscopic substance.
12 D
13 C
Option
Solutions mixed
Any precipitate?
A
CuO(s) and H2SO4(aq)
no precipitate
B
FeSO4(aq) and NaOH(aq)
green precipitate Fe(OH)2
C
NH4Cl(aq) and Ca(NO3)2(aq)
no precipitate
D
Pb(NO3)2(aq) and HCl(aq)
white precipitate PbCl2
Option
Solutions mixed
Any precipitate?
A
BaCl2(aq) and Na2CO3(aq)
white precipitate BaCO3
B
KCl(aq) and AgNO3(aq)
white precipitate AgCl
C
Na2SO4(aq) and MgCl2(aq)
no precipitate
D
NaOH(aq) and NiSO4(aq)
green precipitate Ni(OH)2
79
14 A Options A and C — Chlorides give a white precipitate (AgCl) with silver nitrate solution.
+
–
Ag (aq) Cl (aq)
AgCl(s)
Barium chloride solution gives a white precipitate (BaSO4) with dilute sulphuric
acid.
Ba2+(aq) + SO42–(aq)
BaSO4(s)
15 D Zinc reacts with dilute hydrochloric acid to give zinc chloride solution and hydrogen. Zinc chloride
solution gives a white precipitate (Zn(OH)2) with dilute aqueous ammonia. Zn(OH)2 dissolves in excess
alkali to give a colourless solution.
Zn(s) + 2H+(aq)
Zn2+(aq) + 2OH–(aq)
Zn(OH)2(s) + 4NH3(aq)
Zn2+(aq) + H2(g)
Zn(OH)2(s)
[Zn(NH3)4]2+(aq) + 2OH–(aq)
–1
16 C Molar mass of Ca(OH)2 = [40.1 + 2 x (16.0 + 1.0)] g mol
= 74.1 g mol–1
mass
Number of moles of Ca(OH)2 =
molar mass
5.93 g
=
74.1 g mol–1
= 0.0800
number of moles of Ca(OH)2
volume of solution
0.0800 mol
400.0
dm3
1 000
Molarity of calcium hydroxide solution=
=
= 0.200 mol dm–3
\ the molarity of the calcium hydroxide solution is 0.200 mol dm–3.
(
)
17 D Number of moles of NaOH in 150.0 cm3 of 2.00 mol dm–3 solution
150.0
= 2.00 mol dm–3 x
dm3
1 000
= 0.300 mol
80
Number of moles of NaOH in 80.0 cm3 of 1.20 mol dm–3 solution
80.0
= 1.20 mol dm–3 x
dm3
1 000
= 0.0960 mol
Total number of moles of NaOH in the resulting solution
= (0.300 + 0.0960) mol
= 0.396 mol
Total volume of the resulting solution
= (150.0 + 80.0) cm3
= 230.0 cm3
0.396 mol
230.0
dm3
1 000
Concentration of the resulting solution =
\ the concentration of the resulting solution is 1.72 mol dm–3.
(
)
= 1.72 mol dm–3
18 C pH of solution = –log10(1.58 x 10–12)
= 11.8
19 A Option A — Nitric acid dissociates completely according to the following equation:
H+(aq) + NO3–(aq)
? mol dm–3
HNO3(aq) 0.010 mol dm–3 According to the equation, 1 mole of HNO3 dissociates to give 1 mole of hydrogen ions.
i.e. concentration of hydrogen ions = 0.010 mol dm–3
pH of acid = –log10(0.010)
= –(–2.0)
= 2.0
\ pH of 0.010 mol dm–3 HNO3 is 2.0.
Option B — CH 3COOH is a weak acid while HNO 3 is a strong acid. Hence the concentration of
hydrogen ions in 0.010 mol dm–3 CH3COOH is lower than that in 0.010 mol dm–3 HNO3, i.e.
the pH of CH3COOH is higher than that of HNO3.
Option C — H2SO4 is a dibasic acid while HNO3 is a monobasic acid. Hence the concentration of
hydrogen ions in 0.010 mol dm–3 H2SO4 is higher than that in 0.010 mol dm–3 HNO3, i.e.
the pH of H2SO4 is lower than that of HNO3.
Option D — The pH of 0.010 mol dm–3 NaOH is greater than 7.
20 C (1) and (3) Sodium hydroxide solution is a strong alkali while aqueous ammonia is a weak alkali.
Thus 0.1 mol dm–3 sodium hydroxide solution is more alkaline than 0.1 mol dm–3 aqueous
ammonia, i.e. the pH of sodium hydroxide solution is greater than that of aqueous
ammonia.
(2) The pH of 0.1 mol dm–3 lactice acid is less than 7.
Thus the order of pH of the solutions is: (2) < (1) < (3).
21 D pH of soil before treatment = 4
pH of soil after treatment = 6
Concentration of H+(aq) ions in soil before treatment = 10–4 mol dm–3
Concentration of H+(aq) ions in soil after treatment = 10–6 mol dm–3
Concentration of H+(aq) ions in soil after treatment
Concentration of H+(aq) ions in soil before treatment
10–6 mol dm–3
=
10–4 mol dm–3
1
=
100
\ the concentration of H+(aq) ions in the soil decreased by a factor of 100.
81
22 C pH of acid before dilution = 1
pH of acid after dilution = 3
Concentration of H+(aq) ions in acid before dilution = 10–1 mol dm–3
Concentration of H+(aq) ions in acid after dilution = 10–3 mol dm–3
Concentration of H+(aq) ions in acid after dilution
Concentration of H+(aq) ions in acid before dilution
10–3 mol dm–3
=
10–1 mol dm–3
1
=
100
i.e.the acid is diluted 100 times.
\ diluting 10 cm3 of the acid to 1 000 cm3 with diluted water would cause the pH value to change
from 1 to 3.
23 B Option A — The concentration of hydrogen ions in lemon juice is higher than that in milk.
concentration of hydrogen ions in sea water
concentration of hydrogen ions in laundry detergent
–8
–3
10 mol dm
=
–11
–3
10 mol dm
= 1 000
Option B —
\ the concentration of hydrogen ions in sea water is 1 000 times greater than that in
laundry detergent.
concentration of hydrogen ions in rainwater
concentration of hydrogen ions in sea water
10–5 mol dm–3
=
10–8 mol dm–3
= 1 000
Option C —
\ the concentration of hydrogen ions in rainwater is 1 000 times greater than that in
sea water.
concentration of hydrogen ions in lemon juice
concentration of hydrogen ions in soap solution
–2
–3
10 mol dm
=
–10
–3
10 mol dm
= 108
Option D —
\ the concentration of hydrogen ions in lemon juice is 108 times greater than that in
soap solution.
24 B Sodium chloride is neutral. Adding it to sulphuric acid would NOT affect the pH of the acid.
25 A When we describe acids as strong and weak, we are talking about the extent of their dissociation in
water. When we talk about concentration, we are referring to the amount of an acid in a unit volume
of solution.
Ethanoic acid is a weak acid. 0.1 mol dm–3 solution of ethanoic acid is a dilute solution of a weak acid.
26 D HF is a weak acid while H2SO4 and HClO4 are strong acids. Therefore 0.1 mol dm–3 HF has the lowest
concentration of hydrogen ions. The pH of 0.1 mol dm–3 HF is thus the highest among the three acids.
82
H2SO4 is a dibasic acid while HClO4 is a monobasic acid. Therefore 0.1 mol dm–3 H2SO4 has a higher
concentration of hydrogen ions than 0.1 mol dm–3 HClO4. The pH of 0.1 mol dm–3 H2SO4 is thus lower
than that of the 0.10 mol dm–3 HClO4.
The order of the pH of the three acids is: H2SO4 < HClO4 < HF.
27 C The electrical conductivity of a solution is proportional to the concentration of mobile ions.
–3
Options A and C — H2SO4(aq) is a dibasic acid while HCl(aq) is a monobasic acid. Hence 1.0 mol dm
–3
H2SO4(aq) has a higher concentration of mobile ions than 1.0 mol dm HCl(aq)
does.
Option D — C6H12O6(aq) does NOT conduct electricity.
\ 1.0 mol dm–3 has the greatest electrical conductivity.
28 A Ethanoic acid is a weak acid and hydrochloric acid is a strong acid.
Hencce for acids of the same concentration, ethanoic acid has a lower concentratation of hydrogen
ions than hydrochloric acid does.
Thus the pH of ethanoic acid is higher than that of hydrochloric acid.
29 D Option D — For the neutralization between a strong acid and a strong alkali, the heat released is 57
kJ for 1 mole of water produced.
For neutralization in which either the acid or alkali or both are weak, the heat released is
less than 57 kJ for 1 mole of water produced. The is because some energy is consumed
when the weak acid and weak alkali dissociate to give hydrogen ions and hydroxide ions
before neutralization.
CH3COOH(aq) is a weak acid while HCl(aq) is a strong acid.
Hence the temperature rise of the neutralization between 20 cm of 1 mol dm
–3
3
CH 3COOH(aq) and 1 mol dm NaOH(aq) is smaller than that between 20 cm of
–3
–3
1 mol dm HCl(aq) and 1 mol dm NaOH(aq).
3
–3
30 A All nitrates are soluble in water.
31 A Options C and D — Compounds of potassium and sodium are soluble in water.
32 B Copper(II) sulphate is a soluble salt. To prepare it, mix a dilute acid (dilute sulphuric acid in this case)
with a metal, an insoluble base or an insoluble carbonate. However, copper (metal) CANNOT be used
as it does not react with dilute sulphuric acid.
33 D Option A — Calcium reacts with dilute sulphuric acid to produce insoluble calcium sulphate. Calcium
sulphate forms a protective layer on the surface of calcium. This prevents further reaction
between calcium and dilute sulphuric acid.
Option B — Copper has NO reaction with dilute hydrochloric acid.
Option C — Lead reacts with dilute sulphuric acid to produce insoluble lead(II) sulphate. Lead(II)
sulphate forms a protective layer on the surface of lead. This prevents further reaction
between lead and dilute sulphuric acid.
Option D — Magnesium reacts with dilute hydrochloric acid to give magnesium chloride.
34 C Option C — Potassium chloride is prepared via the titraton of potassium hydroxide solution with dilute
hydrochloric acid. Pipette and burette are required in the process.
83
35 A Molar mass of H2SO4= (2 x 1.0 + 32.1 + 4 x 16.0) g mol–1
= 98.1 mol–1
number of moles of H2SO4
volume of solution
number of moles of H2SO4
=
volume of solution
Molarity of sulphuric acid =
Number of moles of H2SO4= 4.00 mol dm–3 x 2.00 dm3
= 8.00 mol
Mass of H2SO4 required = number of moles of H2SO4 x molar mass of H2SO4
= 8.00 mol x 98.1 g mol–1
= 785 g
4.00 mol dm–3
36 D Molar mass of Na2SO4 = (2 x 23.0 + 32.1 + 4 x 16.0) g mol–1
= 142.1 mol–1
mass
Number of moles of Na2SO4=
molar mass
20.0 g
=
142.1 g mol–1
= 0.141 mol
One mole of Na2SO4 contains 2 moles of Na+ ions.
i.e.number of moles of Na+ ions = 2 x 0.141 mol
= 0.282 mol
Concentration of Na+ ions in solution=
\ the concentration of sodium ions in the solution is 2.82 mol dm–3.
+
number of moles of Na ions
volume of solution
0.282 mol
=
100.0
dm3
1 000
(
)
= 2.82 mol dm–3
37 C Number of moles of potassium phosphate = molarity of solution x volume of solution
250.0
–3
3
= 0.500 mol dm x
dm
1 000
= 0.125 mol
84
One mole of potassium phosphate (K3PO4) contains 4 moles of ions.
\ number of moles of ions = 4 x 0.125 mol
= 0.500 mol
Number of ions = 0.500 mol x 6.02 x 1023 mol–1
= 3.01 x 1023
38 C Suppose V cm3 of NaCl(aq) and V cm3 of ZnCl2(aq) are mixed.
V
dm3
1 000
Number of moles of NaCl in 0.10 mol dm–3 solution = 0.10 mol dm–3 x
Number of moles of ZnCl2 in 0.20 mol dm–3 solution= 0.20 mol dm–3 x
One mole of NaCl contains 1 mole of Cl– ions and one mole of ZnCl2 contains 2 moles of Cl– ions.
Total number of moles of Cl– ions = (0.10 x
Total volume of the mixture = (V + V) cm3
= 2V cm3
Concentration of Cl– ions in the mixture =
= 0.10 x
= 0.20 x
V
mol
1 000
V
dm3
1 000
V
mol
1 000
V
V
+ 2 x 0.20 x
) mol
1 000
1 000
V
= (0.50 x
) mol
1 000
–
total number of moles of Cl ions
total volume of the mixture
(0.5 x
)
V
mol
1 000
=
2V
dm3
1 000
= 0.25 mol dm–3
\ the concentration of chloride ions in the mixture is 0.25 mol dm–3.
39 B Sulphuric acid dissociates completely according to the following equation:
2H+(aq) + SO42–(aq)
? mol dm–3
H2SO4(aq) 0.0250 mol dm–3
According to the equation, 1 mole of H2SO4 dissociates to give 2 moles of hydrogen ions.
i.e.concentration of hydrogen ions = 2 x 0.0250 mol dm–3
= 0.0500 mol dm–3
pH of acid = –log10(0.0500)
= –(–1.30)
= 1.30
\ pH of the acid sample is 1.30.
85
40 D One mole of barium nitrate contains 2 moles of NO3– ions.
0.0250
mol
2
= 0.125 mol
\ number of moles of Ba(NO3)2 =
Molarity of barium nitrate solution =
\ the molarity of the barium nitrate solution is 5.68 x 10–1 mol dm–3.
number of moles of Ba(NO3)2
volume of solution
0.0125 mol
=
22.0
dm3
1 000
(
)
= 0.568 mol dm–3
41 A (MV) before dilution = (MV) after dilution, where M = molarity, V = volume
25.0
V
= 0.50 x
1 000
1 000
3
V = 200 cm
4.0 x
Volume of the final solution = 200 cm3
\ volume of water added = (200 – 25.0) cm3
= 175 cm3
42 B Mass of H2SO4 in 25.0 m3 acid = 41 050 000 g x 92.0%
= 37 800 000 g
Molar mass of H2SO4 = (2 x 1.0 + 32.1 + 4 x 16.0) g mol–1
= 98.1 g mol–1
mass
Number of moles of H2SO4 =
molar mass
37 800 000 g
=
98.1 g mol–1
= 385 000 mol
Molarity of the sulphuric acid=
\ the concentration of the sulphuric acid is 15.4 mol dm–3.
number of moles of H2SO4
molarity of solution
385 000 mol
=
25 000 dm­3
= 15.4 mol dm–3
43 D Option D — Chemicals that absorb atmospheric moisture CANNOT be used to prepare standard
solutions.
86
44 C Na2CO3(aq) + 2HCl(aq) 0.150 mol dm–3 0.250 mol dm–3
25.0 cm3
? cm3
2NaCl(aq) + CO2(g) + H2O(l)
3
Number of moles of Na2CO3 in 25.0 cm solution= molarity of solution x volume of solution
25.0
= 0.150 mol dm–3 x
dm3
1 000
= 0.00375 mol
According to the equation, 1 mole of Na2CO3 requires 2 moles of HCl for complete reaction.
i.e. number of moles of HCl = 2 x 0.00375 mol
= 0.00750 mol
Volume of HCl required=
number of moles of HCl
molarity of solution
0.00750 mol
=
0.250 mol dm­–3
= 0.0300 dm3
= 30.0 cm3
45 C We can represent the dibasic acid X by H2A.
2KOH(aq) + H2A(aq) 0.150 mol dm–3 ? mol dm–3
30.0 cm3
10.0 cm3
K2A(aq) + 2H2O(l)
Number of moles of KOH in 30.0 cm3 solution = molarity of solution x volume of solution
30.0
= 0.150 mol dm–3 x
dm3
1 000
= 0.00450 mol
According to the equation, 1 mole of H2A requires 2 moles of KOH for complete neutralization.
i.e. number of moles of H2A in 10.0 cm3 solution=
Molarity of acid solution =
=
= 0.225 mol dm–3
\ the molarity of the acid soslution is 2.25 x 10–1 mol dm–3.
0.00450
mol
2
= 0.00225 mol
number of moles of H2A
volume of solution
0.00225 mol
10.0
dm3
1 000
(
)
87
46 A We can represent the dibasic acid by H2X.
2NaOH(aq) + H2X(aq) –3
–3
0.100 mol dm ? mol dm
3
3
20.6 cm 25.0 cm
Number of moles of NaOH in 20.6 cm3 solution= molarity of solution x volume of solution
20.6
= 0.100 mol dm–3 x
dm3
1 000
= 0.00206 mol
According to the equation, 1 mole of H2X requires 2 moles of NaOH for complete neutralization.
i.e. number of moles of H2X in 25.0 cm3 solution =
Molarity of acid solution =
=
= 0.0412 mol dm–3
\ the molarity of the acid solution is 0.0412 mol dm–3.
47 B 88
Na2X(aq) + 2H2O(l)
0.00206
mol
2
= 0.00103 mol
number of moles of H2A
volume of solution
0.00103 mol
25.0
dm3
1 000
(
Zn + 2HCl(aq) 2 mol dm–3
20.0 cm3
)
ZnCl2(aq) + H2(g)
Number of moles of HCl in 20.0 cm3 solution= molarity of solution x volume of solution
20.0
= 2 mol dm–3 x
dm3
1 000
= 0.04 mol
According to the equation, 1 mole of Zn will react completely with 2 moles of HCl.
i.e. number of moles of Zn =
\ 0.02 mole of zinc will react completely with the hydrochloric acid.
