3. Solution Guide to Supplementary Exercises

�

Topic 4 Acids and Bases

Part A Unit-basedexercise

Unit 14 Acids and alkalis

Fillintheblanks

� hydrochloric

2 Sulphuric

3 Ethanoic

4 sour

5 red; yellow

6 colourless; red

7 bases

8 dissociate; dissociation

9 hydrogen

�0 mobile ions

�� basicity

�2 monobasic

�3 dibasic

�4 a) salt; hydrogen

b) salt; water; carbondioxide

c) salt; water; carbondioxide

d) salt; water

e) salt; water

�5 salt;water

�6 alkali

�7 bitter

�8 ammonia

�9 precipitates

20 complex

2

2� hygroscopic

22 deliquescent

Trueorfalse

23 F Concentrated sulphuric acid shouldbedilutedby adding the acid towater slowlywhile stirring.

24 T

25 F Solid citric acid has no effect on zinc.

The aqueous solutionof citric acid shows typical properties of an acidwhile solid citric acid does not.

26 F An acid is a hydrogen-containing substance that gives hydrogen ions as the only type of positive ionswhendissolved inwater.

27 T The aqueous solutionof citric acid can conduct electricity.

28 F The basicity of an acid is the maximum number of hydrogen ions produced by one acid molecule. Forexample, ethanoic acid (CH3COOH) ismonobasic becauseonly thehydrogen atom in the –COOH groupcanundergodissociation.

29 T

30 F Phosphoric acid (H3PO4(aq)) is a tribasic acid.

3� T Concentrated sulphuric acid is hygroscopic.

32 T When solid citric acid dissolves inwater, only a fewmolecules dissociate to give ions.

citric acid(s) +water citric acid(aq)

citric acid(aq) H+(aq)+ citrate ion(aq)

Hence an aqueous solutionof citric acid containsmobile citric acidmolecules.

33 T

34 F Phenolphthalein is colourless in acidic solutions.

35 F Manymetal hydroxides are INSOLUBLE inwater.

36 T

3

37 F The discolorations and tight scale buildup that occur in toilet bowls are mostly calcium carbonatedeposits 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 containhydrochloric acid.

38 T Sodium hydroxide absorbs carbon dioxide when exposed to air. Sodium hydrogencarbonate is formedand it undergoes dehydration to give sodium carbonate.

39 F Theproperties of solutions of alkalis dependon thepresenceofmobile hydroxide ions.

40 T Ca2+(aq) +2OH–(aq) Ca(OH)2(s) white precipitate

4� F Iron(II) hydroxidedoes not dissolve in excess dilute aqueous ammonia.

42 F Copper(II) hydroxide does not dissolve in excess dilute sodium hydroxide solution. It dissolves in excessdilute aqueous ammonia to give a deepblue solution.

43 T Fe3+(aq) +3OH–(aq) Fe(OH)3(s) reddishbrownprecipitate

44 F Concentratednitric acid tends to decompose to nitrogendioxidegas andoxygengas.

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 withammonia gas.

H2SO4(l) + 2NH3(g) (NH4)2SO4(s)

Multiplechoicequestions

46 A

47 C Option A — Caustic soda is sodiumhydroxide.

Opiton B — Drain cleansers usually contain sodiumhydroxide.

Option D — Slaked lime is calciumhydroxide.

48 C Option A — Dilute acids reactwith carbonates to give carbondioxidegas.

Option B — Dilute acids have a sour taste.

Option C — Dilute acids conduct electricity due to thepresenceofmobile ions.

Option D — Dilute acids reactwith 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 showNO reactionwith copper.

4

50 B Option C — Hydrochloric acid can turnmethyl orange red.

Option D — Concentratedhydrochloric acid isNOT stored in brownbottles.

5� B Option A — Sulphuric acid is used tomake fertilizers, but it isNOTused as fertilizers.

Option B — Concentrated sulphuric acid is hygroscopic.

Option C — Concentrated sulphuric acid should be diluted by adding the acid to water slowly whilestirring.

Option D — Dilute sulphuric acid showsNOoxidizingproperty.

52 D Option A — Nitric acid isNOT a drying agent.

Option B — Sulphuric acid,NOTnitric acid, is used in car batteries.

Option C — Dilute nitric acid is an oxidizing agent. It reacts with copper to give nitrogen monoxide,NOThydrogen.

Option D — Concentratednitric acid is corrosive.

53 C Aqueous solutionof ethanoic acid can conduct electricity.

54 C Whendilute sulphuric acid reactswith calcium carbonate, insoluble calcium sulphate forms. The calciumsulphate covers the surfaceof calcium carbonate andprevents further reaction.

55 A Option A — Vinegar contains ethanoic acid. It showsNO reactionwith copper.

Option B — Dilute nitric acid is anoxidizing agent. It reacts with copper to give nitrogenmonoxide.

Option C — Copper(II) oxide anddilute nitric acid undergoneutralization.

Option D — Copper(II) hydroxide and vinegar undergoneutralization.

56 A

57 D

58 C Solid citric acid doesNOT reactwithmagnesiumbecause it does not contain hydrogen ions.

The aqueous solutionof citric acid shows theproperties of an acidwhile solid citric acid does not.

59 C Phosphoric acid (H3PO4) is a tribasic acid.

60 A Carbon is a reducing agent. It can reduce FeO(s) to Fe(s).

6� 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 chloridesolution.

Option C — Sodiumhydroxide absorbs carbondioxidewhen exposed to air. Sodiumhydrogencarbonateis 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 andhydroxide ions. However, ammonia does not react with water completely. Only very fewhydroxide ions are formed.

NH3(g) +H2O(l) NH4+(aq)+OH–(aq)

Hence aqueous ammonia contains both ammoniamolecules andhydroxide ions.

Option B — Metals tarnish because they react with the air to form a layer of oxide or sulphide. Acidscan 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 solutionwithmethyl orange.

65 C Dilute sodium hydroxide solution does NOT give a white precipitate with potassium chloride solution.Themixture is a colourless solution.

66 B Aluminium hydroxide dissolves in excess dilute sodium hydroxide solution due to the formation of asoluble complex salt.

Al(OH)3(s) +OH–(aq) [Al(OH)4]–(aq)

aluminate ion

colourless solution

67 B

68 DOption Solution Observation

A ammonium chloride noprecipitate

B iron(II) sulphate greenprecipitate Fe(OH)2

C magnesium chloride white precipitateMg(OH)2

D nickel(II) sulphate greenprecipitateNi(OH)2

69 D Thewhite precipitate ismagnesiumhydroxide.

70 DOption Solution Observation

A calcium nitrate noprecipitate

B copper(II) sulphate pale blueprecipitate

C lead(II) nitrate white precipitate

D zinc sulphate white precipitate

6

7� COption Hydroxide Solubility in NaOH(aq) / NH3(aq)

A Al(OH)3 soluble in excess NaOH(aq)

B Cu(OH)2 soluble in excess NH3(aq)

C Fe(OH)2 insoluble in bothNaOH(aq) andNH3(aq)

D Zn(OH)2 soluble in bothNaOH(aq) andNH3(aq)

72 COption Compound X Addition of NH3(aq) to dilute solution of X

A Fe2(SO4)3 reddishbrownprecipitate Fe(OH)3

B MgCl2 white precipitateMg(OH)2

C NiSO4 greenprecipitateNi(OH)2

D NH4Cl noprecipitate

73 A Concentrated sodiumhydroxide solution is corrosive.

74 B Concentratednitric acid is anoxidizing agent.

75 C

76 A Option A — Concentrated hydrochloric acid and concentrated nitric acid are volatile while concentratedsulphuric acid isNOT.

Option B — Concentrated sulphuric acid isNOT stored in brownbottles.

Option D — Concentrated sulphuric acid is NOT used to dry ammonia gas because the acid wouldreactwith 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 acarbonate, rather than a sulphate.

Solution of X does NOT give a precipitate with dilute sodium hydroxide solution. Hence X probablycontains ammonium ions, rather than iron(II) ions.

\ X is ammonium carbonate.

78 A From test �, it canbededuced thatX is a chloride.

Ag+(aq)+Cl–(aq) AgCl(s)

From test 2, it canbededuced thatX contains iron(II) ions, rather than iron(III) ions.

Fe2+(aq)+2OH–(aq) Fe(OH)2(s) greenprecipitate

Fe3+(aq)+3OH–(aq) Fe(OH)3(s) reddishbrownprecipitate

7

79 BOption Substances Upon mixing the substances

A magnesium and ethanoic acid hydrogengas is produced

B magnesium hydroxide anddilute hydrochloric acid only a salt andwater are produced

C magnesium carbonate anddilute sulphuric acid carbondioxidegas is produced

D magnesium and steam hydrogengas is produced

\ magnesiumhydroxide anddilute hydrochloric acidwouldNOTproduce a gaswhenmixed.

80 C Silver nitrate solutiongives awhite precipitatewith dilute hydrochloric acid.

Ag+(aq) +Cl–(aq) AgCl(s)

8� D NaOH(aq) gives a precipitate Fe2+(aq) ions, butNOTK+(aq) ions.

82 C Zinc canbeobtainedby heating zinc oxidewith carbon.Carbondioxidegas is formed in the process.

zinc oxide+ carbon zinc+ carbondioxide

83 D The aqueous solution of citric acid shows typical properties of an acid while the citric acid crystals donot.

When citric acid crystals (jar 3) are mixed with water (jar 5), the aqueous solution would give carbondioxidegaswith solid hydrogencarbonate (jar �).

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 slightlyacidic.

CO2(g) +H2O(l) H2CO3(aq) carbonic acid

85 D

86 A (�) Acids are covalent compounds when pure. For example, pure sulphuric acid and nitric acid arecolourless liquids consistingofmolecules.

(2) An acidmayNOT contain bothhydrogen andoxygen, e.g.HCl(aq).

(3) Typical acidswouldNOT reactwith copper.

87 B (2) Sulphuric acid is used tomake fertilizers, but it isNOT as fertilizers.

88 B (2) Sulphuric acid,NOThydrochloric acid, is used in car batteries.

89 D (3) When solid citric acid dissolves inwater, only a fewmolecules dissociate to give ions.



8


90 B (2) Vinegar contains ethanoic acid. It showsNO reactionwith copper.

9� D (2) Carbonic acid is formedwhen carbondioxidedissolves inwater.

CO2(g) +H2O(l) H2CO3(aq) carbonic acid

(3) Sulphurous acid is formedwhen sulphur dioxidedissolves inwater.

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

92 D (�) Magnesium reactswith dilute hydrochloric acid to give hydrogengas. (2) and (3) Sodium carbonate and potassium hydrogencarbonate react with dilute hydrochloric acid to

give carbondioxide gas.

93 B (2) Very dilute nitric acid acts like a typical acid. It showsNO reactionwith copper.

94 D All the threemetals (calcium, iron andmagnesium) react with acids to give hydrogen.

95 D (�) Heating ammonium sulphate solutionwith dilute sodiumhydroxide solutiongives ammonia gas.

NH4+(aq)+OH–(aq) NH3(g) +H2O(l)

(2) Iron(II) sulphate solutiongives a greenprecipitatewith dilute sodiumhydroxide solution.

Fe2+(aq)+2OH–(aq) Fe(OH)2(s)

(3) Citric acidwouldundergoneutralization reactionwith dilute sodiumhydroxide solution.

96 D (�) When ammonia gas dissolves in water, it reacts with water to give ammonium ions and hydroxideions. However, ammonia does not react with water completely. Only very few hydroxide ions areformed.


Hence aqueous ammonia contains both ammoniamolecules and ammonium ions.

9

97 B (�) Cu(OH)2 dissolves in excess dilute aqueous ammonia to give a deepblue 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 (�) Heating ammonium chloridewith 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 sodiumhydroxide solution to give a colourless solution.

Zn(OH)2(s) + 2OH–(aq) [Zn(OH)4]2–(aq)

�00A (2) Heating solutions of ammonium chloride and sodiumhydroxide liberates ammonia.

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

�0�A (�) Magnesium chloride solutiongives awhite precipitatewith silver nitrate solution.


(2) Magnesium chloride solutiongives awhite precipitatewith dilute aqueous ammonia.

Mg2+(aq)+2OH–(aq) Mg(OH)2(s)

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

�02B (�) Zinc ismore reactive than silver, but less reactive than sodium.

Zinc granules candisplace silver from silver nitrate solution.

There isNO reactionbetween zinc granules and sodiumnitrate solution.

(2) Both sodiumnitrate solution and silver nitrate solution are colourless.

(3) Only silver nitrate solutiongives awhite precipitatewith dilute hydrochloric acid.


�03A (�) NH3(aq) andNiSO4(aq)would formagreenprecipitatewhenmixed.

Ni2+(aq)+2OH–(aq) Ni(OH)2(s)

(2) NaOH(aq) and FeSO4(aq)would formagreenprecipitatewhenmixed.


(3) There is noobservable changewhenNH3(aq) andNaCl(aq) aremixed.

�0

�04D (�) Ammoniagaswould reactwith dilute sulphuric acid.

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

(2) Carbondioxidegaswouldgive awhite precipitatewith calcium hydroxide solution.

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

(3) Sulphur dioxide gaswould reactwith sodiumhydroxide solution.

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

�05B

�06C Aqueous solutionof ethanoic acid can conduct electricity. Hence ethanoic acid is an electrolyte.

Ethanoic acid is a covalent compoundwhenpure. Pure ethanoic acid consists ofmolecules.

�07D Whendilute sulphuric acid reactswith calcium carbonate, insoluble calcium sulphate forms. The calciumsulphate covers the surfaceof calcium carbonate andprevents further reaction.

�08B Citric acid shows its acidic properties in thepresenceofwater.

When solid citric acid dissolves inwater, only a fewmolecules dissociate to give ions.




�09D

��0D

��C Dilute nitric acid is anoxidizing agent. It reacts with copper to give nitrogenmonoxide,NOThydrogen.

��2A Concentratednitric acid decomposes according to the following chemical equation:

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

��3D Manymetal hydroxides are insoluble inwater.

��4B When ammonia gas dissolves inwater, it reacts withwater to give ammonium ions andhydroxide ions.


Dilute aqueous ammonia conducts electricity due to thepresenceofmobile ions.

��5C Anhydrous calcium chloride is NOT used to dry ammonia gas because calcium chloride would reactwith ammonia gas.

CaCl2(s) + 4NH3(g) CaCl2•4NH3(s)

��

Unit 15 Molarity, pH scale and strengths of acids and alkalis

Fillintheblanks

� 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 acidmolecules

9 ammonium ions; hydroxide ions; ammoniamolecules

�0 concentrated; strong

Trueorfalse

�� F ThepHof a solutiondecreases as the concentrationof hydrogen ions in the solution increases.

�2 T

�3 F ThepH valueof dilute aqueous ammonia is greater than7, but less than�4.

�4 T

�5 F Aqueous solutionof ethanoic acid can conduct electricity. Hence ethanoic acid is an electrolyte.

�6 T When citric acid crystals dissolve in water, only a few molecules dissociate to give ions. Hence citricacid is aweak acid.

�7 F When we describe acids as strong and weak, we are talking about the extent of their dissocation inwater. When we talk about concentration, we are referring to the amount of an acid in a unit volumeof solution.

Forexample,5moldm–3nitric acid is a concentrated solutionofa strongacidwhile0.�moldm–3nitricacid is a dilute solutionof a strong acid.

�8 T

�9 F � mol dm–3 HCl(aq) has a higher concentration of hydrogen ions than � mol dm–3 CH3COOH(aq) does.Hence thepHofHCl(aq) is lower than that ofCH3COOH(aq).

�2

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

NH3(aq) +HCl(aq) NH4Cl(aq)

According to the equations, �mole ofNaOH /NH3 requires �mole ofHCl for complete neutralization.

Hence�0cm3of�moldm–3NaOH(aq) and�0cm3of�moldm–3NH3(aq) require the samenumberofmoles ofHCl for complete neutralization.


2� B Number ofmoles ofKNO3 =molarity ofKNO3 solution x volumeof solution

= 0.25mol dm–3 x40

�000dm3

= �.0 x �0–2mol

22 D Molarmass ofNa2CO3= [2 x 23.0+�2.0+3 x �6.0) gmol–�

= �06.0gmol–�

Number ofmoles ofNa2CO3=mass

molar mass

=�3.8g

�06.0gmol–�

= 0.�30mol

Volume of solution=500

�000dm3

Molarity ofNa2CO3 solution=number ofmoles ofNa2CO3

volumeof solution

= 0.�30mol

( 500�000 ) dm3

= 0.260mol dm–3

\ themolarity of the sodium carbonate solution is 0.260mol dm–3.

23 D Molarmass of (COOH)2•2H2O = [2 x (�2.0+2 x �6.0+�.0) + 2 x (2 x �.0+�6.0)] gmol–�

= �26.0gmol–�

Number ofmoles of (COOH)2•H2Opresent=mass

molarmass

=3.�5g

�26.0gmol–�

= 0.0250mol

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

volumeof solution

= 0.0250mol

( 250.0�000 ) dm3

= 0.�00mol dm–3

�3

24 B Molarmass ofCa(OH)2= [40.�+2 x (�6.0+�.0)] gmol–�

= 74.�gmol–�

Molarity of calciumhydroxide solution=number ofmoles ofCa(OH)2

volumeof solution

0.�50mol dm–3=number ofmoles ofCa(OH)2

( 250.0�000 ) dm3

Number ofmoles ofCa(OH)2 =0.�50mol dm–3 x250.0

�000dm3

= 0.0375mol

Mass ofCa(OH)2=number ofmoles ofCa(OH)2 xmolarmass ofCa(OH)2

= 0.0375mol x 74.�gmol–�

= 2.78g

\ 2.78gof calciumhydroxide are present.

25 D Molarmass of (NH4)2SO4= [2 x (�4.0+4 x �.0) + 32.�+4 x �6.0)] gmol–�

= �32.�gmol–�

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

volumeof solution

�.02mol dm–3 =number ofmoles of (NH4)2SO4

�.50dm3

Number ofmoles of (NH4)2SO4 =�.02mol dm–3 x �.50dm3

= �.53mol

Mass of (NH4)2SO4=number ofmoles of (NH4)2SO4 xmolarmass of (NH4)2SO4

= �.53mol x �32.�gmol–�

= 202g

\ 202gof ammonium sulphate are present.

26 C Molarmass ofKCl= (39.0+35.5) gmol–�

= 74.5gmol–�

Number ofmoles ofKCl present=mass

molarmass

=44.7g

74.5gmol–�

= 0.600mol

Molarity of potassium chloride solution=number ofmoles ofKCl

volumeof solution

2.40mol dm–3 =0.600mol

volumeof solution

Volume of solution =0.600mol

2.40mol dm–3

= 0.250dm3

= 250 cm3

�4

27 B Number ofmoles ofNaOH =molarity of solution x volumeof solution

Option Number of moles of NaOH present

A 3.0mol dm–3 x�00

�000dm3=0.30mol

B 2.5mol dm–3 x200

�000dm3=0.50mol

C �.5mol dm–3 x300

�000dm3=0.45mol

D �.0mol dm–3 x400

�000dm3=0.40mol

\ 200 cm3of 2.5mol dm–3NaOH(aq) contain thegreatest number ofmoles of solute.

28 DOption Number of moles of MgSO4 = molarity of solution x volume of solution

A 0.40mol dm–3 x �.0 dm3=0.40mol

B 0.30mol dm–3 x �.5 dm3=0.45mol

C 0.20mol dm–3 x 2.0 dm3=0.40mol

D 0.�0mol dm–3 x 2.5 dm3=0.25mol

\ 2.5 dm3of 0.�0mol dm–3MgSO4(aq) contain the smallest number ofmoles of solute.

29 D Number ofmoles ofCaCl2 in 250.0 cm3of 0.300mol dm–3 solution

=molarity of solution x volumeof solution

=0.300 mol dm–3 x250.0

�000dm3

=0.0750 mol

Number ofmoles ofCaCl2 in �50.0 cm3of 0.�80mol dm–3 solution


=0.�80 mol dm–3 x�50.0

�000dm3

=0.0270 mol

Total number ofmoles ofCaCl2 in the resulting solution = (0.0750 +0.0270)mol = 0.�02 mol

Total volumeof the resulting solution = (250.0+�50.0) cm3

= 400.0 cm3

Concentrationof the resulting solution= 0.�02mol

( 400.0�000 ) dm3

= 0.255mol dm–3

\ the concentrationof the resulting solution is 2.55 x �0–�mol dm–3.

�5

30 B Number ofmoles ofNaOH in �00.0 cm3of 3.00mol dm–3 solution


=3.00 mol dm–3 x�00.0

�000dm3

=0.300 mol

Number ofmoles ofNaOH in 50.0 cm3of �.20mol dm–3 solution


=�.20 mol dm–3 x50.0

�000dm3

=0.0600 mol

Total number ofmoles ofNaOH in the resulting solution = (0.300 +0.0600)mol = 0.360 mol

Total volumeof the resulting solution = (�00.0+50.0) cm3

= �50.0 cm3

Concentrationof the resulting solution= 0.360mol

( �50.0�000 ) dm3

= 2.40mol dm–3

\ the concentrationof the resulting solution is 2.40mol dm–3.

3� A pHof citric acid solution = –log�0[H+]

= �.85

i.e. log�0[H+] = –�.85

[H+] = �0–�.85

= 0.0�4�mol dm–3

\ concentrationof hydrogen ions in the citric acid solution is �.4� x �0–2mol dm–3.

32 B pHofmilk= –log�0[H+]

= 6.70

i.e. log�0[H+] = –6.70

[H+] = �0–6.70

= 2.00 x �0–7mol dm–3

\ concentrationof hydrogen ions in the sample ofmilk is 2.00 x �0–7mol dm–3.

�6

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, �mole ofH2SO4 dissociates to give 2moles of hydrogen ions.

i.e.concentrationof hydrogen ions=2 x 0.0200mol dm–3

= 0.0400mol dm–3

pHof acid = –log�0(0.0400) = –(–�.40) = �.40

\ pHof the sulphuric acid sample is �.40.

34 A pHof sample of urine= –log�0(7.94 x �0–7) = –(–6.�0) = 6.�0

\ pHof the sample of urine is 6.�0.

35 C pHof sewagebefore treatment=6.75

pHof sewage after treatment=7.00

Concentrationof hydrogen ions in sewagebefore treatment =�0–6.75

= �.78 x �0–7mol dm–3

Concentrationof hydrogen ions in sewage after treatment =�0–7.00mol dm–3

Change in concentrationof hydrogen ions = (�.78 x �0–7 – �0–7)mol dm–3

= 0.78 x �0–7mol dm–3

\ the change in the concentrationof hydrogen ions in the sewage is 7.80 x �0–8mol dm–3.

36 B Limewater is saturated calciumhydroxide solution.

37 D Soft drinks contain carbonic acid.

38 C Option B — Glass cleanser is alkaline. Its pH is greater than7.

Option D — ThepHof seawater is about 8.

39 D Oven cleanser contains sodiumhydroxide.

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 andhydroxide ions.

�7

4� C In general, non-metals react with oxygen to form acidic oxides and metals react with oxygen to formbasic 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 hydroxidesolution.

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

Hence thepHof the solutionof potassiumoxide is higher than that of calciumoxide.

Options B andD—Oxides of carbon and sulphur are acidic. The pH values of the solutions of bothoxides are less than7.

42 D Concentrationof hydrogen ions in solutionX =�0–�

= 0.�mol dm–3

Concentrationof hydrogen ions in solutionY =�0–2

= 0.0�mol dm–3

\ the concentrationof hydrogen ions in solutionX is ten times that in solutionY.

43 D (�) Baking soda solution is hydrogencarbonate solution. Its pH is ~9.

(2) ThepHof lemon juice is ~2.

(3) ThepHofmilk is ~6.

44 D (�) and (2) ThepHof 0.0�0mol dm–3HCl(aq) is higher than that of 0.�00mol dm–3HCl(aq).

(3) and (4) ThepHof 0.�00mol dm–3NaOH(aq) is higher than that of 0.0�0mol dm–3NaOH(aq).

45 A OptionsA,C andD— Concentrated nitric acid, lemon juice and vinegar are acidic. There was nocolour changewhen theywere added to the red solution.

46 D Option A — The concentrationof hydrogen ions in lemon juice is higher than that inwine.

Option B — The concentration of hydrogen ions in lemon juice is higher than that in baking sodasolution.

Option C —concentrationof hydrogen ions inwine

concentrationof hydrogen ions in drain cleanser=

�0–4mol dm–3

�0–�2mol dm–3

\ the concentrationof hydrogen ions is �08 times greater inwine than in drain cleanser.

concentrationof hydrogen ions in distilledwater

concentrationof hydrogen ions in baking soda solution=

�0–7mol dm–3

�0–9mol dm–3

\ the concentration of hydrogen ions is �00 times greater in distilled water than inbaking soda solution.

47 B When the right amounts of an acid (pH < 7) and an alkali (pH > 7) are mixed, they react completelyto produce a salt andwater only (pH=7).

�8

48 A

49 A The electrical conductivity of a solution, and hence the brightness of the bulb, is proportional to theconcentrationofmobile ions.

Option Acid Remark Concentration of ions

A25 cm3of 0.5mol dm–3H2SO4(aq)

H2SO4(aq) is a strong acid


�.5mol dm–3

B25 cm3of 0.5mol dm–3HCl(aq)

HCl(aq) is a strong acid

HCl(aq) H+(aq) +Cl–(aq)�.0mol dm–3

C25 cm3of 0.5mol dm–3CH3COOH(aq)

CH3COOH(aq) is aweak acid

CH3COOH(aq) CH3COO–(aq) +H+(aq)<�.0mol dm–3

D25 cm3of 0.5mol dm–3NH3(aq)

NH3(aq) is aweak alkali

NH3(aq) +H2O(l) NH4+(aq) +OH–(aq)

<�.0mol dm–3

\ 25 cm3 of 0.5 mol dm–3 H2SO4(aq) have the highest concentration of ions and thus will make thebulb thebrightest.

50 C OptionsA, B andD— During the reaction of hydrochloric acid and ethanoic acid with marble chips,marble chipswould react with hydrogen ions in the acids.

Hydrochloric acid is a strong acid that completely dissociates inwater.

Ethanoic acid isweak acid that only partially dissociates inwater.

Therefore � mol dm–3 hydrochloric acid has a higher concentration of hydrogenions than � mol dm–3 ethanoic acid does. The reaction rate between marblechips and ethanoic acid is thus lower and the reaction takes a longer time tocomplete.

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 � mole of CaCO3 forcomplete reaction.

�00 cm3 of � mol dm–3 hydrochloric acid and �00 cm3 of � mol dm–3 ethanoic acidrequire the samenumber ofmoles ofCaCO3 for complete reaction.

Hence themass ofmarble chips left overwouldbe the same for both acids.

5� A Carbonic acid is aweak acid that only partially dissociates inwater.

H2CO3(aq) 2H+(aq)+CO32–(aq)

The concentrationofH2CO3(aq) is thehighest in 0.�0mol dm–3 carbonic acid.

52 C Bothnitric acid andhydrochloric acid aremonobasic strong acids.

HNO3(aq) H+(aq)+NO3–(aq)

HCl(aq) H+(aq)+Cl–(aq)

Therefore the concentration of hydrogen ions in �.0 mol dm–3 nitric acid and �.0 mol dm–3 hydrochloricacid is the same.

�9

Mixing �00 cm3 of the nitric acid with �00 cm3 of the hydrochloric acid would NOT result in a changein the concentrationof hydrogen ions, i.e.wouldNOT 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 inwater. When we talk about concentration, we are referring to the amount of an acid in a unit volumeof solution.

Sulphuric acid is a strong acid.Hence0.�mol dm–3 sulphuric acid is a dilute solutionof a strong acid.

54 B ThepHofNaCl(aq) is 7. Its pH remains constant upondilution.

55 A During the reaction between magnesium and an acid, magnesium reacts with hydrogen ions in theacid.

From thediagrams, it canbededuced thatmagnesium reacts more slowlywith 0.�mol dm–3HA(aq).

The concentrationof hydrogen ions in 0.�mol dm–3HA(aq) is lower than that in 0.�mol dm–3HB(aq).

Hence acidHA is probablyweaker than acidHB.

56 A (3) Washing soda is hydrated sodium carbonate (Na2CO3•�0H2O). 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 theirdissocation in water. When we talk about concentration, we are referring to the amountof an acid in a unit volumeof solution.

For example, 5 mol dm–3 nitric acid is a concentrated solution of a strong acid while0.�mol dm–3 nitric acid is a dilute solutionof a strong acid.

58 A In general, non-metals react with oxygen to form acidic oxides and metals react with oxygen to formbasic oxides.

(�) and (2) Calcium oxide and magnesium oxide are slightly basic. They react with cold water to formhydroxideswhich are slightly soluble. Solutionswith a pH>7 are formed.

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

MgO(s) +H2O(l) Mg(OH)2(s)

59 B (�) Hydrochloric acid is a strong acidwhile ethanoic acid is aweak acid.

(2) The pH is a measure of the concentration of hydrogen ions in a solution. Therefore acid solutionsof the samepH shouldhave the same concentrationof hydrogen ions.

(3) Hydrochloric acid almost completely dissociates in water to give hydrogen ions and chloride ions.Ethanoic acid only partially dissociates inwater, forming very fewhydrogen ions.

Hence hydrochloric acid and ethanoic acid having the same pH value should have differentconcentrations.

60 C (�) Adding � mol dm–3 hydrochloric acid to the ethanoic acid makes the ethanoic acid more acidic,thus decreasing thepHof the ethanoic acid.

(2) and (3) Solid sodium carbonate and magnesium react with the ethanoic acid, thus both increasingthepHof the ethanoic acid.

20

6� A (�) SO2(g) will neutralize the dilute sodium hydroxide solution, making the solution less alkaline. ThusthepHof the solutionwill decrease.

(2) NaCl(s) is neutral. ItwillNOT lower thepHof thedilute sodiumhydroxide solution.

(3) When NH3(g) dissolves in water, it reacts with water to give ammonium ions and hydroxide ions. ItwillNOT lower thepHof thedilute sodiumhydroxide solution.

62 B (�) The electrical conductivity of a solution is proportional to the concentrationofmobile ions.

Hydrochloric acid is a strong acidwhile ethanoic acid is aweak acid.

Therefore � mol dm–3 hydrochloric acid has a higher concentration of mobile ions than ethanoicacid does.

Hence the electrical conductivity of � mol dm–3 hydrochloric acid is higher than that of � mol dm–3ethanoic acid.

(2) CH3COOH(aq)+NaOH(aq) CH3COONa(aq)+H2O(l)

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

According to the equations, � mole of CH3COOH / HCl requires � mole of NaOH for completeneutralization.

\ � mol dm–3 ethanoic acid and � mol dm–3 hydrochloric acid of the same volume require thesamenumber ofmoles ofNaOH for complete neutralization.

