12. solution guide to supplementary exercises

�

Topic 16 Analytical Chemistry

Part A Unit-basedexercise

Unit 53 Qualitative analysis — detecting the presence of inorganic chemical species

Fillintheblanks

�Test Observation Deduction

a)dilute nitric acid; aqueous solution of silver nitrate

— —

b) — orange; green sulphite

c) i) — ii) —

i) reddish brown ii) orange; green

bromine; sulphur dioxide

d) — white precipitate

e) — — chlorine; hypochlorite

f) — A white precipitate —

g) — — sulphate

h) — — ammonium

i) — white; dissolves; colourless —

j) — white —

Trueorfalse

2 F

3 F

4 T

5 F

6 T

7 T

8 F

9 T

�0 T

�� F

2

Multiplechoicequestions

�2 A

�3 C

�4 A

�5 C

�6 A

�7 D

�8 A

�9 A

20 B

2� C

22 D

23 C

24 C

25 B

26 D

27 C

28 D

29 B

30 D

3� B

32 C

33 C

34 B

3

35 A

36 B

37 C

38 D

39 C

40 D

4� B

42 A

43 C

44 D

45 B

46 C

47 A

48 D

49 D

50 A

5� D

52 A

53 A

54 D

55 D

56 B

4

Unit 54 Tests for functional groups, separation and purification of compounds

Fillintheblanks

�Test Observation Deduction

a) — — C=C; C≡C

b) — orange; green aldehyde

c) 2,4-dinitrophenylhydrazine — ketone

d) Tollens’ reagent — aldehyde

e) — — tertiary

f) — —

CHI3;

O

C CH3 ;

OH

C

H

CH3

g) silver nitrate — silver bromide; bromine

h) — Effervescence —

i) — — ester; alcohol

Trueorfalse

2 F

3 T

4 F

5 T

6 T

7 T

8 F

9 T

5


�0 B

�� C

�2 B

�3 D

�4 D

�5 B

�6 A

�7 D

�8 A

�9 C

20 B

2� A

22 B

23 D

24 C

25 D

26 B

27 D

28 C

29 C

30 A

3� C

32 C

6

33 C

34 D

35 A

36 B

37 B

38 A

39 C

40 A

4� C

42 A

43 B

44 D

45 D

Unit 55 Quantitative methods of analysis

Fillintheblanks

� a) instrumental error

b) operative error

c) error of method

2 a) silver nitrate; chromate

b) silver chromate

3 a) starch solution

b) colourless; dark blue

4 a) iodine; iodine; sodium thiosulphate

b) starch solution; dark blue; colourless

7

Trueorfalse

5 T

6 T

7 T

8 F

9 F

�0 F


�� D

�2 B

�3 B

�4 C

�5 C

�6 C

�7 D

�8 B

�9 C

20 A

2� C

22 A

23 A

24 B

25 A

26 B

27 D

8

28 C

29 B

30 A

3� A

32 B

33 D

34 C

35 D

36 A

37 C

38 D

39 C

40 D

4� D

42 C

43 C

Unit 56 Instrumental analytical methods

Fillintheblanks

� a) Tollens’ reagent / acidified aqueous solution of potassium dichromate

b) iodine; sodium hydroxide

c) carbonyl / C=O

d) i) CH3CH2+

ii) CH3CO+

2 a) mixing with aqueous solution of sodium hydrogencarbonate / warming with an alcohol in the presence

of concentrated sulphuric acid

9

b) i) O–H

ii) (�) C–O

(2) C=O

c) i) COOH+

ii) CH3CO+

Trueorfalse

3 F

4 F

5 T

6 T

7 T

8 F

9 T

�0 T

�� F

�2 F


�3 C

�4 A

�5 D

�6 C

�7 D

�8 C

�9 D

20 C

2� C

22 B

�0

23 A

24 D

25 B

26 D

27 C

28 A

29 B

30 B

3� A

32 B

33 A

34 C

35 D

36 B

37 B

38 D

39 D

40 A

4� A

42 C

Unit 57 Contribution of analytical chemistry to our society

Fillintheblanks

�Phase

Paper chromatography

Thin layer chromatography

Gas-liquid chromatography

Mobile — — carrier gas

Stationary water in the paper fibres fine layer of alumina or silica gel —

��

Trueorfalse

2 T

3 F

4 F

5 T

6 T


7 D

8 D

9 B

�0 A

�� B

�2 D

�3 C

�4 C

�5 D

�6 A

�7 A

�8 A

�9 B

20 C

2� D

�2

Part B Topic-based exercise


� A

2 C

3 B

4 B

5 C

6 C

7 C

8 A

9 C

�0 A

�� D

�2 C

�3 A

�4 B

�5 D

�6 D

�7 C

�8 C

�9 B

20 D

2� C

�3

22 A

23 C

24 C

25 D

26 B

27 B

28 A

29 D

30 D

3� B

32 A

33 D

34 C

35 B

36 B

37 A

38 D

39 D

40 B

4� D

42 B

43 B

44 C

45 A

�4

46 C

47 A

48 B

49 B

50 D

5� C

Shortquestions

52 a) i) Dip a clean nichrome wire into concentrated hydrochloric acid. (�)

Dip the nichrome wire into the solid. (�)

Heat the wire in a Bunsen flame. (�)

ii) Green (�)

b) Potassium flame colour would be masked by the strong sodium flame colour. (�)

53 a) White precipitate (�)

insoluble in dilute hydrochloric acid (�)

b) The precipitate would dissolve in dilute hydrochloric acid. (�)

54 a)Silver chloride Silver bromide Silver iodide

Dilute aqueous ammonia

soluble (0.5) insoluble (0.5) insoluble (0.5)

Concentrated aqueous ammonia

soluble (0.5) soluble (0.5) insoluble (0.5)

b) i) White fumes (�)

ii) NH3(g) + HCl(g) NH4Cl(s) (�)

55 a) Any two of the following:

• Ethoxyethane is a good solvent for most carbon compounds. (�)

• Ethoxyethane is immiscible with water. (�)

• Ethoxyethane is volatile. It can easily be removed by distillation. (�)

• Ethoxyethane is chemically unreactive. (�)

b) Avoid naked flame / carry out the evaporation in a fume cupboard. (�)

�5

56Tests Observation

a) Shake a few drops of compound X with aqueous bromine.

The yellow-brown aqueous bromine becomes colourless quickly. (�)

b) Add anhydrous zinc chloride in concentrated hydrochloric acid to compound X and warm.

Cloudiness appears. (�)

c) Add iodine in an aqueous solution of sodium hydroxide to compound X and warm the mixture.

