Class Adm No

Candidate Name:

2024 Mid-Year Exams

Pre-University 3
H2 CHEMISTRY 9729/02
Paper 2 Structured Questions 28 June 2024
2 hours
Candidates answer on the Question paper.
Additional materials: Data Booklet

READ THESE INSTRUCTIONS FIRST

Do not turn over this question paper until you are told to do so.
Write your name, class and admission number on all the work you hand in.
Write in dark blue or black pen.
You may use an HB pencil for any diagrams or graphs.
Do not use staples, paper clips, glue or correction fluid.

Answer all questions.

The use of an approved scientific calculator is expected, where appropriate.
A Data Booklet is provided.

At the end of the examination, fasten all your work securely together.
The number of marks is given in brackets [ ] at the end of each question or part question.

Question 1 2 3 4 5 Total
Marks

15 16 15 16 13 75

This question paper consists of 16 printed pages and 2 blank pages.

 2

Answer all questions.

For
1 Ibuprofen, aspirin, and paracetamol are commonly sold as over-the-counter painkillers that offer
Examiners’
Use
versatile relief for different forms of pain. Table 1.1 shows the physical properties of the painkillers
and some organic compounds.
Table 1.1

melting solubility in solubility in density /

compound Mr
point / °C water / g dm–3 ethanol / g dm–3 g cm–3

206.0 75 0.021 1.2 1.03

ibuprofen
(C13H18O2)

180.0 135 3.0 80 1.40

aspirin
(C9H8O4)

151.0 169 12.8 130 1.26

paracetamol
(C8H9NO2)

94.0 40.5 84.2 highly soluble 1.07

phenol
(C6H6O)

78.5 –112 – – 1.10

ethanoyl chloride
(C2H3OCl)
 3

For
A student was provided with three unlabelled tablets, one from each of the three different Examiners’
Use
painkillers. To identify them, he crushed each of the tablets and dissolved them in ethanol. To a
small portion of each solution, he conducted two simple chemical tests to distinguish between the
painkillers.

(a) (i) Define the term relative molecular mass, Mr. [1]

The weighted average mass of one molecule relative to the 1/12 the mass of a 12C atom.

Marker’s comment:
- Poorly done, many students wrote C12

(ii) Suggest, briefly, two reasons for the difference in solubilities between ibuprofen and
paracetamol in water. [2]

Stronger instantaneous dipole-induced dipole forces of attraction between ibuprofen

molecules due to the larger electron cloud size (Mr = 202.0) than paracetamol
molecules (Mr = 151.0) OR longer non-polar alkyl chain ;

Paracetamol molecules form more extensive hydrogen bonding with water molecules
(due to greater number of electronegative O, N atoms and lone pairs available) than
ibuprofen molecules. ;

Marker’s comment:
- Very poorly done, with many students only giving one reason
- Many students did not make mention to main keyword of electron cloud size / no. of
electrons
- Students who counted the no. of H-bonds did not correctly identify the number

(iii) Suggest the reagents and conditions, stating the observations, for the two simple
chemical tests used to distinguish between ibuprofen, aspirin, and paracetamol. [4]

To 1 cm3 of each solution, add neutral FeCl3(aq). ;

Paracetamol will give a dark / violet colouration.
For aspirin and ibuprofen, no dark / violet colouration OR neutral FeCl3(aq) remains
yellow / orange. ;
OR
To 1 cm3 of each solution, add Br2(aq). ;
For paracetamol, orange Br2(aq) decolourises (and white ppt forms).
For aspirin and ibuprofen, Br2(aq) remains orange. ;
OR
To 1 cm3 of each solution, add Na2CO3(s). ;
For aspirin and ibuprofen, effervescence of a gas that forms white ppt in Ca(OH)2(aq).
For paracetamol, no effervescence is observed. ;

To another 1 cm3 of each solution, add KMnO4(aq) in H2SO4(aq), heat/warm. ;

For ibuprofen, purple KMnO4(aq) decolourised (and effervescence which produces
white ppt in limewater).
For aspirin and paracetamol KMnO4(aq) remains purple. ;

Marker’s comment:
- Many students were still unable to identify the functional groups successfully (e.g.
confused between ketone/aldehyde/ester/acid).