(3) For the neutralization between a strong acid and a strong alkali, the heat released is 57 kJ for �mole ofwater produced.

For neutralization in which either the acid or alkali or both are weak, the heat released is less than57 kJ for �moleofwater produced. The is because someenergy is consumedwhen theweak acidandweak alkali dissociate to give hydrogen ions andhydroxide ions before neutralization.

Ethanoic acid is aweak acidwhile hydrochloric acid is a strong acid.

Hence the temperature rise for the neutralization between � mol dm–3 ethanoic acid and NaOH(aq)is lower than that between�mol dm–3 hydrochloric acid andNaOH(aq).

63 C (�) NaOH(aq) is a strong alkali whileNH3(aq) is aweak alkali.

Hence�mol dm–3NaOH(aq) ismore alkaline than�mol dm–3NH3(aq).

ThepHof �mol dm–3NaOH(aq) is higher than that of �mol dm–3NH3(aq).

(2) The electrical conductivity of a solution is proportional to the concentrationofmobile ions.

NaOH(aq) is a strong alkali whileNH3(aq) is aweak alkali.

Therefore �mol dm–3NaOH(aq) has a higher concentrationofmobile ions than�mol dm–3NH3(aq)does.

Hence the electrical conductivity of � mol dm–3 NaOH(aq) is higher than that of � mol dm–3NH3(aq).

2�

(3) For the neutralization between a strong acid and a strong alkali, the heat released is 57 kJ for �mole ofwater produced.

For neutralization in which either the acid or alkali or both are weak, the heat released is less than57 kJ for �moleofwater produced. The is because someenergy is consumedwhen theweak acidand weak alkali dissociate to give hydrogen ions andhydroxide ions before neutralization.

NaOH(aq) is a strong alkali whileNH3(aq) is aweak acid.

Hence the temperature rise for the neutralization between � mol dm–3 NaOH(aq) and HCl(aq) islarger than that between�mol dm–3NH3(aq) andHCl(aq).

64 D Carbondioxide reactswithwater to form carbonic acid.

CO2(g) +H2O(l) H2CO3(aq)

Carbonic acid is aweak acid. It only partially dissociates inwater.

H2CO3(aq) 2H+(aq)+CO32–(aq)

(�) The resulting solution containsmobile ions and thus conducts electricity better thanwater.

(2) Carbonic acid undergoes neutralizationwith dilute sodiumhydroxide solution.

65 D (2) Aqueous solutionof 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 reactswithmagnesium.

When we describe acids as strong and weak, we are talking about the extent of their dissocation inwater. When we talk about concentration, we are referring to the amount of an acid in a unit volumeof solution.

For example, 5 mol dm–3 nitric acid is a concentrated solution of a strong acid while 0.� mol dm–3nitric acid is a dilute solutionof a strong acid.

68 C Adding � mol dm–3 NaCl(aq) to 50 cm3 of � mol dm–3 HCl(aq) increases the total volume of the acid.Hence the concentrationof hydrogen ions in the acid decreases.

Thus thepHof the acid is affected.

69 D Ethanoic acid is aweak acid.

The basicity of ethanoic acid is �.

22

70 B Sulphuric acid is a strong acid because it almost completely dissociates inwater.

Ethanoic acid is aweak acid because it only partly dissociates inwater.

7� B Solid citric acid doesNOT reactwithmagnesiumbecause it does not contain hydrogen ions.

The aqueous solutionof citric acid shows theproperties of an acidwhile solid citric acid does not.

72 B A solution can conduct electricity due to thepresenceofmobile ions.

When solid citric acid dissolves inwater, someof themolecules dissociate to give ions.


Hence an aqueous solutionof citric acid can conduct electricity.

73 C CH3COOH(aq)+NaOH(aq) CH3COONa(aq)+H2O(l)



\ � mol dm–3 ethanoic acid and � mol dm–3 hydrochloric acid of the same volume require the samenumber ofmoles ofNaOH for complete neutralization.

When ammonia gas dissolves in water, it reacts with water to give ammonium ions and hydroxide ions.However, ammonia does not reactwithwater completely. Only very fewhydroxide ions are formed.


Hence there aremanymobile ammoniamolecules in aqueous ammonia.

74 C Both � mol dm–3 NaOH(aq) and � mol dm–3 NH3(aq) form a reddish brown precipitate withFe2(SO4)3(aq).

75 C CH3COOH(aq)+NaOH(aq) CH3COONa(aq)+H2O(l)



Hence �0 cm3 of � mol dm–3 ethanoic acid and �0 cm3 of � mol dm–3 hydrochloric acid require thesamenumber ofmoles of sodiumhydroxide for complete neutralization.

Unit 16 Salts and neutralization

Fillintheblanks

� neutralization

2 exothermic

3 sodiumhydrogensulphate; sodium sulphate

23

4 calcium; lead(II); barium

5 lead(II); silver

Trueorfalse

6 T

7 F In neutralization, salt and water are the only products. However, potassium carbonate and dilutehydrochloric acid react to give a salt, carbon dioxide gas and water. Therefore the reaction is NOT aneutralization reaction.

8 T

9 F Heat is absorbedduring the sublimationof iodine.Hence this isNOT an exothermic process.

�0 F The reaction between dilute hydrochloric acid and dilute sodium hydroxide solution produces only onesalt, sodium chloride. Sodium chloride is a normal salt.

�� T Sulphur dioxidewould reactwith sodiumhydroxide solution.

SO2(g) + 2NaOH(aq) Na2SO3(aq)+H2O(l)

�2 T Amolecule of sulphuric acid canproduce twohydrogen ions. If dilute sulphuric acid is allowed to reactwith dilute sodium hydroxide solution, two kinds of salt can form. When one of the hydrogen ions isreplaced, the salt formed is sodium hydrogensulphate (NaHSO4). It is called an acid salt. When bothhydrogen ions are replaced, the salt formed is sodium sulphate (Na2SO4). It is called 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

�3 T A molecule of carbonic acid can produce two hydrogen ions. During the reaction between carbonicacid and dilute potassium hydroxide, when one of the hydrogen ions is replaced, the acid saltpotassiumhydrogen carbonate (KHCO3) is formed.

H+

H+

H+

H+CO3

2– CO32–

K+

K+

acid(carbonic acid)

acid salt(potassium hydrogencarbonate)

replacing

(from potassium hydroxide solution)

�4 F ThepHof a salt solutionmayNOT equal 7.

Most acid salts give an acidic solution when dissolved in water. For example, sodium hydrogensulphate(NaHSO4)solutionisacidic.Thisisbecausethehydrogensulphateion(HSO4

–)candissociatetogivehydrogenion.

�5 F Some salts are formed from the neutralization between an acid and an insoluble metal hydroxide / aninsolublemetal oxide.

Forexample,copper(II)oxideisaninsolublemetaloxide.Itreactswithdilutesulphuricacidtogivecopper(II)sulphate andwater.

H2SO4(aq)+CuO(s) CuSO4(aq)+H2O(l)

�6 F A fewacid salts give alkaline solutionswhendissolved inwater.

�7 T

�8 F Silver iodide is yellow in colour.

�9 T Lead(II) nitrate solutiongives awhite precipitatewith 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 inthe reaction.

ZnCl2(aq) +Na2CO3(aq) ZnCO3(s) + 2NaCl(aq)

2� F Whendilute sulphuric acid reactswith lead, insoluble lead(II) sulphate forms. The lead(II) sulphate coversthe surfaceof lead andprevents further reaction.

25

22 F Farmers neutralize acidic soil by adding quicklime. Ammonium sulphate gives an acidic solution whendissolved inwater.

23 T

24 F Sodium hydroxide is highly corrosive. It isNEVERused as an active ingredient in antacids.

25 F A lot of heat is released when concentrated hydrochloric acid reacts with concentrated aqueousammonia. Thiswill cause skin burn.

Furthermore, concentratedaqueousammonia is corrosive.Thehand shouldbewashed immediatelywithplenty ofwater.


26 A Iron(III) hydroxide reacts with dilute hydrochloric acid to give iron(III) chloride and water. The resultingsolution 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 solublecomplex salt.

28 B InsolubleCaCO3 is formed in the reactionbetween solutions ofNa2CO3 andCa(NO3)2.

29 D For the neutralization between a strong acid and a strong alkali, the heat released is 57 kJ for � moleofwater produced.

For neutralization in which either the acid or alkali or both are weak, the heat released is less than 57kJ for � mole of water produced. The is because some energy is consumed when the weak acid andweak alkali dissociate to give hydrogen ions andhydroxide ions before neutralization.

Option Solutions mixed Strength of acid and alkali Temperature rise

—25 cm3of �mol dm–3HCl(aq) and25 cm3of �mol dm–3KOH(aq)

neutralizationbetween a strongacid and a strong alkali

T

A25 cm3of �mol dm–3CH3COOH(aq)and25 cm3of �mol dm–3NH3(aq)

neutralizationbetween aweakacid and aweak alkali

< T

B25 cm3of �mol dm–3CH3COOH(aq)and25 cm3of �mol dm–3NaOH(aq)

neutralizationbetween aweakacid and a strong alkali

< T

C25 cm3of �mol dm–3HNO3(aq) and25 cm3of �mol dm–3NH3(aq)

neutralizationbetween a strongacid and aweak alkali

< T

D25 cm3of �mol dm–3HNO3(aq) and25 cm3of �mol dm–3NaOH(aq)


T

\ the temperature rise for mixing 25 cm3 of � mol dm–3 HNO3(aq) and 25 cm3 of � mol dm–3NaOH(aq) is the same as for mixing 25 cm3 of � mol dm–3 HCl(aq) and 25 cm3 of � mol dm–3KOH(aq).

26

30 C For the neutralization between a strong acid and a strong alkali, the heat released is 57 kJ for � moleofwater produced.

Expt. Acid and alkali mixed

Number of moles of HCl / NaOH mixed

= molarity of solution x volume of solution

Number of moles of water formedHeat

released

�

�00 cm3of �mol dm–3HCl(aq) and�00 cm3of �mol dm–3NaOH(aq)

�mol dm–3 x�00

�000dm3

=0.�mol

HCl(aq) +NaOH(aq)0.�mol 0.�mol

NaCl(aq) +H2O(l)

0.�mol

5.7 kJ

2�00 cm3of 2mol dm–3HCl(aq) and�00 cm3of2mol dm–3NaOH(aq)

2mol dm–3 x�00

�000dm3

=0.2mol

HCl(aq) +NaOH(aq)0.2mol 0.2mol

NaCl(aq) +H2O(l)

0.2mol

��.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 secondmixture, i.e. T� =�

2T2.

3� B For the neutralization between a strong acid and a strong alkali, the heat released is 57 kJ for � moleofwater produced.


Number of moles of HCl / NaOH mixed



released

�

�00 cm3of �mol dm–3HCl(aq) and�00 cm3of �mol dm–3KOH(aq)


�000dm3

=0.�mol

HCl(aq) +KOH(aq)0.�mol 0.�mol

KCl(aq) +H2O(l)

0.�mol

5.7 kJ

250 cm3of �mol dm–3HCl(aq) and50 cm3of�mol dm–3KOH(aq)

�mol dm–3 x50

�000dm3

=0.05mol

HCl(aq) +KOH(aq)0.05mol 0.05mol

KCl(aq) +H2O(l)

0.05mol

2.85 kJ

The first mixture (total volume 200 cm3) is heated up by 5.7 kJ while the second mixture (total volume�00 cm3) isheatedupby2.28kJ.Hence the twomixtures show the same temperature rise, i.e. T�+T2.



Expt. Solutions mixed Strength of acid and alkali Temperature rise

��00 cm3of �mol dm–3HNO3(aq) and�00 cm3of �mol dm–3NaOH(aq)


T�

2�00 cm3of �mol dm–3HNO3(aq) and�00 cm3of �mol dm–3NH3(aq)

neutralizationbetween a strongacid and aweak alkali

T2

\ T� > T2.

27



Mixed Strength of acid and alkali Temperature rise

�00 cm3of 2mol dm–3HNO3(aq)+�00 cm3of 2mol dm–3NaOH(aq)

neutralizationbetween a strong acid anda strong alkali

x

�00 cm3of 2mol dm–3HCl(aq)+�00 cm3of 2mol dm–3NH3(aq)

neutralizationbetween a strong acid andaweak alkali

y < x

�00 cm3of 2mol dm–3HCl(aq)+�00 cm3of 2mol dm–3KOH(aq)

neutralizationbetween a strong acid anda strong alkali

z = x

\ x = z > y

34 C Fe3+(aq)+3OH–(aq) Fe(OH)3(s) reddishbrownprecipitate

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 DOption Solution mixed

A precipitate formed

Ionic equation for reaction involved

Abarium chloride solution and silvernitrate solution

4 Ag+(aq) +Cl–(aq) AgCl(s)

Bbarium chloride solution anddilutesulphuric acid

4 Ba2+(aq) + SO42–(aq) BaSO4(s)

Cbarium chloride solution and sodiumsulphate solution

4 Ba2+(aq) + SO42–(aq) BaSO4(s)

Dbarium chloride solution and sodiumnitrate solution

8 —

\ Sodiumnitrate solutionwouldNOTgive a precipitatewith barium chloride solution.

37 A Option C—Ammonium compounds are soluble inwater.

38 AOption Ions reacted Precipitate formed? Ionic equation involved

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

2–(aq) BaSO4(s)

B Cu2+(aq) andNO3–(aq) noprecipitate —

C Ni2+(aq) andCO32–(aq) a greenprecipitate (NiCO3) Ni2+(aq) +CO3

2–(aq) NiCO3(s)

D Na2+(aq) andOH–(aq) noprecipitate —

28

39 A K2CO3(aq)+Ba(NO3)2(aq) BaCO3(s) + 2KNO3(aq) reactantA white precipitateX

BaCO3(s) + 2HCl(aq) CO2(g) + BaCl2(aq)+H2O(l) white reactant B precipitateX colourless solutionY

40 B Copper(II) oxide reactswith dilute sulphuric acid to give copper(II) sulphate andwater.

CuO(s) +H2SO4(aq) CuSO4(aq)+H2O(l)

Copper(II) sulphate is soluble in water while copper is insoluble. Therefore copper can be separatedfrom themixture by filtration.

4� C Option A — AgNO3(aq) forms awhite precipitate (Ag2CO3)withCO32–(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 awhite precipitatewith both SO42–(aq) ions andCO3

2–(aq) ions.

Ba2+(aq)+ SO42–(aq) BaSO4(s)

Ba2+(aq)+CO32–(aq) BaCO3(s)

Option C — Fe(NO3)3(aq)wouldgive a precipitatewithCO32–(aq) ions, butNOTwith SO4

2–(aq) ions.

Hence Fe(NO3)3(aq) canbeused to separate the anions SO42–(aq) andCO3

2–(aq).

42 D Option A — Calcium carbonate gives gas bubbles with dilute sulphuric acid. However, it is insoluble inwater.

Option B — Lead(II) sulphate does not give gas bubbles with dilute sulphuric acid. It is also insolubleinwater.

Option C — Magnesium chloridedoes not give gas bubbleswith dilute sulphuric acid.

Option D — Potassium carbonate is soluble in water and gives gas bubbles (carbon dioxide) with dilutesulphuric acid.

43 D Option A — Calcium sulphate is insoluble inwater.

Option B — Copper(II) oxide is notwhite in colour. It is also insoluble inwater.

Option C — Iron(II) chloride is notwhite in colour.

Option D — Neutralization occurs when potassium hydroxide is mixed with dilute sulphuric acid. Heatis released.

44 B Copper(II) oxide is black in colour. When copper(II) oxide is mixed with dilute sulphuric acid, a bluesolution results.


29

45 C Lead(II) carbonate gives a colourless gas (carbondioxide)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) withdilute hydrochloric acid.


46 C Zinc reactswith diluteH2SO4(aq) (reactantX) to give a solution containing Zn2+ ions.

Zn(s) +H2SO4(aq) ZnSO4(aq)+H2(g)

The solutiongives a precipitate (ZnCO3(s))withK2CO3(aq) (reactantY).

Zn2+(aq)+CO32–(aq) ZnCO3(s)

47 A Calcium reactswithwater to give calciumhydroxide andhydrogen

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 awhite precipitatewith sodium carbonate solution.

Ca2+(aq)+CO32–(aq) CaCO3(s)

48 B From Test �, it can be deduced that solid X is a sulphate. A sulphate gives a white precipitate (bariumsulphate) with bariumnitrate solution.

Ba2+(aq)+ SO42– BaSO4(s)

From Test 2, it can be deduced that solid X contains aluminium ions rather than calcium ions.Aluminium ions give a white precipitate (zinc hydroxide) with dilute sodium hydroxide solution. Theprecipitate dissolves in excess alkali due to the formationof 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 sulphuricacid.

Pb2+(aq)+ SO42–(aq) PbSO4(s)

50 C OptionsA, B andD— Oxides or hydroxides of copper, iron and zinc are insoluble bases. Sulphates ofcopper, iron and zinc areNOTpreparedby an acid-alkali titration method.

Option C — Sodium sulphate can be prepared by reacting dilute sulphuric acid with dilute sodiumhydroxide solution via an acid-alkali titration method.

5� D

30

As sulphuric acid was added, it removed both barium ions (by precipitation) and hydroxide ions (byneutralization).

Ba(OH)2(aq)+H2SO4(aq) BaSO4(s) + 2H2O(l)

At the equivalencepoint, all thebarium ions andhydroxide ions hadbeen removed.Hence thenumberof ions in themixture falls until the equivalencepoint.

After the equivalence point, the number of ions in the mixture rises as excess sulphuric acid wasadded.

52 C Option A — Ammonium sulphate gives an acidic solution when dissolved in water. Farmers may add itto soilwhichhas become too alkaline.

Option C — Slaked lime (calciumhydroxide) canneutralize the acidic waste.

53 A

54 A

55 B (2) This isNOT a neutralizaton reaction.

In neutralization, salt andwater are theonly products.

56 A (�) Heat is released in neutralization reactions.

(2) Theproduct is sodiumnitrate, a normal salt.

(3) Sodiumnitrate is an ionic compound.

57 C (�) When hydrogen chloride gas dissolves in water, almost all the hydrogen chloride moleculesdissociate to give hydrogen ions and chloride ions. The hydrochloric acid formed contains manlyhydrogen ions (H+(aq)), chloride ions (Cl–(aq)) and water molecules. Hydrochloric acid is a strongacid.


(2) A solution of hydrogen chloride in water can conduct electricity due to the presence of mobileions.

(3) A solution of hydrogen chloride and sodium hydroxide solution undergo a neutralization reactionwhenmixed. The reaction is exothermic.

58 D Iron reactswith dilute sulphuric acid to give a rion(II) sulphate solution andhydrogengas.

Fe(s) +H2SO4(aq) FeSO4(aq)+H2(g)

59 A (�) Heatwouldbe releasedwhen concentrated sulphuric acidwasmixedwithwater.

(2) Dilute sulphuric acid and dilute aqueous ammonia underwent a neutralization reaction when mixed.Heatwould be released.

(3) A white precipitate (barium sulphate) appeared upon mixing dilute sulphuric acid and bariumchloride solution.


3�

60 C (�) There isNO reactionbetween copper anddilute ethanoic acid.

6� A Heat is absorbedduring the evaporationof ethanol.Hence this isNOT an exothermic process.

62 B For the neutralization between a strong acid and a strong alkali, the heat released is 57 kJ for � moleofwater produced.



Number of moles of acid / alkali mixed


Number of moles of water formed

Strength of acid and alkali

Heat released

Temperature rise

20 cm3of �moldm–3HCl(aq) and20 cm3of �moldm–3NaOH(aq)

Number ofmoles ofHCl /NaOH

=�mol dm–3 x20

�000dm3

=0.02mol

H+(aq) + OH–(aq)

0.02mol 0.02mol

H2O(l)

0.02mol

neutralizationbetween astrong acid anda strong alkali

�.�4 kJ Thismixture (totalvolume40 cm3) isheatedupby �.�4kJ.Assume that thetemperature riseis T.

� 20 cm3of �moldm–3HCl(aq) and20 cm3of �moldm–3NH3(aq)

Number ofmoles ofHCl /NH3

=�mol dm–3 x20

�000dm3

=0.02mol

H+(aq) + OH–(aq)

0.02mol 0.02mol

H2O(l)

0.02mol

neutralizationbetween astrong acid anda weak alkali

<�.�4 kJ Thismixture (totalvolume40 cm3)is heatedupby<�.�4 kJ.Hence thetemperature riseis <T.

2 40 cm3of �moldm–3HCl(aq) and40 cm3of �moldm–3NaOH(aq)


=�mol dm–3 x40

�000dm3

=0.04mol

H+(aq) + OH–(aq)

0.04mol 0.04mol

H2O(l)

0.04mol


2.28 kJ Thismixture (totalvolume80 cm3) isheatedupby 2.28kJ.Hence thetemperature riseis T.

3 �0 cm3of 2moldm–3HCl(aq) and�0 cm3of 2moldm–3NaOH(aq)


=2mol dm–3 x�0

�000dm3

=0.02mol

H+(aq) + OH–(aq)

0.02mol 0.02mol

H2O(l)

0.02mol


�.�4 kJ Thismixture (totalvolume20 cm3) isheatedupby �.�4kJ.Hence thetemperature riseis 2T.

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

32

64 DReagent

The white precipitate formed with lead(II) nitrate solution

Ionic equation involved

(�) 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 DSolution

Precipitate formed with silver nitrate solution


(�) 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 DSubstance

White precipitate formed with dilute sulphuric acid


(�) 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 (�) Ammonium iodide is an ionic compound. It contains ammonium ions and iodide ions.

(2) Ammonium iodide solutiongives a precipitate (silver iodide)with silver nitrate solution.

Ag+(aq)+ I–(aq) AgI(s)

68 DSolutions mixed Precipitate formed Ionic equation involved

(�) Pb(NO3)2(aq) andK2SO4(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) andCaCl2(aq) CaCO3 Ca2+(aq) +CO32–(aq) CaCO3(s)

69 A (�) Potassium iodide solution conducts electricity due to the presence of mobile ions (potassium ionsand iodide ions).

(2) Potassium iodide solutiongives a yellowprecipitate (silver iodide)with silver nitrate solution.

Ag+(aq)+ I–(aq) AgI(s)

(3) Potassium iodide solution is colourless.

33

70 B (�) Iron ismore reactive than silver, but less reactive than zinc.

Iron candisplace silver from silver nitrate solution.

There isNO reactionbetween iron and zinc nitrate solution.

(3) Potassium chloride solutiongives awhite 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 aremixed.

7� A The student canobtain calcium chloride crystals from the reactionmixture 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 thewater (usingBunsenburner and evaporatingbasin).

Set the concentrated solution aside to cool and crystallize.

Filter (using filter paper and filter funnel) the crystals from the remaining solution.

72 B (�) Adding excess aqueous ammoniawill leave an alkaline solution after reactingwith the acid.

(2) Adding excess calcium carbonate will not leave an alkaline solution after reacting with the acidbecause calcium carbonate is insoluble inwater.

(3) Sodium chloridedoes not reactwith the acid.

The following equation represents the reaction between dilute aqueous ammonia and dilutehydrochloric acid:

NH3(aq)+HCl(aq) NH4Cl(aq)

Ammonium chloride is produced in the reaction.

73 A (3) Ammonium chloride is soluble in water. NO precipitate is formed in the reaction between diluteaqueous ammonia anddilute hydrochloric acid.

74 C (�) CuO is a base,NOT a salt.

75 D Zinc sulphate can be prepared by reacting dilute sulphuric acid with either zinc, zinc oxide or zinccarbonate.


ZnO(s) +H2SO4(aq) ZnSO4(aq)+H2O(l)

ZnCO3(s) +H2SO4(aq) ZnSO4(aq)+H2O(l) +CO2(g)

76 B (2) The chemical formula of calciumhydrogenphosphate isCaHPO4,NOTCaH2PO4.

77 B (2) Ammoniumethanoate is a normal salt.

It is formed fromethanoic acid, amonobasic acid.Amonobasic acid can formnormal salts only.

(3) Aqueous solution of ammonium ethanoate can conduct electricity due to the presence of mobileions.

34

78 A (�) Antacids contain bases. Theywill react with acidic substances, e.g. vinegar.

79 A

80 C Thenumber ofmoles ofNa+ ions in �0 cm3of �mol dm–3 sodiumhydroxide solution

=�mol dm–3 x�0

�000dm3

=0.0� mol

When �0 cm3 of � mol dm–3 hydrochloric acid are added to �0 cm3 of � mol dm–3 sodium hydroxidesolution,

total volumeof the solutionmixture = (�0+�0) cm3

= 20 cm3

ConcentrationofNa+ ions in the solutionmixture = 0.0�mol

( 20�000 ) dm3

= 0.5mol dm–3

\ the concentrationofNa+ ionswould change.

8� B Sodium chloride is a salt formed fromhydrochloric acid.

Hydrochloric acid is amonobasic acid. It can formnormal salts only.

82 D Some salts are insoluble inwater.

Some salts are formed from the neutralization between an acid and an insoluble metal hydroxide aninsolublemetal oxide.

For example, copper(II) oxide is an insoluble metal oxide. It reacts with dilute sulphuric acid to givecopper(II) sulphate andwater.

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. Ethanoicacid is amonobasic acid. It cannot formacid salt.

84 C There isNO reactionbetween copper anddilute hydrochloric acid.

85 B Ammonium chloride is a salt formed from the reaction between dilute aqueous ammonia and dilutehydrochloric acid.

Ammonium chloride can conduct electricity in aqueous solutiondue to thepresenceofmobile ions.

86 C Nickel(II) carbonate is insoluble inwater.

87 C Sodium hydroxide is corrosive and thusNEVERused in antacids.

88 C A lot of heat is released when vinegar reacts with concentrated sodium hydroxide solution. This willcause skin burn.

The student shouldwash the affected areawith plenty ofwater.

35

Unit 17 Concentration of solutions and volumetric analysis

Fillintheblanks

� a) electronic balance

b) volumetric flask

c) pipette

d) burette

2 standard

3 pipette; burette

4 equivalence

5 methyl orange

6 Phenolphthalein

Trueorfalse

7 F To dilute �00 cm3 of �.0 mol dm–3 hydrochloric acid to 0.�0 mol dm–3, add water until the totalvolume of the solution is � 000 cm3.

8 F The volumetric flask shouldbewashedonlywith distilledwater before use.

Washing the volumetric flask with the acid before use would increase the number of moles of solute itholds,making the concentrationof the sulphuric acid obtainedhigher than calculated.

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

�0 T

�� F Potassium hydroxide absorbs moisture from the air and cannot be weighed accurately. It is unsuitablefor preparing a standard solution.

�2 F A solutionwith an accurately known concentration is a standard solution.

�3 T If we wash the pipette or burette with distilled water only, water droplets remaining on the inside oftheglasswarewill dilute the solution that theglassware is going to contain.

�4 F The conical flask is to hold a specific volume of a solution (usually 25.0 cm3), i.e. a specific amountof the solute. It should NOT be washed with the solution it is to contain before use because theadditional amount of solute remaining in the flaskwill affect the titration results.

�5 F The last dropof the solution in thepipette shouldNOTbeblownout.

36

�6 T The following diagram shows the titration curve for the titration of 20.0 cm3 of 0.� mol dm–3 HCl(aq)with 0.� mol dm–3NaOH(aq).

Phenolphthalein changes colourwithin thepH rangeof the vertical part of the titration curve.

Hencephenolphthalein is a suitable indicator for the titration.

�7 F The followingdiagram shows the titration curve for the titrationof0.�moldm–3CH3COOH(aq)with0.�mol dm–3NaOH(aq).

Methyl orangedoesNOT change colourwithin thepH rangeof the vertical part of the titration curve.

Hencemethyl orange isNOT a suitable indicator for the titration.

�8 F During the titrationof an acid and an alkali, thepHof the solutionmixture mayNOTbe7.00.

For example, during the titration of a weak acid and a strong alkali, the pH of the solution mixture isgreater than7.00.

�9 F During a titration the end point and the equivalence point usually occur at slightly different times.However,wehave to assume that the endpoint is the equivalencepoint and recognize this assumptionas a sourceof error.

37

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

Number of ions moles of NaOH required for complete neutralization

Number ofmoles ofCH3COOH

=�mol dm–3 x20

�000dm3

=0.02mol

CH3COOH(aq) +NaOH(aq) CH3COONa(aq)+H2O(l) 0.02mol ?mol\ 20 cm3of �mol dm–3CH3COOH(aq) require 0.02 mole

ofNaOH for complete neutralization.

Number ofmoles ofH2SO4


�000dm3

=0.0�mol

H2SO4(aq) +2NaOH(aq) Na2SO4(aq) +2H2O(l)0.0�mol ?mol\ �0 cm3of �mol dm–3H2SO4(aq) require 0.02mole of

NaOH for complete neutralization.


2� C Molarmass of (COOH)2•2H2O = [2 x (�2.0+2 x �6.0+�.0) + 2 x (2 x �.0+�6.0)] gmol–�

= �26.0gmol–�

Number ofmoles of (COOH)2•2H2Oused=mass

molarmass

=5.04g

�26.0gmol–�

= 0.0400mol

Concentrationof acid solution=number ofmoles of (COOH)2•2H2O

volumeof solution

= 0.0400mol

( 250.0�000 ) dm3

= 0.�60mol dm–3

22 C Consider � 000 cm3 (i.e. �.00dm3) of the sample.

Mass of � 000 cm3of the sample =�.4�g cm–3 x � 000 cm3

= � 4�0g

Mass ofHNO3 in � 000 cm3of sample =mass of � 000 cm3of sample x percentagebymass ofHNO3 in sample = � 4�0 g x 6�.5% =867 g

Molarmass ofHNO3 = (�.0+�4.0+3 x �6.0) gmol–�

= 63.0gmol–�

Number ofmoles ofHNO3 in �.00dm3of sample=mass

molarmass

=867g

63.0gmol–�

= �3.8mol

38

Molarity of nitric acid =number ofmoles ofHNO3

volumeof solution

=�3.8mol

�.00dm3

= �3.8mol dm–3

\ the concentrationof thenitric acid sample is �3.8mol dm–3.

23 A Mass ofHNO3 in 60.0m3 acid=68000000g x 35.5% = 24�00000g

Molarmass ofHCl = (�.0+35.5) gmol–�

= 36.5gmol–�

Number ofmoles ofHCl=mass

molarmass

=24�00000g

36.5gmol–�

= 660000mol

Molarity of hydrochloric acid =number ofmoles ofHCl

volumeof solution

=660000mol

60000dm3

= ��.0mol dm–3

\ the concentrationof thehydrochloric acid is ��.0mol dm–3.