A bright yellow precipitate forms. (�)

57Tests Observation with X Observation with Y

Aqueous solution of sodiumcarbonate

effervescence occurs (0.5) no observable change (0.5)

2,4-dinitrophenylhydrazine no observable change (0.5) yellow to red precipitate forms (0.5)

Aqueous solution of potassiumdichromate + dilute sulphuricacid

no observable change (0.5)the dichromate solution changes from orange to green (0.5)

58 a)

(� mark for correct set-up; � mark for correct labels; award 0 mark if the set-up is not workable) (2)

b)


�6

59 a) When the aqueous solution of silver nitrate is added to a solution of chloride ions, silver chloride precipitate forms. (�)

When all the chloride ions are precipitated, the first excess silver nitrate solution gives a reddish brown silver chromate precipitate with the chromate indicator. This signals the end point of the titration. (�)

2Ag+(aq) + CrO42–(aq) Ag2CrO4(s) (�)

reddish brown

b) If the solution is too acidic, then part of the indicator will be present as hydrogen chromate ions and more silver ions will be required to form the silver chromate precipitate. (�)

If the solution is too basic, silver oxide may be precipitated. (�)

Structuredquestions

60 a)Test Procedure Observation Deduction

� — — sodium (�)

2 — — oxygen (�)

3 — — chloride (�)

b) Sodium chlorate / NaClO3 (Sodium hypochlorite / NaOCl may be accepted.) (�)

6� a)Test Procedure Observations Deduction

� — — transition metal (0.5)

2 — — ammonia; (0.5) ammonium ions (0.5)

3

— — —

a) — — copper(II) ions (0.5)

b) — — barium sulphate; (0.5) sulphate ions (0.5)

b) Cations – ammonium ion and copper(II) ion

Anion – sulphate ion

62 We need to look at the properties of the cations:

• Ag+(aq) ions form an insoluble chloride. (�)

• PbCl2(s) is insoluble in cold water, but soluble in hot water. (�)

• Al3+(aq) and Zn2+(aq) ions form insoluble hydroxides with dilute aqueous solution of sodium hydroxide and the hydroxides are soluble in excess alkali. (�)

(�)

�7

• Fe2+(aq) ions form an insoluble hydroxide with dilute aqueous solution of sodium hydroxide. (�)

• Al3+(aq) and Fe2+(aq) ions form insoluble hydroxides with dilute aqueous ammonia. (�)

• Zn2+(aq) ions form an insoluble hydroxide with dilute aqueous ammonia and the hydroxide is soluble in excess alkali. (�)

precipitate of

Al(OH)3(s)solution containing

[Zn(NH3)4]2+(aq)

Step 3 — add excess NaOH(aq)

Step 4 — add HCl(aq)

followed by

excess NH3(aq)

Step 1 — add HCl(aq)

precipitate of

AgCl(s)

precipitate of

Fe(OH)2(s)PbCl2(aq)

Step 2 — add hot water

solution containing Ag+(aq), Al3+(aq), Fe2+(aq),

Pb2+(aq), Zn2+(aq)

precipitate of

AgCl(s), PbCl2(s)

solution containing

Al3+(aq), Fe2+(aq), Zn2+(aq)

solution containing

[Al(OH4)]–(aq), [Zn(OH4)]

2–(aq)

(� mark for each step with correct resulting species stated; appropriate steps in a different order is also acceptable) (4)

63 A and E reacted to give a white precipitate (Zn(OH)2(s)) that was soluble in excess A. Thus, A was NH3(aq) and E was ZnCl2(aq). (�)

The white precipitate dissolved due to the formation of complex ions.

Zn2+(aq) + 2OH–(aq) Zn(OH)2(s) (�)

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

A and C reacted to give a white precipitate (Ba(OH)2(s)). Thus, C was BaCl2(aq). (�)

Ba2+(aq) + 2OH–(aq) Ba(OH)2(s) (�)

B gave a white precipitate that was soluble in NH3(aq) with C (BaCl2(aq)). The precipitate was AgCl(s). Thus, B was AgNO3(aq). (�)

Ag+(aq) + Cl–(aq) AgCl(s) (�)

�8

B and F reacted to give a yellow precipitate (AgI(s)). Thus, F was KI(aq). (�)

Ag+(aq) + I–(aq) AgI(s) (�)

C and D reacted to give a white precipitate (BaSO4(s)). Thus, D was Na2SO4(aq). (�)

Ba2+(aq) + SO42–(aq) BaSO4(s) (�)

64 a) Any one of the following:

• Treat with 2 ,4-dinitrophenylhydrazine. (�)

Only compound P gives a yellow to red precipitate. (�)

• Warm with the Tollens’ reagent. (�)

Only compound P gives a silver deposit on the wall of the reaction vessel. (�)

• Warm with acidified aqueous solution of potassium dichromate. (�)

Only compound P turns the dichromate solution from orange to green. (�)

• Add aqueous solution of sodium hydrogencarbonate. (�)

Effervescence occurs for compound Q only. (�)

b) Any one of the following:


Only compound R turns the dichromate solution from orange to green. (�)

• Mix with phosphorus pentachloride. (�)

Only compound R gives steamy fumes of hydrogen chloride. (�)

• Treat with anhydrous zinc chloride in concentrated hydrochloric acid. (�)

Cloudiness appears in 5 minutes for compound R only. (�)

Any other appropriate reaction is also acceptable.

c) Any one of the following:


Only compound T turns the dichromate solution from orange to green. (�)

• Treat with anhydrous zinc chloride in concentrated hydrochloric acid. (�)

Cloudiness appears in � minute for compound U only. (�)

d) Warm with dilute aqueous solution of silver nitrate in ethanol. (�)

Compound V gives a white precipitate. (0.5)

Compound W gives a yellow precipitate. (0.5)

�9

65 a) Mobile phase – the developing solvent (�)

Stationary phase – water in the paper fibres (�)

b)


c) Each amino acid distributes itself between the mobile phase and the stationary phase. (�)

Amino acids that are more soluble in the mobile phase (or developing solvent) travel up more quickly. Thus, the amino acids separate. (�)

d) Rf = 3.� cm5.4 cm

= 0.57

e)

66 a) i) Any one of the following:

• The distance a component travels relative to the solvent. (�)

• Rf = distance travelled by the componentdistance travelled by the solvent

(�)

ii) Retention time is the time between injection and detection of a component / the time a component held in a column. (�)

b) Mobile phase – carrier gas (�)

Stationary phase – liquid coated onto the inside of the column (�)

