AQA A-Level Chemistry Paper 3, June 2025: Question 4

9 marks · Hard difficulty · State/Explain/Numerical

Write the Kc expression, explain why a sample is quenched with ice-cold water before titration, and calculate the value and units of Kc for an equilibrium between iron(III) and iodide ions.

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Question

Question text

04 50.0 cm3 of 0.015 mol dm–3 Fe (SO ) (aq) are mixed with 50.0 cm3 of

24 3

0.015 mol dm–3 KI(aq) at a temperature of 40 °C

The mixture is left for several hours until equilibrium is reached.

2 Fe3+(aq) + 2 I–(aq) ⇌ 2 Fe2+(aq) + I (aq)

A sample of the equilibrium mixture is titrated to determine the amount of I–(aq) in the

equilibrium mixture.

04.1 Give the expression for the equilibrium constant, Kc, for the equilibrium.

[1 mark]

04.2 Before the titration, the sample is ‘quenched’ by adding 50 cm3 of ice-cold water.

Suggest why the sample is ‘quenched’ in this way before the titration.

Explain your answer.

[2 marks]

04.3 The amount of I–(aq) in the equilibrium mixture is found to be 3.00 × 10–4 mol

Calculate the value of Kc at 40 °C for the equilibrium.

2 Fe3+(aq) + 2 I–(aq) ⇌ 2 Fe2+(aq) + I (aq)

Give the units of Kc

Assume that the total volume of the mixture remains constant at 100 cm3

[6 marks]

Units of Kc

Mark scheme

Show the mark scheme

Question Answers Additional comments/Guidelines Mark

[Fe2+]2 [I ] IGNORE state symbols 1

04.1 2

(Kc =) 3+ 2 − 2

[Fe ] [I ] NOT () (1 x AO2)

M1 (dilution and cooling) stops the reaction OR reduces rate M1 ALLOW concentrations/amounts stay

constant 2

04.2

M2 during the titration no more iodide formed OR during the M2 NOT iodide concentration remains constant (2 x AO3)

titration the equilibrium will not shift during titration

– A-LEVEL CHEMISTRY – 7405/3 –

3+ 50 × 0.015 × 2 –3

M1 initial amount of Fe = = 1.5 × 10 mol

1000

AND

– 50 × 0.015 –4

initial amount of I = 1000 = 7.5 × 10 mol

M2 amount of I– reacted = M1 I– – 3 x 10–4 M2 = 4.5 × 10–4 mol

M3 for equilibrium amounts in mol

Fe3+ = (M1 Fe3+ – M2) AND M3 Fe3+ = 1.05 × 10–3 / 0.00105 mol

Fe2+ = M2 AND Fe2+ = 4.5 × 10–4 / 0.00045 mol

M2 I = 2.25 × 10–4 / 0.000225 mol

I = ( ) 2

M4 equilibrium concentrations in mol dm–3

[I–] = 3 x 10–3 AND

M4 [I–] = 3 × 10–3 / 0.003 mol dm–3

M3 Fe3+

[Fe3+] = AND [Fe3+] = 1.05 × 10–2 / 0.015 mol dm–3

0.1 2+ –3 –3 6

1804.3 M3 Fe2+ [Fe ] = 4.5 × 10 / 0.0045 mol dm

2+ –3 –3 (6 x AO2)

[Fe ] = 0.1 AND [I2] = 2.25 × 10 / 0.00225 mol dm

M3 I2

[I2] = 0.1

−3 2 −3

2 (4.5 × 10 ) × 2.25 × 10

(M4 Fe2+) × M4 I M5 = 45.9

22 2

M5 Kc = (1.05 × 10−2) × (3 × 10−3)

3+ 2 - 2

(M4 Fe ) × (M4 I )

(45.9 = 5 marks)

If wrong Kc expression then ECF from 4.1

M6 units = mol–1 dm3 M6 Correct units scores irrespective of anything

else; otherwise units should match Kc

expression used in M5; if Kc expression not

used in M5, then ECF from 4.1

459 = M1, M2, M3 & M5 if mole not concs used

562.5 = M2 to M5 if M1 Fe3+ = 7.5 x 10–4

How to answer it

Equilibrium Constant (K_c) Determination by Titration

📌 What this question tests

This question assesses your ability to handle homogeneous equilibria quantitatively and experimentally:

Writing expressions for equilibrium constants (K_c) including stoichiometric powers.
Explaining practical experimental procedures such as quenching before titration.
Managing multi-step stoichiometry: initial amounts, reacting mole ratios, equilibrium concentrations, K_c calculation, and deriving units.

Question 04.1

Expression for Equilibrium Constant K_c

Reaction: 2Fe³⁺(aq) + 2I⁻(aq) ⇌ 2Fe²⁺(aq) + I₂(aq)

✅ Correct Answer

K_c = [Fe²⁺]² [I₂] / ([Fe³⁺]² [I⁻]²)

1 Mark: Correct species, stoichiometric powers, and square brackets.

🧠 Exam Technique & Notation

Always use square brackets [ ] to denote concentrations. Round brackets ( ) will score zero.
State symbols are not required in equilibrium expressions.
Products go on the numerator (top); reactants go on the denominator (bottom).

Question 04.2

Practical Procedure: Quenching the Reaction

"Suggest why the sample is 'quenched' by adding ice-cold water before titration. Explain your answer."

✅ Correct Answer & Marking Points

Point 1 (Action of quenching):

Dilution and cooling stops the reaction (or dramatically reduces the rate) / keeps amounts or concentrations constant.

Point 2 (Impact on titration):

This prevents the equilibrium from shifting during the titration (or ensures no more iodide reacts / forms while titrating).

2 Marks: 1 mark for stopping the reaction / slowing rate; 1 mark for equilibrium not shifting / iodide amounts remaining fixed during titration.

❌ Common Errors & Examiner Traps

Contradicting dilution: Stating that "iodide concentration remains constant during titration" loses Mark 2. Adding water changes concentration, but it fixes the number of moles (amount) by freezing the equilibrium position!
Incomplete answers: Stating only "it cools the mixture" without linking to collision theory/rate or stopping the reaction.

Question 04.3

Calculation of K_c and Units

Determining K_c from initial volumes/concentrations and equilibrium moles of I⁻

📐 Step-by-Step Calculation

Step 1: Calculate initial amounts (moles) of reactants

Initial Fe³⁺: Note the formula is Fe₂(SO₄)₃, so each mole provides 2 moles of Fe³⁺!
n(Fe³⁺) = (50.0 / 1000) × 0.015 × 2 = 1.50 × 10⁻³ mol
Initial I⁻: From KI:
n(I⁻) = (50.0 / 1000) × 0.015 = 7.50 × 10⁻⁴ mol

Step 2: Calculate moles reacted

Equilibrium moles of I⁻ = 3.00 × 10⁻⁴ mol (given in question).
Amount of I⁻ reacted = 7.50 × 10⁻⁴ - 3.00 × 10⁻⁴ = 4.50 × 10⁻⁴ mol

Step 3: Construct an R.I.C.E. Table (Equilibrium Moles)

Species	2 Fe³⁺(aq)	2 I⁻(aq)	2 Fe²⁺(aq)	I₂(aq)
Initial / mol	1.50 × 10⁻³	7.50 × 10⁻⁴	0	0
Change / mol	-4.50 × 10⁻⁴	-4.50 × 10⁻⁴	+4.50 × 10⁻⁴	+(4.50 × 10⁻⁴ / 2) = +2.25 × 10⁻⁴
Equilibrium / mol	1.05 × 10⁻³	3.00 × 10⁻⁴	4.50 × 10⁻⁴	2.25 × 10⁻⁴

Step 4: Convert to concentrations (Total Volume = 100 cm³ = 0.100 dm³)

[Fe³⁺] = (1.05 × 10⁻³) / 0.100 = 1.05 × 10⁻² mol dm⁻³
[I⁻] = (3.00 × 10⁻⁴) / 0.100 = 3.00 × 10⁻³ mol dm⁻³
[Fe²⁺] = (4.50 × 10⁻⁴) / 0.100 = 4.50 × 10⁻³ mol dm⁻³
[I₂] = (2.25 × 10⁻⁴) / 0.100 = 2.25 × 10⁻³ mol dm⁻³

Step 5: Substitute into the K_c expression

K_c = ([Fe²⁺]² × [I₂]) / ([Fe³⁺]² × [I⁻]²)
K_c = [(4.50 × 10⁻³)² × (2.25 × 10⁻³)] / [(1.05 × 10⁻²)² × (3.00 × 10⁻³)²]
K_c = [2.025 × 10⁻⁵ × 2.25 × 10⁻³] / [1.1025 × 10⁻⁴ × 9.00 × 10⁻⁶]
K_c = (4.55625 × 10⁻⁸) / (9.9225 × 10⁻¹⁰) = 45.9

Step 6: Determine Units

Units = (mol dm⁻³)² × (mol dm⁻³) / [(mol dm⁻³)² × (mol dm⁻³)²]
Units = 1 / (mol dm⁻³) = mol⁻¹ dm³

✅ Final Answers & Mark Allocation

M1: Initial moles of Fe³⁺ (1.50 × 10⁻³) AND I⁻ (7.50 × 10⁻⁴)
M2: Amount of I⁻ reacted = 4.50 × 10⁻⁴ mol
M3: Equilibrium moles for Fe³⁺, Fe²⁺, and I₂
M4: All four equilibrium concentrations calculated
M5: Value of K_c = 45.9 (allow 46)
M6: Units = mol⁻¹ dm³

❌ Top 3 Calculation Traps

The Fe₂(SO₄)₃ multiplier: Forgetting to multiply moles of iron(III) sulfate by 2 gives initial Fe³⁺ = 7.5 × 10⁻⁴ mol. This leads to K_c = 562.5 (costs M1).
Forgetting volume conversion: The powers in numerator (3) and denominator (4) do not cancel out! If you insert moles instead of concentrations, you get K_c = 459 (losing M4).
Stoichiometry of I₂: 2 moles of I⁻ produce 1 mole of I₂. Moles of I₂ formed = (moles I⁻ reacted) / 2.

Topics

Physical Chemistry · 3.1.6 Chemical Equilibria, Le Chatelier's Principle and Kc · 3.1.2 Amount of Substance · 3.1.5 Kinetics

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.