0.04
mol
2
= 0.02 mol
48 D Sr(OH)2(aq) + 2HCl(aq) 0.200 mol dm–3
? mol dm–3
25.0 cm3
20.0 cm3
SrCl2(aq) + 2H2O(l)
3
Number of moles of Sr(OH)2 in 25.0 cm solution= molarity of solution x volume of solution
25.0
= 0.200 mol dm–3 x
dm3
1 000
= 0.00500 mol
According to the equation, 1 mole of Sr(OH)2 requires 2 moles of HCl for complete neutralization.
i.e. number of moles of HCl= 2 x 0.00500 mol
= 0.0100 mol
Molarity of hydrochloric acid =
=
= 0.500 mol dm–3
\ concentration of the hydrochloric acid is 0.500 mol dm–3.
49 A Fe(OH)3(s) + 3HNO3(aq) 1.07 g
0.60 mol dm–3
? cm3
number of moles of HCl
volume of solution
0.0100 mol
20.0
dm3
1 000
(
)
Fe(NO3)3(aq) + 3H2O(l)
Molar mass of Fe(OH)3 = [55.8 + 3 x (16.0 + 1.0)] g mol–1
= 106.8 g mol–1
Number of moles of Fe(OH)3 =
According to the equation, 1 mole of Fe(OH)3 requires 3 moles of HNO3 for complete neutralization.
i.e. number of moles of HNO3 = 3 x 0.0100 mol
= 0.0300 mol
Volume of 0.60 mol dm–3 nitric acid required for complete neutralization
number of moles of HNO3
=
molarity of solution
0.0300 mol
=
0.60 mol dm–3
= 0.050 dm3
= 50 cm3
\ 50 cm3 of nitric acid were required for complete neutralization.
mass
molar mass
1.07 g
=
106.8 g mol–1
= 0.0100 mol
89
50 C Al2O3(s) + 3H2SO4(aq) 2.55 g
? mol dm–3
150.0 cm3
–1
Molar mass of Al2O3 = (2 x 27.0 + 3 x 16.0) g mol
–1
= 102.0 g mol
Number of moles of Al2O3 =
According to the equation, 1 mole of Al2O3 requires 3 moles of H2SO4 for complete neutralization.
i.e. number of moles of H2SO4 = 3 x 0.0250 mol
= 0.0750 mol
Molarity of sulphuric acid=
\ the concentration of the sulphuric acid is 0.500 mol dm–3.
mass
molar mass
2.55 g
=
102.0 g mol–1
= 0.0250 mol
number of moles of H2SO4
volume of solution
0.0750 mol
=
150.0
dm3
1 000
(
)
= 0.500 mol dm–3
51 A Mg(s) + 2HCl(aq) 1.00 g 1.00 mol dm–3
100.0 cm3
90
Al2(SO4)3(aq) + 3H2O(l)
MgCl2(aq) + H2(g)
mass
molar mass
1.00 g
=
24.3 g mol–1
= 0.0412 mol
Number of moles of Mg =
Number of moles of HCl in 100.0 cm3 solution= molarity of solution x volume of solution
100.0
= 1.00 mol dm–3 x
dm3
1 000
= 0.100 mol
According to the equation, 1 mole of Mg reacts with 2 moles of HCl to form 1 mole of H2. During
the reaction, 0.0412 mole of Mg reacted with 0.0824 mole of HCl. Therefore HCl was in excess. The
amount of Mg limited the amount of H2 formed.
i.e. number of moles of H2 formed = 0.0412 mol
Mass of H2 formed = number of moles x molar mass
= 0.0412 mol x 2.00 g mol–1
= 0.0824 g
\ 0.0824 g of hydrogen was formed.
52 D A pipette and a pipette filler are used to deliver a fixed, accurate volume (e.g. 25.0 cm3) of a solution.
A burette is used to deliver various volumes of solution accurately.
53 B 2NaOH(aq) + H2SO4(aq) 0.0400 mol dm–3 ? mol dm–3
25.0 cm3
22.7 cm3
Na2SO4(aq) + 2H2O(l)
Number of moles of NaOH in 25.0 cm3 solution= molarity of solution x volume of solution
25.0
= 0.0400 mol dm–3 x
dm3
1 000
= 0.00100 mol
According to the equation, 2 moles of NaOH require 1 mole of H2SO4 for complete neutralization.
i.e. number of moles of H2SO4 used =
Molarity of sulphuric acid=
\ the concentration of the sulphuric acid is 0.0220 mol dm–3.
0.00100
mol
2
= 5.00 x 10–4 mol
number of moles of H2SO4
volume of solution
5.00 x 10–4 mol
=
22.7
dm3
1 000
(
)
= 0.0220 mol dm–3
54 D
55 A H3PO4(aq) + 2NaOH(aq) 10.0 cm3
0.0240 mol dm–3
250.0 cm3
(used) 25.0 cm3
Average volume of NaOH(aq) required for neutralization=
Number of moles of NaOH in 21.5 cm3 solution= molarity of solution x volume of solution
21.5
= 0.0240 mol dm–3 x
dm3
1 000
= 5.16 x 10–4 mol
According to the equation, 1 mole of H3PO4 requires 2 moles of NaOH for neutralization.
i.e. number of moles of H3PO4 in 25.0 cm3 diluted acid=
Na2HPO4(aq) + 2H2O(l)
21.5 + 21.5 + 21.6
cm3
3
= 21.5 cm3
5.16 x 10–4
mol
2
= 2.58 x 10–4 mol
91
250.0
mol
25.0
mol
Number of moles of H3PO4 in 250.0 cm3 diluted acid = 2.58 x 10–4 x
Number of moles of H3PO4 in 10.0 cm3 original acid = 2.58 x 10–3 mol
Molarity of original phosphoric acid=
\ the molarity of the original phosphoric acid is 0.258 mol dm–3.
= 2.58 x 10–3
number of moles of H3PO4
volume of solution
2.58 x 10–3 mol
=
10.0
dm3
1 000
(
)
= 0.258 mol dm–3
56 C Wash the burette with distilled water and then the alkali it is going to contain. Any water or impurities
in the apparatus will change the concentration of the alkali and this will affect the titration results.
92
57
D
H2X(aq) + 2NaOH(aq) 4.51 g
0.350 mol dm–3
3
250.0 cm 28.0 cm3
Na2X(aq) + 2H2O(l)
(used) 25.0 cm3
Let m g mol–1 be the molar mass of H2X.
Number of moles of H2X in 250.0 cm3 solution=
Number of moles of NaOH in 28.0 cm3 solution= molarity of solution x volume of solution
28.0
= 0.350 mol dm–3 x
dm3
1 000
= 0.00980 mol
According to the equation, 1 mole of H2X requires 2 moles of NaOH for complete neutralization.
i.e. number of moles of H2X in 25.0 cm3 solution =
Number of moles of H2X in 25.0 cm3 solution =
\ the molar mass of H2X is 92.0 g mol–1.
mass
molar mass
4.51 g
=
–1
m g mol
4.51
1
Number of moles of H2X in 25.0 cm3 solution =
x
mol
m
10
0.00980
mol
2
= 0.00490 mol
4.51
1
x
mol = 0.00490 mol
m
10
m = 92.0
58 D Na2CO3(aq) + H2SO4(aq)
5.72 g 1.00 mol dm–3
20.0 cm3
Na2SO4(aq) + CO2(g) + H2O(l)
Molar mass of Na2CO3•nH2O = (106.0 + 18.0n) g mol–1
Number of moles of Na2CO3•nH2O used =
Number of moles of H2SO4 in 20.0 cm3 solution = molarity of solution x volume of solution
20.0
= 1.00 mol dm–3 x
dm3
1 000
= 0.0200 mol
According to the equation, 1 mole of Na2CO3 requires 1 mole of H2SO4 for complete reaction.
i.e. number of moles of Na2CO3•nH2O used = 0.0200 mol
Number of moles of Na2CO3•nH2O used =
\ the value of n is 10.
mass
molar mass
5.72 g
=
(106.0 + 18.0n) g mol–1
5.72
mol = 0.0200 mol
(106.0 + 18.0n)
\ n = 10
59 B From the curve, 45.0 cm3 of sulphuric acid were required to neutralize the sodium hydroxide solution. (The
temperature of the reaction mixture was maximum at the point of complete neutralization.)
H2SO4(aq) + 2NaOH(aq) ? mol dm–3
2.0 mol dm–3
45.0 cm3
50.0 cm3
Na2SO4(aq) + 2H2O(l)
Number of moles of NaOH in 50.0 cm3 solution= molarity of solution x volume of solution
50.0
= 2.0 mol dm–3 x
dm3
1 000
= 0.10 mol
According to the equation, 1 mole of H2SO4 requires 2 moles of NaOH for complete neutralization.
i.e.number of moles of H2SO4 =
Molarity of sulphuric acid=
\ the molarity of the sulphuric acid is 1.1 mol dm–3.
0.10
mol
2
= 0.050 mol
number of moles of H2SO4
volume of solution
0.050 mol
=
45.0
dm3
1 000
(
)
= 1.1 mol dm–3
93
60 A 2NaOH(aq) + H2SO4(aq) 0.200 mol dm–3
15.0 cm3
25.0 cm3
3
Number of moles of NaOH in 25.0 cm solution= molarity of solution x volume of solution
25.0
= 0.200 mol dm–3 x
dm3
1 000
= 0.00500 mol
According to the equation, 2 moles of NaOH react with 1 mole of H2SO4 to form 1 mole of Na2SO4.
i.e.number of moles of Na2SO4 formed =
Total volume of the salt solution = (25.0 + 15.0) cm3
= 40.0 cm3
Concentration of the salt solution=
=
= 0.0625 mol dm–3
\ the molarity of the salt solution obtained is 0.0625 mol dm–3.
0.00500
mol
2
= 0.00250 mol
61 C CaCl2(aq) + Na2CO3(aq)
number of moles of Na2SO4
volume of solution
0.00250 mol
40.0
dm3
1 000
(
)
CaCO3(s) + 2NaCl(aq)
According to the equation, 1 mole of CaCl2 reacts with 1 mole of Na2CO3 to give 1 mole of CaCO3, i.e.
equal volumes of 1 mol dm–3 CaCl2(aq) and 1 mol dm–3 NaCO3(aq) would react.
Volume of CaCl2(aq) and
Na2CO3(aq) reacted
Option
Solutions mixed
A
5 cm of CaCl2(aq) + 25 cm of Na2CO3(aq)
5 cm
B
10 cm3 of CaCl2(aq) + 20 cm3 of Na2CO3(aq)
10 cm3
C
15 cm3 of CaCl2(aq) + 15 cm3 of Na2CO3(aq)
15 cm3
D
20 cm3 of CaCl2(aq) + 10 cm3 of Na2CO3(aq)
10 cm3
3
3
3
The greatest volumes of CaCl2(aq) and Na2CO3(aq) react in Option C. Thus the greatest amount of
precipitate would form.
62 B Pb(NO3)2(aq) + 2NaCl(aq)
94
Na2SO4(aq) + 2H2O(l)
PbCl2(s) + 2NaNO3(aq)
According to the equation, 1 mole of Pb(NO3)2 requires 2 moles of NaCl to give 1 mole of PbCl2.
Option
Reaction
A
10 cm3 of 1 mol dm–3 Pb(NO3)2(aq) react with 20 cm3 of 1 mol dm–3 NaCl(aq)
B
12.5 cm3 of 1 mol dm–3 Pb(NO3)2(aq) react with 25 cm3 of 1 mol dm–3 NaCl(aq)
C
10 cm3 of 1 mol dm–3 Pb(NO3)2(aq) react with 20 cm3 of 1 mol dm–3 NaCl(aq)
D
5 cm3 of 1 mol dm–3 Pb(NO3)2(aq) react with 10 cm3 of 1 mol dm–3 NaCl(aq)
The greatest volumes of Pb(NO3)2(aq) and NaCl(aq) react in Option C. Thus the greatest amount of
precipitate would form.
63 C For the neutralization between a strong acid and a strong alkali, the heat released is 57 kJ for 1 mole
of water produced.
Expt.
Acid and alkali mixed
Number of moles of acid /
alkali mixed
= molarity of solution x
volume of solution
1
50 cm3 of 1 mol dm–3
HNO3(aq) + 50 cm3 of
1 mol dm–3 KOH(aq)
number of moles of HNO3 / KOH
50
= 1 mol dm–3 x
dm3
1 000
= 0.05 mol
HNO3(aq) + KOH(aq)
0.05 mol 0.05 mol
KNO3(aq) + H2O(l)
0.05 mol
2.85 kJ
2
50 cm3 of 2 mol dm–3
HNO3(aq) + 50 cm3 of
2 mol dm–3 KOH(aq)
number of moles of HNO3 / KOH
50
= 2 mol dm–3 x
dm3
1 000
= 0.1 mol
HNO3(aq) + KOH(aq)
0.1 mol 0.1 mol
KNO3(aq) + H2O(l)
0.01 mol
5.7 kJ
Heat
released
The total volumes of the two mixtures are the same. Hence the temperature rise of the first mixture is
1
half that of the second mixture, i.e. T1 =
T2.
2
64 B H2SO4(aq) + 2NaOH(aq)
Number of moles of water
formed
Na2SO4(s) + 2H2O(l)
According to the equation, 1 mole of H2SO4 requires 2 moles of NaOH for complete neutralization, i.e.
V cm3 of 1 mol dm–3 H2SO4(aq) would react with 2V cm3 of 1 mol dm–3 NaOH(aq).
Option
Solutions mixed
Volume of H2SO4(aq) and
NaOH(aq) reacted
A
10 cm3 of H2SO4(aq) + 35 cm3 of NaOH(aq)
10 cm3 of H2SO4(aq) + 20 cm3 of NaOH(aq)
B
15 cm3 of H2SO4(aq) + 30 cm3 of NaOH(aq)
15 cm3 of H2SO4(aq) + 30 cm3 of NaOH(aq)
C
20 cm3 of H2SO4(aq) + 25 cm3 of NaOH(aq)
12.5 cm3 of H2SO4(aq) + 25 cm3 of NaOH(aq)
D
25 cm3 of H2SO4(aq) + 15 cm3 of NaOH(aq)
7.5 cm3 of H2SO4(aq) + 15 cm3 of NaOH(aq)
The greatest volumes of H2SO4(aq) and NaOH(aq) react in Option B. The greatest amount of heat is
released and thus the temperature rise is the greatest.
65 C For the neutralization between a strong acid and a strong alkali, the heat released is 57 kJ for 1 mole
of water produced.
For neutralization in which either the acid or alkali or both are weak, the heat released is less than
57 kJ for 1 mole of water produced. The is because some energy is consumed when the weak acid
and weak alkali dissociate to give hydrogen ions and hydroxide ions before neutralization.
Expt.
Solutions mixed
3
–3
Strength of acid and alkali
Temperature rise
1
100 cm of 1 mol dm CH3COOH(aq)
and 100 cm3 of 1 mol dm–3 NaOH(aq)
neutralization between a weak acid
and a strong alkali
T1
2
100 cm3 of 1 mol dm–3 HCl(aq) and
100 cm3 of 1 mol dm–3 NaOH(aq)
neutralization between a strong acid
and a strong alkali
T2
\ T1 < T2.
95
66 C For the neutralization between a strong acid and a strong alkali, the heat released is 57 kJ for 1 mole
of water produced.
Number of moles of HNO3 /
NaOH mixed
= molarity of solution x
volume of solution
Mixture
3
–3
50 cm of 2 mol dm
3
HNO3(aq) + 50 cm of
–3
2 mol dm NaOH(aq)
2 mol dm–3 x
100 cm3 of 2 mol dm–3
HNO3(aq) + 100 cm3 of
2 mol dm–3 NaOH(aq)
2 mol dm–3 x
100 cm3 of 1 mol dm–3
HNO3(aq) + 100 cm3 of
1 mol dm–3 NaOH(aq)
1 mol dm–3 x
Number of moles of water
formed
Heat Temperature
released
rise
50
dm3
1 000
HNO3(aq) + NaOH(aq)
0.1 mol 0.1 mol
NaNO3(aq) + H2O(l)
0.1 mol
5.7 kJ
x
100
dm3
1 000
HNO3(aq) + NaOH(aq)
0.2 mol 0.2 mol
NaNO3(aq) + H2O(l)
0.2 mol
11.4 kJ
y
100
dm3
1 000
HNO3(aq) + NaOH(aq)
0.1 mol 0.1 mol
NaNO3(aq) + H2O(l)
0.1 mol
5.7 kJ
z
= 0.1 mol
= 0.2 mol
= 0.1 mol
The first mixture (total volume 100 cm3) is heated up by 5.7 kJ while the second mixture (total volume
200 cm3) is heated up by 11.4 kJ. Hence these two mixtures show the same temperature rise, i.e. x = y.
The third mixture (total volume 200 cm3) is heated up by less than 5.7 kJ. Hence its temperature rise is
less than x and y.
\ x = y > z.
67 C For the neutralization between a strong acid and a strong alkali, the heat released is 57 kJ for 1 mole
of water produced.
96
For neutralization in which either the acid or alkali or both are weak, the heat released is less than
57 kJ for 1 mole of water produced. The is because some energy is consumed when the weak acid
and weak alkali dissociate to give hydrogen ions and hydroxide ions before neutralization.