20

c) i) X (�)

It is the first to emerge. / It has the shortest retention time. (�)

ii) Total area of the peaks = �2

x �0 x 4 + �2

x 30 x 4 + �2

x 20 x 8

= �60 (�)

Percentage abundance of component Y = 60�60

x �00%

= 37.5% (�)

67 a) To enable reactants to be heated for a long time. (�)

To prevent any loss of the reaction mixture. (�)

b)

(� mark for correct drawing of separating funnel with tap; � mark for showing the � -bromobutane layer below water; award 0 mark if the funnel is not workable) (2)

c) To dry the � -bromobutane. (�)

d) i)

(� mark for correct set-up; � mark for correct labels; � mark for correct direction of water flow in condenser; award 0 mark if the set-up is not workable) (3)

2�

ii) �00 °C – �04 °C (�)

e) i) AgBr (�)

ii) C4H9Br + H2O C4H9OH + H+ + Br– (�)

iii) As a common solvent for �-bromobutane and silver ions (�)

iv) Any one of the following:

• The reaction is slow at room temperature. (�)

• Heat speeds up the reaction. (�)

• The reactant(s) are flammable. (�)

f) CH3CH2CH2CH2+ / CH2Br+ / CH2CH2CH2Br+ (�)

68 a)

(�)

b) Wear protective gloves. / Carry out in a fume cupboard. (�)

c) The reaction is exothermic / vigorous. (�)

d)

(� mark for Buchner funnel and Buchner flask; � mark for suction applied; � mark for correct labels; award 0 mark if the set-up is not workable) (3)

e) Wash the solid with water. (�)

Add an aqueous solution of silver nitrate to the water obtained. (�)

No creamy precipitate forms. (�)

f) Dissolve the crude sample in minimum amount of hot methanol. (�)

Filter the mixture while hot. (�)

Allow the filtrate to cool and collect the crystals by filtration. (�)

Wash the crystals and dry. (�)

22

69 a) Bromine (�)

b) Aqueous solution of sodium hydroxide (�)

c) –OH group / hydroxyl group (�)

d) Heat with concentrated sulphuric acid / aluminium oxide. (�)

e) i)

C Br or

H

C

H

HH

C

H

H

H C H

H

C

H

HBr

C

H

H

H

(�)

ii)

C OH or

H

C

H

HH

C

H

H

H C H

H

C

H

HOH

C

H

H

H

(�)

iii)

C H

H

C

H

H

C

H

H

(�)

70 a) i) CHI3 (�)

ii) Compound A contains a CH3C

O

group. (�)

iii)

(CH3)2CHC CH3

O

(�)

methylbutanone (�)

b) i) A yellow / red precipitate forms. (�)

ii)N

H

N

NO2

NO2

(CH3)2CH

CH3

C

(�)

c) i)

C4H9 — C — H + 2[Ag(NH3)2]+ + 3OH–

O

C4H9 — C — O– + 2Ag + 4NH3 + 2H2O

O

(�)

ii) Compound A (ketone) cannot be oxidized. (�)

23

7� a) Test Proceduce Observation Deduction

�Ignite �0 drops of X in a crucible.

X burns with a sooty flame.

X has a high carbon content. (�)

2To 2 cm3 of X, add 8 drops of aqueous bromine.

T h e y e l l o w - b r o w n aqueous bromine becomes colourless quickly.

X is unsaturated. (�)

3

To � cm3 of X, add 8 drops of acidified aqueous solution of potassium dichromate and warm.

The dichromate solution changes from orange to green.

X may be an alcohol or aldehyde. (�)

4To 4 d r o p s o f X , add 2 cm 3 o f 2 ,4 -dinitrophenylhydrazine.

An orange precipitate forms.

X may be an aldehyde or a ketone. (�)

5

Mix 4 drops of X with 2 cm3 of Tollens’ reagent in a test tube. Warm in a water bath at 60 °C.

A silver mirror forms on the wall of the test tube.

X is an aldehyde. (�)

b) i) Compound X contains identical groups / atoms on the same side of a double bond. (�)

ii)

H

CHO

HC C

(�)

72 a) i) Suppose we have �00 g of compound X, so there are 58.8 g of carbon, 9.80 g of hydrogen and 3�.4 g of oxygen.

Carbon Hydrogen Oxygen

Mass of element in the compound

58.8 g 9.80 g 3�.4 g

Number of moles of atoms that combine

58.8 g�2.0 g mol–� = 4.90 mol

9.80 g�.0 g mol–� = 9.80 mol

3�.4 g�6.0 g mol–� = �.96 mol (�)

Simplest ratio of atoms

4.90 mol�.96 mol

= 2.509.80 mol�.96 mol

= 5.00�.96 mol�.96 mol

= �.00

Simplest whole number ratio of atoms

2.50 x 2 = 5.00 5.00 x 2 = �0.0 �.00 x 2 = 2.00 (�)

∴ the empirical formula of X is C5H�0O2.

24

ii) Let (C5H�0O2)n be the molecular formula of X.

Relative molecular mass of X = n(5 x �2.0 + �0 x �.0 + 2 x �6.0) = �02n

i.e. �02n = �02.0 n = � (�)

∴ the molecular formula of X is C5H�0O2.

b) Y is propan-2-ol. (�)

Y must contain a CH3

OH

C

H

group (�)

as it gives triiodomethane when warmed with iodine in an aqueous solution of sodium hydroxide.

Z is sodium ethanoate. (�)

Z must contain two (i.e. 5 – 3) carbon atoms. (�)

X is �-methylethyl ethanoate / CH3COOCH(CH3)2 (�)

c) i) Iodine (�)

ii) Hydrolyze with aqueous solution of sodium hydroxide. (0.5)

Acidify with nitric acid. (0.5)

Add aqueous solution of silver nitrate. (0.5)

A yellow precipitate forms. (0.5)

73 a) As compound X has a linear structure, compound Z should also have a linear structure.