24 B Mass ofH2SO4 in 58.0m3 acid=55000000g x 94.0% = 5�700000g

Molarmass ofH2SO4 = (2 x �.0+32.�+4 x �6.0) gmol–�

= 98.�gmol–�

Number ofmoles ofH2SO4=mass

molarmass

=5�700000g

98.�gmol–�

= 527000mol

Molarity of the sulphuric acid =number ofmoles ofH2SO4

volumeof solution

=527000mol

58000dm3

= 9.08mol dm–3

\ themolarity of the sulphuric acid is 9.08mol dm–3.

39

25 D Onemole ofK3PO4 contains 3moles ofK+ ions and�mole of PO43– ions.

Total number ofmole of ions=4 x 0.�00mol = 0.400mol

Total ion concentration in solution=number ofmoles of ions

volumeof solution

= 0.400mol

( 500.0�000 ) dm3

= 0.800mol dm–3

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

26 D Number ofmoles ofNa2CO3present =molarity of solution x volumeof solution = 2.00mol dm–3 x 2.00dm3

= 4.00mol

One mole ofNa2CO3 contains 2moles ofNa+ ions and�mole ofCO32– ions.

Total number of ions present =3 x 4.00mol x 6.02 x �023mol–�

= 7.22 x �024

\ the total number of ions present is 7.22 x �0–4.

27 A pHof acid = –log�0[H+]

= 2.70

i.e. log�0[H+] = –2.70

[H+] = �0–2.70

= �.995 x �0–3mol dm–3

Sulphuric acid dissociates completely according to the following equation:


�.995 x �0–3mol dm–3

Concentrationof sulphate ions =concentrationof hydrogen ions

2

=�.995 x �0–3

2mol dm–3

= 9.98 x �0–4mol dm–3

\ tht concentrationof sulphate ions in the acid is 9.98 x �0–4mol dm–3.

40

28 ASubstance

Number of molesof substance present

Number of ions in one formula unit of substance

Number of molesof ions present

NaCl0.�0mol dm–3 x

�00�000

dm3

=0.0�0mol2 2 x 0.0�0mol=0.020 mol

Fe2(SO4)30.050mol dm–3 x

50�000

dm3

=0.0025mol5 5 x 0.0025mol= 0.0�3 mol

CaCl20.080mol dm–3 x

50�000

dm3

=0.0040mol3 3 x 0.0040mol= 0.0�2 mol

MgSO40.080mol dm–3 x

�00�000

dm3

=0.0080mol2 2 x 0.0080mol= 0.0�6 mol

\ �00 cm3 of 0.�0 mol dm–3 NaCl(aq) contain the greatest number of moles of ions, i.e. the greatestnumber of ions.

29 C Number ofmoles of ZnCl2 in ZnCl2(aq) =molarity of solution x volumeof solution

= 2.00mol dm–3 x�50.0

�000dm3

= 0.300mol

Onemole of ZnCl2 contain 2moles ofCl– ions.

\ number ofmoles ofCl–(aq) ions in ZnCl2(aq) =2 x 0.300mol = 0.600mol

Number ofmoles ofNaCl inNaCl(aq) =molarity of solution x volumeof solution

= �.20mol dm–3 x50.0

�000dm3

= 0.0600mol

Onemole ofNaCl contains �mole ofCl– ions.

\ number ofmoles ofCl–(aq) ions inNaCl(aq)=0.0600mol

Total number ofmoles ofCl–(aq) ions = (0.600+0.0600)mol = 0.660mol

Total volumeof solutionX = (�50.0+50.0) cm3

= 200.0 cm3

ConcentrationofCl–(aq) ions in solutionX= 0.660mol

( 200.0�000 ) dm3

= 3.30mol dm–3

\ the concentrationofCl–(aq) ions in solutionX is 3.30mol dm–3.

4�

30 C Onemole of sodium sulphate (Na2SO4) contains 2moles ofNa+ ions and�mole of SO42– ions.

Number ofmoles ofNa2SO4 in solution=number ofmoles ofNa+ ions

2

=2.00 x �0–2

2mol

= �.00 x �0–2mol

Molarity of the sodium sulphate solution=number ofmoles ofNa2SO4

volumeof solution

= �.00 x �0–2mol

( 20.0�000 ) dm3

= 0.500mol dm–3

\ themolarity of the sodium sulphate solution is 5.00 x �0–�mol dm–3.

3� D Number ofmoles ofNa2CO3 in 2.50mol dm–3 solution=molarity of solution x volumeof solution

= 2.50mol dm–3 x200.0

�000dm3

= 0.500mol

One mole ofNa2CO3 contain 2moles ofNa+ ions.

\ number ofmoles ofNa+(aq) ions in the solution=2 x 0.500mol = �.00mol

Number ofmoles ofNa2CO3 in �.00mol dm–3=molarity of solution x volumeof solution

= �.00mol dm–3 x50.0

�000dm3

= 0.0500mol

Number ofmoles ofNa+(aq) ions in the solution=2 x 0.0500mol = 0.�00mol

Total number ofmoles ofNa+(aq) ions= (�.00+0.�00)mol = �.�0mol

Total volumeof solution = (200.0+50.0) cm3

= 250.0 cm3

ConcentrationofNa+(aq) ions in the resulting solution= �.�0mol

( 250.0�000 ) dm3

= 4.40mol dm–3

\ the concentrationofNa+(aq) ions in the resulting solution is 4.40mol dm–3.

42

32 C Onemole ofmagnesium chloride (MgCl2) contains �mole ofMg2+ ions and2moles ofCl– ions.

Number ofmoles ofMgCl2 in solution=number ofmoles ofCl– ions

2

=2.0 x �0–2

2mol

= �.0 x �0–2mol

Molarity of themagnesium chloride solution=number ofmoles ofMgCl2

volumeof solution

= �.0 x �0–2mol

( 50.0�000 ) dm3

= 0.20mol dm–3

\ themolarity of the solution is 2.0 x �0–�mol dm–3.

33 C Number ofmoles of Fe2(SO4)3 in Fe2(SO4)3(aq) =molarity of solution x volumeof solution

= 0.20mol dm–3 x50

�000dm3

= 0.0�0mol

One mole of Fe2(SO4)3 contain 3moles of SO42– ions.

\ number ofmoles of SO42– ions in Fe2(SO4)3(aq)=3 x 0.0�0mol

= 0.030mol

Number ofmoles ofK2SO4 inK2SO4(aq)=molarity of solution x volumeof solution

= 0.25mol dm–3 x200

�000dm3

= 0.050mol

Onemole ofK2SO4 contains �mole of SO42– ions.

\ number ofmoles of SO42– ions in K2SO4(aq)=0.050mol

Total number ofmoles of SO42– ions in solutionX= (0.030+0.050)mol

= 0.080mol

Total volumeof solutionX = (50+200) cm3

= 250 cm3

Concentrationof SO42–(aq) ions in solutionX= 0.080mol

( 250�000 ) dm3

= 0.32mol dm–3

\ the concentrationof SO42–(aq) ions in solutionX is 0.32mol dm–3.

43

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

5.0 xV

�000 = 0.40 x

250

�000 V =20 cm3

35 A Volume of diluted solution=3.75dm3+250.0 cm3

= 4.00dm3

(MV) before dilution = (MV) after dilution,whereM=molarity, V= volume

3.20 x250.0

�000 =M x 4.00

M =0.200

\ themolarity of thediluted solution is 0.200mol dm–3.


�0 x�20

�000 =2 x

V

�000 V =600

Volume of the final solution=600 cm3

\ volumeofwater added= (600 – �20) cm3

= 480 cm3


0.50 x�00

�000 =0.�0 x

V

�000 V =500 cm3


\ volumeofwater added= (500 – �00) cm3

= 400 cm3

38 B Volume of dilute solutionof acidX= (40+�0) cm3

= 50 cm3


�.0 x40

�000 = M x

50

�000 M =0.80mol dm–3

\ the concentrationof thedilute solutionof acidX is 0.80mol dm–3.

44

39 B Volume of diluted solutionof acidY= (25+25) cm3

= 50 cm3


�.6 x25

�000 = M x

50

�000 M =0.80

\ the concentrationof thedilute solutionof acidY is 0.80mol dm–3.

Hence the concentrationof dilute solutionof acidX equals that of acidY.

During the reaction between magnesium and an acid, magnesium reacts with hydrogen ions in theacid.

From thediagrams, it canbededuced thatmagnesium reactsmore slowlywithdilute solutionof acidX.

Theconcentrationofhydrogen ions indilute solutionofacidX is lower than that indilute solutionofacidY.

As the concentration of dilute solution of acid X equals that of acid Y, it can be concluded that acid Xisweaker than acidY.

40 C Volume of solutionX = (20+80) cm3

= �00 cm3


0.50 x20

�000 =M x

�00

�000 M =0.�0

\ concentrationof acidA in solutionX is 0.�0mol dm–3.

4� A Volume of solutionY = (60+40) cm3

= �00 cm3


0.25 x60

�000 =M x

�00

�000 M =0.�5

\ concentrationof acid B in solutionY is 0.�5mol dm–3.

Option A — During the reaction between magnesium and an acid, magnesium reacts with hydrogenions in the acid.

From thediagrams, it canbededuced thatmagnesium reactsmorequicklywith solutionX.

The concentrationof hydrogen ions in solutionX is higher than that in solutionY.

Although the concentration of acid A in solution X is lower than the concentration ofacid B in solutionY, it canbe concluded that acidA is stronger than acid B.

Option D — As the concentration of acid A in solution X is lower than the concentration of acid Bin solution Y, the number of moles of acid A in solution X is lower than the number ofmoles of acid B in solutionY.

When both solutions of monobasic acids react with magnesium, solution X would giveless hydrogen than solutionY.

45

42 A 2NH3(aq) + H2SO4(aq) (Na4)2SO4(aq) �.60 mol dm–3 ? mol dm–3

25.0 cm3 32.0 cm3

Number ofmoles ofNH3 in �.60 cm3 solution =molarity of solution x volumeof solution

= �.60mol dm–3 x25.0

�000dm3

= 0.0400mol

According to the equation, 2moles ofNH3 require �mole ofH2SO4 for complete neutralization.

Number ofmoles ofH2SO4 in 32.0 cm3 solution=0.0400

2mol

= 0.0200mol

Concentrationof sulphuric acid=number ofmoles ofH2SO4

volumeof solution

= 0.0200mol

( 32.0�000 ) dm3

= 0.625mol dm–3


= 98.�gmol–�

Concentrationof sulphuric acid=98.�gmol–� x 0.625mol dm–3

= 6�.3gdm–3

\ the concentrationof the sulphuric acid is 6�.3gdm–3.

43 A We can represent thedibasic acid solutionbyH2X(aq).

H2X(aq) + 2NaOH(aq) Na2X(aq) + 2H2O(l) 0.0600 mol dm–3 0.�50mol dm–3

25.0 cm3 ? cm3

Number ofmoles ofH2X in 25.0 cm3 solution =molarity of solution x volumeof solution

= 0.0600mol dm–3 x25.0

�000dm3

= 0.00�50mol

According to the equation, �mole ofH2X requires 2moles ofNaOH for complete neutralization.

i.e.number ofmoles ofNaOH=2 x number ofmoles ofH2X = 2 x 0.00�50mol = 0.00300mol

Volume of sodiumhydroxide solution required for complete neutralization

=number ofmoles ofNaOH

molarity of solution

=0.00300mol

0.�50mol dm–3

= 0.0200 dm3

=20.0 cm3

46

44 D K2O(s) + 2HCl(aq) 2KCl(aq) + H2O(l) 6.�2 g 0.50mol dm–3

? cm3

Molarmass ofK2O= (2 x 39.�+�6.0) gmol–�

= 94.2gmol–�

Number ofmoles ofK2O=mass

molarmass

=6.�2g

94.2gmol–�

= 0.0650mol

According to the equation, �mole ofK2O requires 2moles ofHCl for complete neutralization.

i.e.number ofmoles ofHCl=2 x 0.0650mol = 0.�30mol

Volume of hydrochloric acid required for complete neutralization=number ofmoles ofHCl


=0.�30mol

0.50mol dm–3

= 0.26dm3

= 260 cm3

45 B 2Na(s) + 2H2O(l) 2NaOH(aq)+H2(g)......(�) 0.�82 g

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

? cm3

Number ofmoles ofNa=mass

molarmass

=0.�82g

23.0gmol–�

= 0.0079�mol

Number ofmoles ofNaOH formed fromNa=number ofmoles ofNa = 0.0079�mol

According to equation (2), �mole ofNaOH requires �mole ofHCl for complete neutralization.

i.e.number ofmoles ofHCl =0.0079�mol

Volume of hydrochloric acid required for complete neutralization=number ofmoles ofHCl


=0.0079�mol

0.240mol dm–3

= 0.0330dm3

= 33.0 cm3

47

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

50.0 cm3

Molarmass ofCaCO3 = (40.�+�2.0+3 x �6.0) gmol–�

= �00.�gmol–�

Number ofmoles ofCaCO3=mass

molarmass

=7.50g

�00.�gmol–�

= 0.0749mol

Number ofmoles ofHCl in 50.0 cm3 solution=molarity of solution x volumeof solution

= 2.00mol dm–3 x50.0

�000dm3

= 0.�00mol

According to the equation, � mole of CaCO3 reacts with 2 moles of HCl to produce � mole of CO2.During the reaction, 0.�00 mole of HCl reacts with 0.0500 mole of CaCO3. Therefore CaCO3 is inexcess. The amount ofHCl limits the amount ofCO2 evolved.

Number ofmoles ofCO2 evolved=0.�00

2mol

= 0.0500mol

\ 0.0500mole of carbondioxide is evolved.

47 A Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g) �.80 g 0.250mol dm–3

�00.0 cm3

Number ofmoles of Zn=mass

molarmass

=�.80g

65.4gmol–�

= 0.0275mol

Number ofmoles ofHCl in �00.0 cm3 solution=molarity of solution x volumeof solution

= 0.250mol dm–3 x�00.0

�000dm3

= 0.0250mol

According to the equation, � mole of Zn reacts with 2 moles of HCl to produce � mole of H2. Duringthe reaction, 0.0250 mole of HCl reacted with 0.0�25 mole of Zn. Therefore Zn was in excess. Theamount ofHCl limited the amount ofH2 formed.

Number ofmoles ofH2 formed=0.0250

2mol

= 0.0�25mol

Mass ofH2 formed =number ofmoles ofH2 xmolarmass ofH2

= 0.0�25mol x 2.0 gmol–�

= 0.0250g

\ 0.0250gof hydrogenwas formed.

48

48 C CuO(s) + 2HNO3(aq) Cu(NO3)2(aq) + H2O(l) 4.00 g �.40mol dm–3

30.0 cm3

Molarmass ofCuO = (63.5+�6.0) gmol–�

= 79.5gmol–�

Number ofmoles ofCuO=mass

molarmass

=4.00g

79.5gmol–�

= 0.0503mol

Number ofmoles ofHNO3 in 30.0 cm3 solution=molarity of solution x volumeof solution

= �.40mol dm–3 x30.0

�000dm3

= 0.0420mol

According to the equation, � mole of CuO requires 2 moles of HNO3 for complete reaction. During thereaction, 0.0420mole ofHNO3 reactswith 0.02�0mole ofCuO. ThereforeCuO is in excess.

Number ofmoles ofCuO left unreacted= (0.0503 – 0.02�0)mol = 0.0293mol

Mass ofCuO left unreacted=number ofmoles xmolarmass = 0.0293mol x 79.5gmol–�

= 2.33g

\ 2.33gof copper(II) oxide are left unreacted.

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

According to the equation, �mole ofH2SO4 requires 2moles ofNaOH for complete neutralization.

Option Solutions mixed

Number of moles of acid / alkali


Complete neutralization?

A �00 cm3of �mol dm–3H2SO4(aq) and�00 cm3of �mol dm–3NaOH(aq)

number ofmoles ofH2SO4


�000dm3

=0.�mol

number ofmoles ofNaOH


�000dm3

=0.�mol

not enoughNaOH(aq)to neutralizeH2SO4(aq)completely

B �00 cm3of �mol dm–3H2SO4(aq) and�00 cm3of 2mol dm–3NaOH(aq)



�000dm3

=0.�mol


=2mol dm–3 x�00

�000dm3

=0.2mol

complete neutralizationoccurs and a neutralsolution is obtained.

49

C �00 cm3of 2mol dm–3H2SO4(aq) and50 cm3of 2mol dm–3NaOH(aq)


=2mol dm–3 x�00

�000dm3

=0.2mol


=2mol dm–3 x50

�000dm3

=0.�mol


D 200 cm3of 2mol dm–3H2SO4(aq) and�00 cm3of �mol dm–3NaOH(aq)


=2mol dm–3 x200

�000dm3

=0.4mol



�000dm3

=0.�mol


50 D Solid sodiumhydroxide is deliquescent.

5� 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


= 0.200mol dm–3 x27.5

�000dm3

= 0.00550mol

According to the equation, �mole ofHCl requires �mole ofNaOH for complete neutralization.

i.e.number ofmoles ofNaOH =0.00550mol

Molarity of sodiumhydroxide solution=number ofmoles ofNaOH

volumeof solution

= 0.00550mol

( 25.0�000 ) dm3

= 0.220mol 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) �0.0 cm3 0.320mol dm–3

28.8 cm3

250.0 cm3

(used) 25.0 cm3

Number ofmoles ofNaOH in 28.8 cm3 solution= molarity of solution x volumeof solution

= 0.320mol dm–3 x28.8

�000dm3

= 0.00922mol


i.e. number ofmoles ofH2SO4 in 25.0 cm3diluted cleanser=0.00922

2mol

= 0.0046�mol

Number ofmoles ofH2SO4 in 250.0 cm3diluted cleanser=0.0046�mol x250.0 cm3

25.0 cm3

= 0.046�mol

Number ofmoles ofH2SO4 in �0.0 cm3undiluted cleanser=0.046�mol

Molarity ofH2SO4 in undiluted cleanser=number ofmoles ofH2SO4

volumeof solution

= 0.046�mol

( �0.0�000 ) dm3

= 4.6�mol dm–3

\ the concentrationof sulphuric acid in theundiluted toilet cleanser is 4.6�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 0.0200mol dm–3

22.0 cm3

250.0 cm3

(used) 25.0 cm3


= 0.0200mol dm–3 x22.0

�000dm3

= 4.40 x �0–4mol

5�

According to the equation, �mole ofH3PO4 reacts with 2moles ofNaOH.

i.e. number ofmoles ofH3PO4 in 25.0 cm3diluted acid=4.40 x �0–4

2mol

= 2.20 x �0–4mol

Number ofmoles ofH3PO4 in 250.0 cm3diluted acid= 2.20 x �0–4 x250.0

25.0mol

= 2.20 x �0–3mol

Number ofmoles ofH3PO4 in 25.0 cm3original acid=2.20 x �0–3mol

Molarity of original phosphoric acid=number ofmoles ofH3PO4

volumeof solution

= 2.20 x �0–3mol

( 25.0�000 ) dm3

= 0.088mol dm–3

\ the concentrationof theoriginal phosphoric acidwas 8.80 x �0–2mol dm–3.

59 C NaOH(aq) + HA(aq) NaA(aq) + H2O(l) 0.�00 mol dm–3 ?mol dm–3

22.5 cm3 30.0 cm3


= 0.�00mol dm–3 x22.5

�000dm3

= 0.00255mol

According to the equation, �mole ofHA requires �mole ofNaOH for complete neutralization.

i.e.number ofmoles ofHA=0.00225mol

Molarity ofHA(aq) =number ofmoles ofHA

volumeof solution

= 0.00225mol

( 30.0�000 ) dm3

= 0.0750mol dm–3

\ the concentrationofHA(aq) is 7.50 x �0–2mol dm–3.

60 B Only phenolphthalein changes colourwithin thepH rangeof the vertical part of the titration curve.

6� A Wash a pipette and a burette first with distilled water and then with the solution they are going tocontain.

Wash a conical flaskwith distilled water only.

52

62 A A conical flask shouldbewashedwith distilledwater only.

Washing the conical flask with the tablet solution increased the number of moles of solute it held. Thiswould increase the averageburette reading (i.e. the volumeof hydrochloric acid used in the titration).

63 C Therewas problemwithStep 3.

The conical flask shouldNOTbe rinsedwith alkali.

Rinsing the conical flask with the alkali increased the number of moles of solute it held. This wouldincrease the volume of acid used to neutralize the alkali, making the calculated concentration of thealkali too high.

64 A Na2CO3(aq)+2HCl(aq) 2NaCl(aq)+CO2(g) +H2O(l) 3.57g �.20mol dm–3

48.0 cm3

MolarMass ofNa2CO3•H2O= (2 x 23.0+�2.0+3 x �6.0+�8.0n) gmol–�

Number ofmoles ofNa2CO3•H2O=mass

molarmass

=3.57g

(�06.0+�8.0n) gmol–�


= �.20mol dm–3 x48.0

�000dm3

= 0.0576mol

According to the equation, �mole ofNa2CO3 requires 2moles ofHCl for complete reaction.

i.e.number ofmoles ofNa2CO3•H2O=0.0576

2mol

= 0.0288mol

Number ofmoles ofNa2CO3•H2O=3.57

(�06.0+�8.0n)mol= 0.0288mol

n= �

65 C Letn be thebasicity of the acid.

We can represent the acid solutionbyHnX(aq).

HnX(aq) + nNaOH(aq) NanX(aq) + nH2O(l) 0.�25 mol dm–3 0.30mol dm–3

20.0 cm3 25.0 cm3


= 0.30mol dm–3 x25.0

�000dm3

= 0.0075mol

53

Number ofmoles ofHnX in 20.0 cm3 solution =molarity of solution x volumeof solution

= 0.�25mol dm–3 x20.0

�000dm3

= 0.0025mol

According to the equation, �mole ofHnX requires nmoles ofNaOH for complete neutralization.

Number ofmoles ofHnX

Number ofmoles ofNaOH=

�

n

=0.0075mol

0.0025mol n = 3

\ thebasicity of the acid is 3.

66 D We can represent thedibasic acid byH2X.

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

�8.9 cm3

250.0 cm3

(used) 25.0 cm3

Letm gmol–� be themolarmass ofH2X.

Number ofmoles ofH2X in 2.62g solid=mass

molarmass

=2.62g

m gmol–�

Number ofmoles ofH2X in 25.0 cm3 solution=�

�0 x

2.62

mmol

Number ofmoles ofNaOH in �8.9 cm3 solution=molarity of solution x volumeof solution

= 0.220mol dm–3 x�8.9

�000dm3

= 0.004�6mol


i.e. number ofmoles ofH2X=0.004�6

2mol

= 0.00208mol

Number ofmoles ofH2X=�

�0 x

2.62

mmol=0.00208mol

m = �26

\ themolarmass of the acid is �26gmol–�.

54

67 B Number ofmoles ofOH– in 25.0 cm3of �.00mol dm–3NaOH solution


=�.00 mol dm–3 x25.0

�000dm3

=0.0250 mol

68 C

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

28.0 cm3

Number ofmoles ofHCl in 28.0 cm3 solution =molarity of solution x volumeof solution

= 0.550mol dm–3 x28.0

�000dm3

= 0.0�54mol

According to the equation, �mole ofNaOH requires �mole ofHCl for complete neutralization.

i.e.number ofmoles ofOH– in the filtrate =0.0�54mol

\ 0.0�54mole of hydroxide ions is present in the solutionobtained inStage II.

70 C Number ofmoles ofOH– used for precipitation= (0.0250 – 0.0�54)mol = 0.00960mol

Ni2+(aq) +2OH–(aq) Ni(OH)2(s) 0.00960mol

According to the equation, � mole of Ni2+ ions react with 2 moles of OH– ions to give � mole ofNi(OH)2.

\ number ofmoles ofNi2+ ion =0.00960

2mol

= 0.00480mol

Molarity of nickel(II) sulphate solution=number ofmoles ofNi2+ ions

volumeof solution

= 0.00480mol

( 25.0�000 ) dm3

= 0.�92mol dm–3

\ themolarity of thenickel(II) sulphate solution is 0.�92mol dm–3.

7� B Methyl orange is yellow in alkaline solution and red in acidic solutions.

72 D Na2CO3(aq)+H2SO4(aq) Na2SO4(aq)+CO2(g) +H2O(l) 3.20g �.08mol dm–3

(with 24.8 cm3

impurity)

55

Number ofmoles ofH2SO4 in 24.8 cm3 solution =molarity of solution x volumeof solution

= �.08mol dm–3 x24.8

�000dm3

= 0.0268mol

According to the equation, �mole ofNa2CO3 requires �mole ofH2SO4 for complete reaction.

i.e. number ofmoles ofNa2CO3 in the sample=0.0268mol

Molarmass ofNa2CO3 = (2 x 23.0+�2.0+3 x �6.0) gmol–�

= �06.0gmol–�

Mass ofNa2CO3 in the sample =number ofmoles ofNa2CO3 xmolarmass ofNa2CO3 = 0.0268mol x �06.0gmol–�

= 2.84g

\ percentagepurity ofNa2CO3 in the sample=2.84g

3.20g x �00%

= 88.8%

\ thepercentagepurity of sodium carbonate in the sample is 88.8%.

73 C Fe(s) +H2SO4(aq) FeSO4(aq)+H2(g) �.60mol dm–3 ? g �00 cm3

Number ofmoles ofH2SO4 in �00 cm3 solution =molarity of solution x volumeof solution

= �.60mol dm–3 x�00

�000dm3

= 0.�60mol

According to the equation, �mole ofH2SO4 reacts with �mole of Fe to give �mole of FeSO4.

i.e. number ofmoles of FeSO4=0.�60mol

Molarmass of FeSO4•7H2O= [55.8+32.�+4 x �6.0+7 x (2 x �.0+�6.0)] gmol–�

= 277.9gmol–�

Mass of FeSO4•7H2Oobtained=number ofmoles of FeSO4•7H2O xmolarmass of FeSO4•7H2O = 0.�60mol x 277.9gmol–�

= 44.5g

\ 44.5 gof crystals are obtained.

74 C Ca(OH)2(aq) + 2HCl(aq) CaCl2(aq) + 2H2O(l) 0.0200 mol dm–3 0.400mol dm–3

? cm3 25.0 cm3


= 0.400mol dm–3 x25.0

�000dm3

= 0.0�00mol

According to the equation, � mole of Ca(OH)2 requires 2 moles of HCl for complete neutralization,giving�mol ofCaCl2.

56

i.e. number ofmoles ofCa(OH)2 =0.0�00

2mol

= 0.00500mol

Volume ofCa(OH)2 solution=number ofmoles ofCa(OH)2


=0.00500mol

0.0200mol dm–3

= 0.250dm–3

= 250 cm–3

Number ofmoles ofCaCl2 = 0.00500mol

Volume of the resulting solution = (25.0+250) cm3

= 275 cm3

ConcentrationofCaCl2 in the resulting solution=number ofmoles ofCaCl2

volumeof solution

= 0.00500mol

( 275�000 ) dm3

= 0.0�82mol dm–3

\ the concentrationof calcium chloride in the resulting solution is �.82 x �0–2mol dm–3.

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

200.0 cm3

Number ofmoles ofMg=mass

molarmass

=2.00g

24.3gmol–�

= 0.0823mol


= 0.800mol dm–3 x200.0

�000dm3

= 0.�60mol

According to the equation, � mole of Mg reacts with 2 moles of HCl to produce � mole of MgCl2.During the reaction, 0.�60 mole of HCl reacted with 0.0800 mole of Mg. Therefore Mg was in excess.The amount ofHCl limited the amount ofMgCl2 formed.

Number ofmoles ofMgCl2=0.�60

2mol

= 0.0800mol

ConcentrationofMgCl2 in the resulting solution =number ofmoles ofMgCl2

volumeof solution

= 0.0800mol

( 200.0�000 ) dm3

= 0.400mol dm–3

\ the concentrationofmagnesium chloride in the resulting solution is 0.400mol dm–3.

57

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

Cu2+(aq) +CO32–(aq) CuCO3(s)

77 C K2CO3(aq)+MgCl(aq) MgCO3(s) + 2KCl(aq)

According to theequation,�moleofK2CO3 reactswith�moleofMgCl2 toproduce�moleofMgCO3, i.e.equal volumesof �mol dm–3K2CO3(aq) and�mol dm–3MgCl2(aq)would react.

Option Solutions mixedVolume of K2CO3(aq) /

MgCl2(aq) reacted

A �0 cm3of �mol dm–3K2CO3(aq) +30 cm3of �mol dm–3MgCl2(aq) �0 cm3

B �5 cm3of �mol dm–3K2CO3(aq) +25 cm3of �mol dm–3MgCl2(aq) �5 cm3

C 20 cm3of �mol dm–3K2CO3(aq) +20 cm3of �mol dm–3MgCl2(aq) 20 cm3

D 30 cm3of �mol dm–3K2CO3(aq) +�0 cm3of �mol dm–3MgCl2(aq) �0 cm3

The greatest volumes of K2CO3(aq) and MgCl2(aq) react in option C. Thus the greatest amount ofprecipitatewouldbeproduced.

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

�number ofmoles ofCaCO3

=4g

�00.�gmol–�

=0.04mol

number ofmoles ofHCl


�000dm3

=0.�mol

CaCO3(s) + 2HCl(aq)0.04mol 0.�mol

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

2

number ofmoles ofCH3COOH


�000dm3

=0.�mol

CaCO3(s) + 2CHCOOH(aq)0.04mol 0.�mol

(CH3COO)2Ca(aq)+H2O(l) +CO2(g)

According to the equations, � mole of CaCO3 reacts with 2 moles of HCl / CH3COOH to produce �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 ofCaCO3 limits the amount ofCO2produced.

Number ofmoles ofCO2produced=0.04mol

Option A — No calcium carbonate remains in bothbeakers.

Options B andC— Hydrochloric acid and ethanoic acid in both beakers are in excess. They do notreact completely.

Option D — The same number of moles of gas, i.e. the same amount of gas, is produced by bothreactingmixtures.

58

79 C

BeakerNumber of moles of Mg

= mass

molar mass


volume of solutionReaction between Mg and acid

�

number ofmoles ofMg

=�.5g

24.3gmol–�

=0.062mol



�000dm3

=0.�mol

Mg(s) + 2HCl(aq)0.062mol 0.�mol

MgCl2(aq) +H2(g)According to the equation, � moleof Mg reacts with 2 moles of HCltoproduce�moleofH2.Duringthereaction, 0.� mole of HCl reactedwith 0.05 mole of Mg. ThereforeMg was in excess. The amountof HCl limited the amount of H2produced.Number ofmoles ofH2 produced=0.05mol

2



�000dm3

=0.�mol

Mg(s) + H2SO4(aq)0.062mol 0.�mol

MgSO4(aq) +H2(g)According to the equation, � moleof Mg reacts with � mole of H2SO4to produce � mole of H2. Duringthe reaction, 0.062 mole of Mgreacted with 0.062 mole of H2SO4.Therefore H2SO4 was in excess. Theamount of Mg limited the amountofH2 produced.Number ofmoles ofH2 produced=0.062mol

Option A — Themagnesium in Beaker �was in excess.