Mixture
Number of moles of acid /
alkali mixed
= molarity of solution x
volume of solution
100 cm3 of 1 mol dm–3
HCl(aq) + 100 cm3 of
1 mol dm–3 NaOH(aq)
Number of moles of HCl / NaOH
100
= 1 mol dm–3 x
dm3
1 000
= 0.1 mol
HCl(aq) + NaOH(aq)
neutralization
0.1 mol 0.1 mol
between a
5.7 kJ
NaCl(aq) + H2O(l) strong acid and
0.1 mol a strong alkali
x
100 cm3 of 2 mol dm–3
HCl(aq) + 100 cm3 of
2 mol dm–3 NaOH(aq)
Number of moles of HCl / NaOH
100
= 2 mol dm–3 x
dm3
1 000
= 0.2 mol
HCl(aq) + NaOH(aq)
neutralization
0.2 mol 0.2 mol
between a
11.4 kJ
NaCl(aq) + H2O(l) strong acid and
0.2 mol a strong alkali
y
100 cm3 of 1 mol dm–3
HCl(aq) + 100 cm3 of
1 mol dm–3 NH3(aq)
Number of moles of HCl / NH3
100
= 1 mol dm–3 x
dm3
1 000
= 0.1 mol
HCl(aq) + NH4+(aq) + OH–(aq)
neutralization
0.1 mol 0.1 mol
between a
< 5.7 kJ
NH4Cl(aq) + H2O(l) strong acid and
0.1 mol a weak alkali
z
Number of moles of water
Strength of
Heat Temperature
formed
acid and alkali released
rise
All the mixtures have the same total volume (200 cm3).
The heat released in the second case is higher than that released in the first case.
Thus y > x.
The heat released in the third case is less than that released in the first case.
Thus x > z.
\ y > x > z.
68 B For the neutralization between a strong acid and a strong alkali, the heat released is 57 kJ for 1 mole
of water produced.
Number of moles of H2SO4 /
NaOH mixed
= molarity of solution x
volume of solution
Mixture
25 cm3 of 1 mol dm–3
H2SO4(aq)
+
25 cm3 of
2 mol dm–3 NaOH(aq)
3
–3
50 cm of 1 mol dm
H2SO4(aq)
+
50 cm3 of
2 mol dm–3 NaOH(aq)
Number of moles of H2SO4
25
= 1 mol dm–3 x
dm3
1 000
= 0.025 mol
Number of moles of NaOH
25
= 2 mol dm–3 x
dm3
1 000
= 0.05 mol
Number of moles of H2SO4
50
= 1 mol dm–3 x
dm3
1 000
= 0.05 mol
Number of moles of NaOH
50
= 2 mol dm–3 x
dm3
1 000
= 0.1 mol
Number of moles of water
formed
Heat Temperature
released
rise
H2SO4(aq) + 2NaOH(aq)
0.025 mol 0.05 mol
Na2SO4(aq) + 2H2O(l)
0.05 mol
2.85 kJ
w
H2SO4(aq) + 2NaOH(aq)
0.05 mol 0.1 mol
Na2SO4(aq) + 2H2O(l)
0.1 mol
5.7 kJ
x
25 cm3 of 0.5 mol dm–3
H2SO4(aq)
+
25 cm3 of
1 mol dm–3 NaOH(aq)
Number of moles of H2SO4
25
= 0.5 mol dm–3 x
dm3
1 000
H2SO4(aq) + 2NaOH(aq)
= 0.0125 mol
0.0125 mol 0.025 mol
1.425 kJ
Na2SO4(aq) + 2H2O(l)
Number of moles of NaOH
0.025 mol
25
= 1 mol dm–3 x
dm3
1 000
= 0.025 mol
y
50 cm3 of 0.5 mol dm–3
H2SO4(aq)
+
50 cm3 of
1 mol dm–3 NaOH(aq)
Number of moles of H2SO4
50
= 0.5 mol dm–3 x
dm3
1 000
H2SO4(aq) + 2NaOH(aq)
= 0.025 mol
0.025 mol 0.05 mol
Na2SO4(aq) + 2H2O(l)
Number of moles of NaOH
0.05 mol
50
= 1 mol dm–3 x
dm3
1 000
= 0.05 mol
z
2.85 kJ
The first mixture (total volume 50 cm3) is heated up by 2.85 kJ while the second mixture (total volume
100 cm3) is heated up by 5.7 kJ. Hence these two mixtures show the same temperature rise, i.e. w = x.
97
The third mixture (total volume 50 cm3) is heated up by 1.425 kJ while the fourth mixture (total
volume 100 cm3) is heated up by 2.85 kJ. Hence these two mixtures show the same temperature rise, i.e.
y = z.
\ w = x > y = z.
69 C
Indicator
Colour of solution
Possible pH range of solution
Methyl orange
yellow
4.5 – 14
Bromothymol blue
blue
7.5 – 14
Phenolphthalein
colourless
0 – 8.5
\ the pH range of the solution should be 7.5 – 8.5.
70 A The following diagram shows the titration curve for the titration of 20.0 cm3 of 0.100 mol dm–3
CH3COOH(aq) with 0.100 mol dm–3 NaOH(aq).
Q)
FRVJWBMFODF
QPJOU
7PMVNFPG/B0)BR
BEEFEDN
During the titration of a weak acid with a strong alkali, the pH at the equivalence point is greater than 7.
71 C During the titration of a strong acid and a weak alkali, the pH at the equivalence point is less than 7.
72 C Bromocresol green and methyl red change colour within the pH range of the vertical part of the
titration curve.
Hence they are suitable indicators for the titration.
73 A The reaction between two solids should be the slowest.
2.0 g
3.0 min
= 0.67 g min–1
74 D Average reaction rate =
98
75 C Rate = –
=
1
4
d[NH3(g)]
dt
d[O2(g)]
dt
1
5
Rate of consumption of NH3(g) = –
d[NH3(g)]
dt
d[O2(g)]
4
=
dt
5
4
=
(1.80 mol dm–3 s–1)
5
= 1.44 mol dm–3 s–1
76 A Rate = –
=
1
6
1
5
d[O2(g)]
dt
d[H2O(g)]
dt
Rate of formation of H2O(g) =
d[H2O(g)]
dt
d[O2(g)]
6
=
dt
5
6
=
(1.80 mol dm–3 s–1)
5
= 2.16 mol dm–3 s–1
77 D Option D — The intensity of the orange colour of Cr2O72– ions increases as the reaction proceeds.
78 D Option A — The concentration of both reagents in the reaction mixture in Experiments I and II are
different. So they CANNOT be used for comparison.
Option B — The temperature in Experiments I and III are different. So they CANNOT be used for
comparison.
Option C — The temperature in Experiments II and IV are different. So they CANNOT be used for
comparison.
Option D — The total volume of the samples in Experiments III and IV are the same. The only
difference is the volume of the sodium thiosulphate solution used, i.e. the concentration
of sodium thiosulphate in the reaction mixture.
79 B Option A — The concentration of sodium thiosulphate in the reaction mixture in Experiments I and III
are different. So they CANNOT be used for comparison.
Option B — The concentration of both reagents in the reaction mixture in Experiments I and IV are
the same. The only difference is the temperature.
Option C — The concentration of acid in the reaction mixture in Experiments II and III are different. So
they CANNOT be used for comparison.
Option D — The concentration of both reagents in the reaction mixture in Experiments II and IV are
different. So they CANNOT be used for comparison.
99
80 A 2 mol dm–3 H2SO4(aq) has the highest concentration of hydrogen ions.
Hence the initial rate of reaction would be the highest when it is added to the magnesium ribbons.
81 A The reaction represented by curve Y took more time to complete, i.e. the rate of this reaction was
lower.
Option A — Adding water to the acid would decrease its concentration. The initial reaction rate would
decrease.
Option D — As excess hydrochloric acid was used, magnesium would limit the volume of hydrogen
evolved.
Using half piece of magnesium ribbon would decrease the volume of hydrogen evolved.
82 D The rate of a reaction increases when
• the surface area of a solid reactant is increased;
• the concentration of a reactant is increased; and
• the solution is hotter.
83 B The reaction represented by curve Y took more time to complete, i.e. the rate of this reaction was
lower.
Option B — Calcium carbonate powder had a greater surface area than calcium carbonate lumps of
the same mass.
Option D — Calcium carbonate was the limiting reactant as it disappeared in both experiments after
reaction.
The rate of the reaction decreases when the surface area of calcium carbonate is
decreased, i.e. using calcium carbonate lumps instead of calcium carbonate powder.
Using a lower mass of calcium carbonate would decrease the volume of gas formed, i.e.
decrease the pressure.
84 D The tangent to curve II was less steep than that to curve I, i.e. the rate of the reaction represented by
curve II was lower.
100
However, a greater volume of hydrogen was formed in this reaction.
Option A — Increasing the temperature would increase the rate of the reaction.
Option B — Using magnesium powder would increase the rate of the reaction.
Option C — Using a more concentrated sulphuric acid would increase the rate of the reaction.
Option D — Using a less concentrated sulphuric acid would decrease the rate of the reaction.
However, increasing the number of moles of H2SO4 (from 0.1 mole to 0.14 mole) would increase the
volume of hydrogen formed.
85 A The concentration of MnO4– ions in the reaction mixture decreased as the reaction proceeded. The
reaction mixture became lighter in colour gradually. Thus the reaction mixture absorbed less and less
light and so the absorbance went down.
–
The concentration of MnO4 ions in reaction mixture 1 was higher than that in reaction mixture 2.
Hence the initial absorbance of reaction mixture 1 was higher than that of reaction mixture 2 (i.e. Options B,
C and D are INCORRECT).
The concentration of MnO4–(aq) ion in reaction mixture 1 was higher than that in reaction mixture 2.
So, the rate of reaction for reaction mixture 1 was also higher and thus the tangent to its curve would
be steeper.
\ the curves in Option A show the results.
86 B
Beaker
Number of moles of acid
= molarity of solution x volume of solution
A
Number of moles of HCl
45
3
= 1.2 mol dm–3 x
dm
1 000
= 0.054 mol
B
Number of moles of CH3COOH
60
–3
3
= 0.9 mol dm x
dm
1 000
= 0.054 mol
Reaction between CaCO3 and acid
CaCO3(s) + 2HCl(aq)
CaCl2(aq) + H2O(l) + CO2(g)
CaCO3(s) + 2CH3COOH(aq)
(CH3COO)2Ca(aq) + H2O(l) + CO2(g)
–3
Option A — The initial rate of reaction in beaker A is higher as 1.2 mol dm HCl(aq) contains a
–3
higher concentration of hydrogen ions than 0.9 mol dm CH3COOH(aq) does.
Option B — Marble chips disappear in both beakers after reaction. Thus marble chip is the limiting
reagent.
Option C — The average rate of evolution of gas in Beaker A is higher as 1.2 mol dm HCl(aq)
–3
contains a higher concentration of hydrogen ions than 0.9 mol dm CH3COOH(aq) does.
Option D — Equal masses of calcium chloride are formed in both beakers.
However, the volumes of the two reaction mixtures are different.
Hence the concentrations of Ca2+(aq) ions in the two beakers are different.
Equal masses of marble chips would produce the same amount of gas.
–3
87 C The reaction represented by curve Y took less time to complete, i.e. the rate of this reaction was
higher.
Option A — Lowering the temperature would decrease the rate of the reaction.
Option B — Using the same volume of 0.15 mol dm–3 hydrogen peroxide solution instead of 0.12 mol
dm–3 solution (i.e. a greater number of moles of hydrogen peroxide) would increase the
volume of O2 produced.
101
Option C — Using 0.15 mol dm–3 hydrogen peroxide solution instead of 0.12 mol dm–3 solution would
increase the rate of the reaction.
Option D — Using 120 cm3 of 0.10 mol dm–3 hydrogen peroxide solution instead of 100 cm3 would
increase the volume of O2 produced.
3
–3
Number of moles of H2O2 in 100 cm of 0.12 mol dm
100
= 0.12 mol dm–3 x
dm3
1 000
= 0.012 mol
H2O2(aq)
Number of moles of H2O2 in 80 cm3 of 0.15 mol dm–3 H2O2(aq)
80
= 0.15 mol dm–3 x
dm3
1 000
= 0.012 mol
As the number of moles of H2O2 remained the same, the volume of O2 produced would
be the same.
88 B (2) The amount of catalyst used would NOT affect the amount of oxygen produced.
89 D (1) Glass cleanser contains ammonia.
(2) Alkaline solutions are more effective than acidic solutions in creating curls of the hair. The curls are
permanent. Perm solutions are thus alkaline.
(3) Limewater is a saturated solution of calcium hydroxide.
90 A (2) Metals tarnish because they reaect with the air to form a layer of oxide or sulphide. Acids can be
used to remove this layer. One of the oxide layers most difficultly removed is rust. Rust removers
usually contain an acid such as hydrochloric acid or phosphoric acid.
(3) Nitric acid is NOT a drying agent.
91 A (1) Solid citric acid consists of molecules.
(2) Hydrochloric acid is produced in human stomach.
(3) Solid citric acid does NOT react with magnesium because it does not contain hydrogen ions.
92 C (1) Heating solid or solution of ammonium chloride with sodium hydroxide solution liberates ammonia
gas.
NH4Cl(s or aq) + NaOH(aq)
(2) Phenolphthalein gives a colourless solution with acidic solutions.
(3) Dilute sodium hydroxide solution would react with carbon dioxide gas.
NaOH(aq) + CO2(g)
NH3(g) + NaCl(aq) + H2O(l)
NaHCO3(aq)
93 A (1) and (2) Vinegar and grapefruit juice react with magnesium to give hydrogen, a gas which burns
with a ‘pop’ sound.
102
(3) Solid citric acid does NOT react with magnesium because it does not contain hydrogen ions.
94 A (1) Aluminium nitrate solution gives a white precipitate with dilute aqueous ammonia.
3+
–
Al (aq) + 3OH (aq)
Al(OH)3(s)
(2) Iron(III) sulphate solution gives a reddish brown precipitate with dilute aqueous ammonia.
Fe3+(aq) + 3OH–(aq)
Fe(OH)3(s)
95 A (1) and (2) The aqueous solution of citric acid shows typical properties of an acid while solid citric acid
does not.
When solid citric acid was added to an aqueous solution of sodium hydrogencarbonate,
some citric acid molecules dissociated to give hydrogen ions.
citric acid(s) + water
citric acid(aq)
HCO3– ions in sodium hydrogencarbonate reacted with the H+ ions to give carbon dioxide
gas. Effervescence occurred.
HCO3–(aq) + H+(aq)
(3) Solid citric acid would NOT react with solid sodium hydrogencarbonate because it did not contain
hydrogen ions.
citric acid(aq)
H+(aq) + citrate ion(aq)
H2O(l) + CO2(g)
96 A (3) Phenolphthalein is colourless in dilute hydrochloric acid.
97 C (1) When dilute sulphuric acid reacts with calcium carbonate, insoluble calcium sulphate forms. The
calcium sulphate covers the surface of calcium carbonate and prevents further reaction.
(2) Carbonic acid is formed when carbon dioxide dissolves in water.
Cl2(g) + H2O(l)
The aqueous solution can conduct electricity due to the presence of mobile ions.
(3) NaOH(aq) can absorb carbon dioxide.
NaHCO3(aq) is formed which undergoes dehydration to give Na2CO3(s).
NaOH(aq) + CO2(g)
2NaHCO3(aq)
H2CO3(aq)
carbonic acid
NaHCO3(aq)
Na2CO3(s) + H2O(l)
98 A (1) Pure sulphuric acid is a covalent compound. It is a colourless liquid consisting of molecules.
(2) When we describe acids as strong and weak, we are talking about the extent of their dissociation
in water. When we talk about concentration, we are referring to the amount of an acid in a unit
volume of solution.
(3) Concentrated sulphuric acid does NOT give an acid mist in air because it is non-voltatile.
For example, 5 mol dm–3 sulphuric acid is a concentrated solution of a strong acid while 0.1 mol
dm–3 sulphuric acid is a dilute solution of a strong acid.
103
99 A Aqueous ammonia reacts with hydrochloric acid according to the equation below. The salt ammonium
chloride is formed in the reaction.
NH3(aq) + HCl(aq)
NH4Cl(aq)
(1) Neutralization occurs and heat is released. Thus the temperature of the mixture increases.
(2) Ammonium chloride is soluble in water. Hence no precipitate would form.
(3) Ammonium chloride is an ionic compound. The solution mixture conducts electricity due to the
presence of mobile ions from ammonium chloride.
100A (2) Sodium hydroxide solution can absorb sulphur dioxide.
NaOH(aq) + SO2(g)
NaHSO3(aq)
(3) Sulphur dioxide is added to wine as a preservative, NOT a flavour enhancer.
101B (1) We do not use anhydrous calcium chloride to dry ammonia gas because ammonia reacts with it.
CaCl2(s) + 4NH3(g)
CaCl2•4NH3(s)
(3) Concentrataed nitric acid is NOT a drying agent.
102C (2) Pb(NO3)2(aq) + CaCl2(aq)
PbCl2(s) + Ca(NO3)2(aq)
CuCO3(s) + Na2SO4(aq)
(3) Na2CO3(aq) + CuSO4(aq)
103D (1) Magnesium reacts with dilute sulphuric acid while silver has NO reaction.