As compound Z gives cloudiness with the Lucas reagent in 5 minutes, it is probably a secondary alcohol. (�)

As compound Z has no chiral carbon, the –OH group should be attached to the central carbon atom. (�)

C

H

C

H

HH

C

H

H

H C H

H

C

H

HH

C

H

H

OH

H

C

(�)

25

b) Compound Y undergoes hydrogenation to give compound Z which has two more hydrogen atoms, thus compound Y must contain a C=C bond. (�)

Compound X undergoes reduction to give compound Y which is a secondary alcohol, thus compound X is probably a ketone with (�)

the C=O group attached to the central carbon atom. (�)

As compound X exists as a mixture of geometrical isomers, the C=C bond should not be at the end of the carbon chain. (�)

Structure of X:

C

H

C

H

C

H

H

H C H or

H

C

H

HH

C

H

H

O

C C C

H

H

C

H

H

H C H

H

C

H

HH

C

H

H

O

C

(�)

Structure of Y:

C

H

C

H

C

H

H

H C H

H

C

H

HH

C

H

H

OH

H

C C C

H

H

C

H

H

H C H

H

C

H

HH

C

H

H

OH

H

Cor

chiral carbon chiral carbon (�)

74 a) i) (NH4)2C2O4(aq) + 2H+(aq) 2NH4+(aq) + H2C2O4(aq) (�)

ii) Urea decomposes to form hydroxide ions. (�)

The hydroxide ions neutralize the oxalic acid to liberate oxalate ions. (�)

b)

(� mark for Buchner flask and sintered glass filtering crucible; � mark for suction applied; � mark for correct labels; award 0 mark if the set-up is not workable) (3)

26

c) Number of moles of CaC2O4•H2O = 0.462 g�46.� g mol–�

= 3.�6 x � 0–3 mol (�) = number of moles of Ca in mineral

Mass of Ca in mineral = 3.�6 x �0–3 mol x 40.� g mol–�

= 0.�27 g

Percentage by mass of Ca in mineral = 0.�27 g0.599 g

x �00%

= 2�.2% (�)

∴ the percentage by mass of calcium in the mineral is 2�.2%.

d) Any two of the following:

• As no ionic substance is completely insoluble in water, a little calcium oxalate will remain dissolved in the solution. (�)

• A little of the precipitate will be lost as the precipitate is washed. (�)

• Little splashes, inefficient rinsing out of the beaker, and inefficient filtering can cause some loss of the precipitate. (�)

75 a) i) H2O2(aq) O2(g) + 2H+(aq) + 2e– (�)

MnO4–(aq) + 8H+(aq) + 5e– Mn2+(aq) + 4H2O(l) (�)

ii) 2MnO4–(aq) + 5H2O2(aq) + 6H+(aq) 2Mn2+(aq) + 5O2(g) + 8H2O(l) (�)

b) i) Run the aqueous solution of potassium permanganate into the peroxide solution until the first appearance of a persistent pale pink colour. (�)

ii) 2MnO4–(aq) + 5H2O2(aq) + 6H+(aq) 2Mn2+(aq) + 5O2(g) + 8H2O(l)

0.0�80 mol dm–3 ? mol dm–3

�5.7 cm3 25.0 cm3

Number of moles of MnO4– ions in �5.7 cm3 solution = 0.0�80 mol dm–3 x �5.7

� 000 dm3

= 2.83 x �0–4 mol (�)

According to the equation, 2 moles of MnO4– ions react with 5 moles of H2O2.

i.e. number of moles of H2O2 in 25.0 cm3 solution = 52

x 2.83 x �0–4 mol

= 7.08 x �0–4 mol (�)

Concentration of H2O2 solution = 7.08 x �0–4 mol25.0

� 000 dm3

= 2.83 x �0–2 mol dm–3

= 2.83 x �0–2 mol dm–3 x 34.0 g mol–�

= 0.962 g dm–3 (�)

∴ the concentration of the hydrogen peroxide solution is 0.962 g dm–3.

27

76 a) No additional mass was lost during the third heating, indicating that all the water of hydration had been driven off. (�)

b) Mass of water lost = (28.457 – 25.437) g or (28.457 – 25.438) g = 3.02 g

Number of moles of water lost = 3.02 g�8.0 g mol–�

= 0.�68 mol (�)

Mass of anhydrous CaCl2 = (25.437 – 22.347) g or (25.438 – 22.347) g = 3.09 g

Number of moles of anhydrous CaCl2 = 3.09 g��.� g mol–�

= 0.0278 mol (�)

Number of moles of H2ONumber of moles of CaCl2

= 0.�68 mol0.0278 mol

= 6.04 (�)

∴ the value of x is 6.

c) The calculated mass (or moles) of water lost by the hydrate would be too large (�)

because the mass of the solid that was lost would be assumed to be water when it actually included some CaCl2 as well. (�)

d) i) Number of moles of AgCl(s) formed = 9.�8 g�43.4 g mol–�

= 0.0640 mol (�)

Number of moles of CaCl2•xH2O in 7.00 g of sample = 0.06402

mol

= 0.0320 mol (�)

Molar mass of CaCl2•xH2O = 7.00 g0.0320 mol–�

= 2�9 g mol–� (�)

Molar mass of CaCl2•xH2O = (40.� + 2 x 35.5 + �8x) g mol–�

= 2�9 g mol–�

x = 5.99 (�)


ii) Any two of the following:

• A little silver chloride would remain dissolved in the solution. (�)

• A little of the precipitate would be lost as the precipitate was washed. (�)

• Little splashes, inefficient rinsing out of the beaker, and inefficient filtering could cause some loss of the precipitate. (�)

• The precipitate might not be dried completely and so residual moisture could add to its mass. (�)

28

77 a) Run KMnO4(aq) into the contents of the flask until the first appearance of a persistent pale pink colour. (�)

b) 5(COOH)2(aq) + 2MnO4–(aq) + 6H+(aq) �0CO2(g) + 2 Mn2+(aq) + 8H2O(l) (�)

c) 5(COOH)2(aq) + 2MnO4–(aq) + 6H+(aq) �0CO2(g) + 2Mn2+(aq) + 8H2O(l)

0.202 g 0.0280 mol dm–3

22.9 cm3

Number of moles of MnO4– ions in 22.9 cm3 solution = 0.0280 mol dm–3 x 22.9

� 000 dm3

= 6.4� x �0–4 mol (�)

According to the equation, 5 moles of (COOH)2 react with 2 moles of MnO4– ions.

i.e. number of moles of (COOH)2 used = 52

x 6.4� x �0–4 mol

= �.60 x �0–3 mol (�)

Molar mass of (COOH)2•xH2O = 0.202 g�.60 x �0–3 mol

= �26 g mol–� (�) = (2 x �2.0 + 4 x �6.0 + 2 x �.0 + �8x) g mol–�

x = 2 (�)


78 a) A dark blue colour appears at the end point. (�)

When the ascorbic acid has been used up, the next iodine produced by the IO3–(aq) ions will react with

the starch to form a dark blue complex. (�)

b) IO3–(aq) + 5I–(aq) + 6H+(aq) 3I2(aq) + 3H2O(l) (�)

c) IO3–(aq) + 5I–(aq) + 6H+(aq) 3I2(aq) + 3H2O(l)