Option B — The sulphuric acid in Beaker 2was in excess.

Option C — A greater number of moles of hydrogen, i.e. a greater amount of hydrogen, wasproducedby the reactingmixture in Beaker 2.

Option D — Somemagnesium remained in Beaker �, but none remiained in Beaker 2.

80 D Option A — HCl(aq) is amonobasic acidwhileH2SO4(aq) is a dibasic acid.


H2SO4(aq) 2H+(aq)+ SO42–(aq)

� mol dm–3 H2SO4(aq) has a higher concentration of hydrogen ions than � mol dm–3HCl(aq) does.

Hence�mol dm–3H2SO4(aq) has a pH value lower than�mol dm–3HCl(aq) does.

Option B — Methyl orange is red in acids.

Option C — During the reactionof zinc and an acid, zinc reacts with hydrogen ions in the acid.

As � mol dm–3 H2SO4(aq) has a higher concentration of hydrogen ions than � mol dm–3HCl(aq) does, the reaction rate between zinc and � mol dm–3 H2SO4(aq) is higher thanthat between zinc and�mol dm–3H2SO4(aq).

59

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

Reaction between acid and NaOH


=�mol dm–3 x20

�000dm3

=0.02mol

HCl(aq) +NaOH(aq) NaCl(aq) +H2O(l)0.02molAccording to the equation, �mole ofHCl requires�mole ofNaOH for complete neutralization\ number ofmoles ofNaOH required=0.02 mol



�000dm3

=0.0�mol

H2SO4(aq)+2NaOH(aq) Na2SO4(aq)+H2O(l)0.0�molAccording to the equation, �mole ofH2SO4requires 2moles ofNaOH for completeneutralization\ number ofmoles ofNaOH required = 2 x 0.0�mol = 0.02mol

\ 20 cm3 of � mol dm–3 HCl(aq) and �0 cm3 of � mol dm–3 H2SO4(aq) require the samenumber ofmoles ofNaOH for complete neutralization.

8� D

82 B The following diagram shows the titration curve for the titration of 20.0 cm3 of 0.� mol dm–3CH3COOH(aq)with 0.�mol dm–3NaOH(aq).

During the titration of CH3COOH(aq) (a weak acid) with NaOH(aq) (a strong alkali), the pH at theequivalencepoint is greater than7.

83 C OptionsA andD— These two curves are INCORRECT as the initial pH of the 0.� mol dm–3 weakalkali is greater 7.

Option B — This curve shows the pH change as a strong acid is added to a strong alkali. There is amarked change in pH at the equivalencepoint.

60

Option C — The following diagram shows the titration curve for the titration of 20.0 cm3 of0.�mol dm–3NH3(aq)with 0.�mol dm–3HCl(aq).

During the titration of a weak alkali with a strong acid, the pH at the equivalence pointis less than7.

84 D The following diagram shows the titration curve for the titration of 20.0 cm3 of 0.� mol dm–3 KOH(aq)with 0.� mol dm–3CH3COOH(aq).

During the titrationof a strongalkaliwithaweakacid, thepHat theequivalencepoint isgreater than7.

Phenolphthalein changes colourwithin thepH rangeof the vertical part of the titration curve.

Hencephenolphthalein is a suitable indicator for the titration.

85 A ThepHof solutionX is less than7. Thus solutionX shouldbe an acid.

As the pH of the mixture levels off at 7 with excess solid X, it can be concluded that solid X isinsoluble inwater.Otherwise themixturewill be alkaline.

6�

86 DIndicator Colour in solution with a pH of 5

Methyl orange yellow

Bromothylmol blue yellow

Phenolphthalein colourless

\ a yellow colourwouldbeobserved.

87 B The followinggraph shows the relationshipbetween the temperatureof themixture and the volumeofsodium 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 atY. No more heat is produced. The excess alkali added also cools the solution mixture. Therefore thetemperature of the solutionmixture falls fromY to Z.

88 D As sulphuric acid was added, it removed both the barium ions (by precipitation) and hydroxide ions (byneutralization).

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 theelectrical conductivity of themixture fell to almost zero.

After the equivalence point, the number of ions in the mixture rises as excess sulphuric acid wasadded. Hence the electrical conductivity of themixture 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 � mole of H2SO4 for complete neutralization, i.e.2V cm3of �.00mol dm–3NaOH(aq)would react withV cm3of �.00mol dm–3H2SO4(aq).

62

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

A �0.0 cm3ofNaOH(aq)+20.0 cm3ofH2SO4(aq) �0.0 cm3ofNaOH(aq) +5.0 cm3ofH2SO4(aq)

B �5.0 cm3ofNaOH(aq)+�5.0 cm3ofH2SO4(aq) �5.0 cm3ofNaOH(aq) +7.5 cm3ofH2SO4(aq)

C 20.0 cm3ofNaOH(aq)+�0.0 cm3ofH2SO4(aq) 20.0 cm3ofNaOH(aq) +�0.0 cm3ofH2SO4(aq)

D 25.0 cm3ofNaOH(aq)+5.0 cm3ofH2SO4(aq) �0.0 cm3ofNaOH(aq) +5.0 cm3ofH2SO4(aq)

The greatest volumes of NaOH(aq) and H2SO4(aq) react in Option C. The greatest amount of heat isreleased and thus the temperature rise is thegreatest.

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 potassiumsulphate solution. Hence it can be concluded that 6.0 cm3 of lead(II) nitrate solution are required toreact completelywith thepotassium sulphate solution.

Pb(NO3)2(aq) + K2SO4(aq) PbSO4(s) + 2KNO3(aq) ?mol dm–3 �.2mol dm–3

6.0 cm3 �0.0 cm3

Number ofmoles ofK2SO4 in �0.0 cm3 solution=molarity of solution x volumeof solution

= �.2mol dm–3 x�0.0

�000dm3

= 0.0�2mol

According to the equation, �mole ofK2SO4 requires �mole of Pb(NO3)2 for complete reaction.

i.e. number ofmoles of Pb(NO3)2 in 6.0 cm3 solution=0.0�2mol

Molarity of lead(II) nitrate solution=number ofmoles of Pb(NO3)2

volumeof solution

= 0.0�2mol

( 6.0�000 ) dm3

= 2.0mol dm–3

\ the concentrationof lead(II) nitrate solution is 2.0mol dm–3.

9� A (�) Theprecipitate lead(II) sulphate iswhite in colour.

(3) Aburette shouldbeused tomeasure the�.0 cm3portions of the lead(II) nitrate solution.

92 C (�) Number ofmoles of Fe2(SO4)3 =molarity of solution x volumeof solution

= 0.�00mol dm–3 x250.0

�000dm3

= 0.0250mol

63

(2) and (3) One formula unit of Fe2(SO4)3 contains 2 Fe3+ ions and3 SO42– ions.

\ number ofmoles of Fe3+ ions =2 x 0.0250mol = 0.0500mol

\ number ofmoles of SO42– ions=3 x 0.0250mol

= 0.0750mol

93 C (�) Conical flask isNOT required for dilutionprocess.

94 A (�) Wash a pipette firstwith distilled water and thenwith the solution they are going to contain.

(2) and (3) Wash a conical flask and a volumetric flaskwith distilledwater only.

95 A (�) �mole of tribasic acid (H3A) requires 3moles ofNaOH for complete neutralization.

H3A(aq)+3NaOH(aq) Na3A(aq)+3H2O(l)

(2) The number of hydrogen atoms � molecule of an acid contains may be different from the basicityof the acid.

For example, a molecule of ethanoic acid (CH3COOH) contains four hydrogen atoms but only thehydrogen atom in the –COOH group canundergodissociation. Therefore it is amonobasic acid.

96 D The following diagram shows the titration curve for the titration of dilute sodium hydroxide solutionwith dilute hydrochloric acid.

All the indicators change colour within the pH range of the vertical part of the titration curve. Henceall the indicators are suitable for the titration.

64

97 ABeaker Number of moles of H2SO4 = molarity of solution x volume of solution

A �mol dm–3 x�00

�000dm3=0.�mol

B 2mol dm–3 x50

�000dm3=0.�mol

(�) Equal numbers of moles of H2SO4 are present in both beakers. Equal masses of zinc react in bothbeakers.Hence equalmasses of zinc remain in bothbeakers.

(2) Equal amounts of zinc and H2SO4(aq) react in both beakers. Hence equal masses of gas areproduced in both cases.

(3) Equalmasses of zinc sulphate are produced in both cases.

As the volumes of solution in the two beakers are different, zinc sulphate solutions of differentconcentrations are produced.

98 A

ReactionNumber of moles of Mg

= mass

molar mass



�

�g24.3gmol–� = 0.04mol


=2mol dm–3 x50

�000dm3

=0.�mol


MgCl2(aq) +H2(g)

2


=2mol dm–3 x50

�000dm3

=0.�mol

Mg(s) + 2CH3COOH(aq)0.04mol 0.�mol

(CH3COO)2Mg(aq) +H2(g)

According to the equations, � mole of Mg reacts with 2 moles of HCl / CH3COOH to produce � moleofH2.During the reaction, 0.04moleofMg reactedwith0.08moleofHCl /CH3COOH. ThereforeHCl/CH3COOHwas in excess. The amount ofMg limited the amount ofH2 produced.

Number ofmoles ofH2 produced=0.04mol

(�) Magnesium reacted completely in both cases. Hence magnesium disappeared in both reactions atthe end.

(2) Equal amounts of hydrogenwereproduced in both reactions.

(3) HCl(aq) is a strong acidwhileCH3COOH(aq) is aweak acid.

2mol dm–3HCl(aq) had a higher concentrationof hydrogen ions than2mol dm–3CH3COOH(aq).

Hence HCl(aq) reacted more quickly with magnesium than CH3COOH(aq) did. The two reactionstook different times to complete.

65

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 ofmoles ofHCl

=�.2mol dm–3 x50

�000dm3

=0.060mol


=�.0mol dm–3 x60

�000dm3

=0.060mol


=�.0mol dm–3 x60

�000dm3

=0.060mol

HCl(aq) + NaOH(aq)0.060mol 0.060mol

NaCl(aq)+H2O(l)

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

CH3COONa(aq)+H2O(l)

According to the equations, � mole of HCl / CH3COOH requires � mole of NaOH for completeneutralization.

(�) During the reactions, 0.060mole ofHCl /CH3COOH reacts with 0.060mole ofNaOH.

Henceboth acids are completely neutralized.

(2) For the neutralization between a strong acid and a strong alkali, the heat released is 57 kJ for�mole ofwater produced.

For neutralization in which either the acid or alkali or both are weak, the heat released is less than57 kJ for �mole ofwater produced.

HCl(aq) is a strong acid andCH3COOH(aq) is aweak acid.

The heat released in the neutralization between HCl(aq) and NaOH(aq) is higher than that in theneutralization betweenCH3COOH(aq) andNaOH(aq).

Futhermore, the total volume of the reaction mixture in Beaker A is less than that of the reactionmixture in Beaker B.

Hence the temperature rise for the reaction mixture in Beaker A is higher than that for the reactionmixture Beaker B.

(3) Equal numbers ofNa+(aq) ions are present in both reactionmixtures.

However, the total volumesof the reactionmixtures are different.

Hence concentrations ofNa+(aq) ions in the reactionmixtures are different.

�00D Themolarity of a solution is thenumber ofmoles of solute dissolved in � dm3of the solution.

�0�B Ethanedioic acid crystals are suitable for preparing standard acid solutions because it has the followingcharacteristics:

• it is obtainable in a very pure form;

• it has a known chemical formula;

• it dissolves inwater completely at room temperature;

• it is stable anddoes not absorbmoisture from the air; and

• it has a highmolarmass tominimize weighing errors.

66

�02C Solid sodium hydroxide is NOT suitable for preparing a standard solution because it absorbs moisturefrom the air and cannot beweighed accurately.

�03A Wash a burette firstwith distilled water and thenwith the acid it is going to contain.

�04D Rinsing the conical flask with the solution it is about to contain would increase the number of molesof solute it holds.

It is not necessary to dry the conical flask as any water remaining in the flask will NOT change thenumber ofmoles of solute it holds.

�05D H2SO4(aq)+2NaOH(aq) Na2SO4(aq)+2H2O(l)

�00 cm3 of � mol dm–3 sulphuric acid require 200 cm3 of � mol dm–3 sodium hydroxide solution forcomplete neutalization.

Sulphuric acid is a dibasic acid. The number of hydrogen ions in �00 cm3 of � mol dm–3 sulphuric acidis twice as that of hydroxide ions in �00 cm3of �mol dm–3 sodiumhydroxide solution.

�06B The following diagram shows the titration curve for the titration of 20.0 cm3 of 0.� mol dm–3 NH3(aq)with 0.� mol dm–3HCl(aq).

Methyl orange is suitable indicator for the titration because the indicator changes colour within the pHrange of the vertical part of the titration curve.

�07C Number of moles of acid= molarity of solution x volume of solution

Reaction between acid and alkali



�000dm3

=0.0�mol

H2SO4(aq) +2NaOH(aq) Na2SO4(aq) +H2O(l)0.0�molnumber ofmoles ofNaOH required for neutralization=2 x 0.0�mol=0.02mol


=�mol dm–3 x20

�000dm3

=0.02mol

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

\ the acids require the samenumber ofmoles ofNaOH for complete neutralization.

67

Unit 18 Rate of reaction

Fillintheblanks

� concentration; time

2 colorimeter

3 a) surface area

b) concentration

c) temperature

4 catalyst; chemical

Trueorfalse

5 T In the oxidation of oxalate ions by permanganate ions, the intensity of the purple colour ofpermanganate ions decreases as the reactionproceeds.

When we shine light upon the reaction mixture, the absorbance of the reaction mixture is directlyproportional to the colour intensity of the reaction mixture and the concentration of the permanganateions in the reactionmixture.

Hence theprogress of the reaction canbe followedby a colorimeter.

6 T During thedecompositionof hydrogenperoxide solution, oxygengas 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 theprogressof the reactionbymeasuring thepressure inside thevesselwithapressuresensorconnected toa 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 unitvolume. As the reactant particles are consumed gradually, there are fewer particles per unit volume, i.e.the concentrationof reactants falls. So, the reaction slowsdown.

9 F The volumeof a liquid wouldNOT affect the rate of a reaction.

�0 F The surface areaofmarble chips is smaller than that of powderedmarble of the samemass.

Thereactionratebetweendilutehydrochloricacidandmarblechips isslowerthanthatbetweentheacidand powderedmarble.

�� F During the reaction between magnesium and an acid, magnesium would react with hydrogen ions inthe acid.

Hydrochloricacid isastrongacidwhileethanoicacid isaweakacid.Hence2moldm–3hydrochloricacidhas a higher concentrationof hydrogen ions than2mol dm–3 ethanoic acid does.

Thus the reaction between magnesium and 2 mol dm–3 hydrochloric acid is faster than that betweenmagnesiumand2mol dm–3 ethanoic acid.

68

�2 T

�3 T

�4 F Thephysical state of a catalystmaybe the same as thoseof the reactants.

For example, iron acts as a catalyst in the reaction between nitrogen gas and hydrogen gas to produceammonia gas.

�5 T Manganese(IV) oxide can catalyze thedecompositionof hydrogenperoxide.


�6 D Option D — A limestone status damage causedby acid rainmay take years.

�7 D Option D — The reactionbetween two solutions has thegreatest reaction rate.

�8 C Average reaction rate=0.�05mol dm–3

20 s = 5.25 x �0–3mol dm–3 s–�

�9 B

Instantaneous rate of reaction at 30 s =(0.�65 – 0.09)mol dm–3

(60 – 0) s = �.25 x �0–3mol dm–3 s–�

20 A Number ofmoles of Fe2O3 produced in �0.0 seconds =0.30

2mol

= 0.�5mol

Rate of consumptionof Fe2O3 =0.�5mol

�0.0 s = 0.0�5mol s–�

\ the rate of consumptionof Fe2O3 is �.5 x �0–2mol s–�.

69

2� B Rate= –�

2

d[N2O5(g)]

dt

=�

4

d[NO2(g)]

dt

=d[O2(g)]

dt

Instantaneous rate of decompositionofN2O5(g) = –d[N2O5(g)]

dt

=2

4

d[NO2(g)]

dt

=2

4 (3.8 x �0–5mol dm–3 s–�)

= �.9 x �0–5mol dm–3 s–�

22 D Instantaneous rate of decompositionofO2(g) =d[O2(g)]

dt

=�

4

d[NO2(g)]

dt

=�

4 (3.8 x �0–5mol dm–3 s–�)

= 9.50 x �0–6mol dm–3 s–�

23 A

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

25 A The rate of a reaction decreass with time. At the start, there are plenty of reactant particles per unitvolume. As the reactant particles are consumed gradually, there are fewer particles per unit volume, i.e.the concentrationof reactants fall. So, the reaction slowsdown.

26 B In the oxidation of oxalate ions by permanganate ions, the intensity of the purple colour ofpermanganate ions decreases as the reactionproceeds.



27 C Consider a reaction in which a gas is produced. If the reaction vessel is a closed system, the pressureinside the vessel will increase. We can follow the progress of the reaction by measuring the pressureinside the vesselwith 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. Hencethe progress of the reaction CANNOT be followed by using a data-logger with a pressuresensor.

70

28 B Option A — The antacid tablets in Experiments I and II have the same surface area. So, they CANNOTbeused for comparison.

Option B — The only difference between Experiments I and IV is the surface area of the antacidtablets.

Option C — The antacid tablets in Experiments II and III have the same surface area. So, theyCANNOTbeused for comparison.

Option D — The concentration and temperature of the acids in Experiments III and IV are different. So,theyCANNOT beused for comparison.

29 C Option A — The concentration and temperature of the acids in Experiments I and II are different. So,theyCANNOT beused for comparison.

Option C — Theonly differencebetween Experiments II and III is the concentrationof the acids.

Option D — The surface areaof the tablets, concentration and temperatureof the acids in ExperimentsIII and IV are different. So, theyCANNOTbeused for comparison.

30 B The hydrochloric acid was in excess in each case. Hence the amount of calcium carbonate limited theamount of carbon dioxide produced, i.e. the total loss in mass of contents of the reaction flask wasthe same in both cases.

Powdered marble had a greater surface area thanmarble lumpsof the samemass.

The rate of the reaction increasedwhen the surface areaofmarblewas increased.

3� D The initial rate of reactionwoulddecreasewhenhydrochloric acid of a lower concentration is used.

32 D During the reactionbetweenmarble and an acid,marblewould react with hydrogen ions in the acid.

Hydrochloric acid is a strong acidwhile ethanoic acid is aweak acid.

Therefore � mol dm–3 hydrochloric acid has a higher concentration of hydrogen ions than � mol dm–3ethanoic acid.

Hence the initial rate of reactionwoulddecreasewhen ethanoic acid is used.

33 A The initial rate of the reactionwould increasewith an increase in temperature.

34 C The total volumeof hydrogen formed increased as the reactionproceeded.

After some time, no more hydrogen was produced. So, its volume no longer changed. The reactionwasover.

35 B Option A — All the magnesium disappeared after reaction. Magnesium was the limiting reactant. Theamount ofmagnesium limited the amount of hydrogen formed.

Hence increasing the volume of hydrochloric acid used would NOT affect the final volumeof hydrogen formed.

Option B — Using a greater volume of acid of the same concentration had NO effect on the rate ofreaction.

7�

36 D OptionsA andB— During the reaction between sodium carbonate and an acid, sodium carbonatewould reactwith hydrogen ions in the acid.

Sulphuric acid is a dibasic acid while nitric acid is a monobasic acid. Hencesulphuric acid has a higher concentrationof hydrogen ions.

Thus the reaction rate between sodium carbonate and � mol dm–3 sulphuric acidis higher than that between sodium carbonate and�mol dm–3 nitric acid.

Options C andD— The rate of reaction at 40 °C is higher than that at 20 °C.

37 AReaction mixture

Number of moles of Zn

= mass

molar mass

Number of moles of H2SO4


Reaction between Zn and H2SO4

�

5g65.4gmol–� = 0.076mol

2mol dm–3 x�00

�000dm3

=0.2mol

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

ZnSO4(aq) +H2(g)

2 �mol dm–3 x�00

�000dm3

=0.�mol

Zn(s) + H2SO4(aq)0.076mol 0.�mol

ZnSO4(aq) +H2(g)

According to the equation, �mole of Zn reactswith �moleofH2SO4 toproduce�moleofH2.Duringthe reaction, 0.076 mole of Zn reacted with 0.076 mole of H2SO4. Therefore the amount of Zn limitedthe amount ofH2 produced (0.076mole in both cases).

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

As the same amount of gas was produced in both reactions the total mass lost of bothreactionmixtureswould be the same.

Options B,C andD— The rate of reaction between zinc and 2 mol dm–3 sulphuric acid was higherthan that between zinc and�mol dm–3 sulphuric acid.

Hence the reaction between zinc and 2 mol dm–3 sulphuric acid took less timeto complete.

38 C As calcium carbonate disappeared in both experiments after reaction, it could be deduced that calciumcarbonatewas the limiting reactant, i.e. its amount limited the amount of carbondioxide formed.

Curve Y showed a smaller loss in mass, and thus the mass of calcium carbonate used was smaller (i.e.OptionC).

Furthermore, the tangent to curve Y was less steep than that to curve X, i.e. the rate of the reactionrepresented by curveYwas lower.

This agreed with the change stated in Option C. The rate of the reaction would decrease when a lumpof calcium carbonatewas used insteadof calcium carbonate powder.

39 C Curve II showed a greater volume of oxygen formed. Thus the number of moles of hydrogen peroxidedecomposedwasgreater, i.e. adding somehydrogenperoxide solution (OptionC).

Furthermore, the tangent to curve II was less steep than that to curve I, i.e. the rate of decompositionrepresented by curve IIwas lower.

This agreed with the change stated in Option C. Adding some 0.� mol dm–3 hydrogen peroxidesolution to the original � mol dm–3 hydrogen peroxide solution would dilute the original hydrogenperoxide solution. The rate of decompositionwoulddecrease.

72

40 D During the reactionbetweenmagnesiumanddilute hydrochloric acid, hydrogenwas formed.

If the reaction vessel was a closed system, the pressure inside the vessel would increase. We couldfollow the progress of the reaction by measuring the pressure inside the vessel with a pressure sensorconnected to a data-logger interface and a computer.

As magnesium ribbons disappeared in both experiments after reaction, it could be deduced thatmagnesiumwas 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 curveYwashigher.

Option A — Lowering the temperaturewoulddecrease the rate of the reaction.

Option B — Using a greater volumeof hydrochloric acid hadNOeffect on the rate of the reaction.

Option C — Using more magnesium ribbon would increase the amount of hydrogen formed, i.e.increase thepressure change.

Option D — Usingmore concentratedhydrochloric acidwould increase the rate of the reaction.

4� C The second experiment took a longer time to complete, i.e. the rate of reaction of the secondexperiment is lower.

Option A — Using a catalystwould increase the rate of the reaction.

Option B — UsingpowderedMgwould increase the rate of the reaction.

Option C — HCl(aq) is a strong acidwhileCH3COOH(aq) is aweak acid.

Therefore � mol dm–3 HCl(aq) has a higher concentration of hydrogen ions than � moldm–3 ethanoic acid does.

During the reaction between Mg and an acid, Mg would react with hydrogen ions in theacid.

Hence the rate of the reaction between Mg and CH3COOH(aq) (i.e. Option C) would belower than that betweenMgandHCl(aq).

Option D — Increasing the temperaturewould 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

�

8g�00.�gmol–� = 0.08mol


�000dm3

=0.�mol

CaCO3(s) + 2HCl(aq)0.08mol 0.�mol

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

2 2mol dm–3 x�00

�000dm3

=0.2mol

CaCO3(s) + 2HCl(aq)0.08mol 0.2mol

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

According to the equation, �mole ofMg reactswith 2moles ofHCl to produce�mole ofCaCl2.

In Reaction�, 0.�mole ofHCl reactedwith 0.05mole ofCaCO3 to give 0.05mole ofCaCl2.

Thus CaCO3was in excess, i.e.OptionAwas INCORRECT.

73

In Reaction2, 0.08mole ofCaCO3 reactedwith 0.�6mole ofHCl to give 0.08mole ofCaCl2.

ThusHClwas in excess, i.e.OptionBwas INCORRECT.

Option C — Different amounts of CaCl2, and hence calcium chloride solution of differentconcentrations,wereproduced in the two reactions.

Option D — The initial rateof the reactionbetweenCaCO3anda less concentratedHCl(aq) (Reaction�)is smaller than that betweenCaCO3 and amore concentratedHCl(aq) (Reaction 2).

43 B HCl(aq) andCH3COOH(aq) reactwith calcium carbonate according to the following equations:

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


�000dm3=0.�mol

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


�000dm3=0.�mol

Option A — During the reaction between CaCO3 and an acid, CaCO3 reacts with hydrogen ions in theacid.

HCl(aq) is a strong acidwhileCH3COOH(aq) is aweak acid.

� mol dm–3 HCl(aq) has a higher concentration of hydrogen ions than � mol dm–3CH3COOH(aq) does.

Hence the initial rate of reaction in BeakerA is higher than that in Beaker B.

Option B — As equal numbers of moles of HCl and CH3COOH are used, the same amount of marblechipswouldbe consumed in both cases.

Hence equalmasses ofmarble chips remain in the twobeakers after reaction.

OptionsC andD— As excess marble chips are used, HCl(aq) and CH3COOH(aq) are the limitingreagents. Their amounts limit the amount of carbondioxidegas produced.

0.� mole of HCl and 0.� mole of CH3COOH would produce the same amount ofgas.

44 B OptionsA andB—HCl is amonobasic andH2SO4 is a dibasic acid.

Therefore 2 mol dm–3 HCl(aq) has a lower concentration of hydrogen ions than 2mol dm–3 sulphuric acid does.

During the reaction between Zn and an acid, Zn would react with hydrogen ionsin the acid.

Hence the rate of the reaction between Zn and HCl(aq) is lower than thatbetween Zn andH2SO4(aq).

74

Options C andD—

ReactionNumber of moles of Zn

= mass

molar mass


volume of solutionReaction between Zn and acid

�

�0g65.4gmol–� = 0.�5mol


=2mol dm–3 x�00

�000dm3

=0.2mol

Zn(s) + 2HCl(aq)0.�5mol 0.2mol

ZnCl2(aq) +H2(g)

2


=2mol dm–3 x�00

�000dm3

=0.2mol

Zn(s) + H2SO4(aq)0.�5mol 0.2mol

ZnSO4(aq) +H2(g)

According to the first equation, �mole of Zn reacts with 2moles ofHCl.

In Reaction�, 0.2mole ofHCl reactedwith 0.�mole of Zn.

Thus Znwas in excess, i.e.OptionCwas INCORRECT.

According to the second equation, �mole of Zn reactswith �mole of H2SO4.

In Reaction2, 0.�5mole of Zn reactedwith 0.�5mole ofH2SO4.

Thus sulphuric acidwas in excess, i.e.OptionDwas INCORRECT.

45 B The tangent to the curve for experiment I at the start was less steep than that for experiment 2, i.e.the initial rate of reactionof experiment Iwas lower than that of experiment 2.

Thus OptionsA andC are INCORRECT.

The curve for experiment 2 went flat easiler than that for experiment I, i.e. the rate of reaction forexperiment 2 fell to zero easiler.Hence thegraph shown inOptionB is correct.

46 B HCl(aq) andCH3COOH(aq) reactwith calcium carbonate according to the following equations:

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

�.2mol dm–3 x90

�000dm3=0.��mol

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

0.9mol dm–3 x�20

�000dm3=0.��mol

As excess CaCO3 is used, HCl(aq) and CH3COOH(aq) are the limiting reagents. Their amounts limit theamount of carbondioxide gas produced.

0.�� mole of HCl and 0.�� mole of CH3COOH would produce the same amount of gas (i.e. Options AandC are INCORRECT).

During the reactionbetweenCaCO3 and an acid,CaCO3 reactswith hydrogen ions in the acid.

HCl(aq) is a strong acidwhileCH3COOH(aq) is aweak acid.

�.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 � is higher than that in Experiment 2.

\ the curves inOptionB show the results of the experiments.

75

47 D

ExperimentNumber of moles of Zn

= mass

molar mass


volume of solutionReaction between Zn and acid

�

5g65.4gmol–� = 0.076mol



�000dm3

=0.�mol

Zn(s) + 2HCl(aq)0.076mol 0.�mol

ZnCl2(aq) +H2(g)

2



�000dm3

=0.�mol

Zn(s) + H2SO4(aq)0.076mol 0.�mol

ZnSO4(aq) +H2(g)

According to the first equation, �mole of Zn reactswith 2moles ofHCl to produce�mole ofH2.

In Experiment �, 0.�mole ofHCl reactedwith 0.05mole of Zn to produce0.05mole ofH2.

According to the second equation, �mole of Zn reactswith �mole ofH2SO4.

In Experiment 2, 0.076mole of Zn reactedwith 0.076mole ofH2SO4 to produce0.076mole ofH2.

Thus the volume of gas produced in Experiment � was smaller than that produced in Experiment 2 (i.e.OptionsA andB are INCORRECT).

During the reactionbetween Zn and an acid, Zn reactswith hydrogen ions in the acid.

HCl(aq) is amonobasic acidwhileH2SO4(aq) is a dibasic acid.

�mol dm–3HCl(aq) has a lower concentrationof hydrogen ions than�mol dm–3H2SO4(aq) does.

Hence the initial rate of reaction in Experiment � is lower than that in Experiment 2.

\ the curves inOptionD show the results of the experiments.

48 D The concentration of the brown colour iodine in the reaction mixture decreased as the reactionproceeded. The reaction mixture became lighter in colour gradually. Thus the reaction mixture absorbedless and less light and so the absorbancewent down.

The concentrationof I2 in Sample �was the same as that in Sample 2.

Hence the initial absorbance of Sample � was the same as that of Sample 2 (i.e. Options B and C areINCORRECT).

The concentrationof propanone in Sample �was lower than that in Sample 2.

The rate of reaction for Sample �was lower and thus the tangent to its curvewouldbe less steep.

\ the curves inOptionD show the results.

49 B

50 D Option D — Manganese(IV) oxide can act as a catalyst in thedecompositionof hydrogenperoxide.

5� C (�) Following the progress of a reaction using physical property may require the use of expensiveinstructments, e.g. colorimeter.

52 D

76

53 A During the reactionbetweenMgand an acid,Mg reactswith hydrogen ions in the acid.

(�) HCl(aq) is amonobasic acidwhileH2SO4(aq) is a dibasic acid.

2mol dm–3H2SO4(aq) has a higher concentrationof hydrogen ions than2mol dm–3HCl(aq) does.

Hence the initial rate of reaction would increase when using 2 mol dm–3 sulphuric acid instead of 2mol dm–3 hydrochloric acid.