(2) Barium nitrate solution gives a white precipitate with dilute sulphuric acid while potassium nitrate
solution does NOT.
(3) Zinc carbonate gives gas bubbles (carbon dioxide gas) with dilute sulphuric acid while zinc chloride
does NOT.
Ba2+(aq) + SO42–(aq)
BaSO4(s)
104B (1) There is no observable change when zinc bromide solution or zinc iodide solution is mixed with
dilute hydrochloric acid.
(2) Lead(II) nitrate solution gives a white precipitate with dilute hydrochloric acid while magnesium
sulphate solution does NOT.
(3) Both ammonium carbonate and sodium hydrogencarbonate react with dilute hydrochloric acid to
give carbon dioxide gas.
Pb2+(aq) + 2Cl–(aq)
PbCl2(s)
105C (1) Both ammonium nitrate solution and potassium chloride solution are colourless.
104
(2) Ammonia is liberated when ammonium nitrate solution is heated with dilute sodium hydroxide
solution. The ammonia can be tested with moist red litmus paper.
(3) Potassium chloride solution gives a white precipitate with silver nitrate solution while ammonium
nitrate solution does NOT.
NH4+(aq) + OH–(aq)
Ag+(aq) + Cl–(aq)
NH3(g) + H2O(l)
AgCl(s)
106B (1) Both dilute hydrochloric acid and nitric acid react with copper(II) oxide.
(2) Dilute hydrochloric acid gives a white precipitate with silver nitrate solution while dilute nitric acid
does NOT.
(3) Both dilute hydrochloric acid and nitric acid react with sodium hydrogencarbonate solution.
Ag+(aq) + Cl–(aq)
107D
AgCl(s)
Solutions mixed
NaOH(aq) and CuSO4(aq)
(1)
(2)
(3)
Precipitate formed
a pale blue precipitate
NH3(aq) and Mg(NO3)2(aq)
a white precipitate
(NH4)2CO3(aq) and NiCl2(aq)
Ionic equation
2+
Cu (aq) + 2OH–(aq)
2+
–
Mg (aq) + 2OH (aq)
2+
a green precipitate
2–
3
Ni (aq) + CO (aq)
Cu(OH)2(s)
Mg(OH)2(s)
NiCO3(s)
108D (1) Copper(II) oxide and dilute sulphuric acid react to give copper(II) sulphate and water. NO gas is
produced.
CuO(s) + H2SO4(aq)
CuSO4(aq) + H2O(l)
(2) When water is added to solid citric acid, some citric acid molecules dissociate to give hydrogen
ions.
citric acid(s) + water
citric acid(aq)
HCO3 (aq) ions in sodium hydrogencarbonate react with the H ions to give carbon dioxide gas.
HCO3–(aq) + H+(aq)
(3) Ammonium carbonate reacts with dilute hydrochloric acid to give carbon dioxide gas.
citric acid(aq)
H+(aq) + citrate ion(aq)
–
+
H2O(l) + CO2(g)
(NH4)2CO3(s) + 2HCl(aq)
109D (1) CO32–(aq) + Pb2+(aq)
(2) SO42–(aq) + Pb2+(aq)
(3) 2Cl–(aq) + Pb2+(aq)
2NH4Cl(aq) + H2O(l) + CO2(g)
PbCO3(s)
PbSO4(s)
PbCl2(s)
110D Magnesium sulphate is prepared by reacting dilute sulphuric acid with either magnesium, magnesium
carbonate or magnesium oxide.
Mg(s) + H2SO4(aq)
MgCO3(s) + H2SO4(aq)
MgO(s) + H2SO4(aq)
MgSO4(aq) + H2(g)
MgSO4(aq) + H2O(l) + CO2(g)
MgSO4(aq) + H2O(l)
111A (2) and (3) If we wash the pipette or burette with distilled water only, water droplets remaining on
the inside of the glassware will dilute the solution that the glassware is going to contain.
112D (2) and (3) Adding dilute sulphuric acid to limewater and mixing hydrochloric acid with sodium
hydroxide solution are neutralization reactions. Heat is released in these reactions.
105
113A
Number of moles of acid / alkali
= molarity of solution x
volume of solution
Solutions mixed
3
–3
100 cm of 1 mol dm HNO3(aq)
and
3
–3
100 cm of 1 mol dm KOH(aq)
(1)
number of moles of HNO3
100
3
= 1 mol dm–3 x
dm
1 000
= 0.1 mol
number of moles of KOH
100
–3
3
dm
= 1 mol dm x
1 000
= 0.1 mol
(2)
3
–3
100 cm of 1 mol dm H2SO4(aq)
and
3
–3
100 cm of 1 mol dm NaOH(aq)
number of moles of H2SO4
100
–3
3
= 1 mol dm x
dm
1 000
= 0.1 mol
number of moles of NaOH
100
–3
3
dm
= 1 mol dm x
1 000
= 0.1 mol
(3)
100 cm3 of 1 mol dm–3 H2SO4(aq)
and
3
–3
100 cm of 2 mol dm KOH(aq)
number of moles of H2SO4
100
3
= 1 mol dm–3 x
dm
1 000
= 0.1 mol
number of moles of KOH
100
–3
3
dm
= 2 mol dm x
1 000
= 0.2 mol
Complete neutralization?
HNO3(aq) + KOH(aq)
0.1 mol 0.1 mol
KNO3(aq) + H2O(l)
\ complete neutralization
occurs and a neutral
solution is obtained.
H2SO4(aq) + 2NaOH(aq)
0.1 mol 0.1 mol
Na2SO4(aq) + 2H2O(l)
\ not enough NaOH(aq)
to neutralize H2SO4(aq)
completely.
H2SO4(aq) + 2KOH(aq)
0.1 mol 0.2 mol
K2SO4(aq) + 2H2O(l)
\ complete neutralization
occurs and a neutral
solution is obtained.
114B For the neutralization between a strong acid and a strong alkali, the heat released is 57 kJ for 1 mole
of water produced.
For neutralization in which either the acid or alkali or both are weak, the heat released is less than
57 kJ for 1 mole of water produced. The is because some energy is consumed when the weak acid
and weak alkali dissociate to give hydrogen ions and hydroxide ions before neutralization.
Option
Solutions mixed
3
106
–3
Strength of acid and alkali
Temperature rise
—
25 cm of 1 mol dm HCl(aq) and
25 cm3 of 1 mol dm–3 NaOH(aq)
neutralization between a strong acid
and a strong alkali
T
(1)
25 cm3 of 1 mol dm–3 HNO3(aq) and
25 cm3 of 1 mol dm–3 NH3(aq)
neutralization between a strong acid
and a weak alkali
< T
(2)
25 cm3 of 1 mol dm–3 HNO3(aq) and
25 cm3 of 1 mol dm–3 KOH(aq)
neutralization between a strong acid
and a strong alkali
T
(3)
50 cm3 of 1 mol dm–3 CH3COOH(aq) and
50 cm3 of 1 mol dm–3 NaOH(aq)
neutralization between a weak acid
and a strong alkali
< T
\ Mixing 25 cm3 of 1 mol dm–3 HNO3(aq) and 25 cm3 of 1 mol dm–3 KOH(aq) produces a similar
temperature rise as mixing 25 cm3 of 1 mol dm–3 HCl(aq) and 25 cm3 of 1 mol dm–3 NaOH(aq).
115A
Beaker
Number of moles of HCl = molarity of solution x volume of solution
1
–3
x
100
3
dm = 0.1 mol
1 000
–3
x
50
3
dm = 0.1 mol
1 000
1 mol dm
2
2 mol dm
Marble chips react with dilute hydrochloric acid according to the following equation:
CaCO3(s) + 2HCl(aq)
excess
0.1 mol
(1) Equal numbers of moles are present in the two beakers.
Hence equal masses of marble chips are used in the beakers.
Thus equal masses of marble chips remain in the beakers.
(2) The same amount of gas (
(3) Equal amounts of calcium chloride are produced in both cases.
However, the volumes of the reaction mixtures are different.
Hence concentrations of calcium chloride solution in the beakers are different.
116A
Beaker
CaCl2(aq) + H2O(l) + CO2(g)
0.1
mole) is produced in both cases.
2
Number of moles of Mg
mass
=
molar mass
1
2 g
–1
24.3 g mol
= 0.08 mol
2
Number of moles of acid
= molarity of solution x
volume of solution
Reaction between Mg and acid
Number of moles of HCl
90
–3
3
= 1.2 mol dm x
dm
1 000
= 0.11 mol
Mg(s) + 2HCl(aq)
0.08 mol 0.11 mol
MgCl2(aq) + H2(g)
Number of moles of CH3COOH
120
–3
3
= 0.9 mol dm x
dm
1 000
= 0.11 mol
Mg(s) + 2CH3COOH(aq)
0.08 mol 0.11 mol
(CH3COO)2Mg(aq) + H2(g)
According to the equations, 1 mole of Mg reacts with 2 moles of HCl / CH3COOH to produce 1 mole
of H2. During the reaction, 0.11 mole of HCl / CH3COOH reacted with 0.055 mole of Mg.
Therefore Mg was in excess. The amount of HCl / CH3COOH limited the amount of H2 produced.
(1) Mg was in excess. The acids in both beakers reacted completely.
(2) The same amount of gas (
(3) HCl(aq) is a strong acid while CH3COOH(aq) is a weak acid.
1.2 mol dm–3 HCl(aq) has a higher concentration of hydrogen ions than 0.9 mol dm–3 CH3COOH(aq)
does.
Hence the reaction between Mg and HCl(aq) took a shorter time to complete.
0.11
mole) was produced in both cases.
2
107
117B (1) During the acid catalyzed hydrolysis of methyl methanoate, the concentration of ethanoic acid
(CH3COOH) in the reaction mixture increases as the reaction proceeds.
Hence the progress of the reaction can be followed by titration with an alkali.
(3) During the decomposition of a dicarboxylic acid, the concentration of the acid in the reaction
mixture decreases as the reaction proceeds.
Hence the progress of the reaction can be followed by titration with an alkali.
118A (1) The intensity of the yellow-brown colour of Br2(aq) decreases as the reaction proceeds.
Hence the progress of reaction can be followed by using a colorimeter.
(2) The reaction mixture becomes turbid as the sulphur forms.
(3) The reaction does NOT involve any colour change.
A colorimeter can be used to follow the change in light level passing through the reaction mixture
as the turbidity increases.
Hence a colorimeter CANNOT be used to follow the progress of the reaction.
119A (1) Powdered zinc has a greater surface area than a zinc granule of the same mass.
Hence the initial rate of the reaction would increase.
(3) Doubling the volume of the acid used has NO effect on the initial rate of the reaction.
120A HCl(aq) and CH3COOH(aq) react with calcium carbonate according to the following equations:
108
Beaker A CaCO3(s) + 2HCl(aq)
Beaker B CaCO3(s) + 2CH3COOH(aq)
Calcium carbonate remain in both beakers after reaction. Thus HCl(aq) and CH3COOH(aq) are the
limiting reagents. Their amounts limit the amount of carbon dioxide gas produced.
(1) HCl(aq) is a strong acid while CH3COOH(aq) is a weak acid.
1 mol dm–3 HCl(aq) has a higher concentration of hydrogen ions than 1 mol dm–3 CH3COOH(aq)
does.
Hence the initial rate of reaction in Beaker A is greater than that in Beaker B.
(2) As equal numbers of moles of HCl and CH3COOH are used, the same amount of gas is produced
in both cases.
(3) The rate of the reaction between CaCO3 and 1 mol dm–3 HCl(aq) is greater than that between
CaCO3 and 1 mol dm–3 CH3COOH(aq).
1 mol dm–3 x
1 mol dm–3 x
CaCl2(aq) + H2O(l) + CO2(g)
100
dm3 = 0.1 mol
1 000
(CH3COO)2Ca(aq) + H2O(l) + CO2(g)
100
dm3 = 0.1 mol
1 000
Hence the reaction in Beaker A takes a shorter time to complete.
121D
Reaction
Number of moles of Zn
mass
=
molar mass
Number of moles of HCl
= molarity of solution x
volume of solution
2 mol dm–3 x
1
3 g
–1
65.4 g mol
= 0.046 mol
50
3
dm
1 000
Zn(s) + 2HCl(aq)
0.046 mol 0.1 mol
ZnCl2(aq) + H2(g)
50
3
dm
1 000
Zn(s) + 2HCl(aq)
0.046 mol 0.2 mol
ZnCl2(aq) + H2(g)
= 0.1 mol
4 mol dm–3 x
2
Reaction between Zn and HCl
= 0.2 mol
According to the equation, 1 mole of Zn reacts with 2 moles of HCl to produce 1 mole of H2. During
the reaction, 0.046 mole of Zn reacted with 0.092 mole of HCl.
Therefore HCl was in excess. The amount of Zn limited the amount of H2 produced.
(1) All the zinc reacted in both cases.
(2) The hydrogen produced would escape. Thus the mass of each reaction mixture would decrease.
(3) The concentration of HCl(aq) was lower in Reaction 1.
As the same amount of gas (0.046 mol) was produced in both reactions, the losses in mass of the
two reaction mixtures after reaction would be the same.
Thus the initial rate of Reaction 1 was smaller.
122C
Beaker
Number of moles of Mg
mass
=
molar mass
1
1.8 g
–1
24.3 g mol
2
Number of moles of acid
= molarity of solution x
volume of solution
Reaction between Mg and acid
Number of moles of HCl
100
–3
3
dm
= 1 mol dm x
1 000
= 0.1 mol
Mg(s) + 2HCl(aq)
0.074 mol 0.1 mol
MgCl2(aq) + H2(g)
According to the equation, 1 mole
of Mg reacts with 2 moles of HCl
to produce 1 mole of H2.
Hence 0.1 mole of HCl reaccted
with 0.05 mole of Mg to give 0.05
mole of H2.
Thus Mg was in excess.
Number of moles of H2SO4
100
3
= 1 mol dm–3 x
dm
1 000
= 0.1 mol
Mg(s) + H2SO4(aq)
0.074 mol 0.1 mol
MgSO4(aq) + H2(g)
According to the equation, 1 mole
of Mg reacts with 1 mole of H2SO4
to produce 1 mole of H2.
Hence 0.074 mole of Mg reaccted
with 0.074 mole of H2SO4 to give
0.074 mole of H2.
Thus H2SO4(aq) was in excess.
= 0.074 mol
(1) The hydrogen produced would escape. Thus the mass of each reaction mixture would decrease.
Different amounts of hydrogen gas were produced in the two cases.
Hence the losses in mass of the two reaction mixtures after reaction would be different.
109
(2) HCl(aq) is a monobasic acid while H2SO4(aq) is a dibasic acid.
Therefore the concentration of hdyrogen ions in 1 mol dm
–3
dm H2SO4(aq).
Hence the initial rate of reaction in Beaker 1 would be smaller than that in Beaker 2.
(3) Magnesium reacted completely with H2SO4(aq) in Beaker 2.
–3
HCl(aq) was lower than that in 1 mol
123B (1) A catalyst would NOT change the amount of product formed in a reaction.
(3) The physical state of a catalyst may NOT be the same as those of the reactants.
For example, the solid magnese(IV) oxide acts as a catalyst in the decomposition of hydrogen
peroxide solution.
124A
125D Nitric acid is NOT a drying agent.
126A
127B Sodium hydroxide can absorb sulphur dioxide gas because it can react with the gas.
128B Aqueous solution of citric acid can conduct electricity due to the presence of mobile ions.
H+(aq) + citrate ion(aq)
citric acid(aq)
Hence citric acid is an electrolyte.
129D Copper(II) hydroxide does NOT dissolve in excess sodium hydroxide solution.
130A
131C Aqueous solution of ethanoic acid can conduct electricity.
Hence ethanoic acid is an electrolyte.
132C Both ethanoic acid and propanoic acid are monobasic acids as only the hydrogen atom in the –COOH
group of a molecule of each acid can undergo dissociation.
)
)
0
$
$
POMZUIJTIZESPHFO
BUPNDBOVOEFSHP
EJTTPDJBUJPO
0
)
FUIBOPJDBDJE
133D Ammonia is a weak alkali.
110
)
)
)
)
0
$
$
$
)
)
POMZUIJTIZESPHFO
BUPNDBOVOEFSHP
EJTTPDJBUJPO
QSPQBOPJDBDJE
0
)
134D Hydrochloric acid is a strong acid while ethanoic acid is a weak acid.
–3
Therefore 1 mol dm hydrochloric acid contains a higher concentration of hydrogen ions than 1 mol
–3
dm ethanoic acid does.
Hence the acids give a different colour with the same quantity of universal indicator.
135D Ethanoic acid is a weak acid. It only partically dissociates in both concentrated and dilute solutions.
136D Solid citric acid does NOT show typical properties of an acid.
A typical acid does NOT react with copper.
137D Number of moles of H2SO4 in 10 cm3 of 5 mol dm–3 acid
10
= 5 mol dm–3 x
dm3
1 000
= 0.05 mol
Number of moles of H2SO4 in 50 cm3 of 1 mol dm–3 acid
50
= 1 mol dm–3 x
dm3
1 000
= 0.05 mol
The acids contain the same number of moles of H2SO4.
Both acids react with excess zinc granules to produce the same amount of hydrogen.
5 mol dm–3 H2SO4(aq) is a concentrated solution of a strong acid while 1 mol dm–3 H2SO4(aq) is a
dilute solution of a strong acid.
138D HCl(aq) is a strong acid while CH3COOH(aq) is a weak acid.