0.0��8 mol dm–3 ? mol

�7.6 cm3

Number of moles of KIO3 in �7.6 cm3 solution = 0.0��8 mol dm–3 x �7.6� 000

dm3

= 2.08 x �0–4 mol (�)

According to the equation, � mole of IO3– ions reacts with I– ions and H+ ions to give 3 moles of I2.

i.e. number of moles of I2 formed = 3 x 2.08 x �0–4 mol = 6.24 x �0–4 mol (�)

29

Ascorbic acid reacts with iodine according to the following equation:

C6H8O6(aq) + I2(aq) C6H6O6(aq) + 2H+(aq) + 2I–(aq)

? g in � .00 dm3 6.24 x �0–4 mol

25.0 cm3

According to the equation, � mole of ascorbic acid reacts with � mole of I2.

i.e. number of moles of ascorbic acid in 25.0 cm3 sample solution = 6.24 x �0–4 mol (�)

Number of moles of ascorbic acid in �.00 dm3 sample solution

= 6.24 x �0–4 mol25.0

� 000

= 0.0250 mol

Mass of ascorbic acid in �.00 dm3 sample solution = 0.0250 mol x �76.0 g mol–�

= 4.40 g (�)

∴ the mass of ascorbic acid in �.00 dm3 of the sample solution is 4.40 g.

79 a) Cl2(g) + 2KI(aq) I2(aq) + 2KCl(aq) (�)

b) Transfer the solution to a 250.0 cm3 volumetric flask. (�)

Add distilled water to the flask until the bottom of the meniscus reaches the graduated mark of the flask. (�)

c) i) Aqueous solution of sodium thiosulphate (�)

ii) I2(aq) + 2S2O32–(aq) 2I–(aq) + S4O6

2–(aq) (�)

iii) Add Na2S2O3(aq) until the colour of the solution in the conical flask becomes pale yellow. (�)

Add starch solution and continue with the titration until the disappearance of the blue colour. (�)

iv) Cl2(g) + 2KI(aq) I2(aq) + 2KCl(aq)

6.29 x �0–4 mol

25.0 cm3

Number of moles of I2 in 250.0 cm3 = �0 x 6.29 x �0–4 mol = 6.29 x �0–3 mol (�)

According to the equation, 2 moles of KI react with Cl2 to give � mole of I2.

i.e. number of moles of KI in sample = 2 x 6.29 x �0–3 mol = �.26 x �0–2 mol (�)

Mass of KI in sample = �.26 x �0–2 mol x �66.0 g mol–�

= 2.09 g

30

Percentage purity of KI sample = 2.09 g2.�8 g

x �00%

= 95.9% (�)

∴ the percentage purity of the potassium iodide sample is 95.9%.

80 a) i) Cu(s) + 4HNO3(l) Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l) (�)

ii) 2Cu2+(aq) + 4I–(aq) 2Cul(s) + I2(aq) (�)

iii) I2(aq) + 2S2O32–(aq) 2I–(aq) + S4O6

2–(aq) (�)

b) Add Na2S2O3(aq) until the colour of the solution in the flask becomes pale yellow. (�)

Add starch indicator and continue with the titration until the disappearance of the blue colour. (�)

c) I2(aq) + 2S2O32–(aq) 2I–(aq) + S4O6

2–(aq)

25.0 cm3 0.�02 mol dm–3

26.3 cm3

Number of moles of S2O32– ions in 26.3 cm3 solution

= 0.�02 mol dm–3 x 26.3� 000

dm3

= 2.68 x �0–3 mol (�)

According to the equation, � mole of I2 reacts with 2 moles of S2O32– ions.

i.e number of moles of I2 reacted with S2O32– ions

= 2.68 x�0–3

2 mol

= �.34 x �0–3 mol = number of moles of I2 liberated from 25.0 cm3 diluted solution

2Cu2+(aq) + 4I–(aq) I2(aq) + 2CuI(s)

25.0 cm3 �.34 x �0–3 mol

According to the equation, 2 moles of Cu2+ ions react to give � mole of I2.

i.e number of moles of Cu2+ ions in 25.0 cm3 diluted solution = 2 x �.34 x �0–3 mol = 2.68 x �0–3 mol (�)

Number of moles of Cu2+ ions in 250.0 cm3 diluted solution = �0 x 2.68 x�0–3 mol = 2.68 x �0–2 mol (�) = number of moles of Cu in the foil

Mass of Cu in the foil = 2.68 x �0–2 mol x 63.5 g mol–�

= �.70 g

Percentage purity of copper foil = �.70 g�.75 g

x �00%

= 97.�% (�)

∴ the percentage purity of the copper foil is 97.�%.

3�

d) Any one of the following:

• The impurities in the foil do not react with KI(aq) to give I2(aq). (�)

• The impurities in the foil do not react with S2O32–(aq) ions. (�)

8� a) IO3–(aq) + 5I–(aq) + 6H+(aq) 3I2(aq) + 3H2O(l)

8.50 x �0–4 mol dm–3 ? mol dm–3

�00.0 cm3 250.0 cm3

Number of moles of KIO3 in � 00.0 cm3 solution

= 8.50 x �0–4 mol dm–3 x �00.0� 000

dm3

= 8.50 x �0–5 mol (�)

According to the equation, � mole of IO3– ions reacts with I– and H+ ions to give 3 moles of I2.

i.e. number of moles of I2 formed = 3 x 8.50 x �0–5 mol = 2.55 x �0–4 mol

Concentration of I2(aq) = 2.55 x �0–4 mol250.0� 000

dm3

= �.02 x �0–3 mol dm–3 (�)

b) The red colour would interfere with the colour change of the indicator at the end point. / So that the colour of the indicator can be seen. (�)

c) Use starch solution as indicator. (�)

Add aqueous iodine solution until a blue colour is developed. (�)

d) I2(aq) + SO2(aq) + 2H2O(l) 2I–(aq) + SO42–(aq) + 4H+(aq) (�)

e) I2(aq) + SO2(aq) + 2H2O(l) 2I–(aq) + SO42–(aq) + 4H+(aq)

�.02 x �0–3 mol dm–3 ? mol dm–3

��.9 cm3 25.0 cm3

Number of moles of I2 in ��.9 cm3 solution = �.02 x �0–3 mol dm–3 x ��.9� 000

dm3

= �.2� x �0–5 mol (�)

According to the equation, � mole of I2 reacts with � mole of SO2.

i.e. number of moles of SO2 in 25.0 cm3 red wine = �.2� x �0–5 mol

Concentration of SO2 in red wine = �.2� x �0–5 mol25.0

� 000 dm3

= 4.84 x �0–4 mol dm–3 (�)

∴ the concentration of sulphur dioxide in the red wine is 4.84 x �0–4 mol dm–3.