� mol dm–3 HCl(aq) has a higher concentration of hydrogen ions than � mol dm–3 CH3COOH(aq)does.

Hence the initial rate of reaction would decrease when using 2 mol dm–3 ethanoic acid instead of 2mol dm–3 hydrochloric acid.

(3) Using a greater volume of � mol dm–3 hydrochloric acid would NOT affect the initial rate of thereaction.

54 B Zinc reactswithH2SO4(aq) according to the following equation:


Beaker A excess �mol dm–3 x�00

�000dm3=0.�mol

Beaker B excess 2mol dm–3 x50

�000dm3=0.�mol

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

(�) The initial rate of reaction is higher in Beaker B as sulphuric acid of a higher concentration is used.

(2) Thehydrogenproducedwould escape. Thus themass of each reactionmixture woulddecrease.

As zinc is in excess, the amount of acid limits the amount of hydrogenproduced.

Each reaction mixture contains 0.� mole of H2SO4. Hence the same amount of hydrogen wouldform. Thus the total losses inmass of the two reactionmixtures after reaction are the same.

(3) Each reaction mixture contains 0.� mole of H2SO4. Hence the same amount of ZnSO4 would beproduced in each case.However, the total volumesof the reactionmixtures are different.

Thus zinc sulphate solutions of different concentrations are produced in bothbeakers.

55 B

ReactionNumber of moles of Mg

= mass

molar mass



�

�g24.3gmol–� = 0.04mol

2mol dm–3 x50

�000dm3

=0.�mol


MgCl2(aq) +H2(g)

2


4mol dm–3 x50

�000dm3

=0.2mol

Mg(s) + 2HCl(aq)0.04mol 0.2mol

MgCl2(aq) +H2(g)

According to the equation, �mole ofMg reactswith 2moles ofHCl to produce�mole ofH2.

77

In Reaction�, 0.04mole ofMg reactedwith 0.08mole ofHCl to give 0.04mole ofH2.

ThusHClwas in excess.

In Reaction2, 0.04mole ofMg reactedwtih 0.08mol ofHCl to give 0.04mole ofH2.

ThusHCl(aq)was in excess.

(�) Magnesium was the limiting reactant in both reactions. Thus magnesium disappeared in both casesafter reaction.

(2) The same amount of hydrogen (0.04mole)was produced in both reactions.

(3) The initial rate of Reaction2wasgreater as amore concentratedhydrochloric acidwas used.

56 A

57 B Carbondioxidegas is produced in the reactionbetween sodium carbonate anddilute hydrochloric acid.

As the carbondioxidegas escapes, the reactionmixture gets lighter as the reactionproceeds.

58 B In the reaction between iodine and propanone, the intensity of the brown colour of iodine decreasesas the reactionproceeds.

I2(aq) +CH3COCH3(aq) CH3COCH2I(aq)+HI(aq) brown colourless colourless

Hence theprogress of the reaction canbe followedbyusing a colorimeter.

59 A The rate of a reaction decreases with time. At the start, there are plenty of reactant particles per unitvolume. As the reactant particles are consumed gradually, there are fewer particles per unit volume, i.e.the concentrationof reactants fall. So, the reaction slowsdown.

60 D During the reaction between calcium carbonate and an acid, calcium carbonate reacts with hydrogenions in the acid.

Nitric acid is amonobasic acidwhile sulphuric acid is a dibasic acid.

� mol dm–3 nitric acid has a lower concentration of hydrogen ions than � mol dm–3 sulphuric aciddoes.

Hence the rate of reaction between � mol dm–3 nitric acid and calcium carbonate is smaller than thatbetween�mol dm–3 sulphuric acid and calcium carbonate.

6� A

62 D Inmost cases, the rate of a reaction increaseswhen the temperature is increased.

63 C A catalystCANNOT change the amount of product formed in a reaction.

78

Part B Topic-based exercise


� B Option A — Phenolphthalein is colourless in acids.

Option B — Citric acid is aweak acid. It only partially dissociates inwater.


Hence an aqueous solution of citric acid contains both citric acid molecules and hydrogenions.

Option C — An aqueous solution of citric acid reacts with sodium hydrogencarbonate to give carbondioxidegas.

Option D — There isNO reactionbetween an aqueous solutionof citric acid and copper.

2 A Option A — Dilute sodiumhydroxide solution turnsmethyl orange yellow.

Option C — Sodiumhydroxide is a strong alkali.

3 D Zn2+(aq)+2OH–(aq) Zn(OH)2(s)

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

A calcium chloride solution noprecipitate

B chromium(III) sulphate greenprecipitate

C iron(III) sulphate reddishbrownprecipitate

D potassium chloride noprecipitate

5 B Option Solution Addition of Na2CO3(aq) to solution

A aluminium sulphate solution white precipitate

B ammonium nitrate solution noprecipitate

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 blueprecipitate

B Na2CO3(aq) noprecipitate

C Pb(NO3)2(aq) awhite precipitate, insoluble in excess alkali

D ZnCl2(aq) awhite precipitate, soluble in excess alkali

79

7 C Option Solution Reaction with Ba(NO3)2(aq) Reaction with NaOH(aq)

A calcium chloride noprecipitate white precipitateCa(OH)2

B iron(II) chloride noprecipitate greenprecipitate Fe(OH)2

C magnesium sulphate white precipitate BaSO4 white precipitateMg(OH)2

D potassium sulphate white precipitate BaSO4 noprecipitate

8 C Options A andC— Carbonates give a gas (carbondioxide)with dilute hydrochloric acid.

Upon the addition of dilute aqueous ammonia, a solution containing zinc ionsgives awhite precipitate which is soluble in excess alkali.

\ X is zinc carbonate.

Options B andD— Sulphates doNOTgive a gaswith dilute hydrochloric acid.

9 C A solution containing iron(II) ion gives a green precipitate (Fe(OH)2) with dilute sodium hydroxidesolution.

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)

�0 D Barium chloride solution reacts with sodium carbonate soluton to give a white precipitate, bariumcarbonate.

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)

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

�2 D Option Solutions mixed Any precipitate?

A CuO(s) andH2SO4(aq) noprecipitate

B FeSO4(aq) and NaOH(aq) greenprecipitate Fe(OH)2

C NH4Cl(aq) andCa(NO3)2(aq) noprecipitate

D Pb(NO3)2(aq) andHCl(aq) white precipitate PbCl2

�3 C Option Solutions mixed Any precipitate?

A BaCl2(aq) andNa2CO3(aq) white precipitate BaCO3

B KCl(aq) andAgNO3(aq) white precipitateAgCl

C Na2SO4(aq) and MgCl2(aq) noprecipitate

D NaOH(aq) and NiSO4(aq) greenprecipitateNi(OH)2

80

�4 A OptionsA andC— Chlorides give awhite precipitate (AgCl)with silver nitrate solution.

Ag+(aq)Cl–(aq) AgCl(s)

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


�5 D Zinc reacts with dilute hydrochloric acid to give zinc chloride solution and hydrogen. Zinc chloridesolution gives a white precipitate (Zn(OH)2) with dilute aqueous ammonia. Zn(OH)2 dissolves in excessalkali 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)

�6 C Molarmass ofCa(OH)2= [40.�+2 x (�6.0+�.0)] gmol–�

= 74.�gmol–�

Number ofmoles ofCa(OH)2 =mass

molar mass

=5.93g

74.�gmol–�

= 0.0800

Molarity of calciumhydroxide solution=number ofmoles ofCa(OH)2

volumeof solution

= 0.0800mol

( 400.0�000 ) dm3

= 0.200mol dm–3

\ themolarity of the calciumhydroxide solution is 0.200mol dm–3.

�7 D Number ofmoles ofNaOH in �50.0 cm3of 2.00mol dm–3 solution

=2.00 mol dm–3 x�50.0

�000dm3

=0.300 mol

Number ofmoles ofNaOH in 80.0 cm3of �.20mol dm–3 solution

=�.20 mol dm–3 x80.0

�000dm3

=0.0960 mol

Total number ofmoles ofNaOH in the resulting solution = (0.300 +0.0960)mol = 0.396 mol

Total volumeof the resulting solution = (�50.0 +80.0) cm3

= 230.0 cm3

8�

Concentrationof the resulting solution = 0.396mol

( 230.0�000 ) dm3

= �.72mol dm–3

\ the concentrationof the resulting solution is �.72mol dm–3.

�8 C pHof solution = –log�0(�.58 x �0–�2) = ��.8

�9 A Option A — Nitric acid dissociates completely according to the following equation:

HNO3(aq) H+(aq) + NO3–(aq)

0.0�0mol dm–3 ?mol dm–3

According to the equation, �mole ofHNO3dissociates to give �mole of hydrogen ions.

i.e. concentrationof hydrogen ions=0.0�0mol dm–3

pHof acid = –log�0(0.0�0) = –(–2.0) =2.0

\ pHof 0.0�0mol dm–3HNO3 is 2.0.

Option B — CH3COOH is a weak acid while HNO3 is a strong acid. Hence the concentration ofhydrogen ions in0.0�0moldm–3CH3COOH is lower than that in0.0�0moldm–3HNO3, i.e.thepHofCH3COOH is higher than that ofHNO3.

Option C — H2SO4 is a dibasic acid while HNO3 is a monobasic acid. Hence the concentration ofhydrogen ions in 0.0�0 mol dm–3 H2SO4 is higher than that in 0.0�0 mol dm–3 HNO3, i.e.thepHofH2SO4 is lower than that ofHNO3.

Option D — ThepHof 0.0�0mol dm–3NaOH is greater than7.

20 C (�) and (3) Sodiumhydroxide solution is a strong alkaliwhile aqueous ammonia is aweak alkali.

Thus 0.� mol dm–3 sodium hydroxide solution is more alkaline than 0.� mol dm–3 aqueousammonia, i.e. the pH of sodium hydroxide solution is greater than that of aqueousammonia.

(2) ThepHof 0.�mol dm–3 lactice acid is less than7.

Thus theorder of pHof the solutions is: (2) < (�) < (3).

2� D pHof soil before treatment=4

pHof soil after treatment=6

ConcentrationofH+(aq) ions in soil before treatment =�0–4mol dm–3

ConcentrationofH+(aq) ions in soil after treatment =�0–6mol dm–3

ConcentrationofH+(aq) ions in soil after treatment

ConcentrationofH+(aq) ions in soil before treatment=

�0–6mol dm–3

�0–4mol dm–3

=�

�00

\ the concentrationofH+(aq) ions in the soil decreasedby a factor of �00.

82

22 C pHof acid before dilution =�

pH of acid after dilution =3

ConcentrationofH+(aq) ions in acid before dilution=�0–�mol dm–3

ConcentrationofH+(aq) ions in acid after dilution=�0–3mol dm–3

ConcentrationofH+(aq) ions in acid after dilution

ConcentrationofH+(aq) ions in acid before dilution=

�0–3mol dm–3

�0–�mol dm–3

=�

�00

i.e.the acid is diluted�00 times.

\ diluting �0 cm3 of the acid to � 000 cm3 with diluted water would cause the pH value to changefrom� to 3.

23 B Option A — The concentrationof hydrogen ions in lemon juice is higher than that inmilk.

Option B —concentrationof hydrogen ions in seawater

concentrationof hydrogen ions in laundry detergent=

�0–8mol dm–3

�0–��mol dm–3

= � 000

\ the concentration of hydrogen ions in sea water is � 000 times greater than that inlaundry detergent.

Option C —concentrationof hydrogen ions in rainwater

concentrationof hydrogen ions in seawater=

�0–5mol dm–3

�0–8mol dm–3

= � 000

\ the concentration of hydrogen ions in rainwater is � 000 times greater than that inseawater.

Option D —concentrationof hydrogen ions in lemon juice

concentrationof hydrogen ions in soap solution=

�0–2mol dm–3

�0–�0mol dm–3

= �08

\ the concentration of hydrogen ions in lemon juice is �08 times greater than that insoap solution.

24 B Sodium chloride is neutral.Adding it to sulphuric acidwouldNOT affect thepHof the acid.

25 A When we describe acids as strong and weak, we are talking about the extent of their dissociation inwater. When we talk about concentration, we are referring to the amount of an acid in a unit volumeof solution.

Ethanoic acid is aweak acid. 0.�mol dm–3 solutionof ethanoic acid is a dilute solutionof aweak acid.

26 D HF is a weak acid while H2SO4 and HClO4 are strong acids. Therefore 0.� mol dm–3 HF has the lowestconcentrationof hydrogen ions. ThepHof 0.�mol dm–3HF is thus thehighest among the three acids.

H2SO4 is a dibasic acid while HClO4 is a monobasic acid. Therefore 0.� mol dm–3 H2SO4 has a higherconcentration of hydrogen ions than 0.� mol dm–3 HClO4. The pH of 0.� mol dm–3 H2SO4 is thus lowerthan that of the0.�0mol dm–3HClO4.

Theorder of thepHof the three acids is:H2SO4<HClO4<HF.

83

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

Options A andC— H2SO4(aq) is a dibasic acid while HCl(aq) is a monobasic acid. Hence �.0 mol dm–3H2SO4(aq) has a higher concentration of mobile ions than �.0 mol dm–3 HCl(aq)does.

\ �.0mol dm–3 has thegreatest electrical conductivity.

Option D — C6H�2O6(aq) doesNOT conduct electricity.

28 A Ethanoic acid is aweak acid andhydrochloric acid is a strong acid.

Hencce for acids of the same concentration, ethanoic acid has a lower concentratation of hydrogenions thanhydrochloric acid does.

Thus thepHof 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 57kJ for �mole ofwater produced.

For neutralization in which either the acid or alkali or both are weak, the heat released isless than 57 kJ for � mole of water produced. The is because some energy is consumedwhen the weak acid and weak alkali dissociate to give hydrogen ions and hydroxide ionsbefore neutralization.

CH3COOH(aq) is aweak acidwhileHCl(aq) is a strong acid.

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

30 A All nitrates are soluble inwater.

3� A OptionsC andD—Compoundsof potassiumand sodiumare soluble inwater.

32 B Copper(II) sulphate is a soluble salt. To prepare it, mix a dilute acid (dilute sulphuric acid in this case)with a metal, an insoluble base or an insoluble carbonate. However, copper (metal) CANNOT be usedas it does not reactwith dilute sulphuric acid.

33 D Option A — Calcium reacts with dilute sulphuric acid to produce insoluble calcium sulphate. Calciumsulphate forms a protective layer on the surface of calcium. This prevents further reactionbetween calciumanddilute sulphuric acid.

Option B — Copper hasNO reactionwith 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 reactionbetween lead anddilute sulphuric acid.

Option D — Magnesium reactswith dilute hydrochloric acid to givemagnesium chloride.

34 C Option C — Potassium chloride is prepared via the titraton of potassium hydroxide solution with dilutehydrochloric acid. Pipette andburette are required in theprocess.

84

35 A Molarmass ofH2SO4= (2 x �.0+32.�+4 x �6.0) gmol–�

= 98.�mol–�

Molarity of sulphuric acid =number ofmoles ofH2SO4

volumeof solution

4.00mol dm–3 =number ofmoles ofH2SO4

volumeof solution

Number ofmoles ofH2SO4= 4.00mol dm–3 x 2.00dm3

= 8.00mol

Mass ofH2SO4 required=number ofmoles ofH2SO4 xmolarmass ofH2SO4 = 8.00mol x 98.�gmol–�

= 785g

36 D Molarmass ofNa2SO4= (2 x 23.0+32.�+4 x �6.0) gmol–�

= �42.�mol–�

Number ofmoles ofNa2SO4=mass

molarmass

=20.0g

�42.�gmol–�

= 0.�4�mol

One mole ofNa2SO4 contains 2moles ofNa+ ions.

i.e.number ofmoles ofNa+ ions=2 x 0.�4�mol = 0.282mol

ConcentrationofNa+ ions in solution=number ofmoles ofNa+ ions

volumeof solution

= 0.282mol

( �00.0�000 ) dm3

= 2.82mol dm–3

\ the concentrationof sodium ions in the solution is 2.82mol dm–3.

37 C Number ofmoles of potassiumphosphate =molarity of solution x volumeof solution

= 0.500mol dm–3 x250.0

�000dm3

= 0.�25mol

One mole of potassiumphosphate (K3PO4) contains 4moles of ions.

\ number ofmoles of ions =4 x 0.�25mol = 0.500mol

Number of ions =0.500mol x 6.02 x �023mol–�

= 3.0� x �023

85

38 C SupposeV cm3ofNaCl(aq) andV cm3of ZnCl2(aq) aremixed.

Number ofmoles ofNaCl in 0.�0mol dm–3 solution =0.�0mol dm–3 xV

�000dm3

= 0.�0 xV

�000mol

Number ofmoles of ZnCl2 in 0.20mol dm–3 solution=0.20mol dm–3 xV

�000dm3

= 0.20 xV

�000mol

Onemole ofNaCl contains �mole ofCl– ions andonemole of ZnCl2 contains 2moles ofCl– ions.

Total number ofmoles ofCl– ions = (0.�0 xV

�000+2 x 0.20 x

V

�000)mol

= (0.50 xV

�000)mol

Total volumeof themixture = (V+V) cm3

= 2V cm3

ConcentrationofCl– ions in themixture =total number ofmoles ofCl– ions

total volumeof themixture

=(0.5 x

V�000 )mol

2V�000 dm3

= 0.25mol dm–3

\ the concentrationof chloride ions in themixture is 0.25mol dm–3.

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


0.0250 mol dm–3 ?mol dm–3


i.e.concentrationof hydrogen ions=2 x 0.0250mol dm–3

= 0.0500mol dm–3

pH of acid= –log�0(0.0500) = –(–�.30) = �.30

\ pHof the acid sample is �.30.

86

40 D Onemole of bariumnitrate contains 2moles ofNO3– ions.

\ number ofmoles of Ba(NO3)2 =0.0250

2mol

= 0.�25mol

Molarity of bariumnitrate solution=number ofmoles of Ba(NO3)2

volumeof solution

= 0.0�25mol

( 22.0�000 ) dm3

= 0.568mol dm–3

\ themolarity of thebariumnitrate solution is 5.68 x �0–�mol dm–3.

4� A (MV) before dilution = (MV) after dilution,whereM=molarity, V= volume

4.0 x25.0

�000=0.50 x

V

�000 V=200 cm3


\ volumeofwater added = (200 – 25.0) cm3

= �75 cm3

42 B Mass ofH2SO4 in 25.0m3 acid=4�050000g x 92.0% = 37800000g


= 98.�gmol–�

Number ofmoles ofH2SO4 =mass

molarmass

=37800000g

98.�gmol–�

= 385000mol

Molarity of the sulphuric acid=number ofmoles ofH2SO4


=385000mol

25000dm3

= �5.4mol dm–3

\ the concentrationof the sulphuric acid is �5.4mol dm–3.

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

87

44 C Na2CO3(aq) + 2HCl(aq) 2NaCl(aq) + CO2(g) + H2O(l) 0.�50 mol dm–3 0.250mol dm–3

25.0 cm3 ? cm3

Number ofmoles ofNa2CO3 in 25.0 cm3 solution=molarity of solution x volumeof solution

= 0.�50mol dm–3 x25.0

�000dm3

= 0.00375mol

According to the equation, �mole ofNa2CO3 requires 2moles ofHCl for complete reaction.

i.e. number ofmoles ofHCl =2 x 0.00375mol = 0.00750mol

Volume ofHCl required=number ofmoles ofHCl


=0.00750mol

0.250mol dm–3

= 0.0300dm3

= 30.0 cm3

45 C We can represent thedibasic acidXbyH2A.

2KOH(aq) + H2A(aq) K2A(aq) + 2H2O(l) 0.�50 mol dm–3 ?mol dm–3

30.0 cm3 �0.0 cm3

Number ofmoles ofKOH in 30.0 cm3 solution=molarity of solution x volumeof solution

= 0.�50mol dm–3 x30.0

�000dm3

= 0.00450mol

According to the equation, �mole ofH2A requires 2moles ofKOH for complete neutralization.

i.e. number ofmoles ofH2A in �0.0 cm3 solution=0.00450

2mol

= 0.00225mol

Molarity of acid solution=number ofmoles ofH2A

volumeof solution

= 0.00225mol

( �0.0�000 ) dm3

= 0.225mol dm–3

\ themolarity of the acid soslution is 2.25 x �0–�mol dm–3.

88

46 A We can represent thedibasic acid byH2X.

2NaOH(aq) + H2X(aq) Na2X(aq) + 2H2O(l) 0.�00 mol dm–3 ?mol dm–3

20.6 cm3 25.0 cm3

Number ofmoles ofNaOH in 20.6 cm3 solution=molarity of solution x volumeof solution

= 0.�00mol dm–3 x20.6

�000dm3

= 0.00206mol


i.e. number ofmoles ofH2X in 25.0 cm3 solution =0.00206

2mol

= 0.00�03mol

Molarity of acid solution=number ofmoles ofH2A

volumeof solution

= 0.00�03mol

( 25.0�000 ) dm3

= 0.04�2mol dm–3

\ themolarity of the acid solution is 0.04�2mol dm–3.

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

20.0 cm3


= 2mol dm–3 x20.0

�000dm3

= 0.04mol

According to the equation, �mole of Znwill react completely with 2moles ofHCl.

i.e. number ofmoles of Zn =0.04

2mol

= 0.02mol

\ 0.02mole of zincwill react completely with thehydrochloric acid.

89

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 ofmoles of Sr(OH)2 in 25.0 cm3 solution=molarity of solution x volumeof solution

= 0.200mol dm–3 x25.0

�000dm3

= 0.00500mol

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

i.e. number ofmoles ofHCl=2 x 0.00500mol = 0.0�00mol

Molarity of hydrochloric acid=number ofmoles ofHCl

volumeof solution

= 0.0�00mol

( 20.0�000 ) dm3

= 0.500mol dm–3

\ concentrationof thehydrochloric acid is 0.500mol dm–3.

49 A Fe(OH)3(s) + 3HNO3(aq) Fe(NO3)3(aq) + 3H2O(l) �.07 g 0.60mol dm–3

? cm3

Molarmass of Fe(OH)3 = [55.8+3 x (�6.0+�.0)] gmol–�

= �06.8gmol–�

Number ofmoles of Fe(OH)3 =mass

molar mass

=�.07g

�06.8gmol–�

= 0.0�00mol

According to the equation, �mole of Fe(OH)3 requires 3moles ofHNO3 for complete neutralization.

i.e. number ofmoles ofHNO3 =3 x 0.0�00mol = 0.0300mol

Volume of 0.60mol dm–3 nitric acid required for complete neutralization

=number ofmoles ofHNO3


=0.0300mol

0.60 mol dm–3

= 0.050 dm3

=50 cm3

\ 50 cm3of nitric acidwere required for complete neutralization.

90

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

�50.0 cm3

Molarmass ofAl2O3 = (2 x 27.0+3 x �6.0) gmol–�

= �02.0gmol–�

Number ofmoles ofAl2O3 =mass

molarmass

=2.55g

�02.0gmol–�

= 0.0250mol

According to the equation, �mole ofAl2O3 requires 3moles ofH2SO4 for complete neutralization.

i.e. number ofmoles ofH2SO4=3 x 0.0250mol = 0.0750mol

Molarity of sulphuric acid=number ofmoles ofH2SO4

volumeof solution

= 0.0750mol

( �50.0�000 ) dm3

= 0.500mol dm–3

\ the concentrationof the sulphuric acid is 0.500mol dm–3.

5� A Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g) �.00 g �.00mol dm–3

�00.0 cm3

Number ofmoles ofMg=mass

molarmass

=�.00g

24.3gmol–�

= 0.04�2mol


= �.00mol dm–3 x�00.0

�000dm3

= 0.�00mol

According to the equation, � mole of Mg reacts with 2 moles of HCl to form � mole of H2. Duringthe reaction, 0.04�2 mole of Mg reacted with 0.0824 mole of HCl. Therefore HCl was in excess. Theamount ofMg limited the amount ofH2 formed.

i.e. number ofmoles ofH2 formed=0.04�2mol

Mass ofH2 formed =number ofmoles xmolarmass = 0.04�2mol x 2.00gmol–�

= 0.0824g

\ 0.0824gof hydrogenwas formed.

9�

52 D Apipette and a pipette filler are used to deliver a fixed, accurate volume (e.g. 25.0 cm3) of a solution.

Aburette is used to deliver various volumesof 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


= 0.0400mol dm–3 x25.0

�000dm3

= 0.00�00mol

According to the equation, 2moles ofNaOH require �mole ofH2SO4 for complete neutralization.

i.e. number ofmoles ofH2SO4 used =0.00�00

2mol

= 5.00 x �0–4mol


volumeof solution

= 5.00 x �0–4mol

( 22.7�000 ) dm3

= 0.0220mol dm–3


54 D

55 A H3PO4(aq) + 2NaOH(aq) Na2HPO4(aq) + 2H2O(l) �0.0 cm3 0.0240mol dm–3

250.0 cm3

(used) 25.0 cm3

Average volumeofNaOH(aq) required for neutralization=2�.5+2�.5+2�.6

3 cm3

= 2�.5 cm3

Number ofmoles ofNaOH in 2�.5 cm3 solution=molarity of solution x volumeof solution

= 0.0240mol dm–3 x2�.5

�000dm3

= 5.�6 x �0–4mol

According to the equation, �mole ofH3PO4 requires 2moles ofNaOH for neutralization.

i.e. number ofmoles ofH3PO4 in 25.0 cm3diluted acid=5.�6 x �0–4

2mol

= 2.58 x �0–4mol

92

Number ofmoles ofH3PO4 in 250.0 cm3diluted acid =2.58 x �0–4 x250.0

25.0mol

= 2.58 x �0–3mol

Number ofmoles ofH3PO4 in �0.0 cm3original acid=2.58 x �0–3mol

Molarity of original phosphoric acid=number ofmoles ofH3PO4

volumeof solution

= 2.58 x �0–3mol

( �0.0�000 ) dm3

= 0.258mol dm–3

\ themolarity of theoriginal phosphoric acid is 0.258mol dm–3.

56 C Wash the burette with distilled water and then the alkali it is going to contain. Any water or impuritiesin the apparatuswill change the concentrationof the alkali and thiswill affect the titration results.

57 D H2X(aq) + 2NaOH(aq) Na2X(aq) + 2H2O(l) 4.5� g 0.350 mol dm–3

250.0 cm3 28.0 cm3

(used) 25.0 cm3

Letm gmol–� be themolarmass ofH2X.

Number ofmoles ofH2X in 250.0 cm3 solution=mass

molarmass

=4.5�g

m gmol–�

Number ofmoles ofH2X in 25.0 cm3 solution=4.5�

m x

�

�0mol


= 0.350mol dm–3 x28.0

�000dm3

= 0.00980mol


i.e. number ofmoles ofH2X in 25.0 cm3 solution =0.00980

2mol

= 0.00490mol

Number ofmoles ofH2X in 25.0 cm3 solution=4.5�

m x

�

�0mol =0.00490mol

m = 92.0

\ themolarmass ofH2X is 92.0gmol–�.

93

58 D Na2CO3(aq)+H2SO4(aq) Na2SO4(aq)+CO2(g) +H2O(l) 5.72g �.00mol dm–3

20.0 cm3

Molarmass ofNa2CO3•nH2O= (�06.0+�8.0n) gmol–�

Number ofmoles ofNa2CO3•nH2Oused =mass

molarmass

=5.72g

(�06.0+�8.0n) gmol–�

Number ofmoles ofH2SO4 in 20.0 cm3 solution=molarity of solution x volumeof solution

= �.00mol dm–3 x20.0

�000dm3

= 0.0200mol

According to the equation, �mole ofNa2CO3 requires �mole ofH2SO4 for complete reaction.

i.e. number ofmoles ofNa2CO3•nH2Oused=0.0200mol

Number ofmoles ofNa2CO3•nH2Oused=5.72

(�06.0+�8.0n)mol=0.0200mol

\n= �0

\ the valueofn is �0.

59 B From the curve,45.0 cm3of sulphuric acidwere required toneutralize the sodiumhydroxide solution. (Thetemperature of the reactionmixturewasmaximum at thepoint of complete neutralization.)

H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l) ? mol dm–3 2.0mol dm–3

45.0 cm3 50.0 cm3


= 2.0mol dm–3 x50.0

�000dm3

= 0.�0mol


i.e.number ofmoles ofH2SO4 =0.�0

2mol

= 0.050mol


volumeof solution

= 0.050mol

( 45.0�000 ) dm3

= �.�mol dm–3

\ themolarity of the sulphuric acid is �.�mol dm–3.

94

60 A 2NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2H2O(l) 0.200 mol dm–3 �5.0 cm3

25.0 cm3


= 0.200mol dm–3 x25.0

�000dm3

= 0.00500mol

According to the equation, 2moles ofNaOH react with �mole ofH2SO4 to form�mole ofNa2SO4.

i.e.number ofmoles ofNa2SO4 formed=0.00500

2mol

= 0.00250mol

Total volumeof the salt solution = (25.0+�5.0) cm3

= 40.0 cm3

Concentrationof the salt solution=number ofmoles ofNa2SO4

volumeof solution

= 0.00250mol

( 40.0�000 ) dm3

= 0.0625mol dm–3

\ themolarity of the salt solutionobtained is 0.0625mol dm–3.

6� C CaCl2(aq) +Na2CO3(aq) CaCO3(s) + 2NaCl(aq)

According to the equation, �moleofCaCl2 reactswith�moleofNa2CO3 togive�moleofCaCO3, i.e.equal volumesof �mol dm–3CaCl2(aq) and�mol dm–3NaCO3(aq)would react.

Option Solutions mixedVolume of CaCl2(aq) and

Na2CO3(aq) reacted

A 5 cm3ofCaCl2(aq)+25 cm3ofNa2CO3(aq) 5 cm3

B �0 cm3ofCaCl2(aq)+20 cm3ofNa2CO3(aq) �0 cm3

C �5 cm3ofCaCl2(aq)+�5 cm3ofNa2CO3(aq) �5 cm3

D 20 cm3ofCaCl2(aq) +�0 cm3ofNa2CO3(aq) �0 cm3

The greatest volumes of CaCl2(aq) and Na2CO3(aq) react in Option C. Thus the greatest amount ofprecipitatewould form.

62 B Pb(NO3)2(aq)+2NaCl(aq) PbCl2(s) + 2NaNO3(aq)

According to the equation, �mole of Pb(NO3)2 requires 2moles ofNaCl to give �mole of PbCl2.

