AQA A-Level Chemistry Paper 3, 2017: Question 9

1 mark · Easy difficulty · Multiple Choice

Calculate the concentration of hydrogen ions in a solution of a weak monoprotic acid given its acid dissociation constant (Ka).

Practise this question

Question

Multiple choice question 09 asking for the concentration of hydrogen ions, in mol dm⁻³, in a 2.00 × 10⁻³ mol dm⁻³ solution of 2,4,6-trichlorophenol at 298 K, given it is a weak monoprotic acid with Ka = 2.51 × 10⁻⁸ mol dm⁻³. Four options are given: A 5.02 × 10⁻¹¹, B 7.09 × 10⁻⁶, C 1.26 × 10⁻⁵, and D 3.54 × 10⁻³.
Question text

2,4,6-Trichlorophenol is a weak monoprotic acid, with K = 2.51 × 10−8 mol dm−3 at

09 a

298 K.

What is the concentration, in mol dm–3, of hydrogen ions in a 2.00 × 10−3 mol dm−3

solution of 2,4,6-trichlorophenol at 298 K?

[1 mark]

A 5.02 × 10–11

B 7.09 × 10–6

C 1.26 × 10–5

D 3.54 × 10−3

Mark scheme

Show the mark scheme Mark scheme table row showing question number 9 corresponds to correct answer B.

9 B

How to answer it

Calculating [H⁺] for a Weak Monoprotic Acid

📌 What this question tests

This question assesses your ability to apply the acid dissociation constant ( Ka ) expression for a weak monoprotic acid, implement the standard weak acid assumptions ( [H⁺] = [A⁻] and [HA]eqm ≈ [HA]initial ), and rearrange the expression to find hydrogen ion concentration [H⁺] .

Question 09 • 1 Mark

Weak Acid Dissociation Calculation

Determining [H⁺] from Ka and acid concentration

✅ Correct Answer

B: 7.09 × 10⁻⁶ mol dm⁻³

Award 1 mark for selecting option B.

💡 Key Knowledge

  • A weak monoprotic acid dissociates according to:
    HA(aq) ⇌ H⁺(aq) + A⁻(aq)
  • The acid dissociation constant expression is:
    Ka = [H⁺][A⁻] / [HA]
  • Since each HA molecules gives one H⁺ and one A⁻, and negligible dissociation occurs:
    [H⁺] ≈ [A⁻]
    [HA]eqm ≈ [HA]start
  • Simplified formula:
    Ka = [H⁺]² / [HA]  ⇒  [H⁺] = √(Ka × [HA])

📐 Step-by-Step Calculation

  1. Identify given values:
    Ka = 2.51 × 10⁻⁸ mol dm⁻³
    [HA] = 2.00 × 10⁻³ mol dm⁻³
  2. Substitute into rearranged expression:
    [H⁺]² = Ka × [HA]
    [H⁺]² = (2.51 × 10⁻⁸) × (2.00 × 10⁻³) = 5.02 × 10⁻¹¹ mol² dm⁻⁶
  3. Take the square root to find [H⁺]:
    [H⁺] = √(5.02 × 10⁻¹¹) = 7.0852 × 10⁻⁶ mol dm⁻³
  4. Round to 3 significant figures:
    [H⁺] = 7.09 × 10⁻⁶ mol dm⁻³ (matches Option B)

❌ Common Errors & Anatomy of Distractors

  • Distractor A (5.02 × 10⁻¹¹): Forgetting to take the square root at the end ( Ka × [HA] ). This is the most frequent calculator slip.
  • Distractor C (1.26 × 10⁻⁵): Incorrect algebraic rearrangement by dividing instead of multiplying: Ka / [HA] = 2.51 × 10⁻⁸ / 2.00 × 10⁻³.
  • Distractor D (3.54 × 10⁻³): Taking the square root of the inverted division: √(Ka / [HA]) .

🧠 Exam Technique & Examiner Insight

  • Do not be intimidated by complex names: The molecule "2,4,6-trichlorophenol" sounds complicated, but the question explicitly states it is a monoprotic acid. Treat it just like generic weak acid "HA".
  • Sanity-check your answer: A weak acid with concentration 10⁻³ mol dm⁻³ and a tiny Ka (10⁻⁸) should have an [H⁺] significantly smaller than 10⁻³ mol dm⁻³, but definitely not as tiny as 10⁻¹¹ (which would represent a strongly alkaline solution, pH ~ 11!).

Topics

Physical Chemistry · 3.1.12 Acids and Bases

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.