AQA A-Level Chemistry Paper 3, June 2022: Question 14

What this question tests

Understanding the definition and complete dissociation of a strong acid.
Applying stoichiometric mole ratios for a diprotic acid (H₂X → 2H⁺ + X²⁻).
Calculating pH using the logarithmic expression: pH = −log₁₀[H⁺] .
Avoiding common distractor traps in multiple-choice acid-base chemistry questions.

B — 1.40

Mark Scheme Breakdown:
• 1 mark for selecting option B (AO2).

Strong acid: Completely ionises in aqueous solution.
Diprotic acid: Each mole of H₂X releases 2 moles of H⁺ ions in solution:
H₂X(aq) → 2H⁺(aq) + X²⁻(aq)
Therefore: [H⁺] = 2 × [H₂X] .
pH equation: pH = −log₁₀[H⁺] .

Identify the stoichiometric relationship:
The acid H₂X produces 2 moles of H⁺ for every 1 mole of acid dissolved.
[H⁺] = 2 × [H₂X]
Calculate hydrogen ion concentration [H⁺]:
[H⁺] = 2 × 0.020 mol dm⁻³ = 0.040 mol dm⁻³
Calculate the pH:
pH = −log₁₀(0.040) = 1.3979...
Round to appropriate precision:
To 2 decimal places (standard for pH): 1.40

Choosing C (1.70): The most common error. Students treat H₂X as a monoprotic acid (forgetting the 2:1 ratio):
−log₁₀(0.020) = 1.70
Choosing D (2.00): Students divide the concentration by 2 instead of multiplying:
[H⁺] = 0.010 mol dm⁻³ → −log₁₀(0.010) = 2.00
Choosing A (1.00): Guesswork or arithmetic slips assuming [H⁺] = 0.10 mol dm⁻³ .

Always inspect the formula: Look closely at the subscript in H₂X or H₂SO₄. The prefix "di-" or a subscript "2" means you must double the acid concentration to find [H⁺].
Sanity check: Doubling [H⁺] should decrease the pH by approx log₁₀(2) ≈ 0.30 units. Compared to a monoprotic acid of the same concentration (pH = 1.70), a diprotic acid must have a lower pH ( 1.70 − 0.30 = 1.40 ).
Speed tip: In Section A (Multiple Choice), you have around 1.5 minutes per question. Doing −log₁₀(2 × 0.020) directly in your calculator takes under 15 seconds.

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