Option Reaction

A �0 cm3of �mol dm–3 Pb(NO3)2(aq) reactwith 20 cm3of �mol dm–3NaCl(aq)

B �2.5 cm3of � mol dm–3 Pb(NO3)2(aq) reactwith 25 cm3of �mol dm–3NaCl(aq)

C �0 cm3of �mol dm–3 Pb(NO3)2(aq) reactwith 20 cm3of �mol dm–3NaCl(aq)

D 5 cm3of �mol dm–3 Pb(NO3)2(aq) reactwith �0 cm3of �mol dm–3NaCl(aq)

95

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






Heat released

�50 cm3of �mol dm–3HNO3(aq)+50 cm3of�mol dm–3KOH(aq)

number ofmoles ofHNO3 / KOH

=�mol dm–3 x50

�000dm3

=0.05mol

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

KNO3(aq) +H2O(l)0.05mol

2.85 kJ

250 cm3of 2mol dm–3HNO3(aq)+50 cm3of2mol dm–3KOH(aq)

number ofmoles ofHNO3 / KOH

=2mol dm–3 x50

�000dm3

=0.�mol

HNO3(aq) +KOH(aq)0.�mol 0.�mol

KNO3(aq) +H2O(l)0.0�mol

5.7 kJ


half that of the secondmixture, i.e. T� =�

2 T2.

64 B H2SO4(aq)+2NaOH(aq) Na2SO4(s) + 2H2O(l)

According to the equation, � mole of H2SO4 requires 2 moles of NaOH for complete neutralization, i.e.V cm3of �mol dm–3H2SO4(aq)would react with 2V cm3of �mol dm–3NaOH(aq).

Option Solutions mixedVolume of H2SO4(aq) and

NaOH(aq) reacted

A �0 cm3ofH2SO4(aq)+35 cm3ofNaOH(aq) �0 cm3ofH2SO4(aq) +20 cm3ofNaOH(aq)

B �5 cm3ofH2SO4(aq)+30 cm3ofNaOH(aq) �5 cm3ofH2SO4(aq) +30 cm3ofNaOH(aq)

C 20 cm3ofH2SO4(aq)+25 cm3ofNaOH(aq) �2.5 cm3ofH2SO4(aq) +25 cm3ofNaOH(aq)

D 25 cm3ofH2SO4(aq)+�5 cm3ofNaOH(aq) 7.5 cm3ofH2SO4(aq) +�5 cm3ofNaOH(aq)

The greatest volumes of H2SO4(aq) and NaOH(aq) react in Option B. The greatest amount of heat isreleased and thus the temperature rise is thegreatest.


For neutralization in which either the acid or alkali or both are weak, the heat released is less than57 kJ for � mole of water produced. The is because some energy is consumed when the weak acidandweak alkali dissociate to give hydrogen ions andhydroxide ions before neutralization.

Expt. Solutions mixed Strength of acid and alkali Temperature rise

��00 cm3of �mol dm–3CH3COOH(aq)and�00 cm3of �mol dm–3NaOH(aq)

neutralizationbetween aweak acidand a strong alkali

T�

2�00 cm3of �mol dm–3HCl(aq) and�00 cm3of �mol dm–3NaOH(aq)

neutralizationbetween a strong acidand a strong alkali

T2

\ T� < T2.

96


Mixture

Number of moles of HNO3 / NaOH mixed



Heat released

Temperature rise

50 cm3of 2mol dm–3HNO3(aq)+50 cm3of2mol dm–3NaOH(aq)

2mol dm–3 x50

�000dm3

=0.�mol

HNO3(aq) +NaOH(aq)0.�mol 0.�mol

NaNO3(aq) +H2O(l)0.�mol

5.7 kJ x

�00 cm3of 2mol dm–3HNO3(aq) +�00 cm3of2mol dm–3NaOH(aq)

2mol dm–3 x�00

�000dm3

=0.2mol

HNO3(aq) +NaOH(aq)0.2mol 0.2mol

NaNO3(aq) +H2O(l)0.2mol

��.4 kJ y

�00 cm3of �mol dm–3HNO3(aq) +�00 cm3of�mol dm–3NaOH(aq)


�000dm3

=0.�mol

HNO3(aq) +NaOH(aq)0.�mol 0.�mol

NaNO3(aq) +H2O(l)0.�mol

5.7 kJ z

The first mixture (total volume �00 cm3) is heated up by 5.7 kJ while the second mixture (total volume200 cm3) is heatedupby��.4 kJ.Hence these twomixtures show the same temperature rise, i.e. x = y.

The third mixture (total volume 200 cm3) is heated up by less than 5.7 kJ. Hence its temperature rise isless than x and y.

\ x = y > z.


For neutralization in which either the acid or alkali or both are weak, the heat released is less than57 kJ for � mole of water produced. The is because some energy is consumed when the weak acidand weak alkali dissociate to give hydrogen ions andhydroxide ions before neutralization.

Mixture




Strength of acid and alkali

Heat released

Temperature rise

�00 cm3of �mol dm–3HCl(aq)+�00 cm3of�mol dm–3NaOH(aq)

Number ofmoles ofHCl / NaOH

=�mol dm–3 x �00�000

dm3

=0.�mol

HCl(aq) +NaOH(aq)0.�mol 0.�mol

NaCl(aq) +H2O(l)0.�mol

neutralizationbetween a

strong acid anda strong alkali

5.7 kJ x

�00 cm3of 2mol dm–3HCl(aq)+�00 cm3of2mol dm–3NaOH(aq)

Number ofmoles ofHCl / NaOH

=2mol dm–3 x �00�000

dm3

=0.2mol

HCl(aq) +NaOH(aq)0.2mol 0.2mol

NaCl(aq) +H2O(l)0.2mol


strong acid anda strong alkali

��.4 kJ y

�00 cm3of �mol dm–3HCl(aq)+�00 cm3of�mol dm–3NH3(aq)

Number ofmoles ofHCl / NH3

=�mol dm–3 x �00�000

dm3

=0.�mol

HCl(aq) +NH4+(aq) +OH–(aq)

0.�mol 0.� molNH4Cl(aq) +H2O(l)

0.�mol


strong acid anda weak alkali

< 5.7 kJ z

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

97

Theheat released in the second case is higher than that released in the first case.

Thus y > x.

Theheat released in the third case is less than that released in the first case.

Thus x > z.

\ y > x > z.

68 B For the neutralization between a strong acid and a strong alkali, the heat released is 57 kJ for � moleofwater produced.

Mixture

Number of moles of H2SO4 / NaOH mixed



Heat released

Temperature rise

25 cm3of �mol dm–3H2SO4(aq)

+25 cm3of

2mol dm–3NaOH(aq)


=�mol dm–3 x25

�000dm3

=0.025mol

Number ofmoles ofNaOH

=2mol dm–3 x25

�000dm3

=0.05mol

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

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

2.85 kJ w

50 cm3of �mol dm–3H2SO4(aq)

+50 cm3of

2mol dm–3NaOH(aq)


=�mol dm–3 x50

�000dm3

=0.05mol


=2mol dm–3 x50

�000dm3

=0.�mol

H2SO4(aq) +2NaOH(aq)0.05mol 0.�mol

Na2SO4(aq) +2H2O(l)0.�mol

5.7 kJ x

25 cm3of 0.5mol dm–3H2SO4(aq)

+25 cm3of

�mol dm–3NaOH(aq)


=0.5mol dm–3 x25

�000dm3

=0.0�25mol


=�mol dm–3 x25

�000dm3

=0.025mol

H2SO4(aq) +2NaOH(aq)0.0�25mol 0.025mol


�.425 kJ y

50 cm3of 0.5mol dm–3H2SO4(aq)

+50 cm3of

�mol dm–3NaOH(aq)


=0.5mol dm–3 x50

�000dm3

=0.025mol


=�mol dm–3 x50

�000dm3

=0.05mol

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


2.85 kJ z

The first mixture (total volume 50 cm3) is heated up by 2.85 kJ while the second mixture (total volume�00 cm3) is heatedupby5.7 kJ.Hence these twomixtures show the same temperature rise, i.e.w = x.

98

The third mixture (total volume 50 cm3) is heated up by �.425 kJ while the fourth mixture (totalvolume�00 cm3) isheatedupby2.85kJ.Hence these twomixtures show the same temperature rise, i.e.y = z.

\ w = x > y = z.

69 CIndicator Colour of solution Possible pH range of solution

Methyl orange yellow 4.5 – �4

Bromothymol blue blue 7.5 – �4

Phenolphthalein colourless 0 – 8.5

\ thepH rangeof the solution shouldbe7.5 – 8.5.

70 A The following diagram shows the titration curve for the titration of 20.0 cm3 of 0.�00 mol dm–3CH3COOH(aq)with 0.�00mol dm–3NaOH(aq).

During the titrationof aweakacidwitha strongalkali, thepHat theequivalencepoint isgreater than7.

7� C During the titrationof a strong acid and aweak alkali, thepH at the equivalencepoint is less than7.

72 C Bromocresol green and methyl red change colour within the pH range of the vertical part of thetitration curve.

Hence they are suitable indicators for the titration.

73 A The reactionbetween two solids shouldbe the slowest.

74 D Average reaction rate=2.0g

3.0min = 0.67gmin–�

99

75 C Rate= –�

4

d[NH3(g)]

dt

=�

5

d[O2(g)]

dt

Rate of consumptionofNH3(g) = –d[NH3(g)]

dt

=4

5

d[O2(g)]

dt

=4

5 (�.80mol dm–3 s–�)

= �.44mol dm–3 s–�

76 A Rate= –�

5

d[O2(g)]

dt

=�

6

d[H2O(g)]

dt

Rate of formationofH2O(g)=d[H2O(g)]

dt

=6

5

d[O2(g)]

dt

=6

5 (�.80mol dm–3 s–�)

= 2.�6mol dm–3 s–�

77 D Option D — The intensity of theorange colour ofCr2O72– ions increases as the reactionproceeds.

78 D Option A — The concentration of both reagents in the reaction mixture in Experiments I and II aredifferent. So theyCANNOTbeused for comparison.

Option B — The temperature in Experiments I and III are different. So they CANNOT be used forcomparison.

Option C — The temperature in Experiments II and IV are different. So they CANNOT be used forcomparison.

Option D — The total volume of the samples in Experiments III and IV are the same. The onlydifference is the volume of the sodium thiosulphate solution used, i.e. the concentrationof sodium thiosulphate in the reactionmixture.

79 B Option A — The concentration of sodium thiosulphate in the reaction mixture in Experiments I and IIIare different. So theyCANNOTbeused for comparison.

Option B — The concentration of both reagents in the reaction mixture in Experiments I and IV arethe same. Theonly difference is the temperature.

Option C — The concentration of acid in the reaction mixture in Experiments II and III are different. SotheyCANNOT beused for comparison.

Option D — The concentration of both reagents in the reaction mixture in Experiments II and IV aredifferent. So theyCANNOTbeused for comparison.

�00

80 A 2mol dm–3H2SO4(aq) has thehighest concentrationof hydrogen ions.

Hence the initial rate of reactionwouldbe thehighestwhen it is added to themagnesium ribbons.

8� A The reaction represented by curve Y took more time to complete, i.e. the rate of this reaction waslower.

Option A — Adding water to the acid would decrease its concentration. The initial reaction rate woulddecrease.

Option D — As excess hydrochloric acid was used, magnesium would limit the volume of hydrogenevolved.

Usinghalf pieceofmagnesium ribbonwoulddecrease the volumeof hydrogen evolved.

82 D The rate of a reaction increaseswhen

• the surface areaof a solid reactant is increased;

• the concentrationof 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 waslower.

Option B — Calcium carbonate powder had a greater surface area than calcium carbonate lumps ofthe samemass.

The rate of the reaction decreases when the surface area of calcium carbonate isdecreased, i.e. using calcium carbonate lumps insteadof calcium carbonate powder.

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

Using a lower mass of calcium carbonate would decrease the volume of gas formed, i.e.decrease thepressure.

84 D The tangent to curve II was less steep than that to curve I, i.e. the rate of the reaction represented bycurve IIwas lower.

However, a greater volumeof hydrogenwas formed in this reaction.

Option A — Increasing the temperaturewould increase the rate of the reaction.

Option B — Usingmagnesiumpowderwould increase the rate of the reaction.

Option C — Using amore concentrated sulphuric acidwould increase the rate of the reaction.

Option D — Using a less concentrated sulphuric acidwoulddecrease the rate of the reaction.

However, increasing the number of moles of H2SO4 (from 0.� mole to 0.�4 mole) would increase thevolume of hydrogen formed.

�0�

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 lesslight and so the absorbancewent down.

The concentrationofMnO4– ions in reactionmixture �was higher than that in reactionmixture 2.

Hence the initialabsorbanceof reactionmixture�washigher than thatof reactionmixture2 (i.e.OptionsB,C and D are INCORRECT).

The concentrationofMnO4–(aq) ion in reactionmixture �was higher than that in reactionmixture 2.

So, the rate of reaction for reaction mixture � was also higher and thus the tangent to its curve wouldbe steeper.

\ the curves inOptionA show the results.

86 BBeaker

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

Reaction between CaCO3 and acid

A

Number ofmoles ofHCl

=�.2mol dm–3 x45

�000dm3

=0.054mol

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

B


=0.9mol dm–3 x60

�000dm3

=0.054mol

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

Option A — The initial rate of reaction in beaker A is higher as �.2 mol dm–3 HCl(aq) contains ahigher concentrationof hydrogen ions than0.9mol dm–3CH3COOH(aq) does.

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

Equalmasses ofmarble chipswouldproduce the same amount of gas.

Option C — The average rate of evolution of gas in Beaker A is higher as �.2 mol dm–3 HCl(aq)contains a higher concentrationof hydrogen ions than0.9mol dm–3CH3COOH(aq) does.

Option D — Equalmasses of calcium chloride are formed in bothbeakers.

However, the volumesof the two reactionmixtures are different.

Hence the concentrations ofCa2+(aq) ions in the twobeakers are different.

87 C The reaction represented by curve Y took less time to complete, i.e. the rate of this reaction washigher.

Option A — Lowering the temperaturewoulddecrease the rate of the reaction.

Option B — Using the same volume of 0.�5 mol dm–3 hydrogen peroxide solution instead of 0.�2 moldm–3 solution (i.e. a greater number of moles of hydrogen peroxide) would increase thevolumeofO2 produced.

�02

Option C — Using 0.�5 mol dm–3 hydrogen peroxide solution instead of 0.�2 mol dm–3 solution wouldincrease the rate of the reaction.

Number ofmoles ofH2O2 in �00 cm3of 0.�2mol dm–3H2O2(aq)

= 0.�2mol dm–3 x�00

�000dm3

= 0.0�2mol

Number ofmoles ofH2O2 in 80 cm3of 0.�5mol dm–3H2O2(aq)

= 0.�5mol dm–3 x80

�000dm3

= 0.0�2mol

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

Option D — Using �20 cm3 of 0.�0 mol dm–3 hydrogen peroxide solution instead of �00 cm3 wouldincrease the volumeofO2 produced.

88 B (2) The amount of catalyst usedwouldNOT affect the amount of oxygenproduced.

89 D (�) Glass cleanser contains ammonia.

(2) Alkaline solutions are more effective than acidic solutions in creating curls of the hair. The curls arepermanent. Perm solutions are thus alkaline.

(3) Limewater is a saturated solution of calcium hydroxide.

90 A (2) Metals tarnish because they reaect with the air to form a layer of oxide or sulphide. Acids can beused to remove this layer. One of the oxide layers most difficultly removed is rust. Rust removersusually contain an acid such as hydrochloric acid or phosphoric acid.

(3) Nitric acid isNOT a drying agent.

9� A (�) Solid citric acid consists ofmolecules.

(2) Hydrochloric acid is produced in human stomach.

(3) Solid citric acid doesNOT reactwithmagnesiumbecause it does not contain hydrogen ions.

92 C (�) Heating solid or solution of ammonium chloride with sodium hydroxide solution liberates ammoniagas.

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

(2) Phenolphthalein gives a colourless solutionwith acidic solutions.

(3) Dilute sodiumhydroxide solutionwould react with carbondioxidegas.

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

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

(3) Solid citric acid doesNOT reactwithmagnesiumbecause it does not contain hydrogen ions.

�03

94 A (�) Aluminiumnitrate solutiongives awhite precipitatewith dilute aqueous ammonia.

Al3+(aq)+3OH–(aq) Al(OH)3(s)

(2) Iron(III) sulphate solutiongives a reddishbrownprecipitatewith dilute aqueous ammonia.


95 A (�) and (2) The aqueous solution of citric acid shows typical properties of an acid while solid citric aciddoes not.

When solid citric acid was added to an aqueous solution of sodium hydrogencarbonate,some citric acidmolecules dissociated to give hydrogen ions.



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

gas. Effervescenceoccurred.


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

96 A (3) Phenolphthalein is colourless in dilute hydrochloric acid.

97 C (�) When dilute sulphuric acid reacts with calcium carbonate, insoluble calcium sulphate forms. Thecalcium sulphate covers the surfaceof calcium carbonate andprevents further reaction.

(2) Carbonic acid is formedwhen carbondioxidedissolves inwater.

Cl2(g) +H2O(l) H2CO3(aq) carbonic acid

The aqueous solution can conduct electricity due to thepresenceofmobile ions.

(3) NaOH(aq) can absorb carbondioxide.

NaHCO3(aq) is formedwhichundergoes dehydration to giveNa2CO3(s).

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

2NaHCO3(aq) Na2CO3(s) +H2O(l)

98 A (�) Pure sulphuric acid is a covalent compound. It is a colourless liquid consistingofmolecules.

(2) When we describe acids as strong and weak, we are talking about the extent of their dissociationin water. When we talk about concentration, we are referring to the amount of an acid in a unitvolumeof solution.

For example, 5 mol dm–3 sulphuric acid is a concentrated solution of a strong acid while 0.� moldm–3 sulphuric acid is a dilute solutionof a strong acid.

(3) Concentrated sulphuric acid doesNOTgive an acidmist in air because it is non-voltatile.

�04

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

NH3(aq) +HCl(aq) NH4Cl(aq)

(�) Neutralizationoccurs andheat is released. Thus the temperature of themixture increases.

(2) Ammonium chloride is soluble inwater. Hencenoprecipitatewould form.

(3) Ammonium chloride is an ionic compound. The solution mixture conducts electricity due to thepresenceofmobile ions fromammonium chloride.

�00A (2) Sodiumhydroxide solution can absorb sulphur dioxide.

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

(3) Sulphur dioxide is added towine as a preservative, NOT a flavour enhancer.

�0�B (�) Wedonot use anhydrous calcium chloride to dry ammonia gas because ammonia reacts with it.

CaCl2(s) + 4NH3(g) CaCl2•4NH3(s)

(3) Concentrataednitric acid isNOT a drying agent.

�02C (2) Pb(NO3)2(aq)+CaCl2(aq) PbCl2(s) +Ca(NO3)2(aq)

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

�03D (�) Magnesium reactswith dilute sulphuric acidwhile silver hasNO reaction.

(2) Barium nitrate solution gives a white precipitate with dilute sulphuric acid while potassium nitratesolutiondoesNOT.


(3) Zinc carbonate gives gas bubbles (carbon dioxide gas) with dilute sulphuric acid while zinc chloridedoesNOT.

�04B (�) There is no observable change when zinc bromide solution or zinc iodide solution is mixed withdilute hydrochloric acid.

(2) Lead(II) nitrate solution gives a white precipitate with dilute hydrochloric acid while magnesiumsulphate solutiondoesNOT.


(3) Both ammonium carbonate and sodium hydrogencarbonate react with dilute hydrochloric acid togive carbondioxide gas.

�05C (�) Both ammoniumnitrate solution andpotassium chloride solution are colourless.

(2) Ammonia is liberated when ammonium nitrate solution is heated with dilute sodium hydroxidesolution. The ammonia canbe testedwithmoist red litmuspaper.

NH4+(aq)+OH–(aq) NH3(g) +H2O(l)

(3) Potassium chloride solution gives a white precipitate with silver nitrate solution while ammoniumnitrate solutiondoesNOT.


�05

�06B (�) Bothdilute hydrochloric acid andnitric acid react with copper(II) oxide.

(2) Dilute hydrochloric acid gives a white precipitate with silver nitrate solution while dilute nitric aciddoesNOT.


(3) Bothdilute hydrochloric acid andnitric acid react with sodiumhydrogencarbonate solution.

�07DSolutions mixed Precipitate formed Ionic equation

(�) NaOH(aq) andCuSO4(aq) a pale blueprecipitate Cu2+(aq) +2OH–(aq) Cu(OH)2(s)

(2) NH3(aq) andMg(NO3)2(aq) awhite precipitate Mg2+(aq) +2OH–(aq) Mg(OH)2(s)

(3) (NH4)2CO3(aq) andNiCl2(aq) a greenprecipitate Ni2+(aq) +CO32–(aq) NiCO3(s)

�08D (�) Copper(II) oxide and dilute sulphuric acid react to give copper(II) sulphate and water. NO gas isproduced.


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



HCO3–(aq) ions in sodiumhydrogencarbonate react with theH+ ions to give carbondioxide gas.


(3) Ammonium carbonate reactswith dilute hydrochloric acid to give carbondioxidegas.

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

�09D (�) CO32–(aq)+ Pb2+(aq) PbCO3(s)

(2) SO42–(aq)+ Pb2+(aq) PbSO4(s)

(3) 2Cl–(aq)+ Pb2+(aq) PbCl2(s)

��0D Magnesium sulphate is prepared by reacting dilute sulphuric acid with either magnesium, magnesiumcarbonate ormagnesiumoxide.

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

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

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

��A (2) and (3) If we wash the pipette or burette with distilled water only, water droplets remaining onthe insideof theglasswarewill dilute the solution that theglassware is going to contain.

��2D (2) and (3) Adding dilute sulphuric acid to limewater and mixing hydrochloric acid with sodiumhydroxide solution are neutralization reactions.Heat is released in these reactions.

�06

��3A

Solutions mixedNumber of moles of acid / alkali


Complete neutralization?

(�) �00 cm3of �mol dm–3HNO3(aq)and

�00 cm3of �mol dm–3KOH(aq)

number ofmoles ofHNO3


�000dm3

=0.�mol

number ofmoles ofKOH


�000dm3

=0.�mol

HNO3(aq) +KOH(aq)0.�mol 0.�mol

KNO3(aq) +H2O(l)\ complete neutralization

occurs and a neutralsolution is obtained.

(2) �00 cm3of �mol dm–3H2SO4(aq)and

�00 cm3of �mol dm–3NaOH(aq)



�000dm3

=0.�mol



�000dm3

=0.�mol

H2SO4(aq) +2NaOH(aq)0.�mol 0.�mol

Na2SO4(aq) +2H2O(l)\ not enoughNaOH(aq)

to neutralizeH2SO4(aq)completely.

(3) �00 cm3of �mol dm–3H2SO4(aq)and

�00 cm3of 2mol dm–3KOH(aq)



�000dm3

=0.�mol

number ofmoles ofKOH

=2mol dm–3 x�00

�000dm3

=0.2mol

H2SO4(aq) +2KOH(aq)0.�mol 0.2mol

K2SO4(aq) +2H2O(l)\ complete neutralization

occurs and a neutralsolution is obtained.

��4B For the neutralization between a strong acid and a strong alkali, the heat released is 57 kJ for � moleofwater produced.

For neutralization in which either the acid or alkali or both are weak, the heat released is less than57 kJ for � mole of water produced. The is because some energy is consumed when the weak acidand weak alkali dissociate to give hydrogen ions andhydroxide ions before neutralization.

Option Solutions mixed Strength of acid and alkali Temperature rise

—25 cm3of �mol dm–3HCl(aq) and25 cm3of �mol dm–3NaOH(aq)


T

(�)25 cm3of �mol dm–3HNO3(aq) and25 cm3of �mol dm–3NH3(aq)

neutralizationbetween a strong acidand aweak alkali

< T

(2)25 cm3of �mol dm–3HNO3(aq) and25 cm3of �mol dm–3KOH(aq)


T

(3)50 cm3of �mol dm–3CH3COOH(aq) and50 cm3of �mol dm–3NaOH(aq)

neutralizationbetween aweak acidand a strong alkali

< T

\ Mixing 25 cm3 of � mol dm–3 HNO3(aq) and 25 cm3 of � mol dm–3 KOH(aq) produces a similartemperature rise asmixing25 cm3of �mol dm–3HCl(aq) and25 cm3of �mol dm–3NaOH(aq).

�07

��5ABeaker Number of moles of HCl = molarity of solution x volume of solution

��mol dm–3 x

�00�000

dm3=0.�mol

22mol dm–3 x

50�000

dm3=0.�mol

Marble chips reactwith dilute hydrochloric acid according to the following equation:

CaCO3(s) + 2HCl(aq) CaCl2(aq)+H2O(l) +CO2(g) excess 0.�mol

(�) Equal numbers ofmoles are present in the twobeakers.

Hence equalmasses ofmarble chips are used in thebeakers.

Thus equalmasses ofmarble chips remain in thebeakers.

(2) The same amount of gas (0.�

2mole) is produced in both cases.

(3) Equal amounts of calcium chloride are produced in both cases.

However, the volumesof the reactionmixtures are different.

Hence concentrations of calcium chloride solution in thebeakers are different.

��6A


= mass

molar mass



�

2g24.3gmol–� = 0.08mol


=�.2mol dm–3 x90

�000dm3

=0.��mol

Mg(s) + 2HCl(aq)0.08mol 0.��mol

MgCl2(aq) +H2(g)

2


=0.9mol dm–3 x�20

�000dm3

=0.��mol

Mg(s) + 2CH3COOH(aq)0.08mol 0.��mol

(CH3COO)2Mg(aq) +H2(g)

According to the equations, � mole of Mg reacts with 2 moles of HCl / CH3COOH to produce � moleofH2.During the reaction, 0.��mole ofHCl /CH3COOH reactedwith 0.055mole ofMg.

Therefore Mgwas in excess. The amount ofHCl /CH3COOH limited the amount ofH2 produced.

(�) Mgwas in excess. The acids in bothbeakers reacted completely.

(2) The same amount of gas (0.��

2mole)was produced in both cases.


�.2mol dm–3HCl(aq) has a higher concentrationof hydrogen ions than0.9mol dm–3CH3COOH(aq)does.

Hence the reactionbetweenMgandHCl(aq) took a shorter time to complete.

�08

��7B (�) During the acid catalyzed hydrolysis of methyl methanoate, the concentration of ethanoic acid(CH3COOH) in the reactionmixture increases as the reactionproceeds.

Hence theprogress of the reaction canbe followedby titration with an alkali.

(3) During the decomposition of a dicarboxylic acid, the concentration of the acid in the reactionmixture decreases as the reactionproceeds.

Hence theprogress of the reaction canbe followedby titration with an alkali.

��8A (�) The intensity of the yellow-brown colour of Br2(aq) decreases as the reactionproceeds.

Hence theprogress of reaction canbe followedbyusing a colorimeter.

(2) The reactionmixture becomes turbid as the sulphur forms.

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

(3) The reactiondoesNOT involve any colour change.

Hence a colorimeterCANNOT beused to follow theprogress of the reaction.

��9A (�) Powdered zinc has a greater surface area than a zinc granule of the samemass.

Hence the initial rate of the reactionwould increase.

(3) Doubling the volumeof the acid usedhasNOeffect on the initial rate of the reaction.

�20A HCl(aq) andCH3COOH(aq) reactwith calcium carbonate according to the following equations:

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


�000dm3=0.�mol

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


�000dm3=0.�mol

Calcium carbonate remain in both beakers after reaction. Thus HCl(aq) and CH3COOH(aq) are thelimiting reagents. Their amounts limit the amount of carbondioxidegas produced.

(�) HCl(aq) is a strong acidwhileCH3COOH(aq) is aweak acid.

� mol dm–3 HCl(aq) has a higher concentration of hydrogen ions than � mol dm–3 CH3COOH(aq)does.

Hence the initial rate of reaction in BeakerA 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 producedin both cases.

(3) The rate of the reaction between CaCO3 and � mol dm–3 HCl(aq) is greater than that betweenCaCO3 and�mol dm–3CH3COOH(aq).

Hence the reaction in BeakerA takes a shorter time to complete.

�09

�2�D

ReactionNumber of moles of Zn

= mass

molar mass


volume of solutionReaction between Zn and HCl

�

3g65.4gmol–� = 0.046mol

2mol dm–3 x50

�000dm3

=0.�mol

Zn(s) + 2HCl(aq)0.046mol 0.�mol

ZnCl2(aq) +H2(g)

24mol dm–3 x

50�000

dm3

=0.2mol

Zn(s) + 2HCl(aq)0.046mol 0.2mol

ZnCl2(aq) +H2(g)

According to the equation, � mole of Zn reacts with 2 moles of HCl to produce � mole of H2. Duringthe reaction, 0.046mole of Zn reactedwith 0.092mole ofHCl.

Therefore HCl was in excess. The amount of Zn limited the amount ofH2 produced.

(�) All the zinc reacted in both cases.

(2) Thehydrogenproducedwould escape. Thus themass of each reactionmixture woulddecrease.

As the same amount of gas (0.046 mol) was produced in both reactions, the losses in mass of thetwo reactionmixtures after reactionwouldbe the same.

(3) The concentrationofHCl(aq)was lower in Reaction�.

Thus the initial rate of Reaction�was smaller.

�22C


= mass

molar mass



�

�.8g24.3gmol–� = 0.074mol



�000dm3

=0.�mol


MgCl2(aq) +H2(g)According to the equation, � moleof Mg reacts with 2 moles of HClto produce�mole ofH2.Hence 0.� mole of HCl reacctedwith0.05moleofMgtogive0.05mole ofH2.ThusMgwas in excess.

2



�000dm3

=0.�mol

Mg(s) + H2SO4(aq)0.074mol 0.�mol

MgSO4(aq) +H2(g)According to the equation, � moleofMgreactswith�moleofH2SO4to produce�mole ofH2.Hence 0.074 mole of Mg reacctedwith 0.074 mole of H2SO4 to give0.074mole ofH2.ThusH2SO4(aq) was in excess.

(�) Thehydrogenproducedwould escape. Thus themass of each reactionmixture woulddecrease.

Different amounts of hydrogengaswereproduced in the two cases.

Hence the losses inmass of the two reactionmixtures after reactionwouldbedifferent.

��0

(2) HCl(aq) is amonobasic acidwhileH2SO4(aq) is a dibasic acid.

Therefore the concentration of hdyrogen ions in � mol dm–3 HCl(aq) was lower than that in � moldm–3H2SO4(aq).

Hence the initial rate of reaction in Beaker �wouldbe smaller than that in Beaker 2.

(3) Magnesium reacted completelywithH2SO4(aq) in Beaker 2.

�23B (�) A catalystwouldNOT change the amount of product formed in a reaction.

(3) Thephysical state of a catalystmayNOTbe the same as thoseof the reactants.

For example, the solid magnese(IV) oxide acts as a catalyst in the decomposition of hydrogenperoxide solution.

�24A

�25D Nitric acid isNOT a drying agent.

�26A

�27B Sodium hydroxide can absorb sulphur dioxidegas because it can react with thegas.

�28B Aqueous solutionof citric acid can conduct electricity due to thepresenceofmobile ions.


Hence citric acid is an electrolyte.

�29D Copper(II) hydroxidedoesNOTdissolve in excess sodiumhydroxide solution.