[Turn over
 4

For
(b) The student planned to synthesise aspirin from phenol and ethanoyl chloride (in 1 : 2 molar Examiners’
Use
ratio) via an acylation reaction to form an intermediate A, before turning it into aspirin under
controlled conditions in step 2.

The liquid ethanoyl chloride was measured out using a graduated gas syringe and added to
phenol in an inert solvent.

(i) Suggest a catalyst for step 1. [1]

step 1: AlCl3 or FeCl3 catalyst (can accept AlBr3 or FeBr3) ;

(ii) Draw the structure of the intermediate A. [1]

Suggested Intermediate A:

;
Marker’s comment:
- Some students could not appreciate that the question asked for a “reaction
intermediate” (stable structure) and not a “mechanism intermediate” (not so stable
structure).

(iii) State the reagents and conditions for step 2. [1]

step 2: I2 NaOH, warm, followed by dilute acid ;

(iv) Calculate the volume of ethanoyl chloride that is needed to completely react with
4.00 g of phenol to form aspirin using the suggested synthetic pathway in (b)(i). [3]
amount of phenol used = 4.00 ÷ 94.0 = 0.04255 mol ;
mass of ethanoyl chloride required = 0.04255 × 2 × 78.5 = 6.68 g ;
volume of ethanoyl chloride required = 6.68 ÷ 1.10 = 6.07 cm3 (3 s.f.) ;

Marker’s comment:
- Many students did not identify the data that was relevant from the table (e.g. Mr, notice
units of density was g cm-3).
- A significant number of students still multiplied amount by 24.0, not recognising that
the formula only applies for gases since the volume of 1 mol of gas at rtp is 24.0 dm3.
 5

For
(v) The resultant aspirin product was washed, dried, and weighed. Given that 5.57 g of
Examiners’
Use
aspirin was formed, calculate the percentage yield of the reaction. [1]

Theoretical mass of aspirin = 0.04255 × 180.0 = 7.66 g [no marks, LO tested in (b)(ii)]
% yield = (5.57 ÷ 7.66) × 100% = 72.7% ;

Marker’s comment:
- Generally well done.

(vi) Suggest why there are no solubility values for ethanoyl chloride. [1]

Ethanoyl chloride reacts readily with both water and ethanol ;

(in a nucleophilic acyl substitution reaction to form ethanoic acid and ethyl ethanoate
respectively, and HCl).

Marker’s comment:
- Many students did not recall that ethanoyl chloride reacts vigorously with water,
choosing to talk about how it is insoluble instead (ethanoyl chloride is able to form H-
bonds with water). An insoluble substance would still have a value for solubility, either
a very very small number, or at the extreme end, “0”. A “–” implies something else
entirely. Students should also note that liquids and gases have solubility values
associated to them as well.
- Some students recognised the above, but did not talk about ethanol, which is required
by the question (“values”).

[Total: 15]

[Turn over
 6

For
2 (a) Arrange the following molecules in increasing order of gas phase relative basic strength, Examiners’
Use
explaining your answer.

[4]

(weakest base) CH3CONH2 < C6H5-NH2 < CH3CH2NH2 (strongest base) ;

CH3CONH2 is neutral (not basic) as the lone pair on N is delocalised due to p-orbital overlap
with the π electron cloud across the C=O. As O is highly electronegative, the lone pair on N
is not available for donation (to H+ usually). ;

C6H5-NH2 (phenylamine): Lone pair on N is delocalised due to partial p-orbital overlap of N

with π electron cloud of the benzene ring, hence less available for donation and weaker
base ;

compared to CH3CH2NH2 which has an electron-donating alkyl group, increases the electron
density and hence availability of lone pair on N for donation, hence stronger base ;

Marker’s comment:
- Answers of many students have substance but lack keywords required.
- Students are advised to use inequality arrows for ‘ranking’ questions.