1 mol dm–3 HCl(aq) contains a higher concentration of hydrogen ions than 1 mol dm–3 CH3COOH(aq)
does.
Hence the acids have different pH values.
Adding 10 cm3 of 1 mol dm–3 CH3COOH(aq) to 10 cm3 of 1 mol dm–3 HCl(aq) would result in a
change in pH.
139C Both 1 mol dm–3 NaOH(aq) and 1 mol dm–3 NH3(aq) give a precipitate with MgCl2(aq).
140C Our stomach produces hydrochloric acid along with enzymes to digest food. Calcium carbonate gives
carbon dioxide gas when reacting with dilute hydrochloric acid. The gas makes a person uncomfortable.
Therefore calcium carbonate is seldom used in antacids nowadays.
141B Hydrochloric acid is produced by human stomach. Magnesium hydroxide in antacids can neutralize the
excess hydrochloric acid and so the pain can be relieved.
Mg(OH)2(s) + 2HCl(aq)
MgCl2(aq) + 2H2O(l)
111
142B Sodium hydrogensulphate is formed when one of the hydrogen ions in dilute sulphuric acid is replaced
by a sodium ion. It is an acid salt.
Sodium hydrogensulphate solution is acidic. This is because the hydrogensulphate ion can dissociate to
give the hydrogen ion.
HSO4–(aq)
H+(aq) + SO42–(aq)
143C When 50 cm3 of 1 mol dm–3 NaCl(aq) are added to 50 cm3 of 1 mol dm–3 HCl(aq), the concentration
of hydrogen ions decreases because the total volume of the acid has increased.
Hence the pH of the acid would change.
144C For example, 1 mole of sulphuric acid can neutralize two moles of sodium hydroxide.
H2SO4(aq) + 2NaOH(aq)
Na2SO4(aq) + 2H2O(l)
145C When 10 cm3 of 1 mol dm–3 HCl(aq) are added to 10 cm3 of 1 mol dm–3 NaOH, the number of
Na+(aq) ions in the solution mixtue remains unchanged.
However, the volume of the solution has changed.
Hence the concentration of Na+(aq) ions would decrease.
146C Universal indicator is NOT used to detect the end point of a titration because the indicator would NOT
give a sharp colour change at the end point.
147C For the neutralization between a strong acid and a strong alkali, the heat released is 57 kJ for 1 mole
of water produced.
112
Acid and alkali mixed
Number of moles of acid /
alkali mixed
= molarity of solution x
volume of solution
Number of moles of water formed
Heat
released
25 cm3 of 1 mol dm–3
HCl(aq) + 25 cm3 of
1 mol dm–3 NaOH(aq)
number of moles of HCl / NaOH
25
= 1 mol dm–3 x
dm3
1 000
= 0.025 mol
HCl(aq) + NaOH(aq)
0.025 mol 0.025 mol
NaCl(aq) + H2O(l)
0.025 mol
1.425 kJ
25 cm3 of 2 mol dm–3
HCl(aq) + 25 cm3 of
2 mol dm–3 NaOH(aq)
number of moles of HCl / NaOH
25
= 2 mol dm–3 x
dm3
1 000
= 0.05 mol
HCl(aq) + NaOH(aq)
0.05 mol 0.05 mol
NaCl(aq) + H2O(l)
0.05 mol
2.85 kJ
The total volumes of the two mixtures are the same. Hence the temperature rise of the first mixture is
1
half of that of the second mixture, i.e. T1 =
T2.
2
148B Ethanedioic acid crystals can be used to prepare standard solutions because they have the following
characteristics:
• they are obtainable in a very pure form;
• they have a known chemical formula;
• they dissolve in water completely at room temperature;
• they are stable and do not absorb moisture from the air; and
• they have a high molar mass to minimize weighing errors.
149A In the oxidation of oxalate ions by permanganate ions, the intensity of the purple colour of
permanganate ions decreases as the reaction proceeds.
When we shine light upon the reaction mixture, the absorbance of the reaction mixture is directly
proportional to the colour intensity of the reaction mixture and the concentration of the permanganate
ions in the reaction mixture.
Hence the progress of the reaction can be followed by a colorimeter.
150C Hydrochloric acid is a monobasic acid while sulphuric acid is a dibasic acid.
–3
hydrochloric acid has a lower concentration of hydrogen ions than 1 mol dm
–3
Therefore 1 mol dm
sulphuric acid does.
Hence the reaction between 100 cm3 of 1 mol dm–3 hydrochloric acid and zinc granules is slower than
that between 100 cm3 of 1 mol dm–3 sulphuric acid and zinc granules.
151D A catalyst would NOT undergo any permanent chemical changes at the end of a reaction.
Short questions
152
Reaction between
Chemical equation
a) Zinc and dilute hydrochloric acid
Zn(s) + 2HCl(aq)
b) Iron and dilute sulphuric acid
Fe(s) + H2SO4(aq)
c) Zinc carbonate and dilute hydrochloric acid
ZnCO3(s) + 2HCl(aq)
d) Sodium carbonate solution and dilute
sulphuric acid
Na2CO3(aq) + H2SO4(aq)
e) Lead(II) carbonate and dilute nitric acid
ZnCl2(aq) + H2(g)
FeSO4(aq) + H2(g)
ZnCl2(aq) + H2O(l) + CO2(g)
(1)
(1)
(1)
Na2SO4(aq) + H2O(l) + CO2(g)
(1)
PbCO3(s) + 2HNO3(aq)
Pb(NO3)2(aq) + H2O(l) + CO2(g)
(1)
f) Sodium hydrogencarbonate solution and
dilute hydrochloric acid
NaHCO3(aq) + HCl(aq)
NaCl(aq) + H2O(l) + CO2(g)
(1)
g) Sodium hydrogencarbonate solid and dilute
sulphuric acid
2NaHCO3(s) + H2SO4(aq)
Na2SO4(aq) + 2H2O(l) + 2CO2(g)
(1)
h) Magnesium hydroxide solid and dilute nitric
acid
Mg(OH)2(s) + 2HNO3(aq)
Mg(NO3)2(aq) + 2H2O(l)
(1)
113
153
Reaction between
a) Magnesium and sulphuric acid
Mg(s) + 2H+(aq)
b) Sodium carbonate solution and dilute
hydrochloric acid
CO3 (aq) + 2H (aq)
c) Sodium hydrogencarbonate solution and
dilute nitric acid
HCO3 (aq) + H (aq)
d) Iron(III) hydroxide and dilute nitric acid
Fe(OH)3(aq) + 3H (aq)
e) Dilute sodium hydroxide solution and dilute
hydrochloric acid
OH­–(aq) + H+(aq)
154
155
Ionic equation
–
(1)
+
H2O(l) + CO2(g)
(1)
+
H2O(l) + CO2(g)
(1)
+
Reaction between
3+
Fe (aq) + 3H2O(l)
(1)
(1)
H2O(l)
Ionic equation(s)
a) Solution containing calcium ions and dilute
sodium hydroxide solution
Ca (aq) + 2OH (aq)
b) Solution containing aluminium ions and
dilute sodium hydroxide solution
2+
–
Ca(OH)2(s)
(1)
Al (aq) + 3OH (aq)
–
Al(OH)3(s) + OH (aq)
3+
–
Al(OH)3(s)
–
[Al(OH)4] (aq)
(1)
(1)
c) Solution containing magnesium ions and
dilute sodium hydroxide solution
Mg2+(aq) + 2OH–(aq)
Mg(OH)2(s)
(1)
d) Solution containing iron(II) ions and dilute
sodium hydroxide solution
Fe2+(aq) + 2OH–(aq)
Fe(OH)2(s)
(1)
e) Solution containing iron(III) ions and dilute
aqueous ammonia
Fe3+(aq) + 3OH–(aq)
Fe(OH)3(s)
(1)
f) Solution containing copper(II) ions and
dilute aqueous ammonia
Cu2+(aq) + 2OH–(aq)
Cu(OH)2(s) + 4NH3(aq)
Cu(OH)2(s)
[Cu(NH3)4]2+(aq) + 2OH–(aq)
(1)
(1)
g) Solution containing zinc ions and dilute
aqueous ammonia
Zn2+(aq) + 2OH–(aq)
Zn(OH)2(s) + 4NH3(aq)
Zn(OH)2(s)
[Zn(NH3)4]2+(aq) + 2OH–(aq)
(1)
(1)
h) Solution containing ammonium compound
and dilute sodium hydroxide solution (with
heating)
NH4+(aq) + OH–(aq)
NH3(g) + H2O(l)
(1)
Name of ion in solution
Reagent(s) added to the solution
Copper(II) ion
Result
dilute sodium hydroxide solution /
dilute aqueous ammonia
(1)
a blue precipitate
Chloride ion
(1)
dilute nitric acid followed by silver
nitrate solution
a white precipitate
Iron(II) ion / nickel(II) ion
(1)
dilute sodium carbonate solution
a green precipitate
Lead(II) ion / barium ion
(1)
dilute sulphuric acid
a white precipitate
warm with sodium hydroxide solution /
calcium hydroxide solution
(1)
a gas that turns moist red
litmus paper blue is given off
Ammonium ion
114
2–
Mg2+(aq) + H2(g)
156
Reaction
Name of salt
Formula of salt
calcium fluoride
CaF2
(1)
magnesium sulphate
MgSO4
(1)
Dilute sodium hydroxide solution and carbonic
acid
sodium hydrogencarbonate /
sodium carbonate
NaHCO3 / Na2CO3
(1)
Iron(II) carbonate and dilute hydrochloric acid
iron(II) chloride
FeCl2
(1)
Lead(II) oxide and dilute nitric acid
lead(II) nitrate
Pb(NO3)2
(1)
Calcium hydroxide and dilute hydrofluoric acid
Magnesium oxide and dilute sulphuric acid
157a) Gas bubbles are given off. / Magnesium dissolves.
Mg(s) + 2H+(aq)
Mg2+(aq) + H2(g)
(1)
b) Copper(II) oxide dissolves in the acid. / A blue solution forms.
c) Effervescence occurs. / Nickel(II) carbonate dissolves in the acid. / A green solution forms.
d) A white precipitate forms; the precipitate dissolves in excess dilute sodium hydroxide solution to give a
colourless solution.
(1)
Pb2+(aq) + 2OH–(aq)
Pb(OH)2(s) + 2OH–(aq)
e) A white precipitate forms.
f) A white precipitate forms.
g) A green precipitate forms.
h) The solution mixture warms up.
CuO(s) + 2H+(aq)
NiCO3(s) + 2H+(aq)
Cu2+(aq) + H2O(l)
Ni2+(aq) + H2O(l) + CO2(g)
Pb(OH)2(s)
[Pb(OH)4)]2–(aq)
Ca2+(aq) + CO32–(aq)
Pb2+(aq) + 2Cl–(aq)
Fe2+(aq) + 2OH–(aq)
H+(aq) + OH–(aq)
(1)
(1)
(1)
(1)
(1)
(1)
(1)
(1)
CaCO3(s)
(1)
(1)
PbCl2(s)
(1)
(1)
Fe(OH)2(s)
H2O(l)
(1)
(1)
(1)
158a) pH of solution = –log10[H+] = 3.40
i.e.log10[H+] = –3.40
[H+] = 10–3.40
= 3.98 x 10–4 mol dm–3
(1)
(1)
115
116
b) One mole of ZnCl2 contains 2 moles of Cl– ions.
i.e.concentration of Cl– ions = 2 x concentration of ZnCl2 solution
Concentration of Cl– ions in solution =
\ concentration of ZnCl2 solution =
c) One mole of K2CO3 contains 2 moles of K+ ions.
i.e.number of moles of K+ ions from K2CO3(aq) = 2 x molarity of solution x volume of solution
= 2 x 2.00 mol dm–3 x
= 0.400 mol
One mole of KCl contains 1 mole of K+ ions.
i.e.number of moles of K+ ions from KCl(aq) = molarity of solution x volume of solution
= 0.500 mol dm–3 x
= 0.0750 mol
Total number of moles of K+ ions in X = (0.400 + 0.0750) mol
Concentration of K­+ ions in X =
1.80 x 10–2 mol
40.0
dm3
1 000
(
)
= 0.450 mol dm–3
(1)
0.450
mol dm–3
2
= 0.225 mol dm–3
(1)
100.0
dm3
1 000
(1)
150.0
dm3
1 000
(1)
= 0.475 mol
=
+
number of moles of K ions
volume of solution
(
0.475 mol
100.0 + 150.0
1 000
) dm
3
= 1.90 mol dm–3
d) i) (MV) before dilution = (MV) after dilution, where M = molarity, V = volume
8.00 x
\ molarity of the diluted acid is 0.128 mol dm–3.
ii) Molar mass of H2SO4 = (2 x 1.0 + 32.1 + 4 x 16.0) g mol–1
= 98.1 g mol–1
400.0
= M x 25.0
1 000
M = 0.128
Concentration of diluted acid= 0.128 mol dm–3 x 98.1 g mol–1
= 12.6 g dm–3
(1)
(1)
(1)
(1)
Structured questions
159a) When solid citric acid dissolves in water, the molecules dissociate to give hydrogen ions.
(1)
Hydrogencarbonate ions react with hydrogen ions to give carbon dioxide gas. Effervescence occurs. (1)
b) Concentrated sulphuric acid is hygroscopic.
(1)
c) Concentrated sulphuric acid reacts with ammonia gas.
(1)
d) Sulphuric acid is a strong acid. It almost completely dissociates in water to give hydrogen ions.
(1)
Carbonic acid is a weak acid. It only partially dissociates in water, forming very few hydrogen ions. (1)
Hence 0.1 mol dm–3 sulphuric acid has a higher concentration of mobile ions than 0.1 mol dm–3 carbonic
acid.
(1)
Thus the electrical conductivity of 0.1 mol dm­–3 sulphuric acid is higher than that of 0.1 mol dm–3 carbonic
acid.
e) The conical flask should not be washed with the solution it is going to contain because the additional
amount of solute remaining in the flask will affect the titration results.
(1)
160a) Add dilute hydrochloric acid to both solids.
(1)
Effervescence occurs for zinc carbonate. A gas that turns limewater milky is given off.
(1)
The zinc oxide just dissolves in the acid.
(1)
b) Add dilute sodium hydroxide solution / dilute aqueous ammonia to the solutions.
(1)
Magnesium sulphate solution gives a white precipitate
(1)
while there is no observable change for sodium sulphate solution.
(1)
c) Add dilute sulphuric acid to the solutions.
(1)
Barium nitrate solution gives a white precipitate
(1)
while there is no observable change for potassium nitrate solution.
(1)
d) Add dilute hydrochloric acid to the solutions.
(1)
Lead(II) ethanoate solution gives a white precipitate
(1)
while there is no observable change for zinc sulphate solution.
(1)
117
161a) A weak acid is an acid that only partially dissociates in water.
118
(1)
b) pH = –log10[H+] = 1.41
c) H2SO4(aq)
2H+(aq) + SO42–(aq)
0.100 mol dm–3 ? mol dm–3
According to the equation, 1 mole of H2SO4 dissociates to give 2 moles of hydrogen ions.
i.e.concentration of hydrogen ions = 2 x 0.100 mol dm–3 = 0.200 mol dm–3
(1)
pH of acid = –log10(0.200) = 0.700
(1)
d) Any one of the following:
• Measure the pH using universal indicator / pH meter
(1)
The pH of the sulphuric acid is lower than that of the sulphurous acid.
(1)
This shows that the sulphuric acid has a higher concentration of hydrogen ions than the sulphurous
acid does.
(1)
• Test with magnesium / zinc / sodium carbonate
(1)
The sulphuric acid reacts more quickly than the sulphurous acid.
(1)
This shows that the sulphuric acid has a higher concentration of hydrogen ions than the sulphurous
acid does.
(1)
• Measure the electrical conductivity
(1)
The electrical conductivity of the sulphuric acid is higher than that of the sulphurous acid.
(1)
This shows that the sulphuric acid has a higher concentration of mobile ions than the sulphurous acid
does.
(1)
i.e.log10[H+] = –1.41
[H+] = 10–1.41
= 3.89 x 10–2 mol dm–3
(1)
(1)
162a) i) Sodium hydroxide is corrosive.
(1)
(1)
b) i) There is no reaction between copper and dilute hydrochloric acid.
c) i) A lot of heat is released when vinegar reacts with potassium hydroxide solution. This will cause skin
burn.
(1)
d) i) Concentrated sulphuric acid will react with ammonia.
(1)
(1)
ii) Use weak bases such as magnesium hydroxide and aluminium hydroxide.
(1)
ii) Use the reaction between copper(II) oxide / copper(II) hydroxide / copper(II) carbonate and dilute
hydrochloric acid
(1)
ii) Wash the hand with plenty of water.
ii) Use calcium oxide.
(1)
163a) i) HCl(aq)
H+(aq) + Cl–(aq)
ii) CH3COOH(aq)
(1)
H+(aq) + CH3COO–(aq)
(1)
b) Mg(s) + 2HCl(aq)
Number of moles of Mg present=
Number of moles of HCl used= molarity of solution x volume of solution
= 1.0 mol dm
= 0.10 mol
According to the equation, 1 mole of Mg reacts with 2 moles of HCl. During Reaction 1, 0.10 mole of
HCl reacted with 0.050 mole of magnesium. Therefore magnesium was in excess.
(1)
c) i) No more gas bubbles were given off.
(1)
ii) The time required for the completion of Reaction 2 would be longer.