�30A

�3�C Aqueous solutionof ethanoic acid can conduct electricity.

Hence ethanoic acid is an electrolyte.

�32C Both ethanoic acid and propanoic acid are monobasic acids as only the hydrogen atom in the –COOHgroup of amolecule of each acid canundergodissociation.

�33D Ammonia is aweak alkali.

��

�34D Hydrochloric acid is a strong acidwhile ethanoic acid is aweak acid.

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

Hence the acids give a different colourwith the samequantity of universal indicator.

�35D Ethanoic acid is aweak acid. It only partically dissociates in both concentrated anddilute solutions.

�36D Solid citric acid doesNOT show typical properties of an acid.

A typical acid doesNOT reactwith copper.

�37D Number ofmoles ofH2SO4 in �0 cm3of 5mol dm–3 acid

= 5mol dm–3 x�0

�000dm3

= 0.05 mol

Number ofmoles ofH2SO4 in 50 cm3of �mol dm–3 acid

= �mol dm–3 x50

�000dm3

= 0.05 mol

The acids contain the samenumber ofmoles ofH2SO4.

Both acids reactwith 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 � mol dm–3 H2SO4(aq) is adilute solutionof a strong acid.

�38D HCl(aq) is a strong acidwhileCH3COOH(aq) is aweak acid.

� mol dm–3 HCl(aq) contains a higher concentration of hydrogen ions than � mol dm–3 CH3COOH(aq)does.

Hence the acids havedifferent pH values.

Adding �0 cm3 of � mol dm–3 CH3COOH(aq) to �0 cm3 of � mol dm–3 HCl(aq) would result in achange in pH.

�39C Both� mol dm–3NaOH(aq) and�mol dm–3NH3(aq) give a precipitatewithMgCl2(aq).

�40C Our stomach produces hydrochloric acid along with enzymes to digest food. Calcium carbonate givescarbondioxidegaswhen reactingwithdilute hydrochloric acid. Thegasmakes a personuncomfortable.Therefore calcium carbonate is seldomused in antacids nowadays.

�4�B Hydrochloric acid is produced by human stomach. Magnesium hydroxide in antacids can neutralize theexcess hydrochloric acid and so thepain canbe relieved.

Mg(OH)2(s) + 2HCl(aq) MgCl2(aq)+2H2O(l)

��2

�42B Sodium hydrogensulphate is formed when one of the hydrogen ions in dilute sulphuric acid is replacedby a sodium ion. It is an acid salt.

Sodium hydrogensulphate solution is acidic. This is because the hydrogensulphate ion can dissociate togive thehydrogen ion.

HSO4–(aq) H+(aq)+ SO4

2–(aq)

�43C When 50 cm3 of � mol dm–3 NaCl(aq) are added to 50 cm3 of � mol dm–3 HCl(aq), the concentrationof hydrogen ions decreases because the total volumeof the acid has increased.

Hence thepHof the acidwould change.

�44C For example, �mole of sulphuric acid canneutralize twomoles of sodiumhydroxide.

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

�45C When �0 cm3 of � mol dm–3 HCl(aq) are added to �0 cm3 of � mol dm–3 NaOH, the number ofNa+(aq) ions in the solutionmixtue remains unchanged.

However, the volumeof the solutionhas changed.

Hence the concentrationofNa+(aq) ionswoulddecrease.

�46C Universal indicator is NOT used to detect the end point of a titration because the indicator would NOTgive a sharp colour change at the endpoint.

�47C For the neutralization between a strong acid and a strong alkali, the heat released is 57 kJ for � moleofwater produced.

Acid and alkali mixed




released

25 cm3of �mol dm–3HCl(aq)+25 cm3of�mol dm–3NaOH(aq)

number ofmoles ofHCl / NaOH

=�mol dm–3 x25

�000dm3

=0.025mol



�.425 kJ

25 cm3of 2mol dm–3HCl(aq)+25 cm3of2mol dm–3NaOH(aq)

number ofmoles ofHCl / NaOH

=2mol dm–3 x25

�000dm3

=0.05mol



2.85 kJ


half of that of the secondmixture, i.e. T� =�

2 T2.

��3

�48B Ethanedioic acid crystals can be used to prepare standard solutions because they have the followingcharacteristics:

• they are obtainable in a very pure form;

• theyhave a known chemical formula;

• theydissolve inwater completely at room temperature;

• they are stable anddonot absorbmoisture from the air; and

• theyhave a highmolarmass tominimize weighing errors.

�49A In the oxidation of oxalate ions by permanganate ions, the intensity of the purple colour ofpermanganate ions decreases as the reactionproceeds.



�50C Hydrochloric acid is amonobasic acidwhile sulphuric acid is a dibasic acid.

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

Hence the reaction between �00 cm3 of � mol dm–3 hydrochloric acid and zinc granules is slower thanthat between�00 cm3of �mol dm–3 sulphuric acid and zinc granules.

�5�D A catalystwouldNOTundergo anypermanent chemical changes at the endof a reaction.

Shortquestions

�52Reaction between Chemical equation

a) Zinc and dilute hydrochloric acid Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g) (�)

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

c) Zinc carbonate anddilute hydrochloric acid ZnCO3(s) + 2HCl(aq) ZnCl2(aq) + H2O(l) + CO2(g) (�)

d) Sodium carbonate solution anddilutesulphuric acid

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

e) Lead(II) carbonate anddilute nitric acid PbCO3(s) + 2HNO3(aq) Pb(NO3)2(aq) + H2O(l) + CO2(g) (�)

f) Sodium hydrogencarbonate solution anddilute hydrochloric acid

NaHCO3(aq) + HCl(aq) NaCl(aq) + H2O(l) + CO2(g) (�)

g) Sodium hydrogencarbonate solid anddilutesulphuric acid

2NaHCO3(s) + H2SO4(aq) Na2SO4(aq) + 2H2O(l) + 2CO2(g) (�)

h) Magnesiumhydroxide solid anddilute nitricacid

Mg(OH)2(s) + 2HNO3(aq) Mg(NO3)2(aq) + 2H2O(l) (�)

��4

�53Reaction between Ionic equation

a) Magnesiumand sulphuric acid Mg(s) + 2H+(aq) Mg2+(aq) + H2(g) (�)

b) Sodium carbonate solution anddilutehydrochloric acid

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

c) Sodiumhydrogencarbonate solution anddilute nitric acid

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

d) Iron(III) hydroxide anddilute nitric acid Fe(OH)3(aq) + 3H+(aq) Fe3+(aq) + 3H2O(l) (�)

e) Dilute sodiumhydroxide solution anddilutehydrochloric acid

OH–(aq) + H+(aq) H2O(l) (�)

�54Reaction between Ionic equation(s)

a) Solution containing calcium ions anddilutesodiumhydroxide solution

Ca2+(aq) + 2OH–(aq) Ca(OH)2(s) (�)

b) Solution containing aluminium ions anddilute sodiumhydroxide solution

Al3+(aq) + 3OH–(aq) Al(OH)3(s)Al(OH)3(s) + OH–(aq) [Al(OH)4]

–(aq)(�)(�)

c) Solution containing magnesium ions anddilute sodiumhydroxide solution

Mg2+(aq) + 2OH–(aq) Mg(OH)2(s) (�)

d) Solution containing iron(II) ions anddilutesodiumhydroxide solution

Fe2+(aq) + 2OH–(aq) Fe(OH)2(s) (�)

e) Solution containing iron(III) ions anddiluteaqueous ammonia

Fe3+(aq) + 3OH–(aq) Fe(OH)3(s) (�)

f) Solution containing copper(II) ions anddilute aqueous ammonia

Cu2+(aq) + 2OH–(aq) Cu(OH)2(s)Cu(OH)2(s) + 4NH3(aq) [Cu(NH3)4]

2+(aq) + 2OH–(aq)(�)(�)

g) Solution containing zinc ions anddiluteaqueous ammonia

Zn2+(aq) + 2OH–(aq) Zn(OH)2(s)Zn(OH)2(s) + 4NH3(aq) [Zn(NH3)4]

2+(aq) + 2OH–(aq)(�)(�)

h) Solution containing ammonium compoundanddilute sodium hydroxide solution (withheating)

NH4+(aq) + OH–(aq) NH3(g) + H2O(l) (�)

�55Name of ion in solution Reagent(s) added to the solution Result

Copper(II) iondilute sodium hydroxide solution / dilute aqueous ammonia (�)

a blueprecipitate

Chloride ion (�)dilute nitric acid followedby silvernitrate solution

awhite precipitate

Iron(II) ion / nickel(II) ion (�) dilute sodium carbonate solution agreenprecipitate

Lead(II) ion / barium ion (�) dilute sulphuric acid awhite precipitate

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

a gas that turnsmoist redlitmuspaper blue is givenoff

��5

�56Reaction Name of salt Formula of salt

Calciumhydroxide anddilute hydrofluoric acid calcium fluoride CaF2 (�)

Magnesiumoxide anddilute sulphuric acid magnesium sulphate MgSO4 (�)

Dilute sodiumhydroxide solution and carbonicacid

sodium hydrogencarbonate / sodium carbonate

NaHCO3 / Na2CO3 (�)

Iron(II) carbonate anddilute hydrochloric acid iron(II) chloride FeCl2 (�)

Lead(II) oxide anddilute nitric acid lead(II) nitrate Pb(NO3)2 (�)

�57a) Gas bubbles are givenoff. /Magnesiumdissolves. (�)

Mg(s) + 2H+(aq) Mg2+(aq)+H2(g) (�)

b) Copper(II) oxidedissolves in the acid. /Ablue solution forms. (�)

CuO(s) + 2H+(aq) Cu2+(aq)+H2O(l) (�)

c) Effervescenceoccurs. /Nickel(II) carbonate dissolves in the acid. /Agreen solution forms. (�)

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

d) A white precipitate forms; the precipitate dissolves in excess dilute sodium hydroxide solution to give acolourless solution. (�)

Pb2+(aq)+2OH–(aq) Pb(OH)2(s) (�)

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

e) Awhite precipitate forms. (�)

Ca2+(aq) +CO32–(aq) CaCO3(s) (�)

f) Awhite precipitate forms. (�)

Pb2+(aq)+2Cl–(aq) PbCl2(s) (�)

g) A green precipitate forms. (�)

Fe2+(aq) +2OH–(aq) Fe(OH)2(s) (�)

h) The solutionmixturewarmsup. (�)

H+(aq) +OH–(aq) H2O(l) (�)

�58a) pHof solution= –log�0[H+] = 3.40

i.e.log�0[H+]= –3.40 (�)

[H+] = �0–3.40

= 3.98 x �0–4mol dm–3 (�)

��6

b) One mole of ZnCl2 contains 2moles ofCl– ions.

i.e.concentrationofCl– ions=2 x concentrationof ZnCl2 solution

ConcentrationofCl– ions in solution = �.80 x �0–2mol

( 40.0�000 ) dm3

= 0.450mol dm–3 (�)

\ concentrationof ZnCl2 solution = 0.4502

mol dm–3

= 0.225mol dm–3 (�)

c) One mole ofK2CO3 contains 2moles ofK+ ions.

i.e.number ofmoles ofK+ ions fromK2CO3(aq) =2 xmolarity of solution x volumeof solution

= 2 x 2.00mol dm–3 x �00.0�000

dm3

= 0.400mol (�)

One mole ofKCl contains �mole ofK+ ions.

i.e.number ofmoles ofK+ ions fromKCl(aq) =molarity of solution x volumeof solution

= 0.500mol dm–3 x �50.0�000

dm3

= 0.0750mol (�)

Total number ofmoles ofK+ ions inX = (0.400+0.0750)mol

= 0.475mol

ConcentrationofK+ ions inX= number ofmoles ofK+ ionsvolumeof solution

= 0.475mol

( �00.0+�50.0�000 ) dm3

= �.90mol dm–3 (�)

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

8.00 x 400.0�000

=M x 25.0 (�)

M =0.�28 (�)

\ molarity of thediluted acid is 0.�28mol dm–3.

ii) Molarmass ofH2SO4= (2 x �.0+32.�+4 x �6.0) gmol–�

= 98.�gmol–�

Concentrationof diluted acid=0.�28mol dm–3 x 98.�gmol–�

= �2.6gdm–3 (�)

��7

Structuredquestions

�59a) When solid citric acid dissolves inwater, themolecules dissociate to give hydrogen ions. (�)

Hydrogencarbonate ions reactwith hydrogen ions to give carbondioxidegas. Effervescence occurs. (�)

b) Concentrated sulphuric acid is hygroscopic. (�)

c) Concentrated sulphuric acid reactswith ammonia gas. (�)

d) Sulphuric acid is a strong acid. It almost completely dissociates inwater to give hydrogen ions. (�)

Carbonic acid is aweak acid. It only partially dissociates inwater, forming very fewhydrogen ions. (�)

Hence0.�moldm–3sulphuricacidhasahigherconcentrationofmobile ionsthan0.�moldm–3carbonicacid. (�)

Thustheelectricalconductivityof0.�moldm–3sulphuricacidishigherthanthatof0.�moldm–3carbonicacid.

e) The conical flask should not be washed with the solution it is going to contain because the additionalamount of solute remaining in the flaskwill affect the titration results. (�)

�60a) Adddilute hydrochloric acid to both solids. (�)

Effervescenceoccurs for zinc carbonate.Agas that turns limewater milky is givenoff. (�)

The zinc oxide just dissolves in the acid. (�)

b) Add dilute sodiumhydroxide solution / dilute aqueous ammonia to the solutions. (�)

Magnesium sulphate solutiongives awhite precipitate (�)

while there is noobservable change for sodium sulphate solution. (�)

c) Add dilute sulphuric acid to the solutions. (�)

Barium nitrate solutiongives awhite precipitate (�)

while there is noobservable change for potassiumnitrate solution. (�)

d) Add dilute hydrochloric acid to the solutions. (�)

Lead(II) ethanoate solutiongives awhite precipitate (�)

while there is noobservable change for zinc sulphate solution. (�)

��8

�6�a) Aweak acid is an acid that only partially dissociates inwater. (�)

b) pH = –log�0[H+] = �.4�

i.e.log�0[H+] = –�.4� (�)

[H+] = �0–�.4�

= 3.89 x �0–2mol dm–3 (�)

c) H2SO4(aq) 2H+(aq)+ SO42–(aq)

0.�00 mol dm–3 ?mol dm–3


i.e.concentrationof hydrogen ions=2 x 0.�00mol dm–3=0.200mol dm–3 (�)

pH of acid= –log�0(0.200)=0.700 (�)

d) Any oneof the following:

• Measure thepHusinguniversal indicator / pHmeter (�)

ThepHof the sulphuric acid is lower than that of the sulphurous acid. (�)

This shows that the sulphuric acid has a higher concentration of hydrogen ions than the sulphurousacid does. (�)

• Testwithmagnesium / zinc / sodium carbonate (�)

The sulphuric acid reactsmorequickly than the sulphurous acid. (�)

This shows that the sulphuric acid has a higher concentration of hydrogen ions than the sulphurousacid does. (�)

• Measure the electrical conductivity (�)

The electrical conductivity of the sulphuric acid is higher than that of the sulphurous acid. (�)

Thisshowsthatthesulphuricacidhasahigherconcentrationofmobile ionsthanthesulphurousaciddoes. (�)

�62a) i) Sodiumhydroxide is corrosive. (�)

ii) Useweakbases such asmagnesiumhydroxide and aluminiumhydroxide. (�)

b) i) There is no reactionbetween copper anddilute hydrochloric acid. (�)

ii) Use the reaction between copper(II) oxide / copper(II) hydroxide / copper(II) carbonate and dilutehydrochloric acid (�)

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

ii) Wash thehandwith plenty ofwater. (�)

d) i) Concentrated sulphuric acidwill reactwith ammonia. (�)

ii) Use calciumoxide. (�)

��9

�63a) i) HCl(aq) H+(aq)+Cl–(aq) (�)

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

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

Number ofmoles ofMgpresent=mass ofMg

molarmass ofMg

= �.5 g24.3gmol–�

= 0.062mol (�)

Number ofmoles ofHCl used=molarity of solution x volumeof solution

= �.0mol dm–3 x �00�000

dm3

= 0.�0mol (�)

According to the equation, � mole of Mg reacts with 2 moles of HCl. During Reaction �, 0.�0 mole ofHCl reactedwith 0.050mole ofmagnesium. Thereforemagnesiumwas in excess. (�)

c) i) Nomoregas bubblesweregivenoff. (�)

ii) The time required for the completionof Reaction2wouldbe longer. (�)

During the reaction between magnesium and the acids, magnesium would react with hydrogen ionsin the acids.

Hydrochloric acid is a strong acid that completely dissociates inwater. (�)

On theother hand, ethanoic acid is aweak acid that only partially dissociates inwater. (�)

Thereforehydrochloric acid has a higher concentrationof hydrogen ions than ethanoic acid does.(�)

Thereactionratebetweenmagnesiumandethanoicacidisthuslowerandthereactiontakesalongertime to complete.


• Measuring their pH (�)

ThepHof �.0mol dm–3HCl(aq) is lower than that of �.0mol dm–3CH3COOH(aq). (�)

• Measuring their electrical conductivity (�)

The electrical conductivity of �.0 mol dm–3 HCl(aq) is higher than that of �.0 mol dm–3CH3COOH(aq). (�)

e) Sulphuric acid is a dibasic acidwhile hydrochloric acid is amonobasic acid. (�)

Therefore �.0 mol dm–3 sulphuric acid has a higher concentrationof hydrogen ions. (�)

The initial rate of Reaction3 is thus higher than that of Reaction�.

�20

�64a) Pb2+(aq)+ SO42–(aq) PbSO4(s) (�)

b)

(�markforcorrectset-up;�markforlabellingfilterfunnelandfilterpaper;�markforlabellingprecipitateA and solutionB; 0mark if the set-up is notworkable) (3)

c) Cu2+(aq) +2OH–(aq) Cu(OH)2(s) (�)

d) The presenceofNH4+(aq) ions canbe shownbywarming solutionD.An alkaline gaswill evolve. (�)

ThepresenceofK+(aq)ionscannotbeshown.Asinflametest,thelilacflameofpotassiumwillbemaskedby thegolden yellow colour of sodium. (�)

e) Blue (�)

Cu2+(aq) ions are blue in colourwhile theother ions in solutionX are colourless. (�)

�65a) 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) (�)

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

iii) Pb2+(aq)+CO32–(aq) PbCO3(aq) (�)

iv) Pb2+(aq)+2OH–(aq) Pb(OH)2(s) (�)

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

c) Carbon reduction (�)

�2�

�66AwasAl2(SO4)3(aq) as it reactedwithC and E to give awhite precipitate (Al(OH)3): (�)

Al3+(aq)+3OH–(aq) Al(OH)3(s) (�)

CandEarealkalisbecauseAl2(SO4)3(aq)reactedwiththealkalistogiveawhiteprecipitate.Astheprecipitatedissolved when E was in excess, E was NaOH(aq). The white precipitate dissolved due to the formation ofcomplex ions: (�)

Al(OH)3(aq) +OH–(aq) [Al(OH)4]–(aq) (�)

ThusCwasNH3(aq). (�)

BwasHCl(aq) as it underwent neutralizationwith E andheatwas liberated. (�)

HCl(aq) +NaOH(aq) NaCl(aq)+H2O(l) (�)

DwasAgNO3(aq) as it reactedwithCl–(aq) ions (in B) to give awhite precipitate (AgCl): (�)

Ag+(aq)+Cl–(aq) AgCl(s) (�)

�67a) Wash a 25.0 cm3pipette firstwith distilledwater and thenwith the sulphuric acid. (�)

Using a pipette filler, suckup the acid until themeniscus is 2 – 3 cmabove thegraduationmark. (�)

Use the forefinger to control the flow. Release the solution until the meniscus reaches the graduationmark. (�)

Transfer the solution into a conical flask. Allow the tip of the pipette to touch the side of the conicalflask. (�)

b) To obtain consistent results. / Toobtain three titres within 0.� cm3. (�)

c) Average volumeof sodiumhydroxide solution required for neutralization

= 22.2 +22.3+22.23

cm3

=22.2 cm3 (�)

d) 2NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2H2O(l) 0.�80 mol dm–3 ?mol dm–3

22.2 cm3 25.0 cm3


= 0.�80mol dm–3 x22.2

�000dm3

= 0.00400mol (�)


i.e.number ofmoles ofH2SO4 in 25.0 cm3 solution=0.00400

2mol

= 0.00200mol (�)

�22

Molarity of sulphuric acid= number ofmoles ofH2SO4

volumeof solution

= 0.00200mol

( 25.0�000 ) dm3

= 0.0800mol dm–3 (�)

\ the molarity of the sulphuric acid is 0.0800mol dm–3.

e) Thenext titrewouldbe lower. (�)

Some alkali remained in the conical flask. (�)

�68a) Use a burette to contain thehydrochloric acid. (�)

Wash theburettewith distilled water and thenwith thehydrochloric acid. (�)

Add the indicator to the conical flask, and then the acid from the burette until the indicator changesfrom yellow to red. (�)

b) i) Average volumeof hydrochloric acid used= �8.8+�8.8+�8.73

cm3

= �8.8 cm3 (�)

ii) NH3(aq) + HCl(aq) NH4Cl(aq) 9.97g 0.�50mol dm–3

250.0 cm3 �8.8 cm3

(used) 25.0 cm3


= 0.�50mol dm–3 x �8.8�000

dm3

= 0.00282mol

According to the equation, �mole ofNH3 requires �mole ofHCl for complete neutralization.

i.e.number ofmoles ofNH3 in 25.0 cm3of thediluted solution=0.00282mol (�)

Number ofmoles ofNH3 in 250.0 cm3of thediluted solution

=number ofmoles ofNH3 in theoriginal household ammonia solution

=0.00282mol x250.0 cm3

25.0 cm3

=0.0282mol

Mass ofNH3 in theoriginal household ammonia solution =number ofmoles ofNH3 xmolar mass ofNH3

=0.0282mol x �7.0gmol–�

= 0.479g (�)

�23

Percentagebymass ofNH3 in theoriginal household ammonia solution

=0.479g9.97g

x �00%

=4.80% (�)

\ thepercentagebymass of ammonia in thehouseold ammonia solution is 4.80%.

c)

Distilled waterOriginal

household ammonia solution

Diluted household ammonia solution

Standard hydrochloric acid

i) 250.0 cm3 volumetric flask 4 (0.5)

ii) 25.0 cm3pipette fordelivering thedilutedhousehold ammonia solution

4 (0.5)

iii) Conical flask 4 (0.5)

iv) Burette 4 (0.5)

�69a) Molarmass ofKOH= (39.�+�6.0+�.0) gmol–�

= 56.�gmol–�

Number ofmoles ofKOH in �00.0 cm3 solution= massmolarmass

= 7.00g56.�gmol–�

= 0.�25mol (�)

Concentrationof potassiumhydroxide solution = number ofmoles ofKOHvolumeof solution

= 0.�25mol

( �00.0�000 ) dm3

= �.25mol dm–3 (�)

b) i) 25.0 cm3of 6.00mol dm–3 sulphuric acid are needed for thedilution.

Wash a 25.0 cm3pipette firstwith distilledwater and thenwith the acid. (�)

Deliver exactly 25.0 cm3 of the original acid into a 250.0 cm3 volumetric flask using the pipette andpipette filler. (�)

Adddistilled water to the flask until themeniscus reaches thegraduationmark.

Stopper the flask. Turn it upsidedown several times tomix the solutionwell. (�)

ii) 2KOH(aq) + H2SO4(aq) K2SO4(aq) + 2H2O(l) �.00 mol dm–3 0.600mol dm–3

25.0 cm3 ? cm3

�24

Number ofmoles ofKOH in 25.0 cm3 solution=molarity of solution x volumeof solution

= �.00mol dm–3 x 25.0�000

dm3

= 0.0250mol (�)

According to the equation, 2moles ofKOH require �mole ofH2SO4 for complete neutralization.

i.e. number ofmoles ofH2SO4=0.0250

2mol

= 0.0�25mol (�)

Volumeof sulphuric acid required for neutralization= number ofmoles ofH2SO4


=0.0�25mol

0.600mol dm–3

= 0.0208dm3

= 20.8 cm3 (�)

\ the lowest acceptable titre is 20.8 cm3.

c) To check that thepeaches are free frompotassiumhydroxide (�)

because potassiumhydroxide is corrosive. (�)

�70a) Sodium hydroxide solutions absorbmoisture fromair. So theirmasses change. (�)

b) Any two of the following:

• It is obtainable in a very pure form. (�)

• It has a known chemical formula. (�)

• It dissolves inwater completely at room temperature. (�)

• It is stable anddoes not absorbmoisture from the air. (�)

• It has a highmolarmass. (�)

c) Wash theburette firstwith distilled water and thenwith the sodiumhydroxide solution. (�)

Clamp theburette vertically ina stand.Close the stopcock. Fill theburettewithdilute sodiumhydroxidesolution. (�)

Open the stopcock for a few seconds so as to fill the tip of theburettewith solution. (�)

d) Phenolphthalein (�)

From colourless to red / pink (�)

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

250.0 cm3 29.2 cm3

(used) 25.0 cm3

�25

Molarmass of (COOH)2•2H2O = [2 x (�2.0+2 x �6.0+�.0) + 2 x (2 x �.0+�6.0)] gmol–�

= �26.0gmol–�

Number ofmoles of (COOH)2•2H2O in 250.0 cm3 solution=mass

molarmass

= 4.95g�26.0gmol–�

= 0.0393mol (�)

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

250.0 cm3

= 0.00393mol

According to the equation, �mole of (COOH)2 requires 2moles ofNaOH for complete neutralization.

i.e.number ofmoles ofNaOH =2 x 0.00393mol=0.00786mol (�)

Molarity ofNaOH solution=number ofmoles ofNaOH

volumeof solution

= 0.00786mol

( 29.2�000 ) dm3

= 0.269mol dm–3 (�)

\ the molarity of the sodiumhydroxide solution is 0.269mol dm–3.

�7�a) Transfer the solution into a 250.0 cm3 volumetric flask. (�)

Washthebeaker, theglassrodandthefilter funnelwitha littledistilledwaterseveral times.Pourall thewashings into the flask. (�)

Add distilledwater to the flask until themeniscus reaches thegraduationmark.


b) Phenolpthalein (�)

From colourless to red / pink (�)

c) i) Number ofmoles ofNaOH in 26.4 cm3 solution = molarity of solution x volumeof solution

= 0.500mol dm–3 x26.4

�000dm3

= 0.0�32mol (�)

ii) Letn be thebasicity of tartaric acid, sowe can represent the acid byHnX.

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

Number ofmoles ofHnX=mass

molarmass

= 9.90g�50.0gmol–�

= 0.0660mol (�)

�26

Number ofmoles ofHnX in 25.0 cm3 solution=0.0660mol x25.0 cm3

250.0 cm3

= 0.00660mol

Number ofmoles of acid

Number ofmoles ofNaOH=

�n

=0.00660mol0.0�32mol

n = 2 (�)

\ thebasicity of tartaric acid is 2.

d) i) A solutionof accurately known concentration. (�)

ii) Not appropriate as sodiumhydroxide absorbsmoisture in air readily. (�)

e) Use a pHmeter / pH sensor. (�)

�72a) The final volume of the solution would be more than 250.0 cm3 as the solute has volume. / The exactvolume of the solutionwouldnot be known. (�)

b)

(� mark for correct set-up; 0.5mark for each correct label; 0mark if the set-up is notworkable) (3)

c) From yellow to red (�)

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

cm3

= 20.2 cm3 (�)

ii) MOH(aq) + HCl(aq) MCl(aq) + H2O(l) �.26 g 250.0 cm3

(used) 25.0 cm3

Letm gmol–� be themolarmass ofMOH.

�27

Number ofmoles ofMOH in 250.0 cm3 solution= massmolarmass

= �.26gm gmol–�

Number ofmoles ofMOH in 25.0 cm3 solution = �.26m

mol x 25.0 cm3

250.0 cm3

= �.26m

x ��0

mol


= 0.�55mol dm–3 x20.2

�000dm3

= 0.003�3mol (�)

According to the equation, �mole ofMOH requires �mole ofHCl for complete neutralization.

i.e.number ofmoles ofMOH in 25.0 cm3 solution=0.003�3mol (�)

Number ofmoles ofMOH in 25.0 cm3 solution= �.26m

x ��0

mol = 0.003�3mol

m = 40.3 (�)

\ themolarmass of themetal hydroxide is 40.3gmol–�.

�73a) Pipette (�)

b) Wash thepipettewith distilled water and thenwith sodiumhydroxide solution. (�)

c) Wash theburette firstwith distilled water and thenwith the sulphuric acid. (�)

Clamptheburettevertically ina stand.Close the stopcock.Fill theburettewith theacid througha filterfunnel. (�)

Open the stopcock for a few seconds so as to fill the tip of theburettewith acid. (�)


• Methyl orange (�)

from yellow to red (�)

• Phenolphthalein (�)

from red / pink to colourless (�)

e) Extrawater in the conical flaskwill not change thenumber ofmoles of solute it holds. (�)

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

�5.0 cm3 25.0 cm3

�28


= 0.600mol dm–3 x 25.0�000

dm3

= 0.0�50mol (�)

Accordingtotheequation,2molesofNaOHrequire�moleofH2SO4forcompleteneutralization.�moleofNa2SO4 is produced.

i.e.number ofmoles ofH2SO4= 0.0�502

mol

= 0.00750mol (�)


volumeof solution

=0.00750mol

( �5.0�000 ) dm3

= 0.500mol dm–3 (�)


g) Number ofmoles ofNa2SO4 produced=0.00750mol

Concentrationof sodium sulphate in the resulting solution

= number ofmoles ofNa2SO4

volumeof solution

= 0.00750mol

( 25.0+�5.0�000 ) dm3

=0.�88 mol dm–3 (�)

\ the concentrationof sodium sulphate in the resulting solution is 0.�88mol dm–3.

h) Mix 25.0 cm3of dilute sodiumhydroxide solution and�5.0 cm3of dilute sulphuric acid. (�)

Heat the sodium sulphate solutiongently to obtain a concentrated solution.

Set the concentrated solution aside to cool and crystallize. (�)

Filter the crystals from the remaining solution. (�)

Wash the crystalswith a little cold distilledwater.Dry the crystals using filter paper. (�)

�29

�74a)

(�mark for showing theburette andbeaker;�mark for showing thepHmeter and stirrer; 2marks for6 correct labels) (4)

b) An acid that canproduceonehydrogen ionpermolecule. (�)

c) i) 3.2 (�)

ii) pHof HA= –log�0[H+] = 3.2

i.e. log�0[H+] = –3.2 (�)

[H+] = �0–3.2

= 6.3 x �0–4mol dm–3 (�)

d) HA(aq) + NaOH(aq) NaA(aq) + H2O(l) ?mol dm–3 0.�00 mol dm–3

25.0 cm3 28.0 cm3


= 0.�00mol dm–3 x 28.0�000

dm3

= 0.00280mol (�)

According to the equation, �mole ofHA requires �mole ofNaOH for complete neutralization.

i.e.number ofmoles ofHA in 25.0 cm3 solution=0.00280mol (�)

Molarity of acidHA= number ofmoles ofHAvolumeof solution

= 0.00280mol

( 25.0�000 ) dm3

= 0.��2mol dm–3 (�)

\ the concentrationof the acidHA is 0.��2mol dm–3.