(b) Ethylamine, CH3CH2NH2, can be formed from ethane, CH3CH3, via the following steps.
Step 1: Free radical substitution of ethane to form chloroethane.
Step 2: Nucleophilic substitution of chloroethane to ethylamine.
(i) Outline the mechanism for step 1. Draw curly arrows to show the movement of
electrons. [3]
Initiation
1 2
3
UV or heat
Cl ⎯ Cl Cl + Cl

Propagation

4 5

6 7

Termination
CH3CH2 + Cl → CH3CH2Cl (major pdt) 8

Either Cl + Cl → Cl2 /OR/ CH3CH2 + CH3CH2 → CH3CH2CH2CH3 (minor pdt) 9
(3 # points for 1 mark)
 7

Marker’s comment:
- Not required for students to draw the half-arrows in propagation and termination steps
(but aids in understanding)
For
(ii) Suggest the reagents and conditions for step 2. [1] Examiners’
Use

Excess concentrated ethanolic NH3, heat in sealed tube

(c) Phenylamine, C6H5NH2, can be synthesised from benzene.

(i) Suggest the reagents and conditions for steps 1 and 2 and state the type of reaction in
each step. [4]

Step 1:
Electrophilic substitution
Concentrated HNO3, concentrated H2SO4, heat under reflux

Step 2:
Reduction
Sn, concentrated HCl, heat under reflux

Marker’s comment:
- A significant number of students tried to go via chlorobenzene as the reaction
intermediate, not realising that it cannot undergo N.S..
- Many students still wrote (aq) together with conc, but (aq) implies dilute.

(ii) After step 2, a colourless solution containing phenylammonium ions, C6H5NH3+, was
obtained. Upon addition of NaOH(aq), solid crystals of phenylamine, C6H5NH2,
precipitated out of the colourless solution.
Explain the observations in terms of the interactions between particles. [2]

Energy released from the ion-dipole interactions between the phenylammonium ion and
water was sufficient to overcome the ionic bonding (strong electrostatic forces between
oppositely charged ions) in phenylammonium chloride (and the hydrogen bonding
between water molecules). ;

However, the energy released from the formation of hydrogen bonding between
phenylamine and water was insufficient to overcome the extensive instantaneous
dipole-induced dipole forces of attraction between phenylamine molecules (and the
hydrogen bonding between water molecules). ;

Marker’s comment:
- Many students did not recognise that the “observation” referred to precipitation of the
solid.

[Turn over
 8

For
(d) The Hofmann degradation is an organic rearrangement reaction of a primary amide to a Examiners’
Use
primary amine with the loss of one carbon atom. The reagent is heated under reflux with
bromine in sodium hydroxide to transform the primary amide into an isocyanate
intermediate, which is subsequently hydrolysed to a primary amine.

Suggest the amide used to produce the following products via the Hoffmann degradation.
(i)
CH3CH2CONH2

→ CH3CH2NH2

OR

[1]
Marker’s comment:
- Many students drew the correct answer but wrote “H3C” or bonded to wrong C atom

(ii)

[1]

[Total: 16]
 9

For
3 (a) Use of the Data Booklet is relevant for this question. Examiners’
Use
(i) Define the term first electron affinity of fluorine. [1]

The enthalpy change (energy released) when one mole of gaseous F atoms accepts
one mole of electrons to form one mole of gaseous F– ions.

Marker’s comment:
- Poorly done.

(ii) Potassium fluoride, KF, is a main source of fluoride ions for manufacturing and in
chemistry. It is found in the mineral carobbiite.

Use the data in Table 3.1 and relevant bond energy values from the Data Booklet to
complete the Born-Haber cycle below and hence calculate the lattice energy for KF(s).
Table 3.1

enthalpy change of formation, ΔHf, for KF(s) –561 kJ mol–1

enthalpy change of atomisation of potassium, K(s) +90 kJ mol–1
first electron affinity of fluorine –328 kJ mol–1

[5]

Energy / kJ mol–1

K+(g) + F(g) + e–

K+(g) + ½F2(g) + e– 1
/2(+158) –328 K+(g) + F–(g)

+418
K(g) + ½F2(g)
Lattice Energy
K(s) + ½F2(g) +90
0
ΔHf
KF(s)

[1m for each highlighted component]

;;;;
ΔHf = –561 = +90 + (+418) + ½(+158) + (–328) + L.E.
L.E. = –561 – (+90) – (+418) – ½(+158) – (–328) = –820 kJ mol–1 ; (sign and ans)

Marker’s comment:
- Most students still forgot to include the e– / did not draw energy level lower for negative
ΔH value

[Turn over
 10

For
(iii) Explain how the magnitude of lattice energy of MgF2 will compare to that of KF. [2] Examiners’
Use
Charge of Mg2+ is higher and ionic radius of Mg2+ is smaller than K+
Anion is the same ;

𝑞 + .𝑞−
Since |L.E.| ∝ |𝑟 ++𝑟− | (required)
|L.E.| for MgF2 is larger than that of KF. ;

Marker’s comment:
Many candidates are able to apply the correct concept, but left out the comparison of
the anions.
(iv) The enthalpy change of atomisation for calcium is +178 kJ mol–1.
Account for the difference in enthalpy change of atomisation between potassium and
calcium and explain it in terms of structure and bonding. [2]

 Both have giant metallic (lattice) structures.

 Calcium has higher charge and
 more delocalised electrons per atom than potassium,
 more energy required to overcome the stronger metallic bonds.
 More endothermic enthalpy change of atomisation for calcium /OR/ More
exothermic enthalpy change of atomisation for potassium.

5pts: 2m
2-4pts: 1m
0-1pts: 0m

Marker’s comment:
Many candidates did not attempt this question or are unable to apply the correct
concept. Some of the candidates who applied the correct concept used ‘enthalpy
change of atomisation for calcium increases’ instead of ‘more endothermic’.
For
(b) You are provided with the following additional information.
Examiners’
Use
Table 3.2

enthalpy change of hydration of fluoride ions, F–(g) –504 kJ mol–1

enthalpy change of solution of potassium fluoride, KF(s) –15 kJ mol–1
 11

(i) Use your answer in (a)(ii) and the information in Table 3.2 to calculate the enthalpy
change of hydration of K+(g).
[If you did not obtain a value in (a)(ii), use the value of lattice energy for KF(s) to be
–1739 kJ mol–1. This is not the answer for (a)(ii).] [2]

Lattice Energy + ΔHsolution = ΔHhyd (K+(g)) + ΔHhyd (F–(g))

ΔHhyd (K+(g)) = Lattice Energy + ΔHsolution – ΔHhyd (F–(g))

Using –820 answer from (a)(ii),

ΔHhyd (K+(g)) = –820 + (–15) – (–504) = –331 kJ mol–1
If using –1739,
ΔHhyd (K+(g)) = –1739 + (–15) – (–504) = –1250 kJ mol–1

(1m for the relationship between the terms / substitution, 1m for final answer with sign).
Marker’s comment:
Weaker candidates were not able to show the correct relationship to calculate enthalpy
change of hydration.

(ii) Hence, or otherwise, explain the sign of the enthalpy change of hydration of K+(g).
[1]

Formation of ion-dipole attractions between gaseous ions and water molecules releases
energy. Since there is no bond-breaking, the enthalpy change of hydration of an ion is
always exothermic.

Marker’s comment:
This question proved to be demanding. Candidates who applied the correct concept did
not mention the lack of bond-breaking.

(c) Arrange the hydrogen halides HF, HCl and HBr in order of increasing thermal stability.
Explain your answer.
[2]

[Total: 15]

Down Group 17, the H–X bond becomes weaker

Due to increasing atomic radius of the halogen atom X /OR/ longer H–X bond,
less effective orbital overlap between H and X. ;

less energy to break H–X bond /OR/ thermal stability of HX molecule decreases.
order of thermal stability: (lowest) HBr(g) < HCl(g) < HF(g) (highest) ;

Marker’s comment:
Candidates confused the explanation for thermal stability of hydrogen halides with thermal
stability of Group 2 metal carbonates/oxides/nitrates.

[Turn over
 12

For
4 In the automotive transformation to green and sustainable transportation options, rechargeable Examiners’
Use
batteries play a significant role in determining the availability, cost, range, and safety of electric
vehicles (EVs). Lithium-ion batteries (LIBs) are currently the most widely manufactured type of
rechargeable battery. However, as the demand for LIBs has grown exponentially, lithium has
become a limited resource. In addition, lithium mining generates undesirable environmental
impacts such as air and water pollution, land degradation, and groundwater contamination.

To overcome these challenges, sodium-ion batteries (SIBs) are being researched as a viable
alternative. The table below shows a comparison of SIBs with common rechargeable batteries
used in cars today.
Table 4.1

sodium-ion battery lithium-ion battery lead-acid battery

cost per kilowatt-

$40–77 $137 $100–300
hour of capacity

energy density* 75–200 W·h/kg 120–260 W·h/kg 35–40 W·h/kg

no. of cycles before

reaching 80% of 300–1000 3500 900
initial capacity

materials abundant scarce limited

safety high risk high risk moderate risk

optimal temperature
−20 °C to 60 °C 15 °C to 35 °C −20 °C to 60 °C
range for operation

*Energy density is given in Watt-hour per kilogram (W·h/kg),

where 1 Watt = 1 Ampere × 1 Volt

The most promising cathode material for SIBs is Prussian White (PW), which has the formula
Na2Fe[Fe(CN)6], while the anode is typically graphite. The cathode and anode are separated by a
porous membrane, with the electrolyte being diglyme.

diglyme

When the SIB is charged, sodium atoms in the PW electrode are oxidised to sodium ions, which
flow past the membrane to the graphite electrode and before being reduced back to sodium atoms
inside the layers of graphite. When the SIB is being discharged (i.e. being used to power
appliances), the reverse process occurs.
 13

For
(a) (i) Draw a well-labelled diagram to illustrate the set-up for a sodium-ion battery, indicating
Examiners’
Use
the direction of electron-flow during the discharging process. [3]

Labels + drawing
 +,–
 cathode, anode
 direction of electron flow
 identity of both electrodes
 electrolyte
 membrane + voltmeter / load
(6pts: 3m, 4-5pts: 2m, 2-3pts: 1m, 0-1pts: 0m)

Suggested diagram

> >
– +

Anode Cathode
Graphite Prussian White /
Na2Fe[Fe(CN)6]

membrane
diglyme

Also accept:

> >
– +

Anode Cathode
Graphite Prussian White /
Na2Fe[Fe(CN)6]

diglyme
membrane

Marker’s comment:
Many candidates are able to identify the correct electrodes with correct signs and
electron flow. Only some candidates included membrane as part of the set-up.

[Turn over
 14

(ii) Given that the complex ion in Prussian White is [Fe(CN)6]3–, and that the oxidation
number of the Fe ion outside of the complex is +2, state the average oxidation number
of sodium in Prussian White. [1]

+0.5

Marker’s comment:
Majority of the candidates did well in this question.

(iii) During the discharging process, a potential (reversible) side reaction might occur in the
Prussian White electrode.
Suggest an ion-electron equation for this side reaction, and state its standard electrode
potential, EO, value. [2]

[Fe(CN)6]3– + e– → [Fe(CN)6]4– EO = +0.36 V

OR
Fe2+ + 2e– → Fe E O = –0.44 V

Marker’s comment:
This question proved to be challenging. Some candidates gave Fe3+ + e-  Fe2+

(iv) Define the term standard conditions. [1]

Temperature of 298 K, Pressure of 1 bar for any gases,

Concentration of 1 mol dm–3 for any aqueous species.

Marker’s comment:
Majority of the candidates left out the ‘concentration of 1 mol dm-3 for any aqueous
species’.

(v) Suggest why an aqueous electrolyte should not be used. [1]

 During charging, electrolysis of water would occur instead / redox reactions

involving water would interfere in the charging process OR
 Sodium metal would not be produced in aqueous electrolytes OR
 Sodium reacts vigorously / in a highly exothermic reaction with water

Marker’s comment:
Many of the candidates gave vague answers.
For
(b) The lead-acid battery consists of a lead metal electrode and a lead(IV) oxide electrode
Examiners’
Use
immersed in concentrated sulfuric acid. During discharging, both electrodes produce solid
lead(II) sulfate which is coated on the surface of the electrodes.
 15

(i) Write the ion-electron equations for the discharging process of a lead-acid battery, and
hence construct the overall balanced equation. [3]

[O]: Pb + SO42– → PbSO4 + 2e–

[R]: PbO2 + 4H+ + SO42– + 2e–→ PbSO4 + 2H2O
(accept PbO2 + 2H+ + H2SO4 + 2e–→ PbSO4 + 2H2O)
Overall: Pb + PbO2 + 2SO42– + 4H+ → 2PbSO4 + 2H2O
(accept Pb + PbO2 + 2H2SO4 → 2PbSO4 + 2H2O)

OR

[accept if left out SO42– spectator ion: but max 2m]

[O]: Pb → Pb2+ + 2e–
[R]: PbO2 + 4H+ + 2e–→ Pb2+ + 2H2O
Overall: Pb + PbO2 + 4H+ → 2Pb2+ + 2H2O

Marker’s comment:
This question proves to be challenging. Some candidates were able to identify the
correct ion-electron equation for oxidation but left out SO42- spectator ion. Very few
candidates were able to identify the ion-electron equation for reduction. Weaker
candidates used reversible arrows for the equations.

(ii) Explain, in terms of oxidation numbers, why the reverse reaction (the charging process)
of a lead-acid battery is a disproportionation reaction. [2]

Oxidation number of Pb in PbSO4 is increased from +2 to +4 in PbO2 (oxidation) and

simultaneously decreased to 0 in Pb.
Since a single species PbSO4 was simultaneously oxidised and reduced,
disproportionation has occurred.
Marker’s comment:
Many candidates were able to apply the concept but did not include the species after
stating the oxidation state of the element in that species.

[Turn over
 16

(c) Using the information presented and in Table 4.1, suggest which rechargeable battery an
automotive manufacturer should adopt for use in EVs, explaining why it is preferred. [3]

[Total: 16]

SIBs (max 3m)

 Cheapest to manufacture compared to LIBs and LABs (as sodium is a readily
available resource / in large supply.)
 Lowest environmental impact to manufacture as sodium is an abundant resource.
 Higher temperature range of operation for SIBs than LIBs
(Do not look at cycling stability as both are high)

LIBs (max 3m)

 Lithium is most lightweight / lowest density / lower mass per unit volume
 LIBs has highest energy density / stores more energy per unit mass than SIBs &
LABs / longer range
 Longest lifespan / maintains capacity at >80% for the largest number of cycles.

LABs (max 3m)

 Lowest risk of (chemical) fire / safest compared to sodium or lithium leaks
 Higher temperature range of operation for LABs than LIBs
 (to award 3m, student must have justified safety as the top most concern – then the
3rd reason will be the materials for manufacture are quite readily available, - if not,
do not accept this reason.)

Marker’s comment:
Weaker candidates did not compare the chosen rechargeable battery with the other two
batteries.
 17

For
5 Cyanic acid, HOCN, dissociates in water according to the following equation. Examiners’
Use

HOCN ⇌ H+ + –OCN pKa = 3.48

When 25.0 cm3 of 0.100 mol dm–3 cyanic acid was titrated against 0.100 mol dm–3 sodium
hydroxide, the following pH curve was obtained. The experiment was conducted at 25 °C.

Fig. 5.1
pH

volume of NaOH
0 added / cm3

(a) (i) Define the term buffer. [1]

A buffer is a solution which resists pH changes / maintains pH to be relatively constant

when small amounts of acid or base are added.

Marker’s comment:
Candidates did well in this question. Weaker candidates state the species present in a
buffer instead of the definition of a buffer. Some candidates mentioned ‘no change in
pH’.

(ii) On the curve in Fig 5.1, mark out and state the numerical value of:
 the volume at equivalence point,
 the volume at maximum buffer capacity, and
 the pH at maximum buffer capacity. [3]

pH

Maximum x Equivalence
point
buffer
capacity
3.48 x

volume of NaOH
0 added / cm3
12.5 25.0

[Turn over
 18

Marker’s comment:
Majority of the candidates did well in this question. Weaker candidates did not include
the correct labelling and wrong identification of the volume at maximum buffer capacity.

(b) (i) Write down the expression for Ka, stating its units. [2]

[H+ ][ OCN]
−
Ka = ; mol dm–3 ;
[HOCN]

Marker’s comment:
Almost all of the candidates were able to do this question.

(ii) Calculate a value for Ka. [1]

Ka = 10–3.48 = 3.311 ˣ 10–4

(do not penalise for units here)

Marker’s comment:
Many candidates know the relationship between Ka and pKa.
For
(iii) Hence, or otherwise, calculate the initial pH of the cyanic acid solution before titration.
Examiners’
Use
[2]

- 2
[H+ ][ OCN] [H+ ]
Ka = =
[HOCN] [HOCN]

[H+] = √ ( Ka × [HOCN] )

Since H3PO4 is a weak acid, dissociation is to a very small extent, [HOCN] ≈ [HOCN]initial
[H+] ≈ √ ( Ka1 × [HOCN]initial )
≈ √ ( 0.0003311 × 0.100 )
≈ 0.005754 mol dm–3 ;

pH = – lg [H+] = – lg 0.005754 = 2.24 ;

------------
shortcut method: initial pH = [pKa + ( –lg 0.1)] ÷ 2 = 2.24 ;;

Marker’s comment:
Many of the candidates did not attempt this question.

(c) (i) Using your answer in (a)(i), calculate the concentration of the cyanate ions, [–OCN], at
equivalence point. [1]

At equivalence point, all HOCN has been completely reacted to form –OCN ions.

Amount of –OCN = amount of HOCN used = 0.100 × 0.0250 = 0.00250 mol

[–OCN] = 0.00250 ÷ (0.025 + 0.025) = 0.0500 mol dm–3 ;

Marker’s comment:
Majority of the candidates who attempted this question did this correctly.
 19

(ii) Hence, calculate the pH at the equivalence point. [3]

Kb = Kw ÷ Ka
= 10–14 ÷ 10–3.48
= 10–10.52
= 3.02 × 10–11 mol dm–3 ;

[OH–] = √ ( Kb × [–OCN] )
= √ (3.02 × 10–11 × 0.0500)
= 1.229 × 10–6 mol dm–3 ;

pH = 14 – pOH
= 14 – [– lg (1.229 × 10–6) ]
= 14 – 5.91
= 8.09 (3 s.f.) ;

Marker’s comment:
Majority of the candidates did not attempt this question.

[Total: 13]
END

[Turn over