(1)
During the reaction between magnesium and the acids, magnesium would react with hydrogen ions
in the acids.
Hydrochloric acid is a strong acid that completely dissociates in water.
(1)
On the other hand, ethanoic acid is a weak acid that only partially dissociates in water.
(1)
Therefore hydrochloric acid has a higher concentration of hydrogen ions than ethanoic acid does.(1)
The reaction rate between magnesium and ethanoic acid is thus lower and the reaction takes a longer
time to complete.
d) Any one of the following:
• Measuring their pH
• Measuring their electrical conductivity
e) Sulphuric acid is a dibasic acid while hydrochloric acid is a monobasic acid.
Therefore 1.0 mol dm–3 sulphuric acid has a higher concentration of hydrogen ions.
The initial rate of Reaction 3 is thus higher than that of Reaction 1.
MgCl2(aq) + H2(g)
mass of Mg
molar mass of Mg
1.5 g
=
24.3 g mol–1
= 0.062 mol
–3
x
(1)
100
3
dm
1 000
The pH of 1.0 mol dm–3 HCl(aq) is lower than that of 1.0 mol dm–3 CH3COOH(aq).
(1)
(1)
(1)
(1)
The electrical conductivity of 1.0 mol dm –3 HCl(aq) is higher than that of 1.0 mol dm –3
CH3COOH(aq).
(1)
(1)
(1)
119
164a) Pb2+(aq) + SO42–(aq)
PbSO4(s)
(1)
b)
HMBTTSPE
NJYUVSF
GJMUFSQBQFS
QSFDJQJUBUF"
GJMUFSGVOOFM
TPMVUJPO#
120
(1 mark for correct set-up; 1 mark for labelling filter funnel and filter paper; 1 mark for labelling precipitate
A and solution B; 0 mark if the set-up is not workable)
(3)
c) Cu2+(aq) + 2OH–(aq)
d) The presence of NH4+(aq) ions can be shown by warming solution D. An alkaline gas will evolve.
e) Blue
Cu(OH)2(s)
(1)
(1)
The presence of K+(aq) ions cannot be shown. As in flame test, the lilac flame of potassium will be masked
by the golden yellow colour of sodium.
(1)
(1)
Cu2+(aq) ions are blue in colour while the other ions in solution X are colourless.
(1)
165a) A: lead (0.5)
B: lead(II) oxide (0.5)
C: lead(II) nitrate solution (0.5)
D: lead(II) carbonate (0.5)
E: lead(II) hydroxide (0.5)
F: plumbate ion (0.5)
b) i) 2Pb(s) + O2(g)
ii) PbO(s) + 2HNO3(aq)
Pb(NO3)2(aq) + H2O(l)
(1)
iii) Pb2+(aq) + CO32–(aq)
PbCO3(aq)
(1)
iv) Pb (aq) + 2OH (aq)
Pb(OH)2(s)
(1)
v) Pb(OH)2(s) + 2OH–(aq)
c) Carbon reduction
2+
PbO(s)
–
(1)
[Pb(OH)4]2–(aq)
(1)
(1)
166A was Al2(SO4)3(aq) as it reacted with C and E to give a white precipitate (Al(OH)3):
(1)
Al3+(aq) + 3OH–(aq)
C and E are alkalis because Al2(SO4)3(aq) reacted with the alkalis to give a white precipitate. As the precipitate
dissolved when E was in excess, E was NaOH(aq). The white precipitate dissolved due to the formation of
complex ions:
(1)
Al(OH)3(aq) + OH­–(aq)
Thus C was NH3(aq).
(1)
B was HCl(aq) as it underwent neutralization with E and heat was liberated.
(1)
HCl(aq) + NaOH(aq)
(1)
D was AgNO3(aq) as it reacted with Cl–(aq) ions (in B) to give a white precipitate (AgCl):
(1)
Ag+(aq) + Cl–(aq)
(1)
Al(OH)3(s)
(1)
[Al(OH)4]–(aq)
(1)
NaCl(aq) + H2O(l)
AgCl(s)
167a) Wash a 25.0 cm3 pipette first with distilled water and then with the sulphuric acid.
(1)
Using a pipette filler, suck up the acid until the meniscus is 2 – 3 cm above the graduation mark.
(1)
Use the forefinger to control the flow. Release the solution until the meniscus reaches the graduation
mark.
(1)
Transfer the solution into a conical flask. Allow the tip of the pipette to touch the side of the conical
flask.
(1)
b) To obtain consistent results. / To obtain three titres within 0.1 cm3.
c) Average volume of sodium hydroxide solution required for neutralization
d) 2NaOH(aq) + H2SO4(aq) 0.180 mol dm–3
? mol dm–3
3
22.2 cm 25.0 cm3
Number of moles of NaOH in 22.2 cm3 solution = molarity of solution x volume of solution
= 0.180 mol dm–3 x
= 0.00400 mol
According to the equation, 1 mole of H2SO4 requires 2 moles of NaOH for complete neutralization.
i.e.number of moles of H2SO4 in 25.0 cm3 solution=
22.2 + 22.3 + 22.2
3
= 22.2 cm3
=
(1)
cm3
(1)
Na2SO4(aq) + 2H2O(l)
0.00400
mol
2
= 0.00200 mol
22.2
dm3
1 000
(1)
(1)
121
122
number of moles of H2SO4
volume of solution
0.00200 mol
25.0
dm3
1 000
Molarity of sulphuric acid=
=
= 0.0800 mol dm \ the molarity of the sulphuric acid is 0.0800 mol dm .
e) The next titre would be lower.
(1)
(1)
(
)
–3
(1)
–3
Some alkali remained in the conical flask.
168a) Use a burette to contain the hydrochloric acid.
(1)
Wash the burette with distilled water and then with the hydrochloric acid.
(1)
Add the indicator to the conical flask, and then the acid from the burette until the indicator changes
from yellow to red.
(1)
b) i) Average volume of hydrochloric acid used=
ii) NH3(aq) + HCl(aq) –3
9.97 g 0.150 mol dm
3
3
250.0 cm 18.8 cm
3
(used) 25.0 cm
Number of moles of HCl in 18.8 cm solution = molarity of solution x volume of solution
= 0.150 mol dm
= 0.00282 mol
According to the equation, 1 mole of NH3 requires 1 mole of HCl for complete neutralization.
i.e.number of moles of NH3 in 25.0 cm3 of the diluted solution = 0.00282 mol
Number of moles of NH3 in 250.0 cm3 of the diluted solution
= number of moles of NH3 in the original household ammonia solution
= 0.00282 mol x
= 0.0282 mol
Mass of NH3 in the original household ammonia solution
= number of moles of NH3 x molar mass of NH3
= 0.0282 mol x 17.0 g mol–1
= 0.479 g
18.8 + 18.8 + 18.7
3
cm
3
3
= 18.8 cm (1)
NH4Cl(aq)
3
–3
x
18.8
3
dm
1 000
(1)
250.0 cm3
25.0 cm3
(1)
Percentage by mass of NH3 in the original household ammonia solution
=
0.479 g
9.97 g
= 4.80%
\ the percentage by mass of ammonia in the houseold ammonia solution is 4.80%.
c)
x 100%
(1)
Original
Diluted household
Standard
household
ammonia solution hydrochloric acid
ammonia solution
Distilled water
3
i) 250.0 cm volumetric flask
(0.5)
4
3
ii) 25.0 cm pipette for
delivering the diluted
household ammonia solution
(0.5)
4
iii) Conical flask
(0.5)
4
iv) Burette
(0.5)
4
169a) Molar mass of KOH= (39.1 + 16.0 + 1.0) g mol–1
= 56.1 g mol–1
mass
molar mass
7.00 g
=
56.1 g mol–1
= 0.125 mol
3
Number of moles of KOH in 100.0 cm solution =
Concentration of potassium hydroxide solution =
=
= 1.25 mol dm–3
b) i) 25.0 cm of 6.00 mol dm
Wash a 25.0 cm3 pipette first with distilled water and then with the acid.
Deliver exactly 25.0 cm3 of the original acid into a 250.0 cm3 volumetric flask using the pipette and
pipette filler. (1)
Add distilled water to the flask until the meniscus reaches the graduation mark.
Stopper the flask. Turn it upside down several times to mix the solution well.
ii) 2KOH(aq) + H2SO4(aq) 1.00 mol dm–3 0.600 mol dm–3
25.0 cm3
? cm3
3
–3
(1)
number of moles of KOH
volume of solution
0.125 mol
100.0
dm3
1 000
(
)
(1)
sulphuric acid are needed for the dilution.
(1)
(1)
K2SO4(aq) + 2H2O(l)
123
Number of moles of KOH in 25.0 cm3 solution= molarity of solution x volume of solution
= 1.00 mol dm–3 x
= 0.0250 mol
According to the equation, 2 moles of KOH require 1 mole of H2SO4 for complete neutralization.
i.e. number of moles of H2SO4 =
Volume of sulphuric acid required for neutralization=
number of moles of H2SO4
molarity of solution
0.0125 mol
=
0.600 mol dm–3
= 0.0208 dm3
= 20.8 cm3
\ the lowest acceptable titre is 20.8 cm3.
c) To check that the peaches are free from potassium hydroxide
(1)
(1)
25.0
dm3
1 000
0.0250
mol
2
= 0.0125 mol
(1)
because potassium hydroxide is corrosive.
170a) Sodium hydroxide solutions absorb moisture from air. So their masses change.
124
(1)
(1)
(1)
b) Any two of the following:
• It is obtainable in a very pure form.
(1)
• It has a known chemical formula.
(1)
• It dissolves in water completely at room temperature.
(1)
• It is stable and does not absorb moisture from the air.
(1)
• It has a high molar mass.
(1)
c) Wash the burette first with distilled water and then with the sodium hydroxide solution.
Clamp the burette vertically in a stand. Close the stopcock. Fill the burette with dilute sodium hydroxide
solution.
(1)
Open the stopcock for a few seconds so as to fill the tip of the burette with solution.
d) Phenolphthalein
(1)
(1)
e) (COOH)2(aq) + 2NaOH(aq) 4.95 g
? mol dm–3
3
250.0 cm 29.2 cm3
From colourless to red / pink
(used) 25.0 cm3
(COONa)2(aq) + 2H2O(l)
(1)
(1)
Molar mass of (COOH)2•2H2O = [2 x (12.0 + 2 x 16.0 + 1.0) + 2 x (2 x 1.0 + 16.0)] g mol–1
= 126.0 g mol–1
Number of moles of (COOH)2•2H2O in 250.0 cm3 solution=
Number of moles of (COOH)2•2H2O in 25.0 cm3 solution= 0.0393 mol x
According to the equation, 1 mole of (COOH)2 requires 2 moles of NaOH for complete neutralization.
i.e.number of moles of NaOH = 2 x 0.00393 mol = 0.00786 mol
Molarity of NaOH solution=
\ the molarity of the sodium hydroxide solution is 0.269 mol dm–3.
mass
molar mass
4.95 g
=
126.0 g mol–1
= 0.0393 mol
(1)
3
25.0 cm
3
250.0 cm
= 0.00393 mol
(1)
number of moles of NaOH
volume of solution
0.00786 mol
=
29.2
3
dm
1 000
= 0.269 mol dm–3
(
)
(1)
171a) Transfer the solution into a 250.0 cm3 volumetric flask.
(1)
Wash the beaker, the glass rod and the filter funnel with a little distilled water several times. Pour all the
washings into the flask.
(1)
Add distilled water to the flask until the meniscus reaches the graduation mark.
Stopper the flask. Turn it upside down several times to mix the solution well.
b) Phenolpthalein
(1)
(1)
c) i) Number of moles of NaOH in 26.4 cm3 solution = molarity of solution x volume of solution
= 0.500 mol dm–3 x
= 0.0132 mol
ii) Let n be the basicity of tartaric acid, so we can represent the acid by HnX.
HnX(s) + nNaOH(aq)
Number of moles of HnX =
From colourless to red / pink
(1)
26.4
dm3
1 000
(1)
NanX(aq) + nH2O(l)
mass
molar mass
9.90 g
=
150.0 g mol–1
= 0.0660 mol
(1)
125
25.0 cm3
250.0 cm3
Number of moles of HnX in 25.0 cm3 solution = 0.0660 mol x
Number of moles of acid
1
=
Number of moles of NaOH
n
0.00660 mol
=
0.0132 mol
n = 2
\ the basicity of tartaric acid is 2.
d) i) A solution of accurately known concentration.
(1)
(1)
e) Use a pH meter / pH sensor.
= 0.00660 mol
(1)
ii) Not appropriate as sodium hydroxide absorbs moisture in air readily.
(1)
3
172a) The final volume of the solution would be more than 250.0 cm as the solute has volume. / The exact
volume of the solution would not be known.
(1)
b)
EJMVUFIZESPDIMPSJDBDJE
CVSFUUF
DPOJDBMGMBTL
TPMVUJPOPGNFUBMIZESPYJEF
NFUIZMPSBOHF
126
(1 mark for correct set-up; 0.5 mark for each correct label; 0 mark if the set-up is not workable)
c) From yellow to red
d) i) Average volume of hydrochloric acid used=
ii) MOH(aq) + HCl(aq) 1.26 g
250.0 cm3
(used) 25.0 cm3
Let m g mol–1 be the molar mass of MOH.
(3)
(1)
20.2 + 20.3 + 20.2
3
cm
3
3
= 20.2 cm MCl(aq) + H2O(l)
(1)
mass
molar mass
1.26 g
=
m g mol–1
Number of moles of MOH in 250.0 cm3 solution=
Number of moles of HCl in 20.2 cm3 solution = molarity of solution x volume of solution
= 0.155 mol dm–3 x
= 0.00313 mol
According to the equation, 1 mole of MOH requires 1 mole of HCl for complete neutralization.
i.e.number of moles of MOH in 25.0 cm3 solution = 0.00313 mol
(1)
3
Number of moles of MOH in 25.0 cm solution = 1.26 x 1 mol = 0.00313 mol
10
m
m = 40.3
(1)
\ the molar mass of the metal hydroxide is 40.3 g mol–1.
3
3
Number of moles of MOH in 25.0 cm solution = 1.26 mol x 25.0 cm 3
m
250.0 cm
1.26
1
=
x
mol
10
m
20.2
dm3
1 000
(1)
173a) Pipette
(1)
b) Wash the pipette with distilled water and then with sodium hydroxide solution.
(1)
c) Wash the burette first with distilled water and then with the sulphuric acid.
(1)
Clamp the burette vertically in a stand. Close the stopcock. Fill the burette with the acid through a filter
funnel.
(1)
Open the stopcock for a few seconds so as to fill the tip of the burette with acid.
d) Any one of the following:
• Methyl orange
(1)
(1)
• Phenolphthalein
(1)
(1)
e) Extra water in the conical flask will not change the number of moles of solute it holds.
f) H2SO4(aq) + 2NaOH(aq) ? mol dm–3 0.600 mol dm–3
15.0 cm3
25.0 cm3
from yellow to red
from red / pink to colourless
(1)
(1)
Na2SO4(aq) + 2H2O(l)
127
128
Number of moles of NaOH in 25.0 cm3 solution = molarity of solution x volume of solution
= 0.600 mol dm–3 x
= 0.0150 mol
According to the equation, 2 moles of NaOH require 1 mole of H2SO4 for complete neutralization. 1 mole
of Na2SO4 is produced.
i.e.number of moles of H2SO4 = 0.0150 mol
2
= 0.00750 mol
Molarity of sulphuric acid=
\ the concentration of the sulphuric acid is 0.500 mol dm–3.
g) Number of moles of Na2SO4 produced = 0.00750 mol
Concentration of sodium sulphate in the resulting solution
=
=
0.00750 mol
25.0 + 15.0
dm3
1 000
= 0.188 mol dm–3
\ the concentration of sodium sulphate in the resulting solution is 0.188 mol dm–3.
h) Mix 25.0 cm3 of dilute sodium hydroxide solution and 15.0 cm3 of dilute sulphuric acid.
Heat the sodium sulphate solution gently to obtain a concentrated solution.
Set the concentrated solution aside to cool and crystallize.
(1)
Filter the crystals from the remaining solution.
(1)
Wash the crystals with a little cold distilled water. Dry the crystals using filter paper.
(1)
25.0 dm3
1 000
(1)
(1)
number of moles of H2SO4
volume of solution
0.00750 mol
=
15.0
dm3
1 000
(
)
= 0.500 mol dm–3
(1)
number of moles of Na2SO4
volume of solution
(
)
(1)
(1)
174a)
TPEJVNIZESPYJEF
TPMVUJPO
FMFDUSPEFPGQ)
NFUFS
Q)NFUFS
YYY
DNPGBRVFPVTTPMVUJPO
PGNPOPCBTJDBDJE)"
NBHOFUJDTUJSSJOHCBS
NBHOFUJDTUJSSFS
(1 mark for showing the burette and beaker; 1 mark for showing the pH meter and stirrer; 2 marks for
6 correct labels)
(4)
b) An acid that can produce one hydrogen ion per molecule.