32

f) Any one of the following:

• Activated charcoal may adsorb SO2. (�)

• Traces of charcoal may react with I2. (�)

82 a) A CH3CH2CHO (�)

B CH3COCH3 (�)

b) C (CH3)3COH (�)

D CH3CH2CH(OH)CH3 (�)

c) E CH3COOCH3 (�)

F CH3CH2COOH (�)

d) G CH3CH2CH2OH or CH3CH(OH)CH3 (�)

H CH3CH2OCH3 (�)

e)

C COCH3

CH3

CH2CH3

HI

(�)

orC CHO

CH3

CH2CH2CH3

H orC CHO

CH3

CH(CH3)2

H C CH2CHO

CH3

CH2CH3

HJ

(�)

f)

orL C OH

O

C

H

H

H C CH3O

O

H

(�)

M C H

O

C

H

H

HO

(�)

83 a)

butanal

(1)

(1)

C

H

C O

H

HH

C

H

H

H

H

C

methylpropanal

(1)

(1)

C

H

C O

H

CH3

C

H

H

H

33

b) i) 2,4-dinitrophenylhydrazine (�)

Yellow to red precipitate (�)

ii) Any one of the following:

• Tollens’ reagent (�)

Only A and B give a silver mirror when warmed with the reagent. (�)

• Acidified aqueous solution of potassium dichromate (�)

Only A and B turn the dichromate solution from orange to green when warmed with the reagent. (�)

• Iodine in an aqueous solution of sodium hydroxide (i.e. iodoform test) (�)

Only C gives a yellow precipitate when warmed with the reagent. (�)


• Alkenol

e.g. CH2=CHCH2CH2OH (�)

CH3CH=CHCH2OH (�)

• Cyclic alcohol

e.g.

CHOH

CH2

H2C

H2C (�)

d) i) � 680 – � 750 cm–� (�)

ii) 3 230 – 3 670 cm–� (�)

84 a) Warm with iodine in an aqueous solution of sodium hydroxide. (�)

Only compound P gives a bright yellow precipitate. (�)

b) i) CH3CH2COCH2CH3 (�)

ii) [CH3CH2COCH2CH3]+· CH3CH2CO+ + ·CH2CH3

molecular ion m/e = 57 (�)

iii) (�) & (2) Any one of the following:

• m/e = 7� (�)

[CH3CH2CH2COCH3]+· CH3CH2CH2CO+ + ·CH3

molecular ion m/e = 7� (�)

• m/e = 43 (�)

[CH3CH2CH2COCH3]+· CH3CO+ + ·CH2CH2CH3

molecular ion m/e = 43 (�)

34

c) The alcohol obtained from compound P has four different groups attached to a carbon atom. (�)

It is chiral. / Its mirror images are non-superposable. (�)

C

CH3

HOH

C3H7

C

CH3

HOH

C3H7

(�)

The alcohol obtained from compound Q has two identical groups attached to its central carbon atom. (�)

85 a) Reagent Observation

2,4-dinitrophenylhydrazine an orange precipitate forms (�)

Warm with Tollens’ reagent no observable change (�)

Warm with iodine in an aqueous solution of sodium hydroxide

A bright yellow precipitate forms (�)

b) i) The peak m/e = �20 is the molecular ion peak. (�)

ii) C6H5CO+ (�)

c) i) LiAlH4 / ethoxyethane (�)

ii) (�) X: O–H bond (�)

(2) Y: C=O bond (�)

iii) Absorptions in an infrared spectrum are due to vibrations of the bonds present. (�)

The two enantiomers have the same bonds / functional groups. (�)

86 a)

(� mark for correct set-up; � mark for correct direction of water flow in condenser; award 0 mark if the set-up is not workable) (2)

35

b) To convert sodium benzoate to benzoic acid. (�)

c) Because benzoic acid is soluble in hot water. (�)

d) Dissolve the crude sample in the minimum amount of hot water. (�)

Filter the mixture while hot. (�)

Allow the filtrate to cool and collect the crystals by filtration. (�)

Wash the crystals and dry. (�)

e) Determine the melting point of the product and compare the result with literature data. (�)

f) i) Absorption Due to

Ethyl benzoate � 680 – � 750 cm–� C=O bond (�)

Benzoic acid� 680 – � 750 cm–�

2 500 – 3 300 cm–�C=O bondO–H bond

(�)(�)

Ethanol 3 230 – 3 670 cm–� O–H bond (�)

ii) Absence of absorption due to C=O bond (�)

Absorption due to O–H bond not broad (�)

87 a) Functional group present

Chemical test Observation

–C=C– (�) Add aqueous bromine. (�)The yellow-brown aqueous bromine becomes colourless quickly. (�)

–COOH (�)

Any one of the following:• Mix with aqueous solution of sodium hydrogencarbonate. (�)• Warm with ethanol in the presence of concentrated sulphuric acid. (�)

• Effervescence occurs. (�)

• A pleasant / fruity smell is detected. (�)

b) i) CH2OHCH2OH / ethane-�,-2-diol (�)

ii) HOOC–COOH / ethanedioic acid (�)

iii) C3H7COCH3 + 3I2 + 4OH– C3H7COO– + CHI3 + 3I– + 3H2O (�)

iv) (�) B (�)

as it contains a –CHO group. (�)

(2) –OOCCOO– (�)

c) i) The peak at m/e = 86 (�)

ii) The CH3CH2CH2CO+ ion is responsible for the peak at m/e = 7�. (�)

The CH3CO+ ion is responsible for the peak at m/e = 43. (�)

36

88 a)

C CH3

CH3

OH

H3C

(�)

methylpropan-2-ol (�)

b)

C

H

C

CH3

HH

C

H

H

H OH

(�)

methylpropan-�-ol (�)


• Iodine in an aqueous solution of sodium hydroxide (i.e. iodoform test) (�)

Only butan-2-ol gives a yellow precipitate when warmed with the reagent. (�)

• A solution of anhydrous zinc chloride in concentrated hydrochloric acid (i.e. Lucas reagent) (�)

Cloudiness appears in 5 minutes when butan-2-ol is treated with the Lucas reagent. (0.5)

Cloudiness appears in hours when butan-�-ol is treated with the Lucas reagent. (0.5)

d) i) The absorption at 3 230 – 3 670 cm–� (�)

ii) Refer to the fingerprint region. (�)

Match with spectra of known compounds in database. (�)

89 a) i) Any one of the following:

• Warm each chemical with acidified aqueous solution of potassium dichromate. (�)

Only CH3CH2CHO turns the dichromate solution from orange to green. (�)

• Warm each chemical with acidified aqueous solution of potassium permanganate. (�)

Only CH3CH2CHO turns the permanganate solution from purple to colourless. (�)

• Warm each chemical with the Tollens’ reagent. (�)

Only CH3CH2CHO gives a silver mirror on the wall of the reaction vessel. (�)

• Treat each chemical with iodine in an aqueous solution of sodium hydroxide. (�)

Only CH3COCH3 gives a yellow precipitate of iodoform. (�)

ii) Mix each chemical with aqueous solution of sodium hydrogencarbonate. (�)

Effervescence occurs for CH3CH2COOH only. (�)

iii) Treat each chemical with iodine in an aqueous solution of sodium hydroxide. (�)

Only CH3CH2CH2COCH3 gives a yellow precipitate of iodoform. (�)

37

b) i) Only D has a broad absorption at 2 500 – 3 300 cm–�. (�)

ii) Confirm the identity of a compound by comparing the fingerprint region with those of known compounds. (�)

Look for an exact match. (�)

c) i) Any one of the following:

• Major peak due to CH3CH2CO+ (�)

m/e = 57 (�)

• Major peak due to CH3CH2+ (�)

m/e = 29 (�)

ii) • [CH3CH2CO CH2CH3]+· CH3CH2CO+ + ·CH2CH3 (�)

• [CH3CH2CO CH2CH3]+· CH3CH2

+ + ·OCCH2CH3 (�)

90 a)

C

H

C

H

HH

C

H

H

H OH propan-1-ol/ (�)

C

H

C

H

HOH

C

H

H

H OH propan-2-ol/ (�)

b) The dichromate solution changed from orange to green. (�)

c) Theoretical yield of C3H6O = 0.300 x 58.0 g mol–�

= �7.4 g (�)

Precentage yield of C3H6O = 6.09 g�7.4 g

x �00%

= 35.0% (�)

d) i) The presence of C=O bond (�)

ii) The presence of O–H bond in an acid (�)

ii) Propan-�-ol (�)

It was oxidized to propanoic acid besides the carbonyl compound C3H6O. (�)

9� a)

H

OH

H

CC C

HOO

OHHOC + CO2

malic acid

* (1)

H

OH

H

CH C

HO

OHC

lactic acid

38

b) i) The peak at m/e = �34. (�)

ii) The m/e value of the peak marked X is 89. The difference between � 34 and 89 is 45, (�)

so it is probable that a COOH radical is lost from the molecular ion, producing the peak at m/e = 89.

Thus,

H

OH

CC C

H

H +

O

HO is responsible for the peak at m/e = 89. (�)

c) i)

C

H

C

H

O

OH

H

C C

H

C H

H

H

H

H

O

(�)

ii) The wine is more fruity due to the esters. / The wine is less sour as acids are used up. (�)

92 a) i) Place � cm3 of aqueous solution of silver nitrate in a very clean test tube. Add a little dilute aqueous solution of sodium hydroxide. (�)

Add dilute aqueous ammonia dropwise until the precipitate of silver oxide nearly dissolves. (�)

Add � – 2 drops of compound X. Place in a beaker of hot water. (�)

Silver metal deposits on the wall of the test tube. (�)

ii) Silver ions are reduced to metal silver. (�)

X is oxidized to CH3CH2CH2CH2COO–, a carboxylate ion. (�)

b) i) 2,4-dinitrophenylhydrazine (�)

A yellow to red precipitate forms. (�)

ii)N

H

N

NO2

NO2

C4H9

HC

(�)


• Mix with phosphorus pentachloride. (�)

Compound Y gives steamy fumes of hydrogen chloride. (�)

• Warm with ethanoic acid in the presence of concentrated sulphuric acid. (�)

Compound Y reacts to give a product with a pleasant smell. (�

d) i) The molecular ion peak is at m/e = 86. The difference between 86 and 57 is 2 9. (�)

It is probable that a CHO radical is lost from the molecular ion. (�)

39

ii) The base peak is at m/e = 57.

Losing a CHO radical from the molecular ion probably produces a very stable tertiary carbocation. (�)

Thus, this is probably the mass spectrum of compound Z. (�)

93 a) i) Suppose we have � 00 g of compound X, so there are 54.5 g of carbon, 9.�0 g of hydrogen and 36.4 g of oxygen.

Carbon Hydrogen Oxygen

Mass of element in the compound

54.5 g 9.�0 g 36.4 g

Number of moles of atoms that combine

54.5 g�2.0 g mol–� = 4.54 mol

9.�0 g�.0 g mol–� = 9.�0 mol

36.4 g�6.0 g mol–� = 2.28 mol (�)

Simplest ratio of atoms

4.54 mol2.28 mol

= �.999.�0 mol2.28 mol

= 3.992.28 mol2.28 mol

= �.00 (�)

∴ the empirical formula of X is C2H4O.

ii) From the mass spectrum, the molecular ion peak is at m/e = 88. Thus, the relative molecular mass of X is 88.0.

Let (C2H4O)n be the molecular formula of X.

Relative molecular mass of X = n(2 x �2.0 + 4 x �.0 + �6.0) = 44n

i.e. 88 = 44n n = 2 (�)

∴ the molecular formula of X is C4H8O2.

b) O–H bond (in an alcohol) is responsible for the absorption at 3 500 cm–�. (�)

C=O bond is responsible for the absorption at � 700 cm–�. (�)

c) X contains a ketone functional group (�)

but not an aldehyde functional group. (�)

d) CH3CO+ (�)

C3H7+ ions, which also give a peak at m/e = 43, are unlikely to be formed from a ketone with four carbon

atoms. (�)

40

e)

C CH3

H

H

C C

H

H

HO

O

4-hydroxybutanone

(�) (�)

or

C CH3

OH

H

C C

H

H

H

O

3-hydroxybutanone

(�) (�)

94 a) The base peak is the peak with the greatest relative intensity. / The base peak corresponds to the most stable species. (�)

b) The molecular ion peak is at m/e = �36. (0.5)

As there are two oxygen atoms in compound X, the carbon and hydrogen atoms must add up to �04 mass units. (0.5)

As compound X is a mono-substituted aromatic compound, it must contain a C6H5 group, adding up to 77 mass units. (0.5)

(�04 – 77) mass units = 27 mass units. Thus, compound X should also contain two carbon and three hydrogen atoms. (0.5)

The molecular formula of compound X is C8H8O2. (�)

c) CH2+

(�)

d) � 680 – � 750 cm–� due to C=O bond (�)

2 500 – 3 300 cm–� due to O–H bond (�)

e) CH2COOH

(�)

4�

95 The infrared absorption at about � 250 cm–� indicates the presence of a C–O group. (�)

The infrared absorption at about � 750 cm–� indicates the presence of a C=O group. (�)

The absence of a broad absorption band at 2 500 cm–� – 3 300 cm–� indicates that X is not an alcohol or a carboxylic acid. (�)

X is probably an ester. (�)

In the mass spectrum, the molecular ion peak is at m/e = 74.

The peak at m/e = 43 is probably due to the CH3CO+ ion. (�)

The difference between 74 and 43 is 3�. So, it is probable that a OCH3 fragment has been lost from the molecular ion, giving rise to an ion responsible for the peak at m/e = 43. (�)

Thus, compound X should have a –OCH3 group. (�)

A possible structure of compound X is CH3COOCH3. (�)

96 a) SO2(g) + H2O2(aq) SO42–(aq) + 2H+(aq) (�)

Ba2+(aq) + SO42–(aq) BaSO4(s) (�)

b) i) Number of moles of BaSO4(s) formed = 0.0607 g233.4 g mol–�

　 = 2.60 x �0–4 mol (�) = number of moles of SO4

2–(aq) ions in resulting solution

As � mole of SO2(g) reacts with H2O2(aq) to give � mole of SO42–(aq) ions,

i.e. number of moles of SO2(g) in air sample = 2.60 x �0–4 mol

Mass of SO2(g) in air sample = 2.60 x �0–4 mol x 64.� g mol–�

= �.67 x �0–2 g (�)

Concentration in μg m–3 of SO2(g) in air sample = �.67 x �04 μg200.0� 000

m3

= 83 500 μg m–3 (�)

ii) Volume of SO2(g) in air sample = 2.60 x �0–4 mol x 24.0 dm3 mol–�

= 6.24 x �0–3 dm3 (�)

Concentration in ppm of SO2(g) in air sample = 6.24 x �0–3 dm3

200.0 dm3 �06 ppm

= 3�.2 ppm (�)

97 a) Butane-�,4-dioic acid (�)

b) HOCH2CH2CH2CH2OH (�)

42

c) (CH2)4O CO (CH2)2

O

C

O (�)

d) Less waste / less landfill / easier to dispose of / less pollution from burning (�)

e) Any two of the following:

• These plastics are much more expensive. (�)

• These plastics may begin to degrade before the right time, causing inconvenience or even safety concerns. (�)

• These plastic articles look like other plastic articles. If the two types of articles are mixed up in the recycling process, articles made from the recycled plastic may become weaker. (�)

f) i) Base peak (�)

ii) C4H4O3+ (�)

iii) A COOH radical (�)

98 a) Mobile phase – the developing solvent (�)

Stationary phase – fine layer of alumina / silica gel coated onto a glass plate (�)

b) Separation depends on the adsorption of the components onto the stationary phase and the relative solubility of the components in the developing solvent. (�)

Components that adsorb more strongly onto the stationary phase travel up less quickly. Thus, the components separate. (�)

c) i) C4H9+ or CCOOH+ (�)

ii) Any one of the following:

Peak at m/e value of Ion producing it

59 (�) OCOCH3+ (�)

�2� (�) C6H4COOH+ (�)

�37 (�) C6H4(COOH)O+ (�)

�80 (�) C9H8O4+ (molecular ion) (�)

d) OH

NHCOCH3

Y

+

OH

NaOH(aq)

X

NH2

(1)

CH3COONa(1)

43

e) O–H bond (�) 3 230 – 3 670 cm–� (�)

99 CH3CH2CH2CHO

Warm with the Tollens’ reagent. (0.5)

A silver mirror forms on the wall of the reaction vessel. (0.5)

CH3CH2COCH3

Warm with iodine in an aqueous solution of sodium hydroxide. (0.5)

A bright yellow precipitate forms. (0.5)

CH3CH2CH2COOH

Any one of the following:

• Mix with an aqueous solution of sodium hydrogencarbonate. (0.5)

Effervescence occurs. (0.5)

• Heat with ethanol in the presence of concentrated sulphuric acid. (0.5)

A pleasant / fruity smell can be detected. (0.5)

CH3CH2CH2CH2Br

Warm with dilute aqueous solution of silver nitrate in ethanol. (0.5)

A creamy precipitate forms. (0.5)

(CH3)3COH

Treat with anhydrous zinc chloride in concentrated hydrochloric acid. (0.5)

Cloudiness appears quickly. (0.5)

CH3CH2CH2CH2OH

Heat with acidified aqueous solution of potassium dichromate. (0.5)

The dichromate solution changes from orange to green. (0.5)

(3 marks for organization and presentation)

44

�00 Iron(III) ions react with thiocyanate ions to give deep red [Fe(SCN)]2+(aq) ions. (�)

This species absorbs most strongly in the green region of the visible spectrum. Thus, choose a green filter in the colorimeter. (�)

Next prepare a series of standard solutions of [Fe(SCN)]2+(aq) ions. Record the absorbance of each solution. Prepare a calibration graph by plotting absorbance against concentration. (�)

Dissolve a known mass of the foil in concentrated hydrochloric acid to convert the iron to iron(II) ions. (�)

Add aqueous hydrogen peroxide solution / concentrated nitric acid to oxidize the iron(II) ions to iron(III) ions. (�)

Add excess aqueous solution of potassium thiocyanate to the iron(III) ions to give deep red [Fe(SCN)]2+(aq) ions. Measure the absorbance of the solution. Read off the concentration of [Fe(SCN)]2+(aq) ions in the solution from the calibration curve. (�)

Calculate the mass of iron in the aluminium foil and thus the percentage of iron in the foil.


�0� Prepare standard aqueous solution of potassium iodate by dissolving a known mass of KIO3(s) in distilled water and dilute the solution to a known volume. (�)

Add excess aqueous solution of potassium iodide and dilute sulphuric acid to a known volume of the standard IO3

–(aq) to obtain standard iodine solution. (�)

Titrate the standard iodine solution against the aqueous solution of sodium thiosulphate. (�)

When the solution becomes pale yellow, (�)

add starch solution and continue the titration. (�)

The end point is marked by the disappearance of the blue colour. (�)


12. solution guide to supplementary exercises

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