topic acids and bases - stmgssintranet.stmgss.edu.hk/~ted/ccy/answers/cs_suppex... · topic 4 acids...

1

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

2

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. It can absorb water vapour from the air.

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.

33 T

34 F Phenolphthalein is colourless in acidic solutions.

35 F Many metal hydroxides are INSOLUBLE in water.

36 T

3

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.

4

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. Ethanoic acid is an electrolyte.

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 gas.

Option C — Copper(II) oxide and dilute nitric acid undergo neutralization when mixed.

Option D — Copper(II) hydroxide and vinegar undergo neutralization when mixed.

56 A

57 D

58 C Solid citric acid does NOT react with magnesium because it does not contain hydrogen ions.


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.

5

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.

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 Cu(OH)2

C lead(II) nitrate white precipitate Pb(OH)2

D zinc sulphate white precipitate Zn(OH)2

6

71 C Option Hydroxide Solubility in excess 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 excess NaOH(aq) and NH3(aq)

D Zn(OH)2 soluble in both excess NaOH(aq) and NH3(aq)

72 C 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) AgCl(s)

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

7

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.


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 used 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.



8


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.


(3) Sulphurous acid is formed when sulphur dioxide dissolves in water.

SO2(g) + H2O(l) H2SO3(aq) sulphurous acid

92 D (1) Magnesium reacts with dilute sulphuric acid to give hydrogen gas.

(2) and (3) Sodium carbonate and potassium hydrogencarbonate react with dilute sulphuric 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) NH3(g) + H2O(l)

(2) Iron(II) sulphate solution gives a green precipitate with dilute sodium hydroxide solution.

Fe2+(aq) + 2OH–(aq) Fe(OH)2(s)

(3) Citric acid would undergo neutralization reaction with dilute sodium hydroxide solution.

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.


Hence aqueous ammonia contains both ammonia molecules and ammonium ions.

9

97 B (1) Cu(OH)2 dissolves in excess dilute aqueous ammonia to give a deep blue solution.

Cu(OH)2(s) + 4NH3(aq) [Cu(NH3)4]2+(aq) + 2OH–(aq)

(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) [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)

100 A (2) Heating solutions of ammonium chloride and sodium hydroxide liberates ammonia.

NH4Cl(aq) + NaOH(aq) NH3(g) + NaCl(aq) + H2O(l)

101 A (1) Magnesium chloride solution gives a white precipitate with silver nitrate solution.


(2) Magnesium chloride solution gives a white precipitate with dilute aqueous ammonia.

Mg2+(aq) + 2OH–(aq) Mg(OH)2(s)

(3) Copper is less reactive than magnesium. It CANNOT displace magnesium from magnesium chloride solution.

102 B (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.


103 A (1) NH3(aq) and NiSO4(aq) would form a green precipitate when mixed.

Ni2+(aq) + 2OH–(aq) Ni(OH)2(s)

(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.

10

104 D (1) Ammonia gas would react with dilute sulphuric acid.

2NH3(g) + H2SO4(aq) (NH4)2SO4(aq)

(2) Carbon dioxide gas would give a white precipitate with calcium hydroxide solution.

CO2(g) + Ca(OH)2(aq) CaCO3(s) + H2O(l)

(3) Sulphur dioxide gas would react with sodium hydroxide solution.

SO2(g) + NaOH(aq) NaHSO3(aq)

105 B

106 C 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.

107 D 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.

108 B 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.




109 D

110 D

111 C Dilute nitric acid is an oxidizing agent. It reacts with copper to give nitrogen monoxide, NOT hydrogen.

112 A Concentrated nitric acid decomposes according to the following chemical equation:

4HNO3(aq) 4NO2(g) + O2(g) + 2H2O(l)

113 D Many metal hydroxides are insoluble in water.

114 B When ammonia gas dissolves in water, it reacts with water to give ammonium ions and hydroxide ions.


Dilute aqueous ammonia conducts electricity due to the presence of mobile ions.

115 C Anhydrous calcium chloride is NOT used to dry ammonia gas because calcium chloride would react with ammonia gas.

CaCl2(s) + 4NH3(g) CaCl2•4NH3(s)

11

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 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).

12

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.


21 B Number of moles of KNO3 = molarity of KNO3 solution x volume of solution

= 0.25 mol dm–3 x 40

1 000 dm3

= 1.0 x 10–2 mol

22 D Molar mass of Na2CO3 = (2 x 23.0 + 12.0 + 3 x 16.0) g mol–1

= 106.0 g mol–1

Number of moles of Na2CO3 = mass

molar mass

= 13.8 g

106.0 g mol–1

= 0.130 mol

Volume of solution = 500

1 000 dm3

Molarity of Na2CO3 solution = number of moles of Na2CO3

volume of solution

= 0.130 mol

( 5001 000 ) dm3

= 0.260 mol dm–3

∴ the molarity of the sodium carbonate solution is 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

Number of moles of (COOH)2•2H2O present = mass

molar mass

= 3.15 g

126.0 g mol–1

= 0.0250 mol

Molarity of ethanedioic acid solution = number of moles of (COOH)2•2H2O

volume of solution

= 0.0250 mol

( 250.01 000 ) dm3

= 0.100 mol dm–3

∴ the molarity of the ethanedioic acid solution is 0.100 mol dm–3.

13

24 B Molar mass of Ca(OH)2 = [40.1 + 2 x (16.0 + 1.0)] g mol–1

= 74.1 g mol–1

Molarity of calcium hydroxide solution = number of moles of Ca(OH)2

volume of solution

0.150 mol dm–3 = number of moles of Ca(OH)2

( 250.01 000 ) dm3

Number of moles of Ca(OH)2 = 0.150 mol dm–3 x 250.0

1 000 dm3

= 0.0375 mol

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.

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

Molarity of ammonium sulphate solution = number of moles of (NH4)2SO4

volume of solution

1.02 mol dm–3 = number of moles of (NH4)2SO4

1.50 dm3

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.

26 C Molar mass of KCl = (39.0 + 35.5) g mol–1

= 74.5 g mol–1

Number of moles of KCl present = mass

molar mass

= 44.7 g

74.5 g mol–1

= 0.600 mol

Molarity of potassium chloride solution = number of moles of KCl

volume of solution

2.40 mol dm–3 = 0.600 mol

volume of solution

Volume of solution = 0.600 mol

2.40 mol dm–3

= 0.250 dm3

= 250 cm3

∴ the volume of the solution is 250 cm3.

14

27 B Option Number of moles of NaOH = molarity of solution x volume of solution

A 3.0 mol dm–3 x 100

1 000 dm3 = 0.30 mol

B 2.5 mol dm–3 x 200

1 000 dm3 = 0.50 mol

C 1.5 mol dm–3 x 300

1 000 dm3 = 0.45 mol

D 1.0 mol dm–3 x 400

1 000 dm3 = 0.40 mol

∴ 200 cm3 of 2.5 mol dm–3 NaOH(aq) contain the greatest number of moles of solute.

28 D 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–3 x 1.5 dm3 = 0.45 mol

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

∴ 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

= molarity of solution x volume of solution

= 0.300 mol dm–3 x 250.0

1 000 dm3

= 0.0750 mol

Number of moles of CaCl2 in 150.0 cm3 of 0.180 mol dm–3 solution


= 0.180 mol dm–3 x 150.0

1 000 dm3

= 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 = 0.102 mol

( 400.01 000 ) dm3

= 0.255 mol dm–3

∴ the concentration of the resulting solution is 2.55 x 10–1 mol dm–3.

15

30 B Number of moles of NaOH in 100.0 cm3 of 3.00 mol dm–3 solution


= 3.00 mol dm–3 x 100.0

1 000 dm3

= 0.300 mol

Number of moles of NaOH in 50.0 cm3 of 1.20 mol dm–3 solution


= 1.20 mol dm–3 x 50.0

1 000 dm3

= 0.0600 mol

Total number of moles of NaOH in the resulting solution = (0.300 + 0.0600) mol = 0.360 mol


= 150.0 cm3


( 150.01 000 ) dm3

= 2.40 mol dm–3

∴ the concentration of the resulting solution is 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.

16

33 B Sulphuric acid dissociates completely according to the following equation:

H2SO4(aq) 2H+(aq) + SO42–(aq)

0.0200 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 0.0200 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 = 10–6.75

= 1.78 x 10–7 mol dm–3

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.

36 B Limewater is saturated calcium hydroxide solution.

37 D Option A — Baking soda solution is hydrogencarbonate solution. Its pH is about 9.

Option 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)).

NH3(g) + water NH3(aq)

NH3(aq) + H2O(l) NH4+(aq) + OH–(aq)

Thus aqueous ammonia contains ammonia molecules, water molecules and a few ammonium ions and hydroxide ions.

17

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.

CaO(s) + H2O(l) Ca(OH)2(s)

Option C — Potassium oxide is strongly basic. It dissolves in water to form potassium hydroxide solution.

K2O(s) + H2O(l) 2KOH(aq)

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.

42 D Concentration of hydrogen ions in solution X = 10–1 mol dm–3

= 0.1 mol dm–3

Concentration of hydrogen ions in solution Y = 10–2 mol dm–3

= 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 about 9.

(2) The pH of lemon juice is about 2.

(3) The pH of milk is about 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 B, 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 — concentration of hydrogen ions in wine

concentration of hydrogen ions in drain cleanser =

10–4 mol dm–3

10–12 mol dm–3

∴ the concentration of hydrogen ions is 108 times greater in wine than in drain cleanser.

Option D — concentration of hydrogen ions in distilled water

concentration of hydrogen ions in baking soda solution =

10–7 mol dm–3

10–9 mol dm–3

∴ 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).

18

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 Remark Concentration of ions

A25 cm3 of 0.5 mol dm–3 H2SO4(aq)

H2SO4(aq) is a strong acid


1.5 mol dm–3

B25 cm3 of 0.5 mol dm–3 HCl(aq)

HCl(aq) is a strong acid

HCl(aq) H+(aq) + Cl–(aq)1.0 mol dm–3

C25 cm3 of 0.5 mol dm–3 CH3COOH(aq)

CH3COOH(aq) is a weak acid

CH3COOH(aq) CH3COO–(aq) + H+(aq)< 1.0 mol dm–3

D25 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 cm3 of 0.5 mol dm–3 H2SO4(aq) have the highest concentration of mobile ions and thus will make the bulb the brightest.

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) CaCl2(aq) + H2O(l) + CO2(g)

CaCO3(s) + 2CH3COOH(aq) (CH3COO)2Ca(aq) + H2O(l) + CO2(g)

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.

51 A Carbonic acid is a weak acid that only partially dissociates in water.

H2CO3(aq) 2H+(aq) + CO32–(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.

HNO3(aq) H+(aq) + NO3–(aq)

HCl(aq) H+(aq) + Cl–(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.

19

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 dissociation in water. When we talk about concentration, we are referring to the amount of an acid in a unit volume of solution.


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 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 will increase the pH of the ethanoic acid.

20

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 1 mol dm–3 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) CH3COONa(aq) + H2O(l)

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

According to the equations, 1 mole of CH3COOH / 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.

(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. This 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).

63 C (1) NaOH(aq) is a strong alkali while NH3(aq) is a weak alkali.

Hence 1 mol dm–3 NaOH(aq) is more alkaline than 1 mol dm–3 NH3(aq).

The pH of 1 mol dm–3 NaOH(aq) is higher than that of 1 mol dm–3 NH3(aq).

(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).

21




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) 2H+(aq) + CO32–(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.

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.


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.

22

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.


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.


Hence an aqueous solution of citric acid can conduct electricity.

73 C The electrical conductivity of a solution is proportional to the concentration of mobile ions.


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).

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.


Hence there are many mobile ammonia molecules in aqueous ammonia.

74 C Both 1 mol dm–3 NaOH(aq) and 1 mol dm–3 NH3(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 CH3COOH / 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

1 neutralization

2 exothermic

3 sodium hydrogensulphate; sodium sulphate

23

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) + NaOH(aq) NaHSO3(aq)

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 an acid salt. When both hydrogen ions are replaced, the salt formed is sodium sulphate (Na2SO4). It is a normal salt.

H+

H+ H+

H+

H+SO4

2– SO42– SO4

2–

Na+

Na+ Na+

Na+

Na+

acid(sulphuric acid)

acid salt(sodium hydrogensulphate)

normal salt(sodium sulphate)

replacing

(from sodium hydroxide solution)

replacing

(from sodium hydroxide solution)

24

13 T A molecule of carbonic acid can produce two hydrogen ions. During the reaction between carbonic acid and dilute potassium hydroxide solution, the acid salt potassium hydrogen carbonate (KHCO3) forms when one of the hydrogen ions is replaced.

H+

H+

H+

H+CO3

2– CO32–

K+

K+

acid(carbonic acid)

acid salt(potassium hydrogencarbonate)

replacing

(from potassium hydroxide solution)

14 F The pH of a salt solution may NOT be 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 Ammonium compounds are soluble in water.

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.

25

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.


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.


Option Solutions mixed Strength of acid and alkali Temperature rise

—25 cm3 of 1 mol dm–3 HCl(aq) and 25 cm3 of 1 mol dm–3 KOH(aq)

neutralization between a strong acid and a strong alkali

T

A25 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

B25 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

C25 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

D25 cm3 of 1 mol dm–3 HNO3(aq) and 25 cm3 of 1 mol dm–3 NaOH(aq)


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 that for mixing 25 cm3 of 1 mol dm–3 HCl(aq) and 25 cm3 of 1 mol dm–3 KOH(aq).

26

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

Number of moles of HCl / NaOH mixed


Number of moles of water formedHeat

released

1100 cm3 of 1 mol dm–3 HCl(aq) and 100 cm3 of 1 mol dm–3 NaOH(aq)

1 mol dm–3 x 100

1 000 dm3

= 0.1 mol

HCl(aq) + NaOH(aq) 0.1 mol 0.1 mol

NaCl(aq) + H2O(l)

0.1 mol

5.7 kJ


2 mol dm–3 x 100

1 000 dm3

= 0.2 mol


NaCl(aq) + H2O(l)

0.2 mol

11.4 kJ

The total volumes of the two mixtures are the same. Hence the temperature rise of the first mixture is

half that of the second mixture, i.e. T1 = 1

2 T2.

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.


Number of moles of HCl / NaOH mixed



released

1100 cm3 of 1 mol dm–3 HCl(aq) and 100 cm3 of 1 mol dm–3 KOH(aq)

1 mol dm–3 x 100

1 000 dm3

= 0.1 mol

HCl(aq) + KOH(aq) 0.1 mol 0.1 mol

KCl(aq) + H2O(l)

0.1 mol

5.7 kJ

250 cm3 of 1 mol dm–3 HCl(aq) and 50 cm3 of 1 mol dm–3 KOH(aq)

1 mol dm–3 x 50

1 000 dm3

= 0.05 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 2.85 kJ. Hence the two mixtures show the same temperature rise, i.e. T1 = T2.



Expt. Solutions mixed Strength of acid and alkali Temperature rise

1100 cm3 of 1 mol dm–3 HNO3(aq) and 100 cm3 of 1 mol dm–3 NaOH(aq)


T1

2100 cm3 of 1 mol dm–3 HNO3(aq) and 100 cm3 of 1 mol dm–3 NH3(aq)


T2

∴ T1 > T2

27



Mixture Strength of acid and alkali Temperature rise

100 cm3 of 2 mol dm–3 HNO3(aq) + 100 cm3 of 2 mol dm–3 NaOH(aq)


x

100 cm3 of 2 mol dm–3 HCl(aq) + 100 cm3 of 2 mol dm–3 NH3(aq)


y < x

100 cm3 of 2 mol dm–3 HCl(aq) + 100 cm3 of 2 mol dm–3 KOH(aq)


z = x

∴ 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.

Na2CO3(aq) + CuSO4(aq) CuCO3(s) + Na2SO4(aq)

36 D Option Solutions mixed

Precipitate formed?