�30

e) Thymol blue is a suitable indicator (�)

because the indicator changes colourwithin thepH rangeof the vertical part of the titration curve. (�)

�75a) Add the sodiumhydroxide solutionquickly to 27 cm3. Swirl the conical flaskwhen adding the alkali.(�)

Adjust the stopcock of the burette to add one drop of alkali at a time. Swirl the flask after eachaddition. (�)

Continue adding the alkali until the colour of the indicator just changes from colourless to red / pink. (�)

As the titration approaches its end point, wash down any solution sticking to the inside of the conicalflaskwith small amount of distilled water. (�)

b) (CH2COOH)2(aq) + 2NaOH(aq) (CH2COONa)2(aq) + 2H2O(l) ?mol dm–3 0.0980mol dm–3

25.0 cm3 27.6 cm3


= 0.0980mol dm–3 x27.6

�000dm3

= 0.00270mol (�)

According to the equation, � mole of (CH2COOH)2 requires 2 moles of NaOH for completeneutralization.

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

mol

= 0.00�35mol (�)

Concentrationof butanedioic acid= number ofmoles of (CH2COOH)2

volumeof solution

= 0.00�35mol

( 25.0�000 ) dm3

= 0.0540mol dm–3 (�)

\ the concentrationof thebutanedioic acid is 0.0540mol dm–3.

c) Error is lesswith a large titre. (�)

d) i) A solutionof accurately known concentration. (�)

ii) Not appropriate as sodiumhydroxide absorbsmoisture in air readily. (�)

�76a) A strong acid is an acid that almost completely dissociates inwater. (�)

Concentration refers to the number of moles (or amount) of acid in a unit volume of solution / � dm3of solution. (�)

Concentrated anddilute refer to the relative values ofmol dm–3. (�)

�3�

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

ii) Tomake sure that all thehydrochloric acid has been reacted. (�)

iii)

(� mark for correct set-up; � mark for labelling filter funnel and filter paper; � mark for labellingbarium carbonate and solution; 0mark if the set-up is notworkable) (3)

iv) The solubility of barium chloridedecreaseswhen the temperature of the solutiondrops. (�)

Asthesaturatedsolutioncools,thesolventcannotholdall thesolutes.Theextrasolutesseparateoutas crystals. (�)

v) (�)Anhydrousbarium chloridewill form. / Thewater of crystallization will be removed. (�)

(2)Absorb thewater by filter paper. / Place the crystals in a desiccator. (�)

vi) BaCO3(s) + 2HCl(aq) BaCl2(aq) + CO2(g) + H2O(l) �.00mol dm–3

50.0 cm3


= �.00mol dm–3 x 50.0�000

dm3

= 0.0500mol (�)

According to the equation, 2moles ofHCl react with �mole of BaCO3 to give �mole of BaCl2.

i.e.number ofmoles of BaCl2 = 0.05002

mol

= 0.0250mol (�)

Mass of BaCl2•2H2O=number ofmoles xmolarmass = 0.0250mol x 244.3gmol–�

= 6.��g (�)

\ 6.��gof barium chloride crystals canbeobtained.

�32

vii)Anyoneof the following:

• Some crystalsmaybe lostwhen the crystals arewashedwith distilledwater. (�)

• Little splashesmay cause someof the solution andhence the crystals to lose. (�)

• SomeBaCl2may remain in the solution anddonot crystallize out. (�)

�77a)

(�)

b) Mass ofHNO3 in 80m3 acid=86400000g x 70.0% = 60480000g

Number ofmoles ofHNO3 = massmolarmass

= 60480000g63.0gmol–�

= 960000mol (�)

Molarity of nitric acid = number ofmoles ofHNO3

volumeof solution

= 960000mol80000dm3

= �2.0mol dm–3 (�)

c) The�0.0 cm3pipette shouldbeused to deliver the concentratednitric acid.

SupposeV cm3 of 0.240mol dm–3 nitric acid canbeobtained.


�2.0 x �0.0�000

=0.240 xV

�000 V = 500.0

Volume of dilute nitric acid obtained =500.0 cm3

Henceuse the500.0 cm3 volumetric flask for thedilution. (�)

Wash the�0.0 cm3pipette firstwith distilledwater and thenwith theoriginal acid. (�)

Deliver exactly �0.0 cm3 of the original acid into the 500.0 cm3 volumetric flask using the pipette andpipette filler. (�)

Add distilledwater to the flask until themeniscus reaches thegraduationmark.


d) i) pHof acid= –log�0 (0.�00) (�)

= � (�)

�33

ii)

(0.5 mark for curve starting from pH �; 0.5 mark for vertical part of curve between pH 2 – 6; 0.5mark for showing equivalence point at 25 cm3; 0.5 mark for curve finishing between pH 9 – �0,extending to 40 cm3) (2)

iii) Chloro-phenol red (�)

because the indicator changes colourwithin thepH rangeof the vertical part of the titration curve. (�)

�78a) Consider � 000 cm3 (i.e. �.00dm3) of the sample.

Mass of � 000 cm3of the sample=�.27g cm–3 x � 000 cm3

= � 270g

Mass ofH3PO4 in � 000 cm3of sample =mass of � 000 cm3of sample x percentagebymass ofH3PO4 in sample = � 270 x 85.0% =�080 g (�)

Molarmass ofH3PO4 = (3 x �.0+3�.0+4 x �6.0) gmol–�

= 98.0gmol–�

Number ofmoles ofH3PO4 in �.00dm3of sample= massmolarmass

= � 080g98.0gmol–�

= ��.0mol (�)

Molarity of phosphoric acid= number ofmoles ofH3PO4

volumeof solution

= ��.0mol�.00dm3

= ��.0mol dm–3 (�)

b) Wash thehandwith plenty ofwater. (�)

�34

c) Use a burette to contain the sodiumhydroxide solution. (�)

Wash theburettewith distilled water and thenwith the sodiumhydroxide solution. (�)

Add the indicator to the flask, and then the alkali from the burette until the indicator changes fromcolourless to red / pink. (�)

d) i) Average volumeof sodiumhydroxide solutionused= 20.0+20.�+20.03

cm3

= 20.0 cm3 (�)

ii) Number ofmoles ofNaOH required=molarity of solution x volumeof solution

= �.�0mol dm–3 x 20.0�000

dm3

= 0.0220mol (�)

iii) Number ofmoles ofNaOH required=0.0220mol x250.0 cm3

25.0 cm3

= 0.220mol (�)

e) Number ofmoles ofH3PO4 in �0.0 cm3 solution=molarity of solution x volumeof solution

= ��.0mol dm–3 x �0.0�000

dm3

= 0.��0mol (�)

Number ofmoles ofNaOHNumber ofmoles ofH3PO4

= 0.220mol0.��0mol

= 2 (�)

\ equation2best describes the reactionoccurring in the titration. (�)

�79a) Step�—Shouldnot prepare the standard sodiumhydroxide solutionusing themethoddescribed. (�)

This is because sodiumhydroxide is deliquescent andwould absorbmoisture from the air. (�)

Step 2—Shouldnot use ameasuring cylinder to transfer the lemon juice. (�)

This is because ameasuring cylinder cannot give accurate measurements of liquid volumes.(�)

Step 3—Shouldnot rinse theburettewith distilledwater only. (�)

This is because water that remains in the burette would dilute the sodium hydroxidesolution. (�)

Step 5—Shouldnot perform the calculationusingonly one titration result. (�)

This is because errorsmayoccur in the titration. (�)

b) Step� —Standardize the sodiumhydroxide solutionbefore use. (�)

Step 2—Use a pipette to transfer the lemon juice. (�)

Step 3—Rinse theburettewith distilledwater and sodiumhydroxide solutionbefore use. (�)

Step 5—Repeat the titration at least 3 times to obtain consistent results. Use the mean titre for thecalculation. (�)

�35

�80a) As sulphuric acid was added, it removed both the barium ions (by precipitation) and hydroxide ions (byneutralization). (�)

Ba(OH)2(aq)+H2SO4(aq) BaSO4(s) + 2H2O(l) (�)

At theequivalencepoint,all thebarium ionsandhydroxide ionshadbeen removed.Hence theelectricalconductivity of the reactionmixture fell to almost zero. (�)

b) �5.0 cm3 (�)

c) Ba(OH)2(aq) + H2SO4(aq) BaSO4(aq) + 2H2O(l) ?mol dm–3 �.00mol dm–3

�00.0 cm3 �5.0 cm3

Number ofmoles ofH2SO4 in �5.0 cm3 solution=molarity of solution x volumeof solution

= �.00mol dm–3 x �5.0�000

dm3

= 0.0�50mol (�)

According to the equation, �mole of Ba(OH)2 requires �mole ofH2SO4 for complete neutralization.

i.e.number ofmoles of Ba(OH)2 in �00.0 cm3 solution=0.0�50mol (�)

Molarity of bariumhydroxide solution= number ofmoles of Ba(OH)2

volumeof solution

= 0.0�50mol

( �00.0�000 ) dm3

= 0.�50mol dm–3 (�)

\ the concentrationof thebariumhydroxide solution is 0.�50mol dm–3.

d) Only barium sulphate and water are produced in the reaction. Barium sulphate can be obtained byfiltration. (�)

�8�a) Aluminiumhydroxide in the tablet neutralizes the excess hydrochloric acid in the stomach. (�)

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

b) Chewing breaks down the tablets into smaller pieces. This increases the surface area of the tablets andthus increases the reaction rate (brings faster relief of pain). (�)

c) i) Methyl orange: from red to yellow / Phenolphthalein: from colourless to redor pink (2)

ii) NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l) 0.�90mol dm–3

�8.� cm3

Number ofmoles ofNaOH in �8.� cm3 solution=molarity of solution x volumeof solution

= 0.�90mol dm–3 x �8.��000

dm3

= 0.00344mol (�)


i.e. number ofmoles of excessHCl in 25.0 cm3diluted solution=0.00344mol (�)

�36

iii) Number ofmoles ofHCl added in Step 1=molarity of solution x volumeof solution

= �.00mol dm–3 x 50.0�000

dm3

= 0.0500mol (�)

iv) Number ofmoles ofHCl left over after reactionwith drug tablet in Step 1

= 0.00344mol x 250.0 cm3

25.0 cm3

=0.0344mol

Number ofmoles ofHCl reactedwithAl(OH)3 in drug tablet= (0.0500 – 0.0344)mol = 0.0�56mol (�)

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

According to the equation, �mole ofAl(OH)3 requires 3moles ofHCl for complete neutralization.

i.e. number ofmoles ofAl(OH)3 in drug tablet=0.0�56

3mol

= 0.00520mol

Molarmass ofAl(OH)3= (27.0+3 x �.0+3 x �6.0) gmol–�

= 78.0gmol–�

Mass ofAl(OH)3 in drug tablet =number ofmoles xmolarmass = 0.00520mol x 78.0gmol–�

= 0.406g (�)

\ thedrug tablet contains 0.406gof aluminiumhydroxide.

v)Liquid used for washing immediately before use

(�)250.0 cm3 volumetric flask used inStep 2 distilled water (0.5)

(2)25.0 cm3pipette used for delivering thediluted solution inStep 3

diluted solution (0.5)

(3)Conical flask for containing thedilutedsolution inStep 3

distilled water (0.5)

(4)Burette for containing the sodium hydroxidesolution

sodium hydroxide solution (0.5)

�82a) Transfer the resulting solution into a 250.0 cm3 volumetric flask. (�)

Washthebeaker, theglassrodandthefilter funnelwitha littledistilledwaterseveral times.Pourall thewashings into the flask. (�)

Adddistilledwater to the flask until themeniscus reaches thegraduationmark.


b) i) NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l) 0.�25mol dm–3

20.8 cm3

�37


= 0.�25mol dm–3 x 20.8�000

dm3

= 0.00260mol (�)


i.e. number ofmoles of excessHCl in 25.0 cm3diluted solution=0.00260mol (�)

ii) Number ofmoles ofHCl added in Step 1=molarity of solution x volumeof solution

= �.00mol dm–3 x �00.0�000

dm3

= 0.�00mol (�)

iii) Number ofmoles ofHCl left over after reactionwithmagnesium in Step 1

= 0.00260mol x250.0 cm3

25.0 cm3

=0.0260mol

Number ofmoles ofHCl reactedwithmagnesium in Step 1 = (0.�00 – 0.0260)mol = 0.074mol (�)

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

According to the equation, �mole ofMg requires 2moles ofHCl for complete reaction.

i.e. number ofmoles ofMg in sample= 0.0742

mol

= mass ofMgmolarmass ofMg

\ molarmass ofMg= mass ofMg0.074

2 mol

= 0.900g0.074

2 mol

= 24gmol–� (�)

\ the relative atomicmass ofMg is 24.

c) Neutralization is a quick process. (�)

As titration proceeds, the concentration of the acid decreases and hence the rate of reaction betweenthe acid and sodiumhydroxidedecreases. (�)

d) Oxide layerwould cover the surfaceofmagnesium. (�)

Theoxidewould reactwith hydrochloric acid, just asmagnesiumdid. (�)

Theoxidelayerwouldnotcauseanydifferencetothevalueofthetitrebecausemagnesiumoxidewouldreactwith the same volumeof acid asmagnesium. (�)

�38

�83a) Nickel /Ni (�)

b) Ni2+(aq) +CO32–(aq) NiCO3(s) (�)

c) i) Number ofmoles ofK2CO3 in 20.0 cm3 solution=molarity of solution x volumeof solution

= 0.250mol dm–3 x20.0

�000dm3

= 0.00500mol (�)

According to the equation, �mole ofK2CO3 requires �mole ofNiSO4 for complete reaction.

i.e.number ofmoles ofNiSO4=0.00500mol (�)

VolumeofNiSO4 solution=number ofmoles ofNiSO4


= 0.00500mol0.�00mol dm–3

= 0.0500dm3

= 50.0 cm3 (�)

\ 50.0 cm3ofNiSO4 solutionwould react exactly with 20.0 cm3of 0.250mol dm–3K2CO3 solution.

ii) Remove thenickel carbonate by filtration. (�)

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



�84a) Volume of acid (�)

Concentrationof acid (�)

Temperature (�)

b) The hydrochloric acidwas consumed as the reactionproceeded. (�)

Hence the concentrationof the acid in the reactionmixture droppedduring the course of the reaction. (�)

c) i) D (0.5)

ii) A (0.5)

d)

Acid usedRate of reaction

lower same higher

�00 cm3of 2mol dm–3HCl(aq) 4

�00 cm3of 0.5mol dm–3HCl(aq) 4

200 cm3of �mol dm–3HCl(aq) 4

200 cm3of 0.5mol dm–3HCl(aq) 4

�39

�85a)

(� mark for correctly drawn curveof best fit; 2marks for correctly plottedpoints) (3)

b) Loss inmass of contents of beaker for the time period2.0min to 6.0min = (�93.9 – �87.3) g = 6.60 g =mass ofNO(g) formed

Average rate of formationofNO(g)= mass ofNO(g) formedtime

= 6.60g(6.0 – 2.0)min

(�)

= �.65gmin–� (�)

c) Mass ofNO(g) formed in the first 3minutes= (200.0 – �9�.9) g = 8.�0g

Number ofmoles ofNO(g) formed= mass ofNOmolarmass ofNO

= 8.�0g30.0gmol–�

= 0.270mol (�)

According to the equation, 3moles ofCu reacts withHNO3(aq) to give 2moles ofNO(g).

�40

\ number ofmoles ofCu consumed= 32

x 0.270mol

= 0.405mol (�)

Mass ofCu consumed=number ofmoles ofCu xmolarmass ofCu = 0.405mol x 63.5gmol–�

= 25.7g (�)

d) Instantaneous rate of formationofNO(g)= – (�85.0 – �9�.6) g(7.5 – 3.0)min

(�)

= �.47gmin–� (�)

e) i) Increase (�)

ii) Decrease (�)

iii) Increase (�)

�86a) 2H2O2(aq) 2H2O(l) +O2(g) (�)

b) The rates of decompositionof hydrogenperoxide solutionwasA>B>C.

At A, the concentration of hydrogen peroxide solution was high. So, the rate of decomposition washigh. (�)

At B, the concentration of hydrogen peroxide solution was lower than that at A. So, the rate ofdecomposition was lower. (�)

AtC, the rate of decompositionwas zero as all thehydrogenperoxidewasusedup. (�)

c) (�68.235 – �68.�43) g4.0min

(�)

= 0.023 gmin–� (�)

d) i) (�68.235 – �68.�23) g=0.��2g (�)

ii) Number ofmoles ofO2 givenoff= mass ofO2

molarmass ofO2

= 0.��2g32.0gmol–�

= 0.00350mol (�)

According to the equation, 2moles ofH2O2 decompose to give �mole ofO2.

i.e.number ofmoles ofH2O2=2 x 0.00350mol = 0.00700mol (�)

Originalmolarity of hydrogenperoxide solution = number ofmoles ofH2O2

volumeof solution

= 0.00700mol

( 50.0�000 ) dm3

= 0.�40mol dm–3 (�)

�4�

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

ii) The shapeof the curvewouldbe the same. (�)

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

f) Anyoneof the following:

As the change in the mass is very small in this experiment, the use of a data-logger can give moreaccurate results. (�)

The experimental results in the formof graph couldbeobtained immediately. (�)

g) Any oneof the following:

Measure the volumeof gas givenoff. (�)

Measure thepressure of themixture in a closed reaction vessel. (�)

�87a) Colorimeter (�)

b) The intensity of thepurple colour of permanganate ions decreases as the reactionproceeds. (�)

The colorimeter measures the amount of light absorbed by the reaction mixture when a beam of lightpasses through the reactionmixture, i.e. the absorbance. (�)

Theabsorbanceisdirectlyproportionaltothecolourintensityofthereactionmixtureandtheconcentrationof permanganate ions in the reactionmixture. (�)

c)

(�)

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

Determine the slopeof the tangent. (�)

e) i) Rate= – �2

d[MnO4–(aq)]

dt= –

�5

d[C2O4

2–(aq)]dt

\ instantaneous rate of consumptionofC2O42–(aq) ions

=52

(instantaneous rate of consumptionofMnO4–(aq) ions) (�)

ii) Instantaneous rate of consumptionofC2O42–(aq) ions =

52

(3.50 x �0–3mol dm–3 s–�)

= 8.75 x �0–3mol dm–3 s–� (�)

�42

�88a) Whensodiumthiosulphatesolutionreactswithdilutesulphuricacid,ayellowprecipitateofsulphurforms.This changes the light transmittanceof the reactionmixture. (�)

To follow theprogress of the reaction,mark a cross on a pieceof paper. (�)

Put the conical flask containing some sodium thiosulphate solution on top of the paper. Add dilutesulphuric acid to the flask. (�)

Recordthetimet forthesolutionmixturetobecomeopaque, i.e.whenthecrosscanno longerbeseenfrom above. (�)

b) The average rate of reaction from the start to theopaque stage

∝ �time to reach theopaque stage (t)

(�)

c) i)

(�mark for correctly drawn curveof best fit; 2marks for correctly plottedpoints) (3)

ii) 64 – 68 s (�)

d) Change the concentrationof the reactants. (�)

�89a)

XXX.XX

(� mark for correct set-up; 2marks for 4 correct labels; 0mark if the set-up is notworkable) (3)

b) The reaction ratewas higher atX than atY. This is because the reactiongets slower as it proceeds. (�)

c) • The loss inmassof thecontentsof the reaction flask for sampleA isgreater than that for sampleB,i.e. sample A produces more carbon dioxide gas. Therefore sample A has a higher purity of calciumcarbonate than sample B. (�)

�43

• The initial rate for sample B is faster than that for sample A. It is because sample B has a smallerparticle size. (�)

d) Mass ofCO2givenoff = 3.70g

CaCO3(s) + 2HCl(aq) CaCl2(aq)+CO2(g) +H2O(l) ? g 3.70g

Number ofmoles ofCO2=mass ofCO2

molarmass ofCO2

= 3.70g44.0gmol–�

= 0.084�mol (�)

According to the equation, �mole ofCaCO3 reacts with 2moles ofHCl to give �mole ofCO2.

i.e.number ofmoles ofCaCO3 in sampleA=0.084�mol (�)

Mass ofCaCO3 in sampleA=number ofmoles ofCaCO3 xmolarmass ofCaCO3 = 0.084�mol x �00.�gmol–�

= 8.42g

\ percentagebymass ofCaCO3 in sampleA= 8.42g�0.0g

x �00%

= 84.2% (�)

�90a) Agaswas evolved in the reaction. (�)

b)

(� mark for correct set-up; 3marks for 6 correct labels; 0mark if the set-up is notworkable) (4)

c) Number ofmoles ofMgpresent= mass ofMgmolarmass ofMg

= 2.00g24.3gmol–�

= 0.0823mol (�)

Number ofmoles ofH2SO4 present=molarity of solution x volumeof solution

= �.00mol dm–3 x �00.0�000

dm3

= 0.�00mol (�)

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

Accordingtotheequation,�moleofMgreactswith�moleofH2SO4.Duringthereaction,0.0823moleofMg reactedwith 0.0823mol ofH2SO4. ThereforeH2SO4was in excess. (�)

�44

d) As the reactionproceeded, sulphuric acidwas consumedgradually.

AtA, the concentrationof sulphuric acidwas high. So, the rate of reactionwashigh. (�)

At B, the concentrationof sulphuric acidwas lower than that atA. So, the rate of reactionwas lower. (�)

AtC, the rate of reactionwas zero as all themagnesiumwasusedup. (�)

e) i) Sulphuric acid is a dibasic acid while hydrochloric acid is a monobasic acid. Therefore � mol dm–3sulphuric acid has a higher concentrationof hydrogen ions than�mol dm–3 hydrochloric acid. (�)

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

ii)

(�mark for showinga lower rateof reaction, i.e. tangent to curve less steep;�mark for showinga

smaller final pressure, around58

of theoriginal value) (2)

�9�a) 2H2O2(aq) 2H2O(l) +O2(g) (�)

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

c)

(� mark for correctly drawn curveof best fit; 2marks for correctly plottedpoints) (3)

�45

d) 60 – 62 cm3 (�)

e) All hydrogenperoxidehas decomposed. (�)

f)

(�markforshowingtheslowerproductionofoxygen,i.e.tangenttocurvelesssteep;�markforshowinga smaller final volume, about half that of theoriginal curve) (2)

g) Add � g of manganese(IV) oxide to 50 cm3 of hydrogen peroxide solution. Wait until the reactionstops. (�)

Filter to removemanganese(IV) oxide from the solution. (�)

Add ‘fresh’ manganese(IV) oxide to the filtrate. No oxygen is given off. This indicates that the oxygendoes not come frommanganese(IV) oxide. (�)

Add ‘fresh’ hydrogen peroxide solution to the residue (manganese(IV) oxide). Oxygen is given off. Thisindicates that theoxygen comes fromhydrogenperoxide. (�)

h) A gas is givenoff in thedecomposition. (�)

�92a) i) By the reactionbetween sulphuric acid and aqueous ammonia. (�)

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

iii) So the fertilizer canbe taken in by the roots of plants. (�)

b) Used in car batteries (�) /manufacture detergents (�) /manufacture paints (�)

c) i) Aprocess inwhichheat is released to the surroundings. (�)

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

d) A catalyst is a substancewhich alters the rate of a reaction (�)

without itself undergoing anypermanent chemical changes. (�)

�46

e) i)

(�)

ii) Any twoof the following:

Wear safety glasses / protective gloves. (�)

Never addwater to the acid. (�)

In case any acid is split on the skin or clothes,wash the affected areawith plenty ofwater. (�)

iii) Anyoneof the following:

• Keep container tightly closed. (�)

• Store in a dry area. (�)

• Keepaway fromheat sources / combustiblematerials / reducingagents / bases /metals / organicmaterials / oxidizing agents. (�)

f) i) CaO(s) + SO2(g) CaSO3(s) (�)

ii) Neutralization reaction (�)

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

ii) Move the student to awell ventilated area / open space. (�)

�93a) i) [Al(OH)4]– (�)

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

b) Diluting concentrated sulphuric acid / any specific neutralization reaction (�)

c) i) The cleanser is corrosive. (�)

ii) An acidic substance would react with the sodium hydroxide and aluminium in the cleanser, makingthe cleanser ineffective. (�)

Agreat amount of heat andgaswouldbegenerated. (�)

Thismay cause a violent eruption from the drain. The mixing may cause severe injury to the user aswell as damage to porcelain tubs and toilets.

iii)

(�)

d) A strong alkali is an alkali that almost completely dissociates to give hydroxide ions (OH–(aq) inwater. (�)

e) i) By electrolysis of concentrated sodium chloride solution (�)

ii) Manufactureof bleach /manufacture of soaps anddetergents (�)

�47

�94a) SiO2(s) + 6HF(aq) H2SiF6(aq)+2H2O(l) (�)

b) Coat theglasswith layers of beeswax. Then trace thepatternswith ametal needle. (�)

Dip theglass intohydrofluoric acid to attack theunprotectedglass surface. (�)

c) i) An acid that canproduceonehydrogen ionpermolecule. (�)

ii) 0.�mol dm–3 sulphuric acid<0.�mol dm–3 hydrochloric acid<0.�mol dm–3 hydrofluoric acid (�)

Both sulphuric acid andhydrochloric acid are strong acids. Theydissociate completely inwater. (�)

Sulphuric acid is dibasicwhile hydrochloric acid ismonobasic. (�)

Hence 0.� mol dm–3 sulphuric acid contains a higher concentration of hydrogen ions than 0.� moldm–3 hydrochloric acid.

Hydrofluoric acid is aweak acid. It only partially dissociates inwater. (�)

Hence0.�moldm–3hydrofluoricacidhasa lowerconcentrationofhydrogen ions than0.�moldm–3hydrochloric acid.

d) i) Calcium carbonate (�)

ii) Thematerials formakingglass are readily available / abundant in the Earth’s crust. (�)

e) i) Hydrated iron(III) oxide (�)

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

�95Any three of the following:

• Addbarium chloride solution / bariumnitrate solution to the acids. (�)

�mol dm–3H2SO4(aq) gives awhite precipitatewhile �mol dm–3HCl(aq) gives noprecipitate. (�)

• Adddilute nitric acid followedby silver nitrate solution to the acids. (�)

�mol dm–3HCl(aq) gives awhite precipitatewhile �mol dm–3H2SO4(aq) gives noprecipitate. (�)

• Add calcium carbonate to excess acids. (�)

Calcium carbonate disappears after reactionwhen added to �mol dm–3HCl(aq).

Calciumcarbonatedoesnotreactcompletelywith�moldm–3H2SO4(aq)duetotheformationofinsolublecalcium sulphate on its surface. (�)

• Titrate equal volumesof the acidswith sodiumhydroxide solution. (�)

�moldm–3H2SO4(aq)requiresmoresodiumhydroxidesolutiontoreachtheendpointthan�.0moldm–3HCl(aq) asH2SO4(aq) is dibasicwhileHCl(aq) ismonobasic. (�)

(3marks for organization andpresentation)

�48

�96Titrate dilute sodiumhydroxide solutionwith dilute hydrochloric acid until the endpoint is reached. (�)

Mix the same volumes of alkali and acid as in the previous experiment. Do not use any indicator thistime. (�)

Heat the resulting solution to obtain a concentrated solution. (�)

Set the concentrated solution aside to cool and crystallize. (�)




�97Carry out flame test on all the solids. (�)

Only potassium sulphate gives a lilac flame. (�)

Dissolve the remaining three solids separately in water. Then add dilute sodium hydroxide solution untilexcess. (�)

Aluminium sulphate solutiongives awhite precipitatewhich is soluble in excess alkali. (�)

Magnesium sulphate solutiongives awhite precipitatewhichdoes not dissolve in excess alkali. (�)

Ammonium sulphate solutiongives noobservable change. (�)


OR

Dissolve someof each solid inwater separately. Then adddilute sodiumhydroxide solutionuntil excess. (�)

Aluminium sulphate solutiongives awhite precipitatewhich is soluble in excess alkali. (�)

Magnesium sulphate solutiongives awhite precipitatewhichdoes not dissolve in excess alkali. (�)

Ammonium sulphate solution andpotassium sulphate solutiongive noprecipitate. (�)

Warmeachof the two solutions. (�)

The solution containing ammonium sulphate gives a gas that turnsmoist red litmuspaper blue (ammonia).

There is noobservable change for the solution containingpotassium sulphate. (�)


�49

�98Add sodium chloride solution to all the four solutions.

The solution that gives awhite precipitate is silver nitrate solution. (�)

Add silver nitrate solution to the remaining three solutions. (�)

Both calcium chloride solution andpotassium chloride solutiongive awhite precipitate, (�)

while there is noobservable changewith potassium sulphate solution. (�)

Add potassium sulphate solution to calcium chloride solution andpotassium chloride solution. (�)

The solution that gives awhite precipitate is calcium chloride solution. (�)


�99During the hydrolysis, each molecule of ethyl ethanoate hydrolyzed produces one molecules of ethanoicacid. As hydrochloric acid is a catalyst in the reaction, its amount remains the same. Hence the increase intotal amountof acid in the reactionmixture is adirectmeasureof the amountof ethyl ethanoate thathasundergone hydrolysis. (�)

Mix 5.00 cm3 of ethyl ethanoate and �00 cm3 of dilute hydrochloric acid in a reaction flask. Start the stopwatch at the same time. (�)

Immediately withdraw 5.00 cm3 of the reaction mixture. Run the sample into 50 cm3 of water which hasbeen chilled in an ice bath. Titrate with sodium hydroxide solution. Let the volume of sodium hydroxidesolution consumedbeV0. (�)

Withdraw5.00cm3samplesat regular time intervalsandtitratewith thesodiumhydroxidesolution.Let thevolume of sodiumhydroxide solution consumed at time t beVt. (�)

Stopper the reaction flask. Leave for twodays to allow thehydrolysis to complete.

Withdraw a 5.00 cm3 sample and titrate with the sodium hydroxide solution. Let the volume of sodiumhydroxide solution consumedbeV∞. (�)

V∞ is the volume of alkali required to neutralize the hydrochloric acid and the ethanoic acid formed uponcompletehydrolysis.Hence(V∞–V0) isproportionaltotheconcentrationofethylethanoateatthebeginningof the reaction and (V∞ –Vt) is proportional to the concentrationof unreacted ethyl ethanoate at time t. (�)


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3. Solution Guide to Supplementary Exercises