Edexcel A-Level Chemistry Paper 3, June 2024: Question 8

17 marks · Hard difficulty · Calculations

Analyze a flawed titration method to find Ka of ethanoic acid, derive the half-neutralisation expression, calculate the mass of sodium ethanoate needed to form a buffer solution, and calculate the pH of a mixture of sodium hydroxide and sulfuric acid.

Practise this question

Question

A three-part structured chemistry question about acids and bases. Part (a) describes a student's flawed method to determine the acid dissociation constant Ka of ethanoic acid using a weak base (ammonia) and phenolphthalein, with subparts asking to explain why it is invalid, how to amend Step 1, and to show that Ka = 10^-ph at half-neutralisation. Part (b) asks to calculate the mass of sodium ethanoate needed to prepare a buffer solution of a given pH. Part (c) asks to calculate the pH of the resultant solution formed by mixing sodium hydroxide and sulfuric acid.
Question text

8 This question is about acids and bases.

(a) A student devised a method to determine the acid dissociation constant, Ka, of

ethanoic acid, CH3COOH, using a solution of the acid of unknown concentration.

The student’s outline procedure is shown.

Step 1 Titrate 25.0 cm3 of the ethanoic acid solution with a solution of ammonia

of known concentration, using phenolphthalein to find the end-point.

Step 2 Add a further 25.0 cm3 of the same ethanoic acid solution to the mixture

from Step 1.

This gives a solution where the acid has been half-neutralised.

Step 3 Record the pH of the solution from Step 2.

(i) Explain why this procedure would give invalid results.

(3)

(ii) State how Step 1 should be amended to give valid results.

(1)

(iii) Show that when the solution has been half-neutralised, the acid dissociation

constant is given by the expression

K = 10–pH

a

(3)

(b) Calculate the mass of sodium ethanoate needed to be dissolved in 250 cm3 of

0.520 mol dm–3 ethanoic acid to form a buffer solution with pH = 4.48

[K for ethanoic acid = 1.74 × 10–5 mol dm–3]

a

(5)

(c) A student mixes 50.0 cm3 of 0.900 mol dm–3 NaOH(aq) with 20.0 cm3 of

0.400 mol dm–3 H SO (aq).

20 Calculate the pH of the resultant solution.

[Ionic product of water,*P74455A02032*Kw= 1.00×10–14mol2dm–6]

(5)

(Total for Question 8 = 17 marks)

Mark scheme

Show the mark scheme The official mark scheme showing detailed answers and awarding points for each subpart. For 8(a)(i), it awards marks for identifying the use of a weak base, the lack of a rapid pH change, and the inability to determine the end-point. For 8(a)(ii), replacing ammonia with a strong base is required. For 8(a)(iii), steps for deriving Ka = 10^-pH or using the Henderson-Hasselbalch equation are outlined. For 8(b) and 8(c), multi-step calculation mark breakdowns and sample calculations for masses, concentrations, and pH values are provided.

Question

Answer Additional Guidance Mark

Number

8(a)(i) An explanation that makes reference to the following points: (3)

• as the weak acid (is being titrated with) a weak base (1)

• as no rapid change in pH (to find volume at end-point) (1) Accept pH changes gradually around the end

point

Allow there is no vertical section in the titration

curve

• so end-point cannot be determined (1) so a sharp colour change cannot be observed

Allow phenolphthalein would not change colour/

pKIn is too high

Question

Answer Additional Guidance Mark

Number

8(a)(ii) (1)

• (replace ammonia with) a strong base such as NaOH / KOH Allow name or formula of strong base

Allow use of a pH probe/pH meter

Do not award change of indicator

Question

Answer Additional Guidance Mark

Number

8(a)(iii) An answer that makes reference to the following points: (3)

(method 1)

• Ka expression for ethanoic acid (1) K = [CH COO–][H+] or K = [A–][H+]

a 3 a

[CH3COOH] [HA]

• statement that at half-neutralisation concentration [CH COOH] = [CH COO–] or [HA] = [A–]

of anion and acid are equal (1)

• K = [H+] (=10−pH) (1) Allow pK = pH (= − log [H+] = − log K )

a a a

Standalone mark

Do not award Ka = pKa

Penalise omission of square brackets once only

Penalise use of () throughout for [] once only

(method 2)

• Henderson-Hasselbalch expression (1) pH = pK + log ( [A−] ) or pH = pK + log ( [salt] )

a a

[HA] [acid]

or

pK = pH − log ( [A−] )

a

[HA]

• statement that at half-neutralisation concentration

of anion and acid are equal (1)

• pK = pH (so K = [H+] =10−pH) (1)

a a

Question

Answer Additional Guidance Mark

Number

8(b) A calculation that makes reference to the following points: Example of calculation (5)

• (M1) calculation of [H+] (1) 10–4.48 = 3.3113 × 10–5 (mol dm–3)

(1) K = ([H+][CH COONa]) ÷ [CH COOH]

• (M2) rearrangement of Ka expression a 3 3

[CH COONa] = K × [CH COOH] ÷ [H+]

3 a 3

• (M3) calculation of [CH COONa] in buffer (1) [CH COONa] = (1.74 × 10–5 × 0.52) ÷ 3.3113 × 10–5

= 0.27325 (mol dm–3)

OR Use of Henderson Hasselbalch for M1 to M3

• (M1) rearrangement of H−H expression (1) log ([CH3COONa] ÷ [CH3COOH]) = 4.48 – 4.75945 = – 0.27945

• (M2) calculation of [CH COONa] ÷ [CH COOH] (1) [CH COONa] ÷ [CH COOH] = 10– 0.27945= 0.52547

33 3 3

[CH COONa] = 0.52547 × 0.52 = 0.27325 (mol dm–3)

• (M3) calculation of [CH3COONa] (1) 3

---------------------------------------------------------------

• (M4) calculation of Mr of CH3COONa (1) 82

• (M5) calculate the mass of CH3COONa needed (1) = (0.27325 ÷ 4 × 82) = 5.6015 (g) / 5.6 (g)

Allow TE but not from an incorrect rearrangement for M3

Ignore SF except 1 SF

Allow CH COO− for CH COONa in M2 and M3

Correct answer with or without working scores (5)

Question

Answer Additional Guidance Mark

Number

8(c) A calculation that makes reference to the following (5)

points: Example of calculation

• calculate amount of H+(aq) in mol (1) = (20 ÷ 1000) × 0.400 × 2 = 0.016 (mol)

• calculate amount of OH–(aq) in mol (1) = (50.0 ÷ 1000) × 0.900 = 0.045 (mol)

• calculate amount of excess OH–(aq) in mol (1) = 0.045 – 0.016 = 0.029 (mol)

– (1) = 0.029 ÷ (70 ÷ 1000) = 0.41429 (mol dm−3)

• calculate [OH ] in resultant solution

• calculate pH of resultant solution (1) pH = 14 – (– log(0.41429) = 13.617 / 13.6

or

pH = – log (1 × 10−14 ÷ 0.41429) = 13.617 / 13.6

Do not award M5 for a pH less than 7

Ignore SF except 1 SF

Ignore intermediate units even if incorrect

Allow TE throughout

(Total for Question 8 = 17 marks)

How to answer it

Acids, Bases and Buffer Solutions

Edexcel A-Level Chemistry • Exam Study Guide

What this question tests

This multi-part question tests your deep understanding of acid-base titrations, the limitations of titration indicators, derivations involving the acid dissociation constant ( Kₐ ) at half-neutralisation, buffer preparation calculations, and strong acid-strong base neutralization stoichiometric calculations involving excess reagents and Kₘ .

Question 8(a)(i) — Titration Validity

Explain why this procedure would give invalid results (3 marks)

✅ Correct Answer Framework

  • Ethanoic acid is a weak acid being titrated with ammonia, a weak base.
  • There is no rapid change in pH (no vertical inflection section) around the equivalence point.
  • Therefore, the end-point cannot be accurately determined using a standard indicator like phenolphthalein.

❌ Common Errors & Misconceptions

  • Students often state vaguely that "phenolphthalein is the wrong indicator" without explaining the underlying titration curve characteristics.
  • Forgetting that both the acid and the base are weak, meaning titration curves lack the steep pH jump required for visual indicators to work.
Mark breakdown: 1 mark for identifying weak acid + weak base; 1 mark for noting absence of rapid pH change/vertical section; 1 mark for concluding the end-point cannot be determined.

Question 8(a)(ii) — Titration Amendment

State how Step 1 should be amended to give valid results (1 mark)

✅ Correct Answer

Replace ammonia with a strong base (such as NaOH or KOH ). Alternatively, use a pH meter/probe instead of an indicator.

🧠 Exam Technique

Keep your answer direct and concise. State the reagent substitution clearly without over-explaining.

Mark breakdown: 1 mark for stating replacement with a strong base (or mentioning a pH meter/probe).

Question 8(a)(iii) — Deriving Kₐ = 10⁻ᵖᴴ

Show that when the solution has been half-neutralised, Kₐ is given by Kₐ = 10⁻ᵖᴴ (3 marks)

💡 Key Knowledge

The acid dissociation constant expression for ethanoic acid ( HA ) is:

Kₐ = [CH₃COO⁻][H⁺] / [CH₃COOH]

At half-neutralisation, exactly half of the acid has been converted into its salt (ethanoate anions):

[CH₃COOH] = [CH₃COO⁻]

📐 Mathematical Derivation

1. Substitute [CH₃COOH] = [CH₃COO⁻] into the Kₐ expression:

Kₐ = ([CH₃COO⁻][H⁺]) / [CH₃COO⁻]

2. Cancel out the concentration terms:

Kₐ = [H⁺]

3. Take negative logs on both sides, remembering that -log[H⁺] = pH and -log(Kₐ) = pKₐ (or apply [H⁺] = 10⁻ᵖᴴ directly):

Kₐ = 10⁻ᵖᴴ

Mark breakdown: 1 mark for writing the correct Kₐ expression; 1 mark for stating/showing [CH₃COOH] = [CH₃COO⁻] at half-neutralisation; 1 mark for deducing Kₐ = [H⁺] leading to Kₐ = 10⁻ᵖᴴ .

Question 8(b) — Buffer Solution Calculation

Calculate the mass of sodium ethanoate needed to be dissolved in 250 cm³ of 0.520 mol dm⁻³ ethanoic acid to form a buffer solution with pH = 4.48 (Kₐ = 1.74 × 10⁻⁵ mol dm⁻³) (5 marks)

📐 Step-by-Step Calculation

Step 1: Calculate [H⁺] from pH

[H⁺] = 10⁻ᵖᴴ = 10⁻⁴·⁴⁸ = 3.3113 × 10⁻⁵ mol dm⁻³

Step 2: Rearrange the Kₐ expression to find [CH₃COO⁻]

Kₐ = ([CH₃COO⁻][H⁺]) / [CH₃COOH]

[CH₃COO⁻] = (Kₐ × [CH₃COOH]) / [H⁺]

Step 3: Calculate equilibrium concentration of sodium ethanoate

[CH₃COO⁻] = (1.74 × 10⁻⁵ × 0.520) / (3.3113 × 10⁻⁵) = 0.27325 mol dm⁻³

Step 4: Calculate moles of sodium ethanoate in 250 cm³

Moles = concentration × volume (in dm³)

Moles = 0.27325 × (250 / 1000) = 0.06831 mol

*(Alternative approach: scale directly using the 0.250 dm³ volume: 0.27325 ÷ 4 = 0.06831 mol)*

Step 5: Convert moles to mass (Mᵣ of CH₃COONa = 82)

Mass = moles × Mᵣ = 0.06831 × 82 = 5.6015 g

Final Answer: 5.6 g (to 2 sig fig, matching data precision)

❌ Calculation Traps & Common Errors

  • Volume conversion errors: Forgetting to divide 250 cm³ by 1000 to convert to dm³.
  • Molar mass errors: Incorrectly calculating the Mᵣ of sodium ethanoate ( CH₃COONa ) by missing sodium (23 + 24.0 + 32 + 3 = 82).
  • Significant figures: Watch out for rounding intermediate values too early, which leads to rounding errors in the final mass.
Mark breakdown: (M1) Calculation of [H⁺] ; (M2) Rearrangement of Kₐ expression; (M3) Calculation of [CH₃COO⁻] ; (M4) Calculation of Mᵣ (82); (M5) Final calculation of mass.

Question 8(c) — Strong Acid / Strong Base Neutralisation & pH

A student mixes 50.0 cm³ of 0.900 mol dm⁻³ NaOH(aq) with 20.0 cm³ of 0.400 mol dm⁻³ H₂SO₄(aq). Calculate the pH of the resultant solution. (Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶) (5 marks)

📐 Step-by-Step Calculation

Step 1: Calculate initial moles of H⁺ ions

H₂SO₄ is a dibasic acid (produces 2 moles of H⁺ per mole of acid).

Moles H₂SO₄ = (20.0 / 1000) × 0.400 = 0.00800 mol

Moles H⁺ = 0.00800 × 2 = 0.0160 mol

Step 2: Calculate initial moles of OH⁻ ions

NaOH is a monoprotic base.

Moles OH⁻ = (50.0 / 1000) × 0.900 = 0.0450 mol

Step 3: Determine moles of excess OH⁻ after neutralisation

Excess OH⁻ = 0.0450 - 0.0160 = 0.0290 mol

Step 4: Calculate concentration of OH⁻ in the total volume

Total volume = 50.0 cm³ + 20.0 cm³ = 70.0 cm³ = 0.070 dm³

[OH⁻] = 0.0290 / 0.0700 = 0.41429 mol dm⁻³

Step 5: Calculate pOH, then pH using Kw

[H⁺] = Kʷ / [OH⁻] = (1.00 × 10⁻¹⁴) / 0.41429 = 2.4138 × 10⁻¹⁴ mol dm⁻³

pH = -log[H⁺] = 13.6 (or 13.617)

*(Alternatively: find pOH = -log(0.41429) = 0.383, then pH = 14.00 - 0.383 = 13.6)*

❌ Critical Pitfalls in Excess Calculations

  • Stoichiometry trap: Forgetting that H₂SO₄ is diprotic, meaning it releases two H⁺ ions per molecule. Multiplying by 2 is essential!
  • Total volume neglect: Dividing excess moles by the volume of just one solution instead of the combined total volume (70 cm³).
  • Forgetting the final step: Stopping at [OH⁻] or calculating pOH instead of converting successfully to pH.
Mark breakdown: (1) Moles of H⁺ calculated; (1) Moles of OH⁻ calculated; (1) Excess moles of OH⁻ determined correctly; (1) Concentration of [OH⁻] in resultant solution calculated; (1) Final pH calculated correctly (13.6).

Topics

Physical Chemistry · Topic 12: Acid-base Equilibria

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 3, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.