Ionic equation for reaction involved

Abarium chloride solution and silver nitrate solution

✔ Ag+(aq) + Cl–(aq) AgCl(s)

Bbarium chloride solution and dilute sulphuric acid

✔ Ba2+(aq) + SO42–(aq) BaSO4(s)

Cbarium chloride solution and sodium sulphate solution

✔ Ba2+(aq) + SO42–(aq) BaSO4(s)

Dbarium chloride solution and sodium nitrate solution

✘ —

∴ 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? Ionic equation for reaction involved

A Ba2+(aq) and SO42–(aq) a white precipitate (BaSO4) Ba2+(aq) + SO4

2–(aq) BaSO4(s)

B Cu2+(aq) and NO3–(aq) no precipitate —

C Ni2+(aq) and CO32–(aq) a green precipitate (NiCO3) Ni2+(aq) + CO3

2–(aq) NiCO3(s)

D Na+(aq) and OH–(aq) no precipitate —

28

39 A K2CO3(aq) + Ba(NO3)2(aq) BaCO3(s) + 2KNO3(aq) reactant A white precipitate X

BaCO3(s) + 2HCl(aq) CO2(g) + BaCl2(aq) + H2O(l) white reactant B precipitate X 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) Ag2CO3(s)

AgNO3(aq) also forms a white precipitate (Ag2SO4) in concentrated solution of SO42–(aq)

ions.

Option B — Ba(NO3)2(aq) forms a white precipitate with both SO42–(aq) ions and CO3

2–(aq) ions.

Ba2+(aq) + SO42–(aq) BaSO4(s)

Ba2+(aq) + CO32–(aq) BaCO3(s)

Option C — Fe(NO3)3(aq) gives a precipitate with CO32–(aq) ions, but NOT with SO4

2–(aq) ions.

Hence Fe(NO3)3(aq) can be used to separate the anions SO42–(aq) and CO3

2–(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.


29

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.


46 C Zinc reacts with dilute H2SO4(aq) (reactant X) to give a solution containing Zn2+(aq) 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– BaSO4(s)

From Test 2, it can be deduced that solid X contains aluminium ions rather than calcium ions. A solution containing aluminium ions gives a white precipitate (aluminium 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)

Al(OH)3(s) + OH–(aq) [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.

51 D

30

As sulphuric acid was added, it removed both 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 number of ions in the mixture falls until the equivalence point.

After the equivalence point, the number of ions in the mixture rose as excess sulphuric acid was added.

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 solution formed contains mainly hydrogen ions (H+(aq)), chloride ions (Cl–(aq)) and water molecules.


(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 iron(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.

(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.


31

60 C (1) There is NO reaction between copper and dilute ethanoic acid.

61 A (3) Heat is absorbed during the evaporation of ethanol. Hence this is NOT an exothermic process.




Number of moles of acid / alkali mixed


Number of moles of water formed

Strength of acid and alkali

Heat released

Temperature rise


Number of moles of HCl / NaOH

= 1 mol dm–3 x 20

1 000 dm3

= 0.02 mol

H+(aq) + OH–(aq)

0.02 mol 0.02 mol

H2O(l)

0.02 mol


1.14 kJ This mixture (total volume 40 cm3) is heated up by 1.14 kJ.Assume that the temperature rise is T.

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 / NH3

= 1 mol dm–3 x 20

1 000 dm3

= 0.02 mol

H+(aq) + OH–(aq)

0.02 mol 0.02 mol

H2O(l)

0.02 mol


<1.14 kJ This mixture (total volume 40 cm3) is heated up by <1.14 kJ.Hence the temperature rise is <T.

2 40 cm3 of 1 mol dm–3 HCl(aq) and 40 cm3 of 1 mol dm–3 NaOH(aq)


= 1 mol dm–3 x 40

1 000 dm3

= 0.04 mol

H+(aq) + OH–(aq)

0.04 mol 0.04 mol

H2O(l)

0.04 mol


2.28 kJ This mixture (total volume 80 cm3) 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)


= 2 mol dm–3 x 10

1 000 dm3

= 0.02 mol

H+(aq) + OH–(aq)

0.02 mol 0.02 mol

H2O(l)

0.02 mol


1.14 kJ This mixture (total volume 20 cm3) is heated up by 1.14 kJ.Hence the temperature rise is 2T.

∴ the temperature rise for mixing 40 cm3 of 1 mol dm–3 HCl(aq) and 40 cm3 of 1 mol dm–3 NaOH(aq) is the same as that for mixing 20 cm3 of 1 mol dm–3 HCl(aq) and 20 cm3 of 1 mol dm–3 NaOH(aq).

63 D (1) A solution containing iron(III) ions is yellow-brown in colour.

32

64 D Reagent

The white precipitate formed with lead(II) nitrate solution


(1) Dilute sulphuric acid lead(II) sulphate Pb2+(aq) + SO42–(aq) PbSO4(s)

(2) Potassium chloride solution lead(II) chloride Pb2+(aq) + 2Cl–(aq) PbCl2(s)

(3) Sodium carbonate solution lead(II) carbonate Pb2+(aq) + CO32–(aq) PbCO3(s)

65 D Solution

Precipitate formed with silver nitrate solution


(1) KBr(aq) silver bromide Ag+(aq) + Br–(aq) AgBr(s)

(2) Na2CO3(aq) silver carbonate 2Ag+(aq) + CO32–(aq) Ag2CO3(s)

(3) HCl(aq) silver chloride Ag+(aq) + Cl–(aq) AgCl(s)

66 D Substance

White precipitate formed with dilute sulphuric acid

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.

(3) Ammonium iodide solution gives a precipitate (silver iodide) with silver nitrate solution.

Ag+(aq) + I–(aq) AgI(s)

68 D Solutions mixed Precipitate formed Ionic equation for reaction involved

(1) Pb(NO3)2(aq) and K2SO4(aq) PbSO4 Pb2+(aq) + SO42–(aq) PbSO4(s)

(2) NH3(aq) and Fe(NO3)2(aq) Fe(OH)2 Fe2+(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).

(2) Potassium iodide solution gives a yellow precipitate (silver iodide) with silver nitrate solution.

Ag+(aq) + I–(aq) AgI(s)

(3) Potassium iodide solution is colourless.

33

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) AgCI(s)

There is NO observable change when potassium chloride solution and zinc nitrate solution are mixed.

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.

73 A (3) The following equation represents the reaction between dilute aqueous ammonia and dilute hydrochloric acid:


Ammonium chloride is produced in the reaction.

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 Magnesium sulphate can be prepared by reacting dilute sulphuric acid with either magnesium, magnesium oxide or magnesium carbonate.

Mg(s) + H2SO4(aq) MgSO4(aq) + H2(g)

MgO(s) + H2SO4(aq) MgSO4(aq) + H2O(l)

MgCO3(s) + H2SO4(aq) MgSO4(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.

34

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

= 1 mol dm–3 x 10

1 000 dm3

= 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 = 0.01 mol

( 201 000 ) dm3

= 0.5 mol dm–3

∴ the concentration of Na+ ions would change.

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.

The student should wash the affected area with plenty of water.

35

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.

Washing the volumetric flask with the sulphuric acid before use would increase the number of moles of solute it holds, making the concentration of the sulphuric acid obtained higher than calculated.

9 F A volumetric flask is used to prepare a solution of accurately known volume, e.g. 250.0 cm3.

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 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.

36

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).

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).

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 at the equivalence point may NOT be 7.

For example, during the titration of a weak acid with a strong alkali, the pH of the solution mixture at the equivalence point is greater than 7.

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.

37

20 T Number of moles of acid used= molarity of solution x volume of solution

Number of moles of NaOH required for complete neutralization

Number of moles of CH3COOH

= 1 mol dm–3 x 20

1 000 dm3

= 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

= 1 mol dm–3 x 10

1 000 dm3

= 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.


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

= 126.0 g mol–1

Number of moles of (COOH)2•2H2O used = mass

molar mass

= 5.04 g

126.0 g mol–1

= 0.0400 mol

Molarity of acid solution = number of moles of (COOH)2•2H2O

volume of solution

= 0.0400 mol

( 250.01 000 ) dm3

= 0.160 mol dm–3

∴ the molarity of the acid solution is 0.160 mol dm–3.

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

38

Molarity of nitric acid = number of moles of HNO3

volume of solution

= 13.8 mol

1.00 dm3

= 13.8 mol dm–3

∴ the concentration of the nitric acid sample is 13.8 mol dm–3.

23 A Mass of HCl 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 = mass

molar mass

= 24 100 000 g

36.5 g mol–1

= 660 000 mol

Molarity of hydrochloric acid = number of moles of HCl

volume of solution

= 660 000 mol

60 000 dm3

= 11.0 mol dm–3

∴ the concentration of the hydrochloric acid is 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

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 = mass

molar mass

= 51 700 000 g

98.1 g mol–1

= 527 000 mol

Molarity of sulphuric acid = number of moles of H2SO4

volume of solution

= 527 000 mol

58 000 dm3

= 9.08 mol dm–3

∴ the molarity of the sulphuric acid is 9.08 mol dm–3.

39

25 D One mole of K3PO4 contains 3 moles of K+ ions and 1 mole of PO43– ions.

Total number of moles of ions = 4 x 0.100 mol = 0.400 mol

Total ion concentration in solution = number of moles of ions

volume of solution

= 0.400 mol

( 500.01 000 ) dm3

= 0.800 mol dm–3

∴ the total ion concentration in the solution is 0.800 mol dm–3.

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 1024.

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:


1.995 x 10–3 mol dm–3

Concentration of sulphate ions = concentration of hydrogen ions

2

= 1.995 x 10–3

2 mol dm–3

= 9.98 x 10–4 mol dm–3

∴ the concentration of sulphate ions in the acid is 9.98 x 10–4 mol dm–3.

40

28 A

Option SubstanceNumber of moles of substance

present = molarity of solution x volume of solution

Number of moles of ions in one

mole of substance

Number of molesof ions present

A NaCl0.10 mol dm–3 x

1001 000

dm3

= 0.010 mol2 2 x 0.010 mol = 0.020 mol

B Fe2(SO4)30.050 mol dm–3 x

501 000

dm3

= 0.0025 mol5 5 x 0.0025 mol = 0.013 mol

C CaCl20.080 mol dm–3 x

501 000

dm3

= 0.0040 mol3 3 x 0.0040 mol = 0.012 mol

D MgSO40.080 mol dm–3 x

1001 000

dm3

= 0.0080 mol2 2 x 0.0080 mol = 0.016 mol

∴ 100 cm3 of 0.10 mol dm –3 NaCl(aq) contain the greatest number of moles of ions, i.e. the greatest number of ions.

29 C Number of moles of ZnCl2 in ZnCl2(aq) = molarity of solution x volume of solution

= 2.00 mol dm–3 x 150.0

1 000 dm3

= 0.300 mol

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

= 1.20 mol dm–3 x 50.0

1 000 dm3

= 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 = 0.660 mol

( 200.01 000 ) dm3

= 3.30 mol dm–3

∴ the concentration of Cl–(aq) ions in solution X is 3.30 mol dm–3.

41

30 C One mole of sodium sulphate (Na2SO4) contains 2 moles of Na+ ions and 1 mole of SO42– ions.

Number of moles of Na2SO4 in solution = number of moles of Na+ ions

2

= 2.00 x 10–2

2 mol

= 1.00 x 10–2 mol

Molarity of the sodium sulphate solution = number of moles of Na2SO4

volume of solution

= 1.00 x 10–2 mol

( 20.01 000 ) dm3

= 0.500 mol dm–3

∴ the molarity of the sodium sulphate solution is 5.00 x 10–1 mol dm–3.

31 D Number of moles of Na2CO3 in 2.50 mol dm–3 solution = molarity of solution x volume of solution

= 2.50 mol dm–3 x 200.0

1 000 dm3

= 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 solution = molarity of solution x volume of solution

= 1.00 mol dm–3 x 50.0

1 000 dm3

= 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 = 1.10 mol

( 250.01 000 ) dm3

= 4.40 mol dm–3

∴ the concentration of Na+(aq) ions in the resulting solution is 4.40 mol dm–3.

42

32 C One mole of magnesium chloride (MgCl2) contains 1 mole of Mg2+ ions and 2 moles of Cl– ions.

Number of moles of MgCl2 in solution = number of moles of Cl– ions

2

= 2.0 x 10–2

2 mol

= 1.0 x 10–2 mol

Molarity of the magnesium chloride solution = number of moles of MgCl2

volume of solution

= 1.0 x 10–2 mol

( 50.01 000 ) dm3

= 0.20 mol dm–3

∴ the molarity of the solution is 2.0 x 10–1 mol dm–3.

33 C Number of moles of Fe2(SO4)3 in Fe2(SO4)3(aq) = molarity of solution x volume of solution

= 0.20 mol dm–3 x 50

1 000 dm3

= 0.010 mol

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

= 0.25 mol dm–3 x 200

1 000 dm3

= 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 = 0.080 mol

( 2501 000 ) dm3

= 0.32 mol dm–3

∴ the concentration of SO42–(aq) ions in solution X is 0.32 mol dm–3.

43

34 B (MV) before dilution = (MV) after dilution, where M = molarity, V = volume

5.0 x V

1 000 = 0.40 x

250

1 000 V = 20 cm3

∴ 20 cm3 of 5.0 mol dm–3 hydrochloric acid are required.

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 250.0

1 000 = M x 4.00

M = 0.200

∴ the molarity of the diluted solution is 0.200 mol dm–3.


10 x 120

1 000 = 2 x

V

1 000 V = 600

Volume of the final solution = 600 cm3

∴ volume of water added = (600 – 120) cm3

= 480 cm3


0.50 x 100

1 000 = 0.10 x

V

1 000 V = 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


1.0 x 40

1 000 = M x

50

1 000 M = 0.80 mol dm–3

∴ the concentration of the dilute solution of acid X is 0.80 mol dm–3.

44

39 B Volume of diluted solution of acid Y = (25 + 25) cm3

= 50 cm3


1.6 x 25

1 000 = M x

50

1 000 M = 0.80

∴ 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.

40 C Volume of solution X = (20 + 80) cm3

= 100 cm3


0.50 x 20

1 000 = M x

100

1 000 M = 0.10

∴ concentration of acid A in solution X is 0.10 mol dm–3.

41 A Volume of solution Y = (60 + 40) cm3

= 100 cm3


0.25 x 60

1 000 = M x

100

1 000 M = 0.15

∴ 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.

When both solutions of monobasic acids react with magnesium, solution X would give less hydrogen than solution Y.

45

42 A 2NH3(aq) + H2SO4(aq) (NH4)2SO4(aq) 1.60 mol dm–3 ? mol dm–3

25.0 cm3 32.0 cm3

Number of moles of NH3 in 25.0 cm3 solution = molarity of solution x volume of solution

= 1.60 mol dm–3 x 25.0

1 000 dm3

= 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 = 0.0400

2 mol

= 0.0200 mol

Concentration of sulphuric acid = number of moles of H2SO4

volume of solution

= 0.0200 mol

( 32.01 000 ) dm3

= 0.625 mol dm–3


= 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.

43 A We can represent the dibasic acid solution by H2X(aq).

H2X(aq) + 2NaOH(aq) Na2X(aq) + 2H2O(l) 0.0600 mol dm–3 0.150 mol dm–3

25.0 cm3 ? cm3

Number of moles of H2X in 25.0 cm3 solution = molarity of solution x volume of solution

= 0.0600 mol dm–3 x 25.0

1 000 dm3

= 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 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

∴ 20.0 cm3 of sodium hydroxide solution are required.

46

44 D K2O(s) + 2HCl(aq) 2KCl(aq) + H2O(l) 6.12 g 0.50 mol dm–3

? cm3

Molar mass of K2O = (2 x 39.1 + 16.0) g mol–1

= 94.2 g mol–1

Number of moles of K2O = mass

molar mass

= 6.12 g

94.2 g mol–1

= 0.0650 mol

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 = number of moles of HCl


= 0.130 mol

0.50 mol dm–3

= 0.26 dm3

= 260 cm3

∴ 260 cm3 of hydrochloric acid are required.