(1)
c) i) 3.2
(1)
ii) pH of HA = –log10[H+] = 3.2
d) HA(aq) + NaOH(aq) ? mol dm–3 0.100 mol dm–3
25.0 cm3
28.0 cm3
Number of moles of NaOH in 28.0 cm3 solution= molarity of solution x volume of solution
= 0.100 mol dm–3 x
= 0.00280 mol
According to the equation, 1 mole of HA requires 1 mole of NaOH for complete neutralization.
i.e.number of moles of HA in 25.0 cm3 solution = 0.00280 mol
Molarity of acid HA=
\ the concentration of the acid HA is 0.112 mol dm–3.
i.e. log10[H+] = –3.2
[H+] = 10–3.2
= 6.3 x 10–4 mol dm–3
(1)
(1)
NaA(aq) + H2O(l)
28.0
dm3
1 000
(1)
(1)
number of moles of HA
volume of solution
0.00280 mol
=
25.0
dm3
1 000
(
)
–3
= 0.112 mol dm (1)
129
e) Thymol blue is a suitable indicator
(1)
because the indicator changes colour within the pH range of the vertical part of the titration curve. (1)
175a) Add the sodium hydroxide solution quickly to 27 cm3. Swirl the conical flask when adding the alkali.(1)
Adjust the stopcock of the burette to add one drop of alkali at a time. Swirl the flask after each
addition.
(1)
Continue adding the alkali until the colour of the indicator just changes from colourless to red / pink. (1)
As the titration approaches its end point, wash down any solution sticking to the inside of the conical
flask with small amount of distilled water.
(1)
b) (CH2COOH)2(aq) + 2NaOH(aq) ? mol dm–3
0.0980 mol dm–3
25.0 cm3
27.6 cm3
Number of moles of NaOH in 27.6 cm3 solution = molarity of solution x volume of solution
= 0.0980 mol dm–3 x
= 0.00270 mol
According to the equation, 1 mole of (CH 2COOH) 2 requires 2 moles of NaOH for complete
neutralization.
i.e.number of moles of (CH2COOH)2 =
Concentration of butanedioic acid=
=
= 0.0540 mol dm–3
\ the concentration of the butanedioic acid is 0.0540 mol dm .
c) Error is less with a large titre.
(1)
d) i) A solution of accurately known concentration.
(1)
(1)
(CH2COONa)2(aq) + 2H2O(l)
0.00270
mol
2
= 0.00135 mol
(1)
(1)
number of moles of (CH2COOH)2
volume of solution
0.00135 mol
25.0
dm3
1 000
(
)
(1)
–3
ii) Not appropriate as sodium hydroxide absorbs moisture in air readily.
176a) A strong acid is an acid that almost completely dissociates in water.
130
27.6
dm3
1 000
(1)
Concentration refers to the number of moles (or amount) of acid in a unit volume of solution / 1 dm3
of solution.
(1)
Concentrated and dilute refer to the relative values of mol dm–3.
(1)
b) i) BaCO3(s) + 2HCl(aq)
BaCl2(aq) + CO2(g) + H2O(l)
ii) To make sure that all the hydrochloric acid has been reacted.
iii)
(1)
(1)
HMBTTSPE
NJYUVSFPGCBSJVN
DBSCPOBUFBOETPMVUJPO
GJMUFSQBQFS
CBSJVNDBSCPOBUF
GJMUFSGVOOFM
TPMVUJPO
(1 mark for correct set-up; 1 mark for labelling filter funnel and filter paper; 1 mark for labelling
barium carbonate and solution; 0 mark if the set-up is not workable)
(3)
iv) The solubility of barium chloride decreases when the temperature of the solution drops.
v) (1)Anhydrous barium chloride will form. / The water of crystallization will be removed.
(1)
(1)
vi) BaCO3(s) + 2HCl(aq) 1.00 mol dm–3
50.0 cm3
Number of moles of HCl in 50.0 cm3 solution= molarity of solution x volume of solution
= 1.00 mol dm–3 x
= 0.0500 mol
According to the equation, 2 moles of HCl react with 1 mole of BaCO3 to give 1 mole of BaCl2.
i.e.number of moles of BaCl2 =
Mass of BaCl2•2H2O= number of moles x molar mass
= 0.0250 mol x 244.3 g mol–1
= 6.11 g
\ 6.11 g of barium chloride crystals can be obtained.
(1)
As the saturated solution cools, the solvent cannot hold all the solutes. The extra solutes separate out
as crystals.
(1)
(2)Absorb the water by filter paper. / Place the crystals in a desiccator.
BaCl2(aq) + CO2(g) + H2O(l)
0.0500
mol
2
= 0.0250 mol
50.0 dm3
1 000
(1)
(1)
(1)
131
vii)Any one of the following:
• Some crystals may be lost when the crystals are washed with distilled water.
(1)
• Little splashes may cause some of the solution and hence the crystals to lose.
(1)
• Some BaCl2 may remain in the solution and do not crystallize out.
(1)
177a)
PS
132
(1)
b) Mass of HNO3 in 80 m3 acid= 86 400 000 g x 70.0%
= 60 480 000 g
Number of moles of HNO3 =
Molarity of nitric acid =
c) The 10.0 cm3 pipette should be used to deliver the concentrated nitric acid.
Suppose V cm of 0.240 mol dm
(MV) before dilution = (MV) after dilution, where M = molarity, V = volume
12.0 x
Volume of dilute nitric acid obtained = 500.0 cm
Hence use the 500.0 cm3 volumetric flask for the dilution.
(1)
Wash the 10.0 cm3 pipette first with distilled water and then with the original acid.
(1)
Deliver exactly 10.0 cm3 of the original acid into the 500.0 cm3 volumetric flask using the pipette and
pipette filler.
(1)
Add distilled water to the flask until the meniscus reaches the graduation mark.
Stopper the flask. Turn it upside down several times to mix the solution well.
d) i) pH of acid= –log10 (0.100)
(1)
(1)
mass
molar mass
= 60 480 000–1g
63.0 g mol
= 960 000 mol
(1)
number of moles of HNO3
volume of solution
= 960 000 mol
80 000 dm3
= 12.0 mol dm–3
3
–3
(1)
nitric acid can be obtained.
V
10.0 = 0.240 x
1 000
1 000
V = 500.0
3
= 1
(1)
ii)
Q)
7PMVNFNPMENm
BRVFPVTBNNPOJBBEEFEDN
(0.5 mark for curve starting from pH 1; 0.5 mark for vertical part of curve between pH 2 – 6; 0.5
3
mark for showing equivalence point at 25 cm ; 0.5 mark for curve finishing between pH 9 – 10,
3
extending to 40 cm )
(2)
iii) Chloro-phenol red
(1)
because the indicator changes colour within the pH range of the vertical part of the titration curve.
(1)
178a) Consider 1 000 cm3 (i.e. 1.00 dm3) of the sample.
Mass of 1 000 cm3 of the sample= 1.27 g cm–3 x 1 000 cm3
= 1 270 g
Mass of H3PO4 in 1 000 cm3 of sample
= mass of 1 000 cm3 of sample x percentage by mass of H3PO4 in sample
= 1 270 x 85.0%
= 1 080 g
Molar mass of H3PO4 = (3 x 1.0 + 31.0 + 4 x 16.0) g mol–1
= 98.0 g mol–1
Number of moles of H3PO4 in 1.00 dm3 of sample =
Molarity of phosphoric acid=
b) Wash the hand with plenty of water.
mass
molar mass
1 080 g
=
98.0 g mol–1
= 11.0 mol
number of moles of H3PO4
volume of solution
11.0 mol
=
1.00 dm3
= 11.0 mol dm–3
(1)
(1)
(1)
(1)
133
c) Use a burette to contain the sodium hydroxide solution.
(1)
Wash the burette with distilled water and then with the sodium hydroxide solution.
(1)
Add the indicator to the flask, and then the alkali from the burette until the indicator changes from
colourless to red / pink.
(1)
d) i) Average volume of sodium hydroxide solution used=
ii) Number of moles of NaOH required= molarity of solution x volume of solution
20.0
= 1.10 mol dm–3 x
dm3
1 000
= 0.0220 mol
iii) Number of moles of NaOH required = 0.0220 mol x
e) Number of moles of H3PO4 in 10.0 cm3 solution= molarity of solution x volume of solution
10.0
= 11.0 mol dm–3 x
dm3
1 000
= 0.110 mol
Number of moles of NaOH
Number of moles of H3PO4
\ equation 2 best describes the reaction occurring in the titration.
20.0 + 20.1 + 20.0
cm3
3
= 20.0 cm3
(1)
(1)
250.0 cm3
25.0 cm3
= 0.220 mol
0.220 mol
0.110 mol
= 2
(1)
(1)
=
(1)
(1)
179a) Step 1 —Should not prepare the standard sodium hydroxide solution using the method described. (1)
134
Step 2 —Should not use a measuring cylinder to transfer the lemon juice.
Step 3 —Should not rinse the burette with distilled water only.
Step 5 —Should not perform the calculation using only one titration result.
(1)
(1)
b) Step 1 —Standardize the sodium hydroxide solution before use.
(1)
Step 2 —Use a pipette to transfer the lemon juice.
(1)
Step 3 —Rinse the burette with distilled water and sodium hydroxide solution before use.
(1)
Step 5 —Repeat the titration at least 3 times to obtain consistent results. Use the mean titre for the
calculation.
(1)
This is because sodium hydroxide is deliquescent and would absorb moisture from the air. (1)
(1)
This is because a measuring cylinder cannot give accurate measurements of liquid volumes.(1)
(1)
This is because water that remains in the burette would dilute the sodium hydroxide
solution.
(1)
This is because errors may occur in the titration.
180a) As sulphuric acid was added, it removed both the barium ions (by precipitation) and hydroxide ions (by
neutralization).
(1)
Ba(OH)2(aq) + H2SO4(aq)
BaSO4(s) + 2H2O(l)
(1)
At the equivalence point, all the barium ions and hydroxide ions had been removed. Hence the electrical
conductivity of the reaction mixture fell to almost zero.
(1)
b) 15.0 cm3
c) Ba(OH)2(aq) + H2SO4(aq) ? mol dm–3 1.00 mol dm–3
100.0 cm3
15.0 cm3
Number of moles of H2SO4 in 15.0 cm3 solution= molarity of solution x volume of solution
15.0
= 1.00 mol dm–3 x
dm3
1 000
= 0.0150 mol
According to the equation, 1 mole of Ba(OH)2 requires 1 mole of H2SO4 for complete neutralization.
i.e.number of moles of Ba(OH)2 in 100.0 cm3 solution = 0.0150 mol
Molarity of barium hydroxide solution=
\ the concentration of the barium hydroxide solution is 0.150 mol dm–3.
d) Only barium sulphate and water are produced in the reaction. Barium sulphate can be obtained by
filtration.
(1)
(1)
BaSO4(aq) + 2H2O(l)
(1)
(1)
number of moles of Ba(OH)2
volume of solution
0.0150 mol
=
100.0
dm3
1 000
(
)
= 0.150 mol dm–3
(1)
181a) Aluminium hydroxide in the tablet neutralizes the excess hydrochloric acid in the stomach.
(1)
(1)
b) Chewing breaks down the tablets into smaller pieces. This increases the surface area of the tablets and
thus increases the reaction rate (brings faster relief of pain).
(1)
c) i) Methyl orange: from red to yellow / Phenolphthalein: from colourless to red or pink
ii) NaOH(aq) + HCl(aq) 0.190 mol dm–3
18.1 cm3
Number of moles of NaOH in 18.1 cm3 solution= molarity of solution x volume of solution
= 0.190 mol dm–3 x 18.1 dm3
1 000
= 0.00344 mol
According to the equation, 1 mole of HCl requires 1 mole of NaOH for complete neutralization.
i.e. number of moles of excess HCl in 25.0 cm3 diluted solution = 0.00344 mol
Al(OH)3(s) + 3HCl(aq)
AlCl3(aq) + 3H2O(l)
(2)
NaCl(aq) + H2O(l)
(1)
(1)
135
iii) Number of moles of HCl added in Step 1= molarity of solution x volume of solution
50.0
–3
3
= 1.00 mol dm x
dm
1 000
= 0.0500 mol
iv) Number of moles of HCl left over after reaction with drug tablet in Step 1
3
250.0 cm
= 0.00344 mol x
3
25.0 cm
= 0.0344 mol
Number of moles of HCl reacted with Al(OH)3 in drug tablet= (0.0500 – 0.0344) mol
= 0.0156 mol
Al(OH)3(s) + 3HCl(aq)
According to the equation, 1 mole of Al(OH)3 requires 3 moles of HCl for complete neutralization.
i.e. number of moles of Al(OH)3 in drug tablet=
Molar mass of Al(OH)3= (27.0 + 3 x 1.0 + 3 x 16.0) g mol–1
= 78.0 g mol–1
Mass of Al(OH)3 in drug tablet = number of moles x molar mass
= 0.00520 mol x 78.0 g mol–1
= 0.406 g
\ the drug tablet contains 0.406 g of aluminium hydroxide.
v)
(1)
(1)
AlCl3(aq) + 3H2O(l)
0.0156
mol
3
= 0.00520 mol
(1)
Liquid used for washing immediately before use
3
distilled water
(0.5)
(2) 25.0 cm pipette used for delivering the
diluted solution in Step 3
diluted solution
(0.5)
(3) Conical flask for containing the diluted
solution in Step 3
distilled water
(0.5)
sodium hydroxide solution
(0.5)
(1) 250.0 cm volumetric flask used in Step 2
3
(4) Burette for containing the sodium hydroxide
solution
182a) Transfer the resulting solution into a 250.0 cm3 volumetric flask.
136
(1)
Wash the beaker, the glass rod and the filter funnel with a little distilled water several times. Pour all the
washings into the flask.
(1)
Add distilled water to the flask until the meniscus reaches the graduation mark.
Stopper the flask. Turn it upside down several times to mix the solution well.
b) i) NaOH(aq) + HCl(aq) 0.125 mol dm–3
20.8 cm3
NaCl(aq) + H2O(l)
(1)
Number of moles of NaOH in 20.8 cm3 solution= molarity of solution x volume of solution
= 0.125 mol dm–3 x
= 0.00260 mol
According to the equation, 1 mole of HCl requires 1 mole of NaOH for complete neutralization.
i.e. number of moles of excess HCl in 25.0 cm3 diluted solution = 0.00260 mol
ii) Number of moles of HCl added in Step 1= molarity of solution x volume of solution
= 1.00 mol dm–3 x 100.0 dm3
1 000
= 0.100 mol
iii) Number of moles of HCl left over after reaction with magnesium in Step 1
= 0.00260 mol x
= 0.0260 mol
Number of moles of HCl reacted with magnesium in Step 1
= (0.100 – 0.0260) mol
= 0.074 mol
Mg(aq) + 2HCl(aq)
According to the equation, 1 mole of Mg requires 2 moles of HCl for complete reaction.
i.e. number of moles of Mg in sample=
\ molar mass of Mg=
=
= 24 g mol–1
\ the relative atomic mass of Mg is 24.
c) Neutralization is a quick process.
d) Oxide layer would cover the surface of magnesium.
(1)
The oxide would react with hydrochloric acid, just as magnesium did.
(1)
The oxide layer would not cause any difference to the value of the titre because magnesium oxide would
react with the same volume of acid as magnesium.
(1)
20.8
dm3
1 000
(1)
(1)
(1)
3
250.0 cm
3
25.0 cm
(1)
MgCl2(aq) + H2(g)
=
mass of Mg
0.074
mol
2
0.074
mol
2
mass of Mg
molar mass of Mg
0.900 g
0.074
mol
2
(1)
(1)
As titration proceeds, the concentration of the acid decreases and hence the rate of reaction between
the acid and sodium hydroxide decreases.
(1)
137
183a) Nickel / Ni
(1)
b) Ni2+(aq) + CO32–(aq)
c) i) Number of moles of K2CO3 in 20.0 cm3 solution= molarity of solution x volume of solution
20.0
= 0.250 mol dm–3 x
dm3
1 000
= 0.00500 mol
According to the equation, 1 mole of K2CO3 requires 1 mole of NiSO4 for complete reaction.
i.e.number of moles of NiSO4 = 0.00500 mol
Volume of NiSO4 solution=
number of moles of NiSO4
molarity of solution
0.00500 mol
=
0.100 mol dm–3
= 0.0500 dm3
= 50.0 cm3
\ 50.0 cm3 of NiSO4 solution would react exactly with 20.0 cm3 of 0.250 mol dm–3 K2CO3 solution.
ii) Remove the nickel carbonate by filtration.
Heat the potassium sulphate solution to evaporate about half of the water. Set the concentrated
solution aside to cool and crystallize.
(1)
Filter the crystals from the remaining solution.
(1)
Wash the crystals with a little cold distilled water. Dry the crystals using filter paper.
(1)
NiCO3(s)
(1)
(1)
(1)
(1)
184a) Volume of acid
(1)
Concentration of acid
(1)
Temperature
(1)
b) The hydrochloric acid was consumed as the reaction proceeded.
c) i) D (0.5)
(0.5)
d)
(1)
Hence the concentration of the acid in the reaction mixture dropped during the course of the reaction.
