OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2023: Question 10

1 mark · Medium difficulty · Multiple Choice

Calculate the percentage dissociation of a methanoic acid solution given its concentration and acid dissociation constant Ka.

Practise this question

Question

Multiple choice question 10 asking for the percentage dissociation of a 0.015 mol dm-3 solution of methanoic acid, HCOOH, with Ka = 1.60 x 10^-4 mol dm-3. Four options are provided: A 0.016%, B 1.1%, C 1.82%, D 10.3%. An answer box is shown at the bottom with 1 mark allocated.
Question text

10 What is the percentage dissociation of a 0.015 mol dm–3 solution of methanoic acid, HCOOH

(K = 1.60 × 10–4 mol dm–3)?

a

A 0.016%

B 1.1%

C 1.82%

D 10.3%

Your answer [1]

Mark scheme

Show the mark scheme Mark scheme showing question number 10 with the correct answer option D, worth 1 mark.

10 D 1 AO2.6

How to answer it

Calculating Percentage Dissociation of a Weak Acid

What this question tests

This question tests your understanding of weak acid equilibria, acid dissociation constants ( K_a ), and the ability to calculate hydrogen ion concentration to determine percentage dissociation. You must confidently manipulate the K_a expression, apply root-solving techniques for weak acids, and use ratios to find percentages.

Question 10 Multiple Choice

Exam Breakdown

✅ Correct Answer: D (10.3%)

Option D is correct because applying the weak acid approximation leads to a hydrogen ion concentration of 1.549 × 10⁻³ mol dm⁻³ , which when divided by the initial concentration and multiplied by 100 gives 10.3% .

💡 Key Knowledge

  • Weak acids only partially dissociate in aqueous solution.
  • The expression for acid dissociation constant is K_a = ([H⁺][HCOO⁻]) / [HCOOH] .
  • Approximation used: [H⁺] = [HCOO⁻] and [HCOOH]equilibrium ≈ [HCOOH]initial .

🧠 Exam Technique

Don't panic when you see percentage dissociation! Treat it as a two-stage problem: first find [H⁺] using the rearranged K_a formula, then calculate the percentage using the initial concentration.

❌ Common Errors

  • Square root omission: Forgetting to square root the product of K_a × [HA] , leading to wildly incorrect dimensions (Option A).
  • Inverted division: Dividing initial concentration by [H⁺] instead of the other way around.

📐 Step-by-Step Calculation Guide

  1. Write the expression for K_a:
    K_a = [H⁺]² / [HCOOH]
  2. Rearrange to make [H⁺]² the subject:
    [H⁺]² = K_a × [HCOOH]
    [H⁺]² = (1.60 × 10⁻⁴) × 0.015 = 2.4 × 10⁻⁶
  3. Calculate [H⁺]:
    [H⁺] = √(2.4 × 10⁻⁶) = 1.549 × 10⁻³ mol dm⁻³
  4. Calculate Percentage Dissociation:
    Percentage Dissociation = ([H⁺] / [HCOOH]initial) × 100
    Percentage Dissociation = (1.549 × 10⁻³ / 0.015) × 100 = 10.326%
  5. Round appropriately:
    Match the data given in the question (3 significant figures) to yield 10.3%.
Examiner Note: This question discriminates well across ability tiers. High-scoring candidates swiftly set up the calculation without getting confused by the terminology "percentage dissociation", recognising it simply as ([ dissociated moles ] / [ total moles ]) × 100 .

Topics

Module 5: Physical chemistry and transition elements · 5.1 Rates, equilibrium and pH

Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.