45 B 2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)......(1) 0.182 g

NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)......(2) 0.240 mol dm–3

? cm3

Number of moles of Na = mass

molar mass

= 0.182 g

23.0 g mol–1

= 0.00791 mol

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 = number of moles of HCl


= 0.00791 mol

0.240 mol dm–3

= 0.0330 dm3

= 33.0 cm3

∴ 33.0 cm3 of hydrochloric acid are required.

47

46 B CaCO3(s) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g) 7.50 g 2.00 mol dm–3

50.0 cm3

Molar mass of CaCO3 = (40.1 + 12.0 + 3 x 16.0) g mol–1

= 100.1 g mol–1

Number of moles of CaCO3 = mass

molar mass

= 7.50 g

100.1 g mol–1

= 0.0749 mol

Number of moles of HCl in 50.0 cm3 solution = molarity of solution x volume of solution

= 2.00 mol dm–3 x 50.0

1 000 dm3

= 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 = 0.100

2 mol

= 0.0500 mol

∴ 0.0500 mole of carbon dioxide is evolved.

47 A Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g) 1.80 g 0.250 mol dm–3

100.0 cm3

Number of moles of Zn = mass

molar mass

= 1.80 g

65.4 g mol–1

= 0.0275 mol


= 0.250 mol dm–3 x 100.0

1 000 dm3

= 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 = 0.0250

2 mol

= 0.0125 mol

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.

48

48 C CuO(s) + 2HNO3(aq) Cu(NO3)2(aq) + H2O(l) 4.00 g 1.40 mol dm–3

30.0 cm3

Molar mass of CuO = (63.5 + 16.0) g mol–1

= 79.5 g mol–1

Number of moles of CuO = mass

molar mass

= 4.00 g

79.5 g mol–1

= 0.0503 mol

Number of moles of HNO3 in 30.0 cm3 solution = molarity of solution x volume of solution

= 1.40 mol dm–3 x 30.0

1 000 dm3

= 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.

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

Number of moles of acid / alkali


Complete neutralization?

A 100 cm3 of 1 mol dm–3 H2SO4(aq) and 100 cm3 of 1 mol dm–3 NaOH(aq)

number of moles of H2SO4

= 1 mol dm–3 x 100

1 000 dm3

= 0.1 mol

number of moles of NaOH

= 1 mol dm–3 x 100

1 000 dm3

= 0.1 mol

not enough NaOH(aq) to neutralize H2SO4(aq) completely

B 100 cm3 of 1 mol dm–3 H2SO4(aq) and 100 cm3 of 2 mol dm–3 NaOH(aq)


= 1 mol dm–3 x 100

1 000 dm3

= 0.1 mol


= 2 mol dm–3 x 100

1 000 dm3

= 0.2 mol

complete neutralization occurs and a neutral solution is obtained

49

C 100 cm3 of 2 mol dm–3 H2SO4(aq) and 50 cm3 of 2 mol dm–3 NaOH(aq)


= 2 mol dm–3 x 100

1 000 dm3

= 0.2 mol


= 2 mol dm–3 x 50

1 000 dm3

= 0.1 mol


D 200 cm3 of 2 mol dm–3 H2SO4(aq) and 100 cm3 of 1 mol dm–3 NaOH(aq)


= 2 mol dm–3 x 200

1 000 dm3

= 0.4 mol


= 1 mol dm–3 x 100

1 000 dm3

= 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) NaCl(aq) + H2O(l) 0.200 mol dm–3 ? mol dm–3

27.5 cm3 25.0 cm3


= 0.200 mol dm–3 x 27.5

1 000 dm3

= 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 = number of moles of NaOH

volume of solution

= 0.00550 mol

( 25.01 000 ) dm3

= 0.220 mol dm–3

∴ molarity of the sodium hydroxide solution is 0.220 mol dm–3.

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

50

56 B H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l) 10.0 cm3 0.320 mol dm–3

28.8 cm3

250.0 cm3

(used) 25.0 cm3

Number of moles of NaOH in 28.8 cm3 solution = molarity of solution x volume of solution

= 0.320 mol dm–3 x 28.8

1 000 dm3

= 0.00920 mol


i.e. number of moles of H2SO4 in 25.0 cm3 diluted cleanser = 0.00920

2 mol

= 0.00460 mol

Number of moles of H2SO4 in 250.0 cm3 diluted cleanser = 0.00460 mol x 250.0 cm3

25.0 cm3

= 0.0460 mol

Number of moles of H2SO4 in 10.0 cm3 undiluted cleanser = 0.0460 mol

Molarity of H2SO4 in undiluted cleanser = number of moles of H2SO4

volume of solution

= 0.0460 mol

( 10.01 000 ) dm3

= 4.60 mol dm–3

∴ the concentration of sulphuric acid in the undiluted toilet cleanser is 4.60 mol dm–3.

57 A Phenolphthalein is colourless in an acid and red / pink in an alkali.

58 D H3PO4(aq) + 2NaOH(aq) Na2HPO4(aq) + 2H2O(l) 25.0 cm3 0.0200 mol dm–3

22.0 cm3

250.0 cm3

(used) 25.0 cm3


= 0.0200 mol dm–3 x 22.0

1 000 dm3

= 4.40 x 10–4 mol

51

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 = 4.40 x 10–4

2 mol

= 2.20 x 10–4 mol

Number of moles of H3PO4 in 250.0 cm3 diluted acid = 2.20 x 10–4 mol x 250.0 cm3

25.0 cm3

= 2.20 x 10–3 mol

Number of moles of H3PO4 in 25.0 cm3 original acid = 2.20 x 10–3 mol

Molarity of original phosphoric acid = number of moles of H3PO4

volume of solution

= 2.20 x 10–3 mol

( 25.01 000 ) dm3

= 0.0880 mol dm–3

∴ the concentration of the original phosphoric acid was 8.80 x 10–2 mol dm–3.

59 C NaOH(aq) + HA(aq) NaA(aq) + H2O(l) 0.100 mol dm–3 ? mol dm–3

22.5 cm3 30.0 cm3


= 0.100 mol dm–3 x 22.5

1 000 dm3

= 0.00225 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) = number of moles of HA

volume of solution

= 0.00225 mol

( 30.01 000 ) dm3

= 0.0750 mol dm–3

∴ the concentration of HA(aq) is 7.50 x 10–2 mol dm–3.

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.

52

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 required to neutralize the alkali, making the calculated concentration of the alkali too high.

64 A Na2CO3(aq) + 2HCl(aq) 2NaCl(aq) + CO2(g) + H2O(l) 3.57 g 1.20 mol dm–3

48.0 cm3

Molar mass of Na2CO3•nH2O = (2 x 23.0 + 12.0 + 3 x 16.0 + 18.0n) g mol–1

= (106.0 + 18.0n) g mol–1

Number of moles of Na2CO3•nH2O = mass

molar mass

= 3.57 g

(106.0 + 18.0n) g mol–1


= 1.20 mol dm–3 x 48.0

1 000 dm3

= 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•nH2O = 0.0576

2 mol

= 0.0288 mol

Number of moles of Na2CO3•nH2O = 3.57

(106.0 + 18.0n) mol = 0.0288 mol

n = 1

∴ the value of n is 1.

65 C Let n be the basicity of the acid.

We can represent the acid solution by HnX(aq).

HnX(aq) + nNaOH(aq) NanX(aq) + nH2O(l) 0.125 mol dm–3 0.30 mol dm–3

20.0 cm3 25.0 cm3


= 0.30 mol dm–3 x 25.0

1 000 dm3

= 0.0075 mol

53

Number of moles of HnX in 20.0 cm3 solution = molarity of solution x volume of solution

= 0.125 mol dm–3 x 20.0

1 000 dm3

= 0.0025 mol

According to the equation, 1 mole of HnX requires n moles of NaOH for complete neutralization.

Number of moles of HnX

Number of moles of NaOH =

1

n

= 0.0025 mol

0.0075 mol n = 3

∴ the basicity of the acid is 3.

66 D We can represent the dibasic acid by H2X.

H2X(aq) + 2NaOH(aq) Na2X(aq) + 2H2O(l) 2.62 g 0.220 mol dm–3

18.9 cm3

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 = mass

molar mass

= 2.62 g

m g mol –1

Number of moles of H2X in 25.0 cm3 solution = 1

10 x

2.62

m mol


= 0.220 mol dm–3 x 18.9

1 000 dm3

= 0.00416 mol


i.e. number of moles of H2X = 0.00416

2 mol

= 0.00208 mol

Number of moles of H2X = 1

10 x

2.62

m mol = 0.00208 mol

m = 126

∴ the molar mass of the acid is 126 g mol–1.

54

67 B Number of moles of OH– ions in 25.0 cm3 of 1.00 mol dm–3 NaOH solution


= 1.00 mol dm–3 x 25.0

1 000 dm3

= 0.0250 mol

68 C

69 A NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l) ? mol 0.550 mol dm–3

28.0 cm3


= 0.550 mol dm–3 x 28.0

1 000 dm3

= 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– ions 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– ions used for precipitation = (0.0250 – 0.0154) mol = 0.00960 mol

Ni2+(aq) + 2OH–(aq) Ni(OH)2(s) 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.

∴ number of moles of Ni2+ ion = 0.00960

2 mol

= 0.00480 mol

Molarity of nickel(II) sulphate solution = number of moles of Ni2+ ions

volume of solution

= 0.00480 mol

( 25.01 000 ) dm3

= 0.192 mol dm–3

∴ the molarity of the nickel(II) sulphate solution is 0.192 mol dm–3.

71 B Methyl orange is yellow in an alkaline solution and red in an acidic solution.

72 D Na2CO3(aq) + H2SO4(aq) Na2SO4(aq) + CO2(g) + H2O(l) 3.20 g 1.08 mol dm–3

(with 24.8 cm3

impurity)

55

Number of moles of H2SO4 in 24.8 cm3 solution = molarity of solution x volume of solution

= 1.08 mol dm–3 x 24.8

1 000 dm3

= 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

∴ percentage purity of Na2CO3 in the sample = 2.84 g

3.20 g x 100%

= 88.8%

∴ the percentage purity of sodium carbonate in the sample is 88.8%.

73 C Fe(s) + H2SO4(aq) FeSO4(aq) + H2(g) 1.60 mol dm–3 ? g 100 cm3

Number of moles of H2SO4 in 100 cm3 solution = molarity of solution x volume of solution

= 1.60 mol dm–3 x 100

1 000 dm3

= 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) CaCl2(aq) + 2H2O(l) 0.0200 mol dm–3 0.400 mol dm–3

? cm3 25.0 cm3


= 0.400 mol dm–3 x 25.0

1 000 dm3

= 0.0100 mol

According to the equation, 1 mole of Ca(OH)2 requires 2 moles of HCl for complete neutralization, giving 1 mole of CaCl2.

56

i.e. number of moles of Ca(OH)2 = 0.0100

2 mol

= 0.00500 mol

Volume of Ca(OH)2 solution = number of moles of Ca(OH)2


= 0.00500 mol

0.0200 mol dm–3

= 0.250 dm3

= 250 cm3

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 = number of moles of CaCl2

volume of solution

= 0.00500 mol

( 2751 000 ) dm3

= 0.0182 mol dm–3

∴ the concentration of calcium chloride in the resulting solution is 1.82 x 10–2 mol dm–3.

75 A Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g) 2.00 g 0.800 mol dm–3

200.0 cm3

Number of moles of Mg = mass

molar mass

= 2.00 g

24.3 g mol–1

= 0.0823 mol


= 0.800 mol dm–3 x 200.0

1 000 dm3

= 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 = 0.160

2 mol

= 0.0800 mol

Concentration of MgCl2 in the resulting solution = number of moles of MgCl2

volume of solution

= 0.0800 mol

( 200.01 000 ) dm3

= 0.400 mol dm–3

∴ the concentration of magnesium chloride in the resulting solution is 0.400 mol dm–3.

57

76 D Copper(II) sulphate solution and sodium carbonate solution react to give a green precipitate (copper(II) carbonate).

Cu2+(aq) + CO32–(aq) CuCO3(s)

77 C K2CO3(aq) + MgCl2(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 completely.

Option Solutions mixedVolume of K2CO3(aq) /

MgCl2(aq) reacted

A 10 cm3 of 1 mol dm–3 K2CO3(aq) + 30 cm3 of 1 mol dm–3 MgCl2(aq) 10 cm3

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 cm3 of 1 mol dm–3 K2CO3(aq) + 10 cm3 of 1 mol dm–3 MgCl2(aq) 10 cm3

The greatest volumes of K2CO3(aq) and MgCl2(aq) react in Option C. Thus the greatest amount of precipitate would be produced.

78 D

BeakerNumber of moles of

CaCO3 = mass

molar mass

Number of moles of acid= molarity of solution x

volume of solutionReaction between CaCO3 and acid

1number of moles of CaCO3

= 4 g

100.1 g mol–1

= 0.04 mol

number of moles of HCl

= 1 mol dm–3 x 100

1 000 dm3

= 0.1 mol

CaCO3(s) + 2HCl(aq) 0.04 mol 0.1 mol

CaCl2(aq) + H2O(l) + CO2(g)

2

number of moles of CH3COOH

= 1 mol dm–3 x 100

1 000 dm3

= 0.1 mol

CaCO3(s) + 2CH3COOH(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.

58

79 C

BeakerNumber of moles of Mg

= mass

molar mass


volume of solutionReaction between Mg and acid

1

number of moles of Mg

= 1.5 g

24.3 g mol–1

= 0.062 mol


= 1 mol dm–3 x 100

1 000 dm3

= 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

2


= 1 mol dm–3 x 100

1 000 dm3

= 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

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.



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.

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 HCl(aq).

59

Option D — Number of moles of acid = molarity of solution x volume of solution

Reaction between acid and NaOH(aq)


= 1 mol dm–3 x 20

1 000 dm3

= 0.02 mol

HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)0.02 mol


∴ number of moles of NaOH required = 0.02 mol


= 1 mol dm–3 x 10

1 000 dm3

= 0.01 mol

H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + H2O(l)0.01 mol


∴ number of moles of NaOH required = 2 x 0.01 mol = 0.02 mol

∴ 20 cm3 of 1 mol dm–3 HCl(aq) and 10 cm3 of 1 mol dm–3 H2SO4(aq) require the same number of moles of NaOH for complete neutralization.

81 D

82 B The following diagram shows the titration curve for the titration of 20.0 cm3 of 0.1 mol dm–3 CH3COOH(aq) with 0.1 mol dm–3 NaOH(aq).

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 than 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.

60

Option C — The following diagram shows the titration curve for the titration of 20.0 cm3 of 0.1 mol dm–3 NH3(aq) with 0.1 mol dm–3 HCl(aq).

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).

During the titration of KOH(aq) (a strong alkali) with CH3COOH(aq) (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 Y is less than 7. Thus solution Y should be an acid.

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.

61

86 D Indicator Colour in solution with a pH of 5

Methyl orange yellow

Bromothymol 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.

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 rose 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).

62

Option Solutions mixed Volumes of NaOH(aq) and H2SO4(aq) reacted

A 10.0 cm3 of NaOH(aq) + 20.0 cm3 of H2SO4(aq) 10.0 cm3 of NaOH(aq) and 5.0 cm3 of H2SO4(aq)

B 15.0 cm3 of NaOH(aq) + 15.0 cm3 of H2SO4(aq) 15.0 cm3 of NaOH(aq) and 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) and 10.0 cm3 of H2SO4(aq)

D 25.0 cm3 of NaOH(aq) + 5.0 cm3 of H2SO4(aq) 10.0 cm3 of NaOH(aq) and 5.0 cm3 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 deduced 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) PbSO4(s) + 2KNO3(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

= 1.2 mol dm–3 x 10.0

1 000 dm3

= 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 = number of moles of Pb(NO3)2

volume of solution

= 0.012 mol

( 6.01 000 ) dm3

= 2.0 mol dm–3

∴ the concentration of lead(II) nitrate solution is 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

= 0.100 mol dm–3 x 250.0

1 000 dm3

= 0.0250 mol

63

(2) and (3) One mole of Fe2(SO4)3 contains 2 moles of Fe3+ ions and 3 moles of SO42– ions.

∴ number of moles of Fe3+ ions = 2 x 0.0250 mol = 0.0500 mol

∴ number of moles of SO42– ions = 3 x 0.0250 mol

= 0.0750 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 it is 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.

96 D The following diagram shows the titration curve for the titration of 20.0 cm3 of 0.1 mol dm–3 NaOH(aq) with 0.1 mol dm–3 HCl(aq).

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.

64

97 A Beaker Number of moles of H2SO4 = molarity of solution x volume of solution

A 1 mol dm–3 x 100

1 000 dm3 = 0.1 mol

B 2 mol dm–3 x 50

1 000 dm3 = 0.1 mol

Sulphuric acid and zinc react according to the following equation:

H2SO4(aq) + Zn(s) ZnSO4(aq) + H2(g) 0.1 mol excess

According to the equation, 1 mole of H2SO4 reacts with 1 mole of Zn to produce 1 mole of H2 and 1 mole of ZnSO4. As Zn is in excess, the amount of H2SO4 limits the amounts of products formed.

i.e. number of moles of Zn used = 0.1 mol

number of moles of H2 produced = 0.1 mol

number of moles of ZnSO4 produced = 0.1 mol

(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 H2SO4 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

ReactionNumber of moles of Mg

= mass

molar mass



1

1 g24.3 g mol–1 = 0.04 mol


= 2 mol dm–3 x 50

1 000 dm3

= 0.1 mol


MgCl2(aq) + H2(g)

2


= 2 mol dm–3 x 50

1 000 dm3

= 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.

65

(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

BeakerNumber of moles of acid= molarity of solution x

volume of solution

Number of moles of alkali= molarity of solution x

volume of solution

Reaction between acid and alkali

A number of moles of HCl

= 1.2 mol dm–3 x 50

1 000 dm3

= 0.060 molnumber of moles of NaOH

= 1.0 mol dm–3 x 60

1 000 dm3

= 0.060 mol


NaCl(aq) + H2O(l)

B number of moles of CH3COOH

= 1.0 mol dm–3 x 60

1 000 dm3

= 0.060 mol

CH3COOH(aq) + NaOH(aq) 0.060 mol 0.060 mol

CH3COONa(aq) + H2O(l)

According to the equations, 1 mole of HCl / CH3COOH 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.

Hence both acids are completely neutralized.


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 greater 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 in Beaker B.

Hence the temperature rise for the reaction mixture in Beaker A is higher than that in 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 the concentrations of Na+(aq) ions in the reaction mixtures are different.

100 D The molarity of a solution is the number of moles of solute dissolved in 1 dm3 of the solution.

101 B 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.

66

102 C Solid sodium hydroxide is NOT suitable for preparing a standard solution because it absorbs moisture from the air and thus cannot be weighed accurately.

103 A Wash a burette first with distilled water and then with the acid it is going to contain.

104 D Rinsing the conical flask with the solution it is about to contain would increase the number of moles of solute it holds. This would lead to an error.

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.

105 D 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.

106 B The following diagram shows the titration curve for the titration of 20.0 cm3 of 0.1 mol dm–3 NH3(aq) with 0.1 mol dm–3 HCl(aq).

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.

107 C Number of moles of acid= molarity of solution x volume of solution

Reaction between acid and alkali


= 1 mol dm–3 x 10

1 000 dm3

= 0.01 mol

H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + H2O(l)0.01 molnumber of moles of NaOH required for neutralization= 2 x 0.01 mol= 0.02 mol


= 1 mol dm–3 x 20

1 000 dm3

= 0.02 mol

CH3COOH(aq) + NaOH(aq) CH3COONa(aq) + H2O(l)0.02 molnumber of moles of NaOH required for neutralization= 0.02 mol

∴ the acids require the same number of moles of NaOH for complete neutralization.

67

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) 2H2O(l) + O2(g)

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.

10 F The surface area of marble chips is smaller than that of powdered marble of the same mass.

The reaction 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 magnesium and 2 mol dm–3 ethanoic acid.

68

12 T

13 T

14 F The physical state of a catalyst may be the same as those of the reactants.

For example, the hydrolysis of ethyl ethanoate can be catalyzed by dilute hydrochloric acid.

15 T Manganese(IV) oxide can catalyze the decomposition of hydrogen peroxide.


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.

18 C Average reaction rate = 0.105 mol dm–3

20 s = 5.25 x 10–3 mol dm–3 s–1

19 B

Instantaneous rate of reaction at 30 s = (0.165 – 0.09) mol dm–3

(60 – 0) s = 1.25 x 10–3 mol dm–3 s–1

20 A Number of moles of Fe2O3 consumed in 10.0 seconds = 0.30

2 mol

= 0.15 mol

Rate of consumption of Fe2O3 = 0.15 mol

10.0 s = 0.015 mol s–1

∴ the rate of consumption of Fe2O3 is 1.5 x 10–2 mol s–1.

69

21 B Rate = – 1

2

d[N2O5(g)]

dt

= 1

4

d[NO2(g)]

dt

= d[O2(g)]

dt

Instantaneous rate of decomposition of N2O5(g) = – d[N2O5(g)]

dt

= 2

4

d[NO2(g)]

dt

= 2

4 (3.8 x 10–5 mol dm–3 s–1)

= 1.9 x 10–5 mol dm–3 s–1

22 D Instantaneous rate of decomposition of O2(g) = d[O2(g)]

dt

= 1

4

d[NO2(g)]

dt

= 1

4 (3.8 x 10–5 mol dm–3 s–1)

= 9.50 x 10–6 mol dm–3 s–1

23 A CaCO3(s) and HCl(aq) react according to the following equation:

CaCO3(s) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g)

Number of moles of HCl consumed in 10.0 minutes = 0.200 mol min–1 x 10.0 min = 2.00 mol

According to the equation, 1 mole of CaCO3 reacts with 2 moles of HCl.

i.e. number of moles of CaCO3 consumed = 2.00

2 mol

= 1.00 mol

Molar mass of CaCO3 = (40.1 + 12.0 + 3 x 16.0) g mol–1

= 100.1 g mol–1

Mass of CaCO3 consumed = number of moles of CaCO3 x molar mass of CaCO3

= 1.00 mol x 100.1 g mol–1

= 100.1 g

Mass of CaCO3 remaining = (145.0 – 100.1) g = 44.9 g

∴ there are 44.9 g of CaCO3(s) remaining.

24 C The sealed flask was a closed system. Therefore the mass of the flask plus its contents remained the same.

70

25 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.

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.



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.

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 B — The only difference between Experiments I and IV is the surface area of the antacid tablets. 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. equal mass of gas was produced in both cases. Thus 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 was used.

71

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 does.

Hence the initial rate of reaction would decrease when ethanoic acid was 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.

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 1 mol dm–3 sulphuric acid has a higher concentration of hydrogen ions than 1 mol dm–3 nitric acid does.

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 Reaction mixture

Number of moles of Zn

= mass

molar mass



Reaction between Zn and H2SO4

1

5 g65.4 g mol–1 = 0.076 mol

2 mol dm–3 x 100

1 000 dm3

= 0.2 mol

Zn(s) + H2SO4(aq) 0.076 mol 0.2 mol

ZnSO4(aq) + H2(g)

2 1 mol dm–3 x 100

1 000 dm3

= 0.1 mol


ZnSO4(aq) + H2(g)

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 of H2 produced in both cases).

72

Option A — The hydrogen produced would escape. Thus the mass of each reaction mixture would decrease.

As the same amount of gas was produced in both reactions, the total mass lost of both reaction mixtures would be the same.

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.

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 gas formed.

The carbon dioxide formed would escape. Thus the masses of both reaction mixtures would decrease.

Curve Y showed a smaller loss in mass, and thus the mass of carbon dioxide gas formed was smaller. It could be deduced 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. some hydrogen peroxide solution was added (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 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 1 mol dm–3 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.

73

Option D — Using more concentrated hydrochloric acid would increase the rate of the reaction.

However, the amount of hydrogen formed (i.e. the pressure change) would remain the same.

41 C The second experiment took a longer time to complete, i.e. the rate of reaction of the second experiment was 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

ReactionNumber of moles of CaCO3

= mass

molar mass

Number of moles of HCl= molarity of solution x

volume of solutionReaction between CaCO3 and HCl

1

8 g100.1 g mol–1 = 0.08 mol

1 mol dm–3 x 100

1 000 dm3

= 0.1 mol



2 2 mol dm–3 x 100

1 000 dm3

= 0.2 mol



According to the equation, 1 mole of Mg reacts with 2 moles of HCl to produce 1 mole of CaCl2.

Option A — 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.

Option B — 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 CaCl2, and hence calcium chloride solutions of different concentration, 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).

74

43 B HCl(aq) and CH3COOH(aq) react with calcium carbonate according to the following equations:

Beaker A CaCO3(s) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g)

number of moles of HCl = 1 mol dm–3 x 100

1 000 dm3 = 0.1 mol

Beaker B CaCO3(s) + 2CH3COOH(aq) (CH3COO)2Ca(aq) + H2O(l) + CO2(g)

number of moles of CH3COOH = 1 mol dm–3 x 100

1 000 dm3 = 0.1 mol

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.

Hence equal masses of marble chips remain in the two beakers after reaction.

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.

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(aq) is a monobasic acid while H2SO4(aq) 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 reacts with hydrogen ions in the acid.

Hence the rate of the reaction between Zn and HCl(aq) (Reaction 1) is lower than that between Zn and H2SO4(aq) (Reaction 2).

Options C and D —

ReactionNumber of moles of Zn

= mass

molar mass


volume of solutionReaction between Zn and acid

1

10 g65.4 g mol–1 = 0.15 mol


= 2 mol dm–3 x 100

1 000 dm3

= 0.2 mol

Zn(s) + 2HCl(aq) 0.15 mol 0.2 mol

ZnCl2(aq) + H2(g)

2


= 2 mol dm–3 x 100

1 000 dm3

= 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 reacts with 0.1 mole of Zn.

75

Thus Zn is in excess, i.e. Option C is INCORRECT.

According to the second equation, 1 mole of Zn reacts with 1 mole of H2SO4.

In Reaction 2, 0.15 mole of Zn reacts with 0.15 mole of H2SO4.

Thus sulphuric acid is in excess, i.e. Option D is INCORRECT.

45 B Options A and C — At the start, the tangent to the curve for Experiment 1 was less steep than that for Experiment 2, i.e. the initial rate of reaction of Experiment 1 was lower than that of Experiment 2.

Thus Options A and C are INCORRECT.

Option B — The curve for Experiment 2 went flat earlier than that for Experiment 1, i.e. the rate of reaction for Experiment 2 fell to zero earlier.

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:

Experiment 1 CaCO3(s) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g)

number of moles of HCl = 1.2 mol dm–3 x 90

1 000 dm3 = 0.11 mol

Experiment 2 CaCO3(s) + 2CH3COOH(aq) (CH3COO)2Ca(aq) + H2O(l) + CO2(g)

number of moles of CH3COOH = 0.9 mol dm–3 x 120

1 000 dm3 = 0.11 mol

Options A and C — 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.

Option B — 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.

Thus the curves in Option B show the results of the experiments.

47 D

ExperimentNumber of moles of Zn

= mass

molar mass


volume of solutionReaction between Zn and acid

1

5 g65.4 g mol–1 = 0.076 mol


= 1 mol dm–3 x 100

1 000 dm3

= 0.1 mol


ZnCl2(aq) + H2(g)

2


= 1 mol dm–3 x 100

1 000 dm3

= 0.1 mol


ZnSO4(aq) + H2(g)

76

Options A and B — 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 reacts 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 reacts with 0.076 mole of H2SO4 to produce 0.076 mole of H2.

Thus the volume of gas produced in Experiment 1 is smaller than that produced in Experiment 2, i.e. Options A and B are INCORRECT.

Option D — 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.

Thus 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.

Options B and C — 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.

Option D — 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.

Thus 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

53 A During the reaction between magnesium and an acid, magnesium 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.

77



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 2 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)

As zinc granules remain in both beaker after reaction, it can be deduced that zinc is in excess in both cases.

Beaker Zinc Number of moles of H2SO4 = molarity of solution x volume of solution

1

excess

1 mol dm–3 x 100

1 000 dm3 = 0.1 mol

2 2 mol dm–3 x 50

1 000 dm3 = 0.1 mol

(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.

Thus zinc sulphate solutions of different concentrations are produced in both beakers.

55 B

ReactionNumber of moles of Mg

= mass

molar mass



1

1 g24.3 g mol–1 = 0.04 mol

2 mol dm–3 x 50

1 000 dm3

= 0.1 mol


MgCl2(aq) + H2(g)

24 mol dm–3 x

501 000

dm3

= 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 mole of HCl to give 0.04 mole of H2.

Thus HCl(aq) was in excess.

78

(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) CH3COCH2I(aq) + HI(aq) brown colourless colourless

Hence the progress of the reaction can be followed by using a colorimeter.

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 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.

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.

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.

79

Part B Topic-based exercise


1 B Option A — Phenolphthalein is colourless in acids.

Option B — Citric acid is a weak acid. It only partially dissociates in water.


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.

Option C — Sodium hydroxide is a strong alkali.

3 D Option D — Zn2+(aq) + 2OH–(aq) Zn(OH)2(s)

4 D Option Solution Addition of NH3(aq) to solution

A calcium chloride no precipitate

B chromium(III) sulphate green precipitate

C iron(III) sulphate reddish brown precipitate

D potassium chloride no precipitate

5 B 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

6 C 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

80

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) BaCO3(s)

Barium carbonate is soluble in dilute nitric acid because it can react with the acid.

BaCO3(s) + 2H+(aq) Ba2+(aq) + H2O(l) + CO2(g)

11 C Option C — Concentrated sulphuric acid is a hygroscopic substance.

12 D 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

13 C 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

81

14 A Options A and C — Chlorides give a white precipitate (AgCl) with silver nitrate solution.


Barium chloride solution gives a white precipitate (BaSO4) with dilute sulphuric acid.


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) + H2(g)

Zn2+(aq) + 2OH–(aq) Zn(OH)2(s)

Zn(OH)2(s) + 4NH3(aq) [Zn(NH3)4]2+(aq) + 2OH–(aq)

16 C 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 = mass

molar mass

= 5.93 g

74.1 g mol–1

= 0.0800 mol

Molarity of calcium hydroxide solution = number of moles of Ca(OH)2

volume of solution

= 0.0800 mol

( 400.01 000 ) dm3

= 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

= 2.00 mol dm–3 x 150.0

1 000 dm3

= 0.300 mol

Number of moles of NaOH in 80.0 cm3 of 1.20 mol dm–3 solution

= 1.20 mol dm–3 x 80.0

1 000 dm3

= 0.0960 mol

Total number of moles of NaOH in the resulting solution = (0.300 + 0.0960) mol = 0.396 mol


= 230.0 cm3

82


( 230.01 000 ) dm3

= 1.72 mol dm–3

∴ the concentration of the resulting solution is 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:

HNO3(aq) H+(aq) + NO3–(aq)

0.010 mol dm–3 ? 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 — CH3COOH is a weak acid while HNO3 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 lactic 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.

83

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 distilled 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.

Option B — concentration of hydrogen ions in sea water

concentration of hydrogen ions in laundry detergent =

10–8 mol dm–3

10–11 mol dm–3

= 1 000

∴ the concentration of hydrogen ions in sea water is 1 000 times greater than that in laundry detergent.

Option C — 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

∴ the concentration of hydrogen ions in rainwater is 1 000 times greater than that in sea water.

Option D — concentration of hydrogen ions in lemon juice

concentration of hydrogen ions in soap solution =

10–2 mol dm–3

10–10 mol dm–3

= 108

∴ 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.

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 does. 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.

84

27 C The electrical conductivity of a solution is proportional to the concentration of mobile ions.

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–3 HCl(aq) does.

∴ 1.0 mol dm–3 H2SO4(aq) has the greatest electrical conductivity.

Option D — C6H12O6(aq) does NOT conduct electricity.

28 A Ethanoic acid is a weak acid and hydrochloric acid is a strong acid.

Hence for acids of the same concentration, ethanoic acid has a lower concentration 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.


CH3COOH(aq) is a weak acid while HCl(aq) is a strong acid.

Hence the temperature rise of the neutralization between 20 cm3 of 1 mol dm–3 CH3COOH(aq) and 1 mol dm–3 NaOH(aq) is smaller than that between 20 cm3 of 1 mol dm–3 HCl(aq) and 1 mol dm–3 NaOH(aq).

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 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.

85

35 A Molar mass of H2SO4 = (2 x 1.0 + 32.1 + 4 x 16.0) g mol–1

= 98.1 g mol–1


volume of solution

4.00 mol dm–3 = number of moles of H2SO4

2.00 dm3

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

∴ 785 g of pure sulphuric acid are required.

36 D Molar mass of Na2SO4 = (2 x 23.0 + 32.1 + 4 x 16.0) g mol–1

= 142.1 g mol–1

Number of moles of Na2SO4 = mass

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 = number of moles of Na+ ions

volume of solution

= 0.282 mol

( 100.01 000 ) dm3

= 2.82 mol dm–3

∴ the concentration of sodium ions in the solution is 2.82 mol dm–3.

37 C Number of moles of potassium phosphate = molarity of solution x volume of solution

= 0.500 mol dm–3 x 250.0

1 000 dm3

= 0.125 mol

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

∴ 3.01 x 1023 ions are present.

86

38 C Suppose V cm3 of NaCl(aq) and V cm3 of ZnCl2(aq) are mixed.

Number of moles of NaCl in 0.10 mol dm–3 solution = 0.10 mol dm–3 x V

1 000 dm3

= 0.10 x V

1 000 mol

Number of moles of ZnCl2 in 0.20 mol dm–3 solution = 0.20 mol dm–3 x V

1 000 dm3

= 0.20 x V

1 000 mol

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 V

1 000 + 2 x 0.20 x

V

1 000) mol

= (0.50 x V

1 000) mol

Total volume of the mixture = (V + V) cm3

= 2V cm3

Concentration of Cl– ions in the mixture = total number of moles of Cl– ions

total volume of the mixture

= (0.5 x

V1 000 ) mol

2V1 000 dm3

= 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:


0.0250 mol dm–3 ? mol dm–3


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.

87

40 D One mole of barium nitrate contains 2 moles of NO3– ions.

∴ number of moles of Ba(NO3)2 = 0.0250

2 mol

= 0.0125 mol

Molarity of barium nitrate solution = number of moles of Ba(NO3)2

volume of solution

= 0.0125 mol

( 22.01 000 ) dm3

= 0.568 mol dm–3

∴ the molarity of the barium nitrate solution is 5.68 x 10–1 mol dm–3.

41 A (MV) before dilution = (MV) after dilution, where M = molarity, V = volume

4.0 x 25.0

1 000 = 0.50 x

V

1 000 V = 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


= 98.1 g mol–1

Number of moles of H2SO4 = mass

molar mass

= 37 800 000 g

98.1 g mol–1

= 385 000 mol

Molarity of the sulphuric acid = number of moles of H2SO4

volume of solution

= 385 000 mol

25 000 dm 3

= 15.4 mol dm–3

∴ the concentration of the sulphuric acid is 15.4 mol dm–3.

43 D Option D — Chemicals that absorb atmospheric moisture CANNOT be used to prepare standard solutions.

88

44 C Na2CO3(aq) + 2HCl(aq) 2NaCl(aq) + CO2(g) + H2O(l) 0.150 mol dm–3 0.250 mol dm–3

25.0 cm3 ? cm3

Number of moles of Na2CO3 in 25.0 cm3 solution = molarity of solution x volume of solution

= 0.150 mol dm–3 x 25.0

1 000 dm3

= 0.00375 mol

According to the equation, 1 mole of Na2CO3 requires 2 moles of HCl for complete reaction.


Volume of hydrochloric acid required = number of moles of HCl


= 0.00750 mol

0.250 mol dm –3

= 0.0300 dm3

= 30.0 cm3

∴ 30.0 cm3 of hydrochloric acid are required.

45 C We can represent the dibasic acid X by H2A.

2KOH(aq) + H2A(aq) K2A(aq) + 2H2O(l) 0.150 mol dm–3 ? mol dm–3

30.0 cm3 10.0 cm3

Number of moles of KOH in 30.0 cm3 solution = molarity of solution x volume of solution

= 0.150 mol dm–3 x 30.0

1 000 dm3

= 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 = 0.00450

2 mol

= 0.00225 mol

Molarity of acid solution = number of moles of H2A

volume of solution

= 0.00225 mol

( 10.01 000 ) dm3

= 0.225 mol dm–3

∴ the molarity of the acid solution is 2.25 x 10–1 mol dm–3.

89

46 A We can represent the dibasic acid by H2X.

2NaOH(aq) + H2X(aq) Na2X(aq) + 2H2O(l) 0.100 mol dm–3 ? mol dm–3

20.6 cm3 25.0 cm3


= 0.100 mol dm–3 x 20.6

1 000 dm3

= 0.00206 mol


i.e. number of moles of H2X in 25.0 cm3 solution = 0.00206

2 mol

= 0.00103 mol

Molarity of acid solution = number of moles of H2X

volume of solution

= 0.00103 mol

( 25.01 000 ) dm3

= 0.0412 mol dm–3

∴ the molarity of the acid solution is 0.0412 mol dm–3.

47 B Zn + 2HCl(aq) ZnCl2(aq) + H2(g) 2 mol dm–3

20.0 cm3


= 2 mol dm–3 x 20.0

1 000 dm3

= 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 = 0.04

2 mol

= 0.02 mol

∴ 0.02 mole of zinc will react completely with the hydrochloric acid.

90

48 D Sr(OH)2(aq) + 2HCl(aq) SrCl2(aq) + 2H2O(l) 0.200 mol dm–3 ? mol dm–3

25.0 cm3 20.0 cm3

Number of moles of Sr(OH)2 in 25.0 cm3 solution = molarity of solution x volume of solution

= 0.200 mol dm–3 x 25.0

1 000 dm3

= 0.00500 mol

According to the equation, 1 mole of Sr(OH)2 requires 2 moles of HCl for complete neutralization.


Molarity of hydrochloric acid = number of moles of HCl

volume of solution

= 0.0100 mol

( 20.01 000 ) dm3

= 0.500 mol dm–3

∴ the concentration of the hydrochloric acid is 0.500 mol dm–3.

49 A Fe(OH)3(s) + 3HNO3(aq) Fe(NO3)3(aq) + 3H2O(l) 1.07 g 0.60 mol dm–3

? cm3

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 = mass

molar mass

= 1.07 g

106.8 g mol–1

= 0.0100 mol

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


= 0.0300 mol

0.60 mol dm–3

= 0.050 dm3

= 50 cm3

∴ 50 cm3 of nitric acid are required for complete neutralization.

91

50 C Al2O3(s) + 3H2SO4(aq) Al2(SO4)3(aq) + 3H2O(l) 2.55 g ? mol dm–3

150.0 cm3

Molar mass of Al2O3 = (2 x 27.0 + 3 x 16.0) g mol–1

= 102.0 g mol–1

Number of moles of Al2O3 = mass

molar mass

= 2.55 g

102.0 g mol–1

= 0.0250 mol

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


volume of solution

= 0.0750 mol

( 150.01 000 ) dm3

= 0.500 mol dm–3


51 A Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g) 1.00 g 1.00 mol dm–3

100.0 cm3

Number of moles of Mg = mass

molar mass

= 1.00 g

24.3 g mol–1

= 0.0412 mol


= 1.00 mol dm–3 x 100.0

1 000 dm3

= 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.

92

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) Na2SO4(aq) + 2H2O(l) 0.0400 mol dm–3 ? mol dm–3

25.0 cm3 22.7 cm3


= 0.0400 mol dm–3 x 25.0

1 000 dm3

= 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 = 0.00100

2 mol

= 5.00 x 10–4 mol


volume of solution

= 5.00 x 10–4 mol

( 22.71 000 ) dm3

= 0.0220 mol dm–3


54 D

55 A H3PO4(aq) + 2NaOH(aq) Na2HPO4(aq) + 2H2O(l) 10.0 cm3 0.0240 mol dm–3

250.0 cm3

(used) 25.0 cm3

Average volume of NaOH(aq) required for neutralization = 21.5 + 21.5 + 21.6

3 cm3

= 21.5 cm3


= 0.0240 mol dm–3 x 21.5

1 000 dm3

= 5.16 x 10–4 mol

According to the equation, 1 mole of H3PO4 would react with 2 moles of NaOH.

i.e. number of moles of H3PO4 in 25.0 cm3 diluted acid = 5.16 x 10–4

2 mol

= 2.58 x 10–4 mol

93

Number of moles of H3PO4 in 250.0 cm3 diluted acid = 2.58 x 10–4 mol x 250.0 cm3

25.0 cm3

= 2.58 x 10–3 mol

Number of moles of H3PO4 in 10.0 cm3 original acid = 2.58 x 10–3 mol

Molarity of original phosphoric acid = number of moles of H3PO4

volume of solution

= 2.58 x 10–3 mol

( 10.01 000 ) dm3

= 0.258 mol dm–3

∴ the molarity of the original phosphoric acid is 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 thus affect the titration results.

57 D H2X(aq) + 2NaOH(aq) Na2X(aq) + 2H2O(l) 4.51 g 0.350 mol dm–3

28.0 cm3

250.0 cm3

(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 = mass

molar mass

= 4.51 g

m g mol–1

Number of moles of H2X in 25.0 cm3 solution = 4.51

m x

1

10 mol


= 0.350 mol dm–3 x 28.0

1 000 dm3

= 0.00980 mol


i.e. number of moles of H2X in 25.0 cm3 solution = 0.00980

2 mol

= 0.00490 mol

Number of moles of H2X in 25.0 cm3 solution = 4.51

m x

1

10 mol = 0.00490 mol

m = 92.0

∴ the molar mass of H2X is 92.0 g mol–1.

94

58 D Na2CO3(aq) + H2SO4(aq) Na2SO4(aq) + CO2(g) + H2O(l) 5.72 g 1.00 mol dm–3

20.0 cm3

Molar mass of Na2CO3•nH2O = (106.0 + 18.0n) g mol–1

Number of moles of Na2CO3•nH2O used = mass

molar mass

= 5.72 g

(106.0 + 18.0n) g mol–1


= 1.00 mol dm–3 x 20.0

1 000 dm3

= 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 = 5.72

(106.0 + 18.0n) mol = 0.0200 mol

∴ n = 10

∴ the value of n is 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) Na2SO4(aq) + 2H2O(l) ? mol dm–3 2.0 mol dm–3

45.0 cm3 50.0 cm3


= 2.0 mol dm–3 x 50.0

1 000 dm3

= 0.10 mol


i.e. number of moles of H2SO4 = 0.10

2 mol

= 0.050 mol


volume of solution

= 0.050 mol

( 45.01 000 ) dm3

= 1.1 mol dm–3


95

60 A 2NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2H2O(l) 0.200 mol dm–3 15.0 cm3

25.0 cm3


= 0.200 mol dm–3 x 25.0

1 000 dm3

= 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 = 0.00500

2 mol

= 0.00250 mol

Total volume of the salt solution = (25.0 + 15.0) cm3

= 40.0 cm3

Concentration of the salt solution = number of moles of Na2SO4

volume of solution

= 0.00250 mol

( 40.01 000 ) dm3

= 0.0625 mol dm–3

∴ the molarity of the salt solution obtained is 0.0625 mol dm–3.

61 C CaCl2(aq) + Na2CO3(aq) 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 Na2CO3(aq) would react.

Option Solutions mixedVolume of CaCl2(aq) and

Na2CO3(aq) undergoing reaction

A 5 cm3 of CaCl2(aq) + 25 cm3 of Na2CO3(aq) 5 cm3

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

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) 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, i.e. V cm3 of 1 mol dm–3 Pb(NO3)2(aq) would react with 2V cm3 of 1 mol dm–3 NaCl(aq).

Option Solutions mixedVolume of Pb(NO3)2(aq) and NaCl(aq) undergoing

reaction

A 10 cm3 of Pb(NO3)2(aq) + 30 cm3 of NaCl(aq) 10 cm3 of Pb(NO3)2(aq) react with 20 cm3 of NaCl(aq)

B 15 cm3 of Pb(NO3)2(aq) + 25 cm3 of NaCl(aq) 12.5 cm3 of Pb(NO3)2(aq) react with 25 cm3 of NaCl(aq)

C 20 cm3 of Pb(NO3)2(aq) + 20 cm3 of NaCl(aq) 10 cm3 of Pb(NO3)2(aq) react with 20 cm3 of NaCl(aq)

D 30 cm3 of Pb(NO3)2(aq) + 10 cm3 of NaCl(aq) 5 cm3 of Pb(NO3)2(aq) react with 10 cm3 of NaCl(aq)

96

The greatest volumes of Pb(NO3)2(aq) and NaCl(aq) react in Option B. Thus the greatest amount of precipitate would form.






Heat released

150 cm3 of 1 mol dm–3 HNO3(aq) + 50 cm3 of 1 mol dm–3 KOH(aq)

number of moles of HNO3 / KOH

= 1 mol dm–3 x 50

1 000 dm3

= 0.05 mol

HNO3(aq) + KOH(aq) 0.05 mol 0.05 mol

KNO3(aq) + H2O(l)0.05 mol

2.85 kJ

250 cm3 of 2 mol dm–3 HNO3(aq) + 50 cm3 of 2 mol dm–3 KOH(aq)

number of moles of HNO3 / KOH

= 2 mol dm–3 x 50

1 000 dm3

= 0.1 mol


KNO3(aq) + H2O(l)0.1 mol

5.7 kJ

The total volumes of the two mixtures were the same. Hence the temperature rise of the first mixture

was half that of the second mixture, i.e. T1 = 1

2 T2.

64 B H2SO4(aq) + 2NaOH(aq) 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 mixedVolume of H2SO4(aq) and

NaOH(aq) undergoing reaction

A 10 cm3 of H2SO4(aq) + 35 cm3 of NaOH(aq) 10 cm3 of H2SO4(aq) react with 20 cm3 of NaOH(aq)

B 15 cm3 of H2SO4(aq) + 30 cm3 of NaOH(aq) 15 cm3 of H2SO4(aq) react with 30 cm3 of NaOH(aq)

C 20 cm3 of H2SO4(aq) + 25 cm3 of NaOH(aq) 12.5 cm3 of H2SO4(aq) react with 25 cm3 of NaOH(aq)

D 25 cm3 of H2SO4(aq) + 15 cm3 of NaOH(aq) 7.5 cm3 of H2SO4(aq) react with 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.



Expt. Solutions mixed Strength of acid and alkali Temperature rise

1100 cm3 of 1 mol dm–3 CH3COOH(aq) + 100 cm3 of 1 mol dm–3 NaOH(aq)


T1

2100 cm3 of 1 mol dm–3 HCl(aq) + 100 cm3 of 1 mol dm–3 NaOH(aq)


T2

∴ T1 < T2

97


Mixture

Number of moles of HNO3 / NaOH mixed



Heat released

Temperature rise


2 mol dm–3 x 50

1 000 dm3

= 0.1 mol

HNO3(aq) + NaOH(aq) 0.1 mol 0.1 mol

NaNO3(aq) + H2O(l)0.1 mol

5.7 kJ x


2 mol dm–3 x 100

1 000 dm3

= 0.2 mol



11.4 kJ y


1 mol dm–3 x 100

1 000 dm3

= 0.1 mol



5.7 kJ z

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 5.7 kJ. Hence its temperature rise is less than x and y.

∴ x = y > z



Mixture




Strength of acid and alkali

Heat released

Temperature rise



= 1 mol dm–3 x 1001 000

dm3

= 0.1 mol


NaCl(aq) + H2O(l)0.1 mol

neutralization between a

strong acid and a strong alkali

5.7 kJ x



= 2 mol dm–3 x 1001 000

dm3

= 0.2 mol




strong acid and a strong alkali

11.4 kJ 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

= 1 mol dm–3 x 1001 000

dm3

= 0.1 mol

HCl(aq) + NH4+(aq) + OH–(aq)

0.1 mol 0.1 mol NH4Cl(aq) + H2O(l)

0.1 mol


strong acid and a weak alkali

< 5.7 kJ z

All the mixtures have the same total volume (200 cm3).

98

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


Mixture

Number of moles of H2SO4 / NaOH mixed



Heat released

Temperature rise

25 cm3 of 1 mol dm–3 H2SO4(aq)

+25 cm3 of

2 mol dm–3 NaOH(aq)


= 1 mol dm–3 x 25

1 000 dm3

= 0.025 mol

Number of moles of NaOH

= 2 mol dm–3 x 25

1 000 dm3

= 0.05 mol

H2SO4(aq) + 2NaOH(aq) 0.025 mol 0.05 mol

Na2SO4(aq) + 2H2O(l)0.05 mol

2.85 kJ w

50 cm3 of 1 mol dm–3 H2SO4(aq)

+ 50 cm3 of



= 1 mol dm–3 x 50

1 000 dm3

= 0.05 mol


= 2 mol dm–3 x 50

1 000 dm3

= 0.1 mol



5.7 kJ x

25 cm3 of 0.5 mol dm–3 H2SO4(aq)

+ 25 cm3 of



= 0.5 mol dm–3 x 25

1 000 dm3

= 0.0125 mol


= 1 mol dm–3 x 25

1 000 dm3

= 0.025 mol



1.425 kJ y

50 cm3 of 0.5 mol dm–3 H2SO4(aq)

+ 50 cm3 of



= 0.5 mol dm–3 x 50

1 000 dm3

= 0.025 mol


= 1 mol dm–3 x 50

1 000 dm3

= 0.05 mol



2.85 kJ z

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.

99

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).

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 with 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.

74 D Average reaction rate = 2.0 g

3.0 min = 0.67 g min–1

100

75 C Rate = – 1

4

d[NH3(g)]

dt

= – 1

5

d[O2(g)]

dt

Rate of consumption of NH3(g) = – d[NH3(g)]

dt

= 4

5

d[O2(g)]

dt

= 4

5 (1.80 mol dm–3 s–1)

= 1.44 mol dm–3 s–1

76 A Rate = – 1

5

d[O2(g)]

dt

= 1

6

d[H2O(g)]

dt

Rate of formation of H2O(g) = d[H2O(g)]

dt

= – 6

5

d[O2(g)]

dt

= 6

5 (1.80 mol dm–3 s–1)

= 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 samples in Experiments I and II are different. So they CANNOT be used for comparison.

Option B — The temperatures in Experiments I and III are different. So they CANNOT be used for comparison.

Option C — The temperatures 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 samples in Experiments I and III are different. So they CANNOT be used for comparison.

Option B — The concentration of both reagents in the samples in Experiments I and IV are the same. The only difference is the temperature.

Option C — The concentration of acid in the samples in Experiments II and III are different. So they CANNOT be used for comparison.

Option D — The concentration of both reagents in the samples in Experiments II and IV are different. So they CANNOT be used for comparison.

101

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.

Thus Option D is INCORRECT.

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.

The changes is pressure were the same in both experiments. It could be deduced that equal volumes of gas were produced.

Option B — Calcium carbonate powder had a greater surface area than calcium carbonate lumps of the same mass.

The rate of the reaction decreases when the surface area of calcium carbonate is decreased, i.e. when using calcium carbonate lumps instead of calcium carbonate powder.

Option D — Calcium carbonate was the limiting reactant as it disappeared in both experiments after reaction.

Using a lower mass of calcium carbonate would decrease the volume of gas formed, i.e. decrease the pressure. This Option D is INCORRECT.

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.

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.

102

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.

Option A — 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.

Thus the curves in Option A show the results.

Options B, C and D — 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.

∴ Options B, C and D are INCORRECT.

86 B Beaker

Number of moles of acid = molarity of solution x volume of solution

Reaction between CaCO3 and acid

A


= 1.2 mol dm–3 x 45

1 000 dm3

= 0.054 mol

CaCO3(s) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g)

B


= 0.9 mol dm–3 x 60

1 000 dm3

= 0.054 mol

CaCO3(s) + 2CH3COOH(aq) (CH3COO)2Ca(aq) + H2O(l) + CO2(g)

Option A — The initial rate of reaction in beaker A is higher as 1.2 mol dm–3 HCl(aq) contains a higher concentration of hydrogen ions than 0.9 mol dm–3 CH3COOH(aq) does.

Option B — Marble chips disappear in both beakers after reaction. Thus marble chip is the limiting reagent.

Equal masses of marble chips would produce the same amount of gas.

Option C — The average rate of evolution of gas in Beaker A is higher as 1.2 mol dm–3 HCl(aq) contains a higher concentration of hydrogen ions than 0.9 mol dm–3 CH3COOH(aq) does.

Option D — Equal masses of marble chips react in the two beakers. Thus equal amounts of Ca2+(aq) ions are formed in the beakers.

However, the volumes of the two reaction mixtures are different.

Hence the concentrations of Ca2+(aq) ions in the two beakers are different.

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.

103

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.

Number of moles of H2O2 in 100 cm3 of 0.12 mol dm–3 solution

= 0.12 mol dm–3 x 100

1 000 dm3

= 0.012 mol

Number of moles of H2O2 in 80 cm3 of 0.15 mol dm–3 solution

= 0.15 mol dm–3 x 80

1 000 dm3

= 0.012 mol

As the number of moles of H2O2 remained the same, the volume of O2 produced would be the same.

Thus the curves in Option C show the results.

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.

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 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.

(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) NH3(g) + NaCl(aq) + H2O(l)

(2) Phenolphthalein gives a colourless solution with acidic solutions.

(3) Dilute sodium hydroxide solution would react with carbon dioxide gas.

NaOH(aq) + CO2(g) NaHCO3(aq)

104

93 A (1) and (2) Vinegar and grapefruit juice react with magnesium to give hydrogen, a gas which burns with a ‘pop’ sound.

(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.

Al3+(aq) + 3OH–(aq) Al(OH)3(s)

(2) Iron(III) sulphate solution gives a reddish brown precipitate with dilute aqueous ammonia.


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.



HCO3– ions in sodium hydrogencarbonate reacted with the H+ ions to give carbon dioxide

gas. Effervescence occurred.


(3) Solid citric acid would NOT react with solid sodium hydrogencarbonate because it did not contain hydrogen ions.

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.


The aqueous solution can conduct electricity due to the presence of mobile ions.

(3) Sodium hydroxide solution can absorb carbon dioxide.

Sodium hydrogencarbonate is formed which undergoes dehydration to give sodium carbonate.

NaOH(aq) + CO2(g) NaHCO3(aq)

2NaHCO3(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.

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.

(3) Concentrated sulphuric acid does NOT give an acid mist in air because it is non-volatile.

105

99 A Aqueous ammonia reacts with hydrochloric acid according to the equation below. The salt ammonium chloride is formed in the reaction.


(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.

100 A (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 as a flavour enhancer.

101 B (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) Concentrated nitric acid is NOT a drying agent.

102 C (2) Pb(NO3)2(aq) + CaCl2(aq) PbCl2(s) + Ca(NO3)2(aq)

(3) Na2CO3(aq) + CuSO4(aq) CuCO3(s) + Na2SO4(aq)

103 D (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.

104 B (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.

105 C (1) Both ammonium nitrate solution and potassium chloride solution are colourless.

(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.

NH4+(aq) + OH–(aq) NH3(g) + H2O(l)

There is NO observable change for potassium chloride solution.

(3) Potassium chloride solution gives a white precipitate with silver nitrate solution while ammonium nitrate solution does NOT.


106

106 B (1) Both dilute hydrochloric acid and nitric acid can 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 can react with sodium hydrogencarbonate solution.

107 D Solutions mixed Precipitate formed Ionic equation for reaction involved

(1) NaOH(aq) and CuSO4(aq) a pale blue precipitate Cu2+(aq) + 2OH–(aq) Cu(OH)2(s)

(2) NH3(aq) and Mg(NO3)2(aq) a white precipitate Mg2+(aq) + 2OH–(aq) Mg(OH)2(s)

(3) (NH4)2CO3(aq) and NiCl2(aq) a green precipitate Ni2+(aq) + CO32–(aq) NiCO3(s)

108 D (1) Copper(II) oxide and dilute sulphuric acid react to give copper(II) sulphate and water. NO gas is produced.


(2) When water is added to solid citric acid, some citric acid molecules dissociate to give hydrogen ions.



HCO3– ions in sodium hydrogencarbonate react with the H+ ions to give carbon dioxide gas.


(3) Ammonium carbonate reacts with dilute hydrochloric acid to give carbon dioxide gas.

(NH4)2CO3(s) + 2HCl(aq) 2NH4Cl(aq) + H2O(l) + CO2(g)

109 D (1) Pb2+(aq) + CO32–(aq) PbCO3(s)

(2) Pb2+(aq) + SO42–(aq) PbSO4(s)

(3) Pb2+(aq) + 2Cl–(aq) PbCl2(s)

110 D Magnesium sulphate is prepared by reacting dilute sulphuric acid with either magnesium, magnesium carbonate or magnesium oxide.


MgCO3(s) + H2SO4(aq) MgSO4(aq) + H2O(l) + CO2(g)

MgO(s) + H2SO4(aq) MgSO4(aq) + H2O(l)

111 A (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.

112 D (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.

107

113 B

Solutions mixedNumber of moles of acid / alkali


Complete neutralization?

(1) 100 cm3 of 1 mol dm–3 HNO3(aq) and

100 cm3 of 1 mol dm–3 KOH(aq)

number of moles of HNO3

= 1 mol dm–3 x 100

1 000 dm3

= 0.1 mol

number of moles of KOH

= 1 mol dm–3 x 100

1 000 dm3

= 0.1 mol


KNO3(aq) + H2O(l)

∴ complete neutralization occurs and a neutral solution is obtained.

(2) 100 cm3 of 1 mol dm–3 H2SO4(aq) and

100 cm3 of 1 mol dm–3 NaOH(aq)


= 1 mol dm–3 x 100

1 000 dm3

= 0.1 mol


= 1 mol dm–3 x 100

1 000 dm3

= 0.1 mol


Na2SO4(aq) + 2H2O(l)

∴ not enough NaOH(aq) to neutralize H2SO4(aq) completely.

(3) 100 cm3 of 1 mol dm–3 H2SO4(aq) and

100 cm3 of 2 mol dm–3 KOH(aq)


= 1 mol dm–3 x 100

1 000 dm3

= 0.1 mol

number of moles of KOH

= 2 mol dm–3 x 100

1 000 dm3

= 0.2 mol

H2SO4(aq) + 2KOH(aq) 0.1 mol 0.2 mol

K2SO4(aq) + 2H2O(l)

∴ complete neutralization occurs and a neutral solution is obtained.



Option Solutions mixed Strength of acid and alkali Temperature rise

—25 cm3 of 1 mol dm–3 HCl(aq) and 25 cm3 of 1 mol dm–3 NaOH(aq)


T

(1)25 cm3 of 1 mol dm–3 HNO3(aq) and 25 cm3 of 1 mol dm–3 NH3(aq)


< T

(2)25 cm3 of 1 mol dm–3 HNO3(aq) and 25 cm3 of 1 mol dm–3 KOH(aq)


T

(3)50 cm3 of 1 mol dm–3 CH3COOH(aq) and 50 cm3 of 1 mol dm–3 NaOH(aq)


< 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).

108

115 A Beaker Number of moles of HCl = molarity of solution x volume of solution

11 mol dm–3 x

1001 000

dm3 = 0.1 mol

22 mol dm–3 x

501 000

dm3 = 0.1 mol

Marble chips react with dilute hydrochloric acid according to the following equation:

CaCO3(s) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g) excess 0.1 mol

(1) Equal numbers of moles of HCl 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 (0.1

2 mole) is produced in both cases.

(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.

116 A


= mass

molar mass



1

2 g24.3 g mol–1 = 0.08 mol

Number of moles of HCl

= 1.2 mol dm–3 x 90

1 000 dm3

= 0.11 mol


MgCl2(aq) + H2(g)

2

Number of moles of CH3COOH

= 0.9 mol dm–3 x 120

1 000 dm3

= 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 (0.11

2 mole) was produced in both cases.


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.

109

117 B (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.

118 A (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.

A colorimeter can be used to follow the change in light level passing through the reaction mixture as the turbidity increases.

(3) The reaction does NOT involve any colour change.

Hence a colorimeter CANNOT be used to follow the progress of the reaction.

119 A (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.

120 A HCl(aq) and CH3COOH(aq) react with calcium carbonate according to the following equations:

Beaker A CaCO3(s) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g)

number of moles of HCl = 1 mol dm–3 x 100

1 000 dm3 = 0.1 mol

Beaker B CaCO3(s) + 2CH3COOH(aq) (CH3COO)2Ca(aq) + H2O(l) + CO2(g)

number of moles of CH3COOH = 1 mol dm–3 x 100

1 000 dm3 = 0.1 mol

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.



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).

Hence the reaction in Beaker A takes a shorter time to complete.

110

121 D

ReactionNumber of moles of Zn

= mass

molar mass


volume of solutionReaction between Zn and HCl

1

3 g65.4 g mol–1 = 0.046 mol

2 mol dm–3 x 50

1 000 dm3

= 0.1 mol


ZnCl2(aq) + H2(g)

24 mol dm–3 x

501 000

dm3

= 0.2 mol


ZnCl2(aq) + H2(g)

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.


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.

(3) The concentration of HCl(aq) was lower in Reaction 1.

Thus the initial rate of Reaction 1 was smaller.

122 C


= mass

molar mass



1

1.8 g24.3 g mol–1 = 0.074 mol

Number of moles of HCl

= 1 mol dm–3 x 100

1 000 dm3

= 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 reacted with 0.05 mole of Mg to give 0.05 mole of H2.

Thus Mg was in excess.

2


= 1 mol dm–3 x 100

1 000 dm3

= 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 reacted with 0.074 mole of H2SO4 to give 0.074 mole of H2.

Thus H2SO4(aq) was in excess.

111


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.

(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 HCl(aq) was lower than that in 1 mol dm–3 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.

123 B (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.

124 A

125 D Nitric acid is NOT a drying agent.

126 A

127 B Sodium hydroxide can absorb sulphur dioxide gas because it can react with the gas.

128 B Aqueous solution of citric acid can conduct electricity due to the presence of mobile ions.


Hence citric acid is an electrolyte.

129 D Copper(II) hydroxide does NOT dissolve in excess sodium hydroxide solution.

130 A

131 C Aqueous solution of ethanoic acid can conduct electricity.

Hence ethanoic acid is an electrolyte.

132 C 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.

112

133 D Ammonia is a weak alkali.

134 D Hydrochloric acid is a strong acid while ethanoic acid is a weak acid.

Therefore 1 mol dm–3 hydrochloric acid contains a higher concentration of hydrogen ions than 1 mol dm–3 ethanoic acid does.

Hence the acids give a different colour with the same quantity of universal indicator.

135 D Ethanoic acid is a weak acid. It only partically dissociates in both concentrated and dilute solutions.

136 D Solid citric acid does NOT show typical properties of an acid.

A typical acid does NOT react with copper.

137 D Number of moles of H2SO4 in 10 cm3 of 5 mol dm–3 acid

= 5 mol dm–3 x 10

1 000 dm3

= 0.05 mol

Number of moles of H2SO4 in 50 cm3 of 1 mol dm–3 acid

= 1 mol dm–3 x 50

1 000 dm3

= 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.

138 D 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.

139 C Both 1 mol dm–3 NaOH(aq) and 1 mol dm–3 NH3(aq) give a precipitate with MgCl2(aq).

140 C 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.

141 B 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)

113

142 B 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) + SO4

2–(aq)

143 C 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.

144 C For example, 1 mole of sulphuric acid can neutralize two moles of sodium hydroxide.

H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l)

145 C 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 mixture remains unchanged.

However, the volume of the solution has changed.

Hence the concentration of Na+(aq) ions would decrease.

146 C 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.


Acid and alkali mixed




released


number of moles of HCl / NaOH

= 1 mol dm–3 x 25

1 000 dm3

= 0.025 mol



1.425 kJ


number of moles of HCl / NaOH

= 2 mol dm–3 x 25

1 000 dm3

= 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

half of that of the second mixture, i.e. T1 = 1

2 T2.

114

148 B 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.

149 A In the oxidation of oxalate ions by permanganate ions, the intensity of the purple colour of permanganate ions decreases as the reaction proceeds.



150 C Hydrochloric acid is a monobasic acid while sulphuric acid is a dibasic acid.

Therefore 1 mol dm–3 hydrochloric acid has a lower concentration of hydrogen ions than 1 mol dm–3 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.

151 D 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) ZnCl2(aq) + H2(g) (1)

b) Iron and dilute sulphuric acid Fe(s) + H2SO4(aq) FeSO4(aq) + H2(g) (1)

c) Zinc carbonate and dilute hydrochloric acid ZnCO3(s) + 2HCl(aq) ZnCl2(aq) + H2O(l) + CO2(g) (1)

d) Sodium carbonate solution and dilute sulphuric acid

Na2CO3(aq) + H2SO4(aq) Na2SO4(aq) + H2O(l) + CO2(g) (1)

e) Lead(II) carbonate and dilute nitric acid 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)

115

153 Reaction between Ionic equation

a) Magnesium and sulphuric acid Mg(s) + 2H+(aq) Mg2+(aq) + H2(g) (1)

b) Sodium carbonate solution and dilute hydrochloric acid

CO32–(aq) + 2H+(aq) H2O(l) + CO2(g) (1)

c) Sodium hydrogencarbonate solution and dilute nitric acid

HCO3–(aq) + H+(aq) H2O(l) + CO2(g) (1)

d) Iron(III) hydroxide and dilute nitric acid Fe(OH)3(aq) + 3H+(aq) Fe3+(aq) + 3H2O(l) (1)

e) Dilute sodium hydroxide solution and dilute hydrochloric acid

OH –(aq) + H+(aq) H2O(l) (1)

154 Reaction between Ionic equation(s)

a) Solution containing calcium ions and dilute sodium hydroxide solution

Ca2+(aq) + 2OH–(aq) Ca(OH)2(s) (1)

b) Solution containing aluminium ions and dilute sodium hydroxide solution

Al3+(aq) + 3OH–(aq) Al(OH)3(s)Al(OH)3(s) + OH–(aq) [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)Cu(OH)2(s) + 4NH3(aq) [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)Zn(OH)2(s) + 4NH3(aq) [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)

155 Name of ion in solution Reagent(s) added to the solution Result

Copper(II) iondilute 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

Ammonium ionwarm with sodium hydroxide solution /calcium hydroxide solution (1)

a gas that turns moist red litmus paper blue is given off

116

156 Reaction Name of salt Formula of salt

Calcium hydroxide and dilute hydrofluoric acid calcium fluoride CaF2 (1)

Magnesium oxide and dilute sulphuric acid 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)

157 a) Gas bubbles are given off. / Magnesium dissolves. (1)

Mg(s) + 2H+(aq) Mg2+(aq) + H2(g) (1)

b) Copper(II) oxide dissolves in the acid. / A blue solution forms. (1)

CuO(s) + 2H+(aq) Cu2+(aq) + H2O(l) (1)

c) Effervescence occurs. / Nickel(II) carbonate dissolves in the acid. / A green solution forms. (1)

NiCO3(s) + 2H+(aq) Ni2+(aq) + H2O(l) + CO2(g) (1)

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) (1)

Pb(OH)2(s) + 2OH–(aq) [Pb(OH)4)]2–(aq) (1)

e) A white precipitate forms. (1)

Ca2+(aq) + CO32–(aq) CaCO3(s) (1)

f) A white precipitate forms. (1)

Pb2+(aq) + 2Cl–(aq) PbCl2(s) (1)

g) A green precipitate forms. (1)

Fe2+(aq) + 2OH–(aq) Fe(OH)2(s) (1)

h) The solution mixture warms up. (1)

H+(aq) + OH–(aq) H2O(l) (1)

158 a) pH of solution = –log10[H+] = 3.40

i.e. log10[H+] = –3.40 (1)

[H+] = 10–3.40

= 3.98 x 10–4 mol dm–3 (1)

117

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 = 1.80 x 10–2 mol

( 40.01 000 ) dm3

= 0.450 mol dm–3 (1)

∴ concentration of ZnCl2 solution = 0.4502

mol dm–3

= 0.225 mol dm–3 (1)

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 100.01 000

dm3

= 0.400 mol (1)

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 150.01 000

dm3

= 0.0750 mol (1)

Total number of moles of K+ ions in X = (0.400 + 0.0750) mol

= 0.475 mol

Concentration of K + ions in X = number of moles of K+ ionsvolume of solution

= 0.475 mol

( 100.0 + 150.01 000 ) dm3

= 1.90 mol dm–3 (1)

d) i) (MV) before dilution = (MV) after dilution, where M = molarity, V = volume

8.00 x 400.01 000

= M x 25.0 (1)

M = 0.128 (1)

∴ 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

Concentration of diluted acid = 0.128 mol dm–3 x 98.1 g mol–1

= 12.6 g dm–3 (1)

118

Structured questions

159 a) 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)

160 a) 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)

119

161 a) A weak acid is an acid that only partially dissociates in water. (1)

b) pH = –log10[H+] = 1.41

i.e. log10[H+] = –1.41 (1)

[H+] = 10–1.41

= 3.89 x 10–2 mol dm–3 (1)

c) H2SO4(aq) 2H+(aq) + SO42–(aq)

0.100 mol dm–3 ? mol dm–3


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)

162 a) i) Sodium hydroxide is corrosive. (1)

ii) Use weak bases such as magnesium hydroxide and aluminium hydroxide. (1)

b) i) There is no reaction between copper and dilute hydrochloric acid. (1)

ii) Use the reaction between copper(II) oxide / copper(II) hydroxide / copper(II) carbonate and dilute hydrochloric acid (1)

c) i) A lot of heat is released when vinegar reacts with potassium hydroxide solution. This will cause skin burn. (1)

ii) Wash the hand with plenty of water. (1)

d) i) Concentrated sulphuric acid will react with ammonia. (1)

ii) Use calcium oxide. (1)

120

163 a) i) HCl(aq) H+(aq) + Cl–(aq) (1)

ii) CH3COOH(aq) H+(aq) + CH3COO–(aq) (1)

b) Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)

Number of moles of Mg present = mass of Mg

molar mass of Mg

= 1.5 g24.3 g mol–1

= 0.062 mol (1)

Number of moles of HCl used = molarity of solution x volume of solution

= 1.0 mol dm–3 x 1001 000

dm3

= 0.10 mol (1)

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.


• Measuring their pH (1)

The pH of 1.0 mol dm–3 HCl(aq) is lower than that of 1.0 mol dm–3 CH3COOH(aq). (1)

• Measuring their electrical conductivity (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)

e) Sulphuric acid is a dibasic acid while hydrochloric acid is a monobasic acid. (1)

Therefore 1.0 mol dm–3 sulphuric acid has a higher concentration of hydrogen ions. (1)

The initial rate of Reaction 3 is thus higher than that of Reaction 1.

121

164 a) Pb2+(aq) + SO42–(aq) PbSO4(s) (1)

b)

(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) Cu(OH)2(s) (1)

d) The presence of NH4+(aq) ions can be shown by warming solution D. An alkaline gas will evolve. (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)

e) Blue (1)

Cu2+(aq) ions are blue in colour while the other ions in solution X are colourless. (1)

165 a) 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) PbO(s) (1)

ii) PbO(s) + 2HNO3(aq) Pb(NO3)2(aq) + H2O(l) (1)

iii) Pb2+(aq) + CO32–(aq) PbCO3(aq) (1)

iv) Pb2+(aq) + 2OH–(aq) Pb(OH)2(s) (1)

v) Pb(OH)2(s) + 2OH–(aq) [Pb(OH)4]2–(aq) (1)

c) Carbon reduction (1)

122

166 A 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) Al(OH)3(s) (1)

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) [Al(OH)4]–(aq) (1)

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) NaCl(aq) + H2O(l) (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) AgCl(s) (1)

167 a) 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. (1)

c) Average volume of sodium hydroxide solution required for neutralization

= 22.2 + 22.3 + 22.23

cm3

= 22.2 cm3 (1)

d) 2NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2H2O(l) 0.180 mol dm–3 ? mol dm–3

22.2 cm3 25.0 cm3


= 0.180 mol dm–3 x 22.2

1 000 dm3

= 0.00400 mol (1)


i.e. number of moles of H2SO4 in 25.0 cm3 solution = 0.00400

2 mol

= 0.00200 mol (1)

123


volume of solution

= 0.00200 mol

( 25.01 000 ) dm3

= 0.0800 mol dm–3 (1)


e) The next titre would be lower. (1)

Some alkali remained in the conical flask. (1)

168 a) 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 = 18.8 + 18.8 + 18.73

cm3

= 18.8 cm3 (1)

ii) NH3(aq) + HCl(aq) NH4Cl(aq) 9.97 g 0.150 mol dm–3

250.0 cm3 18.8 cm3

(used) 25.0 cm3


= 0.150 mol dm–3 x 18.81 000

dm3

= 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 (1)

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 250.0 cm3

25.0 cm3

= 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 (1)

124

Percentage by mass of NH3 in the original household ammonia solution

= 0.479 g9.97 g

x 100%

= 4.80% (1)

∴ the percentage by mass of ammonia in the houseold ammonia solution is 4.80%.

c)

Distilled waterOriginal

household ammonia solution

Diluted household ammonia solution

Standard hydrochloric acid

i) 250.0 cm3 volumetric flask ✔ (0.5)

ii) 25.0 cm3 pipette for delivering the diluted household ammonia solution

✔ (0.5)

iii) Conical flask ✔ (0.5)

iv) Burette ✔ (0.5)

169 a) Molar mass of KOH = (39.1 + 16.0 + 1.0) g mol–1

= 56.1 g mol–1

Number of moles of KOH in 100.0 cm3 solution = massmolar mass

= 7.00 g56.1 g mol–1

= 0.125 mol (1)

Concentration of potassium hydroxide solution = number of moles of KOHvolume of solution

= 0.125 mol

( 100.01 000 ) dm3

= 1.25 mol dm–3 (1)

b) i) 25.0 cm3 of 6.00 mol dm–3 sulphuric acid are needed for the dilution.

Wash a 25.0 cm3 pipette first with distilled water and then with the acid. (1)

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. (1)

ii) 2KOH(aq) + H2SO4(aq) K2SO4(aq) + 2H2O(l) 1.00 mol dm–3 0.600 mol dm–3

25.0 cm3 ? cm3

125

Number of moles of KOH in 25.0 cm3 solution = molarity of solution x volume of solution

= 1.00 mol dm–3 x 25.01 000

dm3

= 0.0250 mol (1)

According to the equation, 2 moles of KOH require 1 mole of H2SO4 for complete neutralization.


2 mol

= 0.0125 mol (1)

Volume of sulphuric acid required for neutralization = number of moles of H2SO4


= 0.0125 mol

0.600 mol dm–3

= 0.0208 dm3

= 20.8 cm3 (1)

∴ the lowest acceptable titre is 20.8 cm3.

c) To check that the peaches are free from potassium hydroxide (1)

because potassium hydroxide is corrosive. (1)

170 a) Sodium hydroxide solutions absorb moisture from air. So their masses change. (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. (1)

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. (1)

d) Phenolphthalein (1)

From colourless to red / pink (1)

e) (COOH)2(aq) + 2NaOH(aq) (COONa)2(aq) + 2H2O(l) 4.95 g ? mol dm–3

250.0 cm3 29.2 cm3

(used) 25.0 cm3

126

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 = mass

molar mass

= 4.95 g126.0 g mol–1

= 0.0393 mol (1)

Number of moles of (COOH)2•2H2O in 25.0 cm3 solution = 0.0393 mol x 25.0 cm3

250.0 cm3

= 0.00393 mol

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 (1)

Molarity of NaOH solution = number of moles of NaOH

volume of solution

= 0.00786 mol

( 29.21 000 ) dm3

= 0.269 mol dm–3 (1)

∴ the molarity of the sodium hydroxide solution is 0.269 mol dm–3.

171 a) 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)



b) Phenolpthalein (1)

From colourless to red / pink (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 26.4

1 000 dm3

= 0.0132 mol (1)

ii) Let n be the basicity of tartaric acid, so we can represent the acid by HnX.

HnX(s) + nNaOH(aq) NanX(aq) + nH2O(l)

Number of moles of HnX = mass

molar mass

= 9.90 g150.0 g mol–1

= 0.0660 mol (1)

127

Number of moles of HnX in 25.0 cm3 solution = 0.0660 mol x 25.0 cm3

250.0 cm3

= 0.00660 mol

Number of moles of acid

Number of moles of NaOH =

1n

= 0.00660 mol0.0132 mol

n = 2 (1)

∴ the basicity of tartaric acid is 2.

d) i) A solution of accurately known concentration. (1)

ii) Not appropriate as sodium hydroxide absorbs moisture in air readily. (1)

e) Use a pH meter / pH sensor. (1)

172 a) The final volume of the solution would be more than 250.0 cm3 as the solute has volume. / The exact volume of the solution would not be known. (1)

b)

(1 mark for correct set-up; 0.5 mark for each correct label; 0 mark if the set-up is not workable) (3)

c) From yellow to red (1)

d) i) Average volume of hydrochloric acid used = 20.2 + 20.3 + 20.23

cm3

= 20.2 cm3 (1)

ii) MOH(aq) + HCl(aq) MCl(aq) + H2O(l) 1.26 g 250.0 cm3

(used) 25.0 cm3

Let m g mol–1 be the molar mass of MOH.

128

Number of moles of MOH in 250.0 cm3 solution = massmolar mass

= 1.26 gm g mol–1

Number of moles of MOH in 25.0 cm3 solution = 1.26m

mol x 25.0 cm3

250.0 cm3

= 1.26m

x 110

mol


= 0.155 mol dm–3 x 20.2

1 000 dm3

= 0.00313 mol (1)

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)

Number of moles of MOH in 25.0 cm3 solution = 1.26m

x 110

mol = 0.00313 mol

m = 40.3 (1)

∴ the molar mass of the metal hydroxide is 40.3 g mol–1.

173 a) 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. (1)


• Methyl orange (1)

from yellow to red (1)

• Phenolphthalein (1)

from red / pink to colourless (1)

e) Extra water in the conical flask will not change the number of moles of solute it holds. (1)

f) H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l) ? mol dm–3 0.600 mol dm–3

15.0 cm3 25.0 cm3

129


= 0.600 mol dm–3 x 25.01 000

dm3

= 0.0150 mol (1)

According to the equation, 2 moles of NaOH require 1 mole of H2SO4 for complete neutralization. 1 mole of Na2SO4 is produced.


mol

= 0.00750 mol (1)


volume of solution

= 0.00750 mol

( 15.01 000 ) dm3

= 0.500 mol dm–3 (1)


g) Number of moles of Na2SO4 produced = 0.00750 mol

Concentration of sodium sulphate in the resulting solution

= number of moles of Na2SO4

volume of solution

= 0.00750 mol

( 25.0 + 15.01 000 ) dm3

= 0.188 mol dm–3 (1)

∴ 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. (1)

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)

130

174 a)

(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

i.e. log10[H+] = –3.2 (1)

[H+] = 10–3.2

= 6.3 x 10–4 mol dm–3 (1)

d) HA(aq) + NaOH(aq) NaA(aq) + H2O(l) ? mol dm–3 0.100 mol dm–3

25.0 cm3 28.0 cm3


= 0.100 mol dm–3 x 28.01 000

dm3

= 0.00280 mol (1)

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 (1)

Molarity of acid HA = number of moles of HAvolume of solution

= 0.00280 mol

( 25.01 000 ) dm3

= 0.112 mol dm–3 (1)

∴ the concentration of the acid HA is 0.112 mol dm–3.

131

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)

175 a) 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) (CH2COONa)2(aq) + 2H2O(l) ? mol dm–3 0.0980 mol dm–3

25.0 cm3 27.6 cm3


= 0.0980 mol dm–3 x 27.6

1 000 dm3

= 0.00270 mol (1)

According to the equation, 1 mole of (CH2COOH)2 requires 2 moles of NaOH for complete neutralization.

i.e. number of moles of (CH2COOH)2 = 0.002702

mol

= 0.00135 mol (1)

Concentration of butanedioic acid = number of moles of (CH2COOH)2

volume of solution

= 0.00135 mol

( 25.01 000 ) dm3

= 0.0540 mol dm–3 (1)

∴ the concentration of the butanedioic acid is 0.0540 mol dm–3.

c) Error is less with a large titre. (1)

d) i) A solution of accurately known concentration. (1)

ii) Not appropriate as sodium hydroxide absorbs moisture in air readily. (1)

176 a) A strong acid is an acid that almost completely dissociates in water. (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)

132

b) i) BaCO3(s) + 2HCl(aq) BaCl2(aq) + CO2(g) + H2O(l) (1)

ii) To make sure that all the hydrochloric acid has been reacted. (1)

iii)

(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. (1)

As the saturated solution cools, the solvent cannot hold all the solutes. The extra solutes separate out as crystals. (1)

v) (1) Anhydrous barium chloride will form. / The water of crystallization will be removed. (1)

(2) Absorb the water by filter paper. / Place the crystals in a desiccator. (1)

vi) BaCO3(s) + 2HCl(aq) BaCl2(aq) + CO2(g) + H2O(l) 1.00 mol dm–3

50.0 cm3


= 1.00 mol dm–3 x 50.01 000

dm3

= 0.0500 mol (1)

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 = 0.05002

mol

= 0.0250 mol (1)

Mass of BaCl2•2H2O = number of moles x molar mass = 0.0250 mol x 244.3 g mol–1

= 6.11 g (1)

∴ 6.11 g of barium chloride crystals can be obtained.

133

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)

177 a)

(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 = massmolar mass

= 60 480 000 g63.0 g mol–1

= 960 000 mol (1)

Molarity of nitric acid = number of moles of HNO3

volume of solution

= 960 000 mol80 000 dm3

= 12.0 mol dm–3 (1)

c) The 10.0 cm3 pipette should be used to deliver the concentrated nitric acid.

Suppose V cm3 of 0.240 mol dm–3 nitric acid can be obtained.


12.0 x 10.01 000

= 0.240 x V

1 000 V = 500.0

Volume of dilute nitric acid obtained = 500.0 cm3

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)



d) i) pH of acid = –log10 (0.100) (1)

= 1 (1)

134

ii)

(0.5 mark for curve starting from pH 1; 0.5 mark for vertical part of curve between pH 2 – 6; 0.5 mark for showing equivalence point at 25 cm3; 0.5 mark for curve finishing between pH 9 – 10, extending to 40 cm3) (2)

iii) Chloro-phenol red (1)

because the indicator changes colour within the pH range of the vertical part of the titration curve. (1)

178 a) 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 (1)

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 = massmolar mass

= 1 080 g98.0 g mol–1

= 11.0 mol (1)

Molarity of phosphoric acid = number of moles of H3PO4

volume of solution

= 11.0 mol1.00 dm3

= 11.0 mol dm–3 (1)

b) Wash the hand with plenty of water. (1)

135

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 = 20.0 + 20.1 + 20.03

cm3

= 20.0 cm3 (1)

ii) Number of moles of NaOH required = molarity of solution x volume of solution

= 1.10 mol dm–3 x 20.01 000

dm3

= 0.0220 mol (1)

iii) Number of moles of NaOH required = 0.0220 mol x 250.0 cm3

25.0 cm3

= 0.220 mol (1)

e) Number of moles of H3PO4 in 10.0 cm3 solution = molarity of solution x volume of solution

= 11.0 mol dm–3 x 10.01 000

dm3

= 0.110 mol (1)

Number of moles of NaOHNumber of moles of H3PO4

= 0.220 mol0.110 mol

= 2 (1)

∴ equation 2 best describes the reaction occurring in the titration. (1)

179 a) Step 1 — Should not prepare the standard sodium hydroxide solution using the method described. (1)

This is because sodium hydroxide is deliquescent and would absorb moisture from the air. (1)

Step 2 — Should not use a measuring cylinder to transfer the lemon juice. (1)

This is because a measuring cylinder cannot give accurate measurements of liquid volumes. (1)

Step 3 — Should not rinse the burette with distilled water only. (1)

This is because water that remains in the burette would dilute the sodium hydroxide solution. (1)

Step 5 — Should not perform the calculation using only one titration result. (1)

This is because errors may occur in the titration. (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)

136

180 a) 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 (1)

c) Ba(OH)2(aq) + H2SO4(aq) BaSO4(aq) + 2H2O(l) ? mol dm–3 1.00 mol dm–3

100.0 cm3 15.0 cm3


= 1.00 mol dm–3 x 15.01 000

dm3

= 0.0150 mol (1)

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 (1)

Molarity of barium hydroxide solution = number of moles of Ba(OH)2

volume of solution

= 0.0150 mol

( 100.01 000 ) dm3

= 0.150 mol dm–3 (1)

∴ 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)

181 a) Aluminium hydroxide in the tablet neutralizes the excess hydrochloric acid in the stomach. (1)

Al(OH)3(s) + 3HCl(aq) AlCl3(aq) + 3H2O(l) (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 (2)

ii) NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l) 0.190 mol dm–3

18.1 cm3


= 0.190 mol dm–3 x 18.11 000

dm3

= 0.00344 mol (1)


i.e. number of moles of excess HCl in 25.0 cm3 diluted solution = 0.00344 mol (1)

137

iii) Number of moles of HCl added in Step 1 = molarity of solution x volume of solution

= 1.00 mol dm–3 x 50.01 000

dm3

= 0.0500 mol (1)

iv) Number of moles of HCl left over after reaction with drug tablet in Step 1

= 0.00344 mol x 250.0 cm3

25.0 cm3

= 0.0344 mol

Number of moles of HCl reacted with Al(OH)3 in drug tablet = (0.0500 – 0.0344) mol = 0.0156 mol (1)

Al(OH)3(s) + 3HCl(aq) AlCl3(aq) + 3H2O(l)

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 = 0.0156

3 mol

= 0.00520 mol

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 (1)

∴ the drug tablet contains 0.406 g of aluminium hydroxide.

v) Liquid used for washing immediately before use

(1) 250.0 cm3 volumetric flask used in Step 2 distilled water (0.5)

(2) 25.0 cm3 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)

(4) Burette for containing the sodium hydroxide solution

sodium hydroxide solution (0.5)

182 a) Transfer the resulting 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)



b) i) NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l) 0.125 mol dm–3

20.8 cm3

138


= 0.125 mol dm–3 x 20.81 000

dm3

= 0.00260 mol (1)


i.e. number of moles of excess HCl in 25.0 cm3 diluted solution = 0.00260 mol (1)

ii) Number of moles of HCl added in Step 1 = molarity of solution x volume of solution

= 1.00 mol dm–3 x 100.01 000

dm3

= 0.100 mol (1)

iii) Number of moles of HCl left over after reaction with magnesium in Step 1

= 0.00260 mol x 250.0 cm3

25.0 cm3

= 0.0260 mol

Number of moles of HCl reacted with magnesium in Step 1 = (0.100 – 0.0260) mol = 0.074 mol (1)

Mg(aq) + 2HCl(aq) MgCl2(aq) + H2(g)

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 = 0.0742

mol

= mass of Mgmolar mass of Mg

∴ molar mass of Mg = mass of Mg0.074

2 mol

= 0.900 g0.074

2 mol

= 24 g mol–1 (1)

∴ the relative atomic mass of Mg is 24.

c) Neutralization is a quick process. (1)

As titration proceeds, the concentration of the acid decreases and hence the rate of reaction between the acid and sodium hydroxide decreases. (1)

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)

139

183 a) Nickel / Ni (1)

b) Ni2+(aq) + CO32–(aq) NiCO3(s) (1)

c) i) Number of moles of K2CO3 in 20.0 cm3 solution = molarity of solution x volume of solution

= 0.250 mol dm–3 x 20.0

1 000 dm3

= 0.00500 mol (1)

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 (1)

Volume of NiSO4 solution = number of moles of NiSO4


= 0.00500 mol0.100 mol dm–3

= 0.0500 dm3

= 50.0 cm3 (1)

∴ 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. (1)

Heat the potassium sulphate solution to evaporate about half of the water. Set the concentrated solution aside to cool and crystallize. (1)



184 a) Volume of acid (1)

Concentration of acid (1)

Temperature (1)

b) The hydrochloric acid was consumed as the reaction proceeded. (1)

Hence the concentration of the acid in the reaction mixture dropped during the course of the reaction. (1)

c) i) D (0.5)

ii) A (0.5)

d)

Acid usedRate of reaction

lower same higher

100 cm3 of 2 mol dm–3 HCl(aq) ✔

100 cm3 of 0.5 mol dm–3 HCl(aq) ✔

200 cm3 of 1 mol dm–3 HCl(aq) ✔

200 cm3 of 0.5 mol dm–3 HCl(aq) ✔

140

185 a)

(1 mark for correctly drawn curve of best fit; 2 marks for correctly plotted points) (3)

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) = mass of NO(g) formedtime

= 6.60 g(6.0 – 2.0) min

(1)

= 1.65 g min–1 (1)

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 = mass of NOmolar mass of NO

= 8.10 g30.0 g mol–1

= 0.270 mol (1)

According to the equation, 3 moles of Cu reacts with HNO3(aq) to give 2 moles of NO(g).

141

∴ number of moles of Cu consumed = 32

x 0.270 mol

= 0.405 mol (1)

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 (1)

d) Instantaneous rate of formation of NO(g) = – (185.0 – 191.6) g(7.5 – 3.0) min

(1)

= 1.47 g min–1 (1)

e) i) Increase (1)

ii) Decrease (1)

iii) Increase (1)

186 a) 2H2O2(aq) 2H2O(l) + O2(g) (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) (168.235 – 168.143) g4.0 min

(1)

= 0.023 g min–1 (1)

d) i) (168.235 – 168.123) g = 0.112 g (1)

ii) Number of moles of O2 given off = mass of O2

molar mass of O2

= 0.112 g32.0 g mol–1

= 0.00350 mol (1)

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 (1)

Original molarity of hydrogen peroxide solution = number of moles of H2O2

volume of solution

= 0.00700 mol

( 50.01 000 ) dm3

= 0.140 mol dm–3 (1)

142

e) i) As a catalyst. / To increase the rate of decomposition. (1)

ii) The shape of the curve would be the same. (1)

The rate of decomposition is independent of the amount of catalyst added. (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. (1)

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)

187 a) Colorimeter (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)

(1)

d) Draw a tangent to the curve at time = 0. (1)

Determine the slope of the tangent. (1)

e) i) Rate = – 12

d[MnO4–(aq)]

dt = –

15

d[C2O4

2–(aq)]dt

∴ instantaneous rate of consumption of C2O42–(aq) ions

= 52

(instantaneous rate of consumption of MnO4–(aq) ions) (1)

ii) Instantaneous rate of consumption of C2O42–(aq) ions =

52

(3.50 x 10–3 mol dm–3 s–1)

= 8.75 x 10–3 mol dm–3 s–1 (1)

143

188 a) 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. (1)

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

∝ 1time to reach the opaque stage (t)

(1)

c) i)


ii) 64 – 68 s (1)

d) Change the concentration of the reactants. (1)

189 a)

XXX.XX

(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)

144

• 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) CaCl2(aq) + CO2(g) + H2O(l) ? g 3.70 g

Number of moles of CO2 = mass of CO2

molar mass of CO2

= 3.70 g44.0 g mol–1

= 0.0841 mol (1)

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 (1)

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 = 8.42 g10.0 g

x 100%

= 84.2% (1)

190 a) A gas was evolved in the reaction. (1)

b)

(1 mark for correct set-up; 3 marks for 6 correct labels; 0 mark if the set-up is not workable) (4)

c) Number of moles of Mg present = mass of Mgmolar mass of Mg

= 2.00 g24.3 g mol–1

= 0.0823 mol (1)

Number of moles of H2SO4 present = molarity of solution x volume of solution

= 1.00 mol dm–3 x 100.01 000

dm3

= 0.100 mol (1)


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)

145

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. (1)

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. (1)

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)

Hence the initial rate of the reaction between magnesium and hydrochloric acid is lower than that between magnesium and sulphuric acid. (1)

ii)

(1 mark for showing a lower rate of reaction, i.e. tangent to curve less steep; 1 mark for showing a

smaller final pressure, around 58

of the original value) (2)

191 a) 2H2O2(aq) 2H2O(l) + O2(g) (1)

b) As a catalyst. / Increase the rate of decomposition. (1)

c)


146

d) 60 – 62 cm3 (1)

e) All hydrogen peroxide has decomposed. (1)

f)

(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. (1)

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)

192 a) i) By the reaction between sulphuric acid and aqueous ammonia. (1)

ii) H2SO4(aq) + 2NH3(aq) (NH4)2SO4(aq) (1)

iii) So the fertilizer can be taken in by the roots of plants. (1)

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)

ii) Diluting concentrated sulphuric acid / any specific neutralization reaction (1)

d) A catalyst is a substance which alters the rate of a reaction (1)

without itself undergoing any permanent chemical changes. (1)

147

e) i)

(1)

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) CaSO3(s) (1)

ii) Neutralization reaction (1)

g) i) Attack the respiratory system / lungs. (1)

ii) Move the student to a well ventilated area / open space. (1)

193 a) i) [Al(OH)4]– (1)

ii) 2Al(s) + 2OH–(aq) + 6H2O(l) 2[Al(OH)4]–(aq) + 3H2(g) (1)

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. (1)

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)

(1)

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)

ii) Manufacture of bleach / manufacture of soaps and detergents (1)

148

194 a) 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)

Dip the glass into hydrofluoric acid to attack the unprotected glass surface. (1)

c) i) An acid that can produce one hydrogen ion per molecule. (1)

ii) 0.1 mol dm–3 sulphuric acid < 0.1 mol dm–3 hydrochloric acid < 0.1 mol dm–3 hydrofluoric acid (1)

Both sulphuric acid and hydrochloric acid are strong acids. They dissociate completely in water. (1)

Sulphuric acid is dibasic while hydrochloric acid is monobasic. (1)

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. (1)

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)

ii) The materials for making glass are readily available / abundant in the Earth’s crust. (1)

e) i) Hydrated iron(III) oxide (1)

ii) Fe2O3(s) + 6HF(aq) 2FeF3(aq) + 3H2O(l) (1)

195 Any three of the following:

• Add barium chloride solution / barium nitrate solution to the acids. (1)

1 mol dm–3 H2SO4(aq) gives a white precipitate while 1 mol dm–3 HCl(aq) gives no precipitate. (1)

• Add dilute nitric acid followed by silver nitrate solution to the acids. (1)

1 mol dm–3 HCl(aq) gives a white precipitate while 1 mol dm–3 H2SO4(aq) gives no precipitate. (1)

• Add calcium carbonate to excess acids. (1)

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)

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)

(3 marks for organization and presentation)

149

196 Titrate dilute sodium hydroxide solution with dilute hydrochloric acid until the end point is reached. (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)




197 Carry 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. (1)


150

198 Add 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)


199 During 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)


topic acids and bases - stmgssintranet.stmgss.edu.hk/~ted/ccy/answers/cs_suppex... · topic 4 acids...

Documents