(1)
ii) A Rate of reaction
Acid used
lower
same
3
–3
100 cm of 2 mol dm HCl(aq)
3
–3
100 cm of 0.5 mol dm HCl(aq)
3
–3
200 cm of 0.5 mol dm
HCl(aq)
higher
4
4
3
–3
200 cm of 1 mol dm HCl(aq)
138
(1)
4
4
185a)
.BTTPGCFBLFSBOEDPOUFOUTH
m
H
m
NJO
5JNFNJO
(1 mark for correctly drawn curve of best fit; 2 marks for correctly plotted points)
b)
Loss in mass of contents of beaker for the time period 2.0 min to 6.0 min
= (193.9 – 187.3) g
= 6.60 g
= mass of NO(g) formed
Average rate of formation of NO(g)=
c) Mass of NO(g) formed in the first 3 minutes= (200.0 – 191.9) g
= 8.10 g
Number of moles of NO(g) formed=
According to the equation, 3 moles of Cu reacts with HNO3(aq) to give 2 moles of NO(g).
mass of NO(g) formed
time
6.60 g
=
(6.0 – 2.0) min
= 1.65 g min–1
mass of NO
molar mass of NO
8.10 g
=
30.0 g mol–1
= 0.270 mol
(3)
(1)
(1)
(1)
139
\ number of moles of Cu consumed =
Mass of Cu consumed= number of moles of Cu x molar mass of Cu
= 0.405 mol x 63.5 g mol–1
= 25.7 g
d) Instantaneous rate of formation of NO(g)= –
(1)
(1)
e) i) Increase
(1)
ii) Decrease
(1)
iii) Increase
(1)
(1)
(185.0 – 191.6) g
(7.5 – 3.0) min
= 1.47 g min–1
186a) 2H2O2(aq)
140
3
x 0.270 mol
2
= 0.405 mol
2H2O(l) + O2(g)
(1)
(1)
b) The rates of decomposition of hydrogen peroxide solution was A > B > C.
At A, the concentration of hydrogen peroxide solution was high. So, the rate of decomposition was
high.
(1)
At B, the concentration of hydrogen peroxide solution was lower than that at A. So, the rate of
decomposition was lower.
(1)
At C, the rate of decomposition was zero as all the hydrogen peroxide was used up.
(1)
c)
(1)
(168.235 – 168.143) g 4.0 min
= 0.023 g min–1
d) i) (168.235 – 168.123) g = 0.112 g
ii) Number of moles of O2 given off=
According to the equation, 2 moles of H2O2 decompose to give 1 mole of O2.
i.e.number of moles of H2O2 = 2 x 0.00350 mol
= 0.00700 mol
Original molarity of hydrogen peroxide solution =
(1)
(1)
mass of O2
molar mass of O2
0.112 g
=
32.0 g mol–1
= 0.00350 mol
(1)
(1)
number of moles of H2O2
volume of solution
0.00700 mol
=
50.0
dm3
1 000
(
)
= 0.140 mol dm–3
(1)
e) i) As a catalyst. / To increase the rate of decomposition.
(1)
ii) The shape of the curve would be the same.
(1)
(1)
f) Any one of the following:
As the change in the mass is very small in this experiment, the use of a data-logger can give more
accurate results.
(1)
The experimental results in the form of graph could be obtained immediately.
g) Any one of the following:
Measure the volume of gas given off.
(1)
Measure the pressure of the mixture in a closed reaction vessel.
(1)
The rate of decomposition is independent of the amount of catalyst added.
(1)
(1)
b) The intensity of the purple colour of permanganate ions decreases as the reaction proceeds.
(1)
The colorimeter measures the amount of light absorbed by the reaction mixture when a beam of light
passes through the reaction mixture, i.e. the absorbance.
(1)
The absorbance is directly proportional to the colour intensity of the reaction mixture and the concentration
of permanganate ions in the reaction mixture.
(1)
c)
d) Draw a tangent to the curve at time = 0.
"CTPSCBODF
187a) Colorimeter
5JNF
(1)
Determine the slope of the tangent.
–
(1)
(1)
2–
4
d[C2O (aq)]
1
1 d[MnO4 (aq)]
= –
dt
5
2
dt
\ instantaneous rate of consumption of C2O42–(aq) ions
e) i) Rate = –
5
(instantaneous rate of consumption of MnO4–(aq) ions)
2
5
ii) Instantaneous rate of consumption of C2O42–(aq) ions =
(3.50 x 10–3 mol dm–3 s–1)
2
= 8.75 x 10–3 mol dm–3 s–1
=
(1)
(1)
141
188a) When sodium thiosulphate solution reacts with dilute sulphuric acid, a yellow precipitate of sulphur forms.
This changes the light transmittance of the reaction mixture.
(1)
To follow the progress of the reaction, mark a cross on a piece of paper.
Put the conical flask containing some sodium thiosulphate solution on top of the paper. Add dilute
sulphuric acid to the flask.
(1)
Record the time t for the solution mixture to become opaque, i.e. when the cross can no longer be seen
from above.
(1)
b) The average rate of reaction from the start to the opaque stage
c) i)
∝ 1
time to reach the opaque stage (t)
(1)
(1)
5JNFUT
5FNQFSBUVSFž$
ii) 64 – 68 s
d) Change the concentration of the reactants.
189a)
(1 mark for correctly drawn curve of best fit; 2 marks for correctly plotted points)
(3)
(1)
(1)
DPUUPOXPPM
EJMVUF
IZESPDIMPSJD
BDJE
DBMDJVNDBSCPOBUF
XXX.XX
FMFDUSPOJD
CBMBODF
142
(1 mark for correct set-up; 2 marks for 4 correct labels; 0 mark if the set-up is not workable)
(3)
b) The reaction rate was higher at X than at Y. This is because the reaction gets slower as it proceeds. (1)
c) • The loss in mass of the contents of the reaction flask for sample A is greater than that for sample B,
i.e. sample A produces more carbon dioxide gas. Therefore sample A has a higher purity of calcium
carbonate than sample B.
(1)
• The initial rate for sample B is faster than that for sample A. It is because sample B has a smaller
particle size.
(1)
d) Mass of CO2 given off = 3.70 g
CaCO3(s) + 2HCl(aq)
? g
Number of moles of CO2=
According to the equation, 1 mole of CaCO3 reacts with 2 moles of HCl to give 1 mole of CO2.
i.e.number of moles of CaCO3 in sample A = 0.0841 mol
Mass of CaCO3 in sample A = number of moles of CaCO3 x molar mass of CaCO3
= 0.0841 mol x 100.1 g mol–1
= 8.42 g
\ percentage by mass of CaCO3 in sample A =
CaCl2(aq) + CO2(g) + H2O(l)
3.70 g
mass of CO2
molar mass of CO2
3.70 g
=
44.0 g mol–1
= 0.0841 mol
(1)
8.42 g
10.0 g
= 84.2%
(1)
x 100%
(1)
190a) A gas was evolved in the reaction.
(1)
b)
TVDUJPOGMBTL
EBUBMPHHFS
JOUFSGBDF
QSFTTVSF
TFOTPS
NBHOFTJVN
SJCCPO
DPNQVUFS
EJMVUF
TVMQIVSJDBDJE
(1 mark for correct set-up; 3 marks for 6 correct labels; 0 mark if the set-up is not workable)
c) Number of moles of Mg present=
Number of moles of H2SO4 present= molarity of solution x volume of solution
= 1.00 mol dm
= 0.100 mol
Mg(s) + H2SO4(aq)
According to the equation, 1 mole of Mg reacts with 1 mole of H2SO4. During the reaction, 0.0823 mole
of Mg reacted with 0.0823 mol of H2SO4. Therefore H2SO4 was in excess.
(1)
mass of Mg
molar mass of Mg
2.00 g
=
24.3 g mol–1
= 0.0823 mol
–3
x
(4)
(1)
100.0
3
dm
1 000
(1)
MgSO4(aq) + H2(g)
143
d) As the reaction proceeded, sulphuric acid was consumed gradually.
At A, the concentration of sulphuric acid was high. So, the rate of reaction was high.
At B, the concentration of sulphuric acid was lower than that at A. So, the rate of reaction was lower.
(1)
At C, the rate of reaction was zero as all the magnesium was used up.
e) i) Sulphuric acid is a dibasic acid while hydrochloric acid is a monobasic acid. Therefore 1 mol dm–3
sulphuric acid has a higher concentration of hydrogen ions than 1 mol dm–3 hydrochloric acid. (1)
ii)
(1)
(1)
Hence the initial rate of the reaction between magnesium and hydrochloric acid is lower than that
between magnesium and sulphuric acid.
(1)
1SFTTVSF
DVSWF9
DVSWF:
5JNF
(1 mark for showing a lower rate of reaction, i.e. tangent to curve less steep; 1 mark for showing a
5
smaller final pressure, around
of the original value)
(2)
8
191a) 2H2O2(aq)
2H2O(l) + O2(g)
(1)
b) As a catalyst. / Increase the rate of decomposition.
c)
(1)
7PMVNFPGPYZHFODN
5JNFT
144
(1 mark for correctly drawn curve of best fit; 2 marks for correctly plotted points)
(3)
d) 60 – 62 cm3
(1)
e) All hydrogen peroxide has decomposed.
(1)
f)
7PMVNFPGPYZHFODN
5JNFT
(1 mark for showing the slower production of oxygen, i.e. tangent to curve less steep; 1 mark for showing
a smaller final volume, about half that of the original curve)
(2)
g) Add 1 g of manganese(IV) oxide to 50 cm3 of hydrogen peroxide solution. Wait until the reaction
stops.
(1)
Filter to remove manganese(IV) oxide from the solution.
Add ‘fresh’ manganese(IV) oxide to the filtrate. No oxygen is given off. This indicates that the oxygen
does not come from manganese(IV) oxide.
(1)
Add ‘fresh’ hydrogen peroxide solution to the residue (manganese(IV) oxide). Oxygen is given off. This
indicates that the oxygen comes from hydrogen peroxide.
(1)
h) A gas is given off in the decomposition.
(1)
(1)
192a) i) By the reaction between sulphuric acid and aqueous ammonia.
(1)
ii) H2SO4(aq) + 2NH3(aq)
(1)
iii) So the fertilizer can be taken in by the roots of plants.
b) Used in car batteries (1) / manufacture detergents (1) / manufacture paints (1)
c) i) A process in which heat is released to the surroundings.
(1)
(1)
d) A catalyst is a substance which alters the rate of a reaction
(1)
(1)
(NH4)2SO4(aq)
ii) Diluting concentrated sulphuric acid / any specific neutralization reaction
without itself undergoing any permanent chemical changes.
(1)
145
e) i)
ii) Any two of the following:
Wear safety glasses / protective gloves.
(1)
Never add water to the acid.
(1)
In case any acid is split on the skin or clothes, wash the affected area with plenty of water. (1)
iii) Any one of the following:
• Keep container tightly closed.
(1)
• Store in a dry area.
(1)
• Keep away from heat sources / combustible materials / reducing agents / bases / metals / organic
materials / oxidizing agents.
(1)
f) i) CaO(s) + SO2(g)
g) i) Attack the respiratory system / lungs.
(1)
(1)
(1)
CaSO3(s)
(1)
ii) Neutralization reaction
(1)
ii) Move the student to a well ventilated area / open space.
193a) i) [Al(OH)4]–
146
ii) 2Al(s) + 2OH–(aq) + 6H2O(l)
(1)
2[Al(OH)4]–(aq) + 3H2(g)
b) Diluting concentrated sulphuric acid / any specific neutralization reaction
(1)
c) i) The cleanser is corrosive.
(1)
ii) An acidic substance would react with the sodium hydroxide and aluminium in the cleanser, making
the cleanser ineffective.
(1)
A great amount of heat and gas would be generated.
This may cause a violent eruption from the drain. The mixing may cause severe injury to the user as
well as damage to porcelain tubs and toilets.
iii)
d) A strong alkali is an alkali that almost completely dissociates to give hydroxide ions (OH–(aq) in water. (1)
e) i) By electrolysis of concentrated sodium chloride solution
(1)
(1)
ii) Manufacture of bleach / manufacture of soaps and detergents
(1)
(1)
(1)
194a) SiO2(s) + 6HF(aq)
H2SiF6(aq) + 2H2O(l)
(1)
b) Coat the glass with layers of beeswax. Then trace the patterns with a metal needle.
(1)
(1)
c) i) An acid that can produce one hydrogen ion per molecule.
ii) 0.1 mol dm–3 sulphuric acid < 0.1 mol dm–3 hydrochloric acid < 0.1 mol dm–3 hydrofluoric acid
Both sulphuric acid and hydrochloric acid are strong acids. They dissociate completely in water. (1)
Sulphuric acid is dibasic while hydrochloric acid is monobasic.
Hence 0.1 mol dm–3 sulphuric acid contains a higher concentration of hydrogen ions than 0.1 mol
dm–3 hydrochloric acid.
Hydrofluoric acid is a weak acid. It only partially dissociates in water.
Hence 0.1 mol dm–3 hydrofluoric acid has a lower concentration of hydrogen ions than 0.1 mol dm–3
hydrochloric acid.
d) i) Calcium carbonate
(1)
(1)
e) i) Hydrated iron(III) oxide
Dip the glass into hydrofluoric acid to attack the unprotected glass surface.
ii) The materials for making glass are readily available / abundant in the Earth’s crust.
ii) Fe2O3(s) + 6HF(aq)
(1)
(1)
(1)
(1)
(1)
2FeF3(aq) + 3H2O(l)
(1)
195Any three of the following:
• Add barium chloride solution / barium nitrate solution to the acids.
• Add dilute nitric acid followed by silver nitrate solution to the acids.
• Add calcium carbonate to excess acids.
Calcium carbonate disappears after reaction when added to 1 mol dm–3 HCl(aq).
Calcium carbonate does not react completely with 1 mol dm–3 H2SO4(aq) due to the formation of insoluble
calcium sulphate on its surface.
(1)
• Titrate equal volumes of the acids with sodium hydroxide solution.
1 mol dm–3 H2SO4(aq) requires more sodium hydroxide solution to reach the end point than 1.0 mol dm–3
HCl(aq) as H2SO4(aq) is dibasic while HCl(aq) is monobasic.
(1)
1 mol dm–3 H2SO4(aq) gives a white precipitate while 1 mol dm–3 HCl(aq) gives no precipitate.
1 mol dm–3 HCl(aq) gives a white precipitate while 1 mol dm–3 H2SO4(aq) gives no precipitate.
(1)
(1)
(1)
(1)
(1)
(1)
(3 marks for organization and presentation)
147
196Titrate dilute sodium hydroxide solution with dilute hydrochloric acid until the end point is reached. 148
(1)
Mix the same volumes of alkali and acid as in the previous experiment. Do not use any indicator this
time. (1)
Heat the resulting solution to obtain a concentrated solution.
(1)
Set the concentrated solution aside to cool and crystallize.
(1)
Filter the crystals from the remaining solution.
(1)
Wash the crystals with a little cold distilled water. Dry the crystals using filter paper.
(1)
(3 marks for organization and presentation)
197Carry out flame test on all the solids.
(1)
Only potassium sulphate gives a lilac flame.
(1)
Dissolve the remaining three solids separately in water. Then add dilute sodium hydroxide solution until
excess. (1)
Aluminium sulphate solution gives a white precipitate which is soluble in excess alkali.
(1)
Magnesium sulphate solution gives a white precipitate which does not dissolve in excess alkali.
(1)
Ammonium sulphate solution gives no observable change.
(1)
OR
Dissolve some of each solid in water separately. Then add dilute sodium hydroxide solution until excess. (1)
Aluminium sulphate solution gives a white precipitate which is soluble in excess alkali.
(1)
Magnesium sulphate solution gives a white precipitate which does not dissolve in excess alkali.
(1)
Ammonium sulphate solution and potassium sulphate solution give no precipitate.
(1)
Warm each of the two solutions.
(1)
The solution containing ammonium sulphate gives a gas that turns moist red litmus paper blue (ammonia).
There is no observable change for the solution containing potassium sulphate.
(3 marks for organization and presentation)
(1)
(3 marks for organization and presentation)
198Add sodium chloride solution to all the four solutions.
The solution that gives a white precipitate is silver nitrate solution.
(1)
Add silver nitrate solution to the remaining three solutions.
(1)
Both calcium chloride solution and potassium chloride solution give a white precipitate,
(1)
while there is no observable change with potassium sulphate solution.
(1)
Add potassium sulphate solution to calcium chloride solution and potassium chloride solution.
(1)
The solution that gives a white precipitate is calcium chloride solution.
(1)
(3 marks for organization and presentation)
199During the hydrolysis, each molecule of ethyl ethanoate hydrolyzed produces one molecules of ethanoic
acid. As hydrochloric acid is a catalyst in the reaction, its amount remains the same. Hence the increase in
total amount of acid in the reaction mixture is a direct measure of the amount of ethyl ethanoate that has
undergone hydrolysis.
(1)
Mix 5.00 cm3 of ethyl ethanoate and 100 cm3 of dilute hydrochloric acid in a reaction flask. Start the stop
watch at the same time.
(1)
Immediately withdraw 5.00 cm3 of the reaction mixture. Run the sample into 50 cm3 of water which has
been chilled in an ice bath. Titrate with sodium hydroxide solution. Let the volume of sodium hydroxide
solution consumed be V0.
(1)
Withdraw 5.00 cm3 samples at regular time intervals and titrate with the sodium hydroxide solution. Let the
volume of sodium hydroxide solution consumed at time t be Vt.
(1)
Stopper the reaction flask. Leave for two days to allow the hydrolysis to complete.
Withdraw a 5.00 cm3 sample and titrate with the sodium hydroxide solution. Let the volume of sodium
hydroxide solution consumed be V∞.
(1)
V∞ is the volume of alkali required to neutralize the hydrochloric acid and the ethanoic acid formed upon
complete hydrolysis. Hence (V∞ – V0) is proportional to the concentration of ethyl ethanoate at the beginning
of the reaction and (V∞ – Vt) is proportional to the concentration of unreacted ethyl ethanoate at time t.
(1)
(3 marks for organization and presentation)
149