AQA A-Level Chemistry Paper 3, 2024: Question 18

Kw vs Temperature: Which statement is correct?

What this question tests

Skills & knowledge assessed (AO3)

Using Kᵥ (Kᵥ = [H⁺][OH⁻]) to find ion concentrations in pure water.
Recognising that in pure water [H⁺] = [OH⁻] .
Interpreting how changing temperature changes Kᵥ and therefore pH of neutral water.
Spotting common conceptual traps: “neutral means pH 7” is only true at 25 °C.

Part (multiple choice) — Temperature dependence of Kᵥ

Mark allocation: 1 mark (AO3). You must choose the single correct statement.
Given: Kᵥ = 6.40 × 10⁻¹⁵ mol² dm⁻⁶ at 18 °C; Kᵥ = 1.00 × 10⁻¹⁴ mol² dm⁻⁶ at 25 °C.

✅ Correct answer (from mark scheme)

A — The concentration of hydroxide ions in water at 18 °C is 8.00 × 10⁻⁸ mol dm⁻³.

Mark scheme: Q18 = A (1 mark).

🧠 Exam technique (how to secure the mark fast)

For pure water: [H⁺] = [OH⁻] , so Kᵥ = [H⁺]² .
Therefore [OH⁻] = √Kᵥ at that temperature.
Only calculate what the statement asks (here it asks [OH⁻] at 18 °C).

📐 Calculation (step-by-step)

We want: [OH⁻] in pure water at 18 °C.

Start: Kᵥ = [H⁺][OH⁻]

In pure water: [H⁺] = [OH⁻] = x, so Kᵥ = x².

Set up: x = √Kᵥ = √(6.40 × 10⁻¹⁵)
Split into number and power: √6.40 × √(10⁻¹⁵)
√6.40 ≈ 2.53
√(10⁻¹⁵) = 10⁻⁷⋅⁵ = 10⁻⁸ × √10 ≈ 10⁻⁸ × 3.162
Multiply: 2.53 × 3.162 × 10⁻⁸ ≈ 8.00 × 10⁻⁸

Answer: [OH⁻] = 8.00 × 10⁻⁸ mol dm⁻³

Units check: Kᵥ has units mol² dm⁻⁶, so √Kᵥ has units mol dm⁻³ (correct for concentration).

💡 Key knowledge (why temperature matters)

Kᵥ increases from 18 °C to 25 °C (6.40 × 10⁻¹⁵ → 1.00 × 10⁻¹⁴).
If Kᵥ increases, then in pure water both [H⁺] and [OH⁻] increase (because both equal √Kᵥ).
So pH of pure water changes with temperature (neutral water is not always pH 7).

❌ Common errors (what loses marks)

Forgetting the square root: using Kᵥ directly as [OH⁻] instead of √Kᵥ.
Assuming neutral always means pH 7: at temperatures other than 25 °C, neutral pH ≠ 7 because Kᵥ changes.
Mixing up direction of acidity: if [H⁺] increases with temperature, pH decreases (more acidic), even though the solution can still be neutral because [H⁺] = [OH⁻].
Units ignored: concentrations must be mol dm⁻³; Kᵥ is mol² dm⁻⁶.

🧠 Examiner-style commentary (how top answers think)

Stronger candidates immediately recognise [H⁺] = [OH⁻] in pure water and jump to [OH⁻] = √Kᵥ .
Weaker responses often pick options based on memorised “pH 7 = neutral” rather than using the data given (Kᵥ at two temperatures).
This is a 1-mark question: the exam rewards a clean, correct numerical inference rather than a long explanation.

Quick check of the other options (why they are wrong)

Option B

Claim: Dissociation of water into ions is exothermic.

Why it’s wrong using the data: Kᵥ increases when temperature increases. Equilibria shift to favour the endothermic direction when temperature rises, so dissociation is endothermic, not exothermic.

Option C

Claim: pH is the same at 25 °C and 18 °C.

Why it’s wrong: different Kᵥ values mean different [H⁺] (= √Kᵥ), so pH must differ.

Option D

Claim: Water becomes less acidic as temperature is raised.

Why it’s wrong: higher Kᵥ means higher [H⁺] in pure water, so pH decreases (more acidic), even though it remains neutral in the sense that [H⁺] = [OH⁻].

💡 Take-home message

“Neutral” means [H⁺] = [OH⁻] , not “pH = 7”. Since Kᵥ changