Edexcel A-Level Chemistry Paper 1, June 2017: Question 9

7 marks · Medium difficulty · Calculations

Write ionic equations for buffer action involving phosphate ions and calculate the pH of a buffer solution formed by mixing sodium hydroxide and ethanoic acid.

Practise this question

Question

Question text

9 This question is about buffer solutions.

(a) A buffer solution is formed from disodium hydrogenphosphate, containing HPO2– ions,

and sodium dihydrogenphosphate, containing H PO– ions.

Write the ionic equations involving HPO2– and H PO– ions to show how this

42 4

solution acts as a buffer solution.

(2)

(b) Another buffer solution was formed by mixing 20.0 cm3 of sodium hydroxide solution

of concentration 0.100 mol dm–3 with 25.0 cm3 of ethanoic acid of concentration

0.150 mol dm–3.

CH3COOH + NaOH r CH3COONa + H2O

Calculate the pH of this buffer solution.

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

(5)

(Total for Question 9 = 7 marks)

Mark scheme

Show the mark scheme

Question

Answer Additional Guidance Mark

Number

9(a) Penalise non-ionic equations, e.g. (2)

using NaOH or HCl once only

Equations must show reaction of

ions with H+ / H O+ and OH−

Allow ⇌

Ignore state symbols

HPO 2− + H+ → H PO −

42 4

HPO 2− + H O+ → H PO − + H O (1)

43 2 4 2

Allow H PO − → HPO 2− + H+ and

− − 2− 2 4 4

H2PO4 + OH → HPO4 + H2O (1) + −

H + OH → H2O

Question

Answer Additional Guidance Mark

Number

9(b) Example of calculation (5)

calculation of the amount of NaOH / salt amount of NaOH = amount of salt formed

(1) =0.100 x 20.0/1000 = 0.00200

calculation of initial amount of acid (1) initial amount of acid = 0.150 x 25.0/1000

= 0.00375

calculation of the amount of acid left (1) amount of acid left = 0.00375 − 0.00200

= 0.00175

calculation of [H+] (1) total volume = 20.0 + 25.0 = 45.0 (cm3)

[salt] = 0.00200 x 1000/45.0 = 0.0444 (mol dm―3)

[acid] = 0.00175 x 1000/45.0 = 0.0389(mol dm―3)

K = [H+][salt] so [H+] = K [acid]

a a

[acid] [salt]

[H+] = 1.74 x 10―5 x 0.0389/0.0444

= 1.52446 x 10―5 (mol dm―3)

Allow use of moles instead of concentrations

calculation of pH (1) pH = −log[H+] = −log(1.52446 x 10―5)

= 4.817 / 4.82 / 4.8

Allow TE for each step

Ignore SF except 1 SF

Correct answer without working score (5)

Allow alternative methods, for example

pH = pKa – log [acid]

[salt]

pH = − log 1.74 x 10―5 − log 0.0389

0.0444

pH = 4.817 / 4.82 / 4.8 scores M4 and M5

pH = pKa + log[salt]

[acid]

pH = −log1.74 x 10―5 + log 0.0444

0.0389

pH = 4.817 / 4.82 / 4.8 scores M4 and M5

(Total for Question 9 = 7 marks)

How to answer it

Buffer Solutions Exam Guide

📋 What this question tests

This question assesses your understanding of acidic and basic buffer systems. You are tested on writing precise ionic equations to show how conjugate acid-base pairs neutralise added hydrogen or hydroxide ions, alongside performing complex multi-step stoichiometric calculations involving strong base / weak acid titrations to find the resulting pH.

Question Part (a)

Mechanism of Buffer Action

Writing ionic equations for buffer neutralisation (2 marks)

✅ Correct Answer

For the first mark (neutralising added acid):

HPO₄²⁻ + H⁺ → H₂PO₄⁻

(Or using H₃O⁺ → H₂PO₄⁻ + H₂O)

For the second mark (neutralising added alkali):

H₂PO₄⁻ + OH⁻ → HPO₄²⁻ + H₂O

💡 Key Knowledge

A buffer must contain a reserve of both a weak acid component and its conjugate base component.
Added H⁺ ions react with the conjugate base component ( HPO₄²⁻ ).
Added OH⁻ ions react with the weak acid component ( H₂PO₄⁻ ).

❌ Common Errors

Including spectator ions like Na⁺ in the equations (non-ionic equations lose marks).
Writing equations where the buffer species decompose rather than react with added H⁺ / OH⁻ .

🧠 Exam Technique

Always check the prompt for the word ionic. Leave out spectator ions completely, ensure charges balance on both sides, and show clear addition of H⁺ or OH⁻ .

🎯 Mark breakdown: 1 mark for the acid-neutralising equation; 1 mark for the alkali-neutralising equation.

Question Part (b)

Buffer Calculation from Mixing

Calculating the pH of a buffer formed by mixing NaOH and CH₃COOH (5 marks)

📐 Step-by-Step Calculation

Find moles of NaOH (limiting reactant) / salt formed:
0.100 × (20.0 / 1000) = 0.00200 mol
Find initial moles of ethanoic acid (excess):
0.150 × (25.0 / 1000) = 0.00375 mol
Calculate moles of acid left after reaction:
0.00375 - 0.00200 = 0.00175 mol
Calculate [H⁺] using K_a expression:
Total volume = 20.0 + 25.0 = 45.0 cm³
[Salt] = 0.00200 × (1000 / 45.0) = 0.0444 mol dm⁻³
[Acid] = 0.00175 × (1000 / 45.0) = 0.0389 mol dm⁻³
[H⁺] = K_a × ([acid] / [salt]) = 1.74 × 10⁻⁵ × (0.0389 / 0.0444) = 1.52 × 10⁻⁵ mol dm⁻³
(Note: molar ratios of concentrations can also be used directly since volume cancels out).
Calculate pH:
pH = -log(1.52 × 10⁻⁵) = 4.82 (Accept 4.81 - 4.84)

❌ Common Calculation Traps

Forgetting to subtract the reacted moles from the initial acid to find the remaining acid.
Using the total volume to find concentrations is good practice, though moles ratio works because volumes cancel out in the K_a expression.
Inverting the acid-to-salt ratio in the K_a rearrangement.

🧠 Top-Level Exam Strategy

Always layout stoichiometric calculations using a clear "Initial moles → Change → Moles remaining" structure. This guarantees error-carried-forward (TE) marks even if a minor arithmetic slip occurs earlier.

🎯 Mark breakdown: 1 mark for moles of NaOH/salt; 1 mark for initial moles of acid; 1 mark for moles of acid remaining; 1 mark for [H⁺] calculation; 1 mark for final pH.

Topics

Physical Chemistry · Topic 12: Acid-base Equilibria

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 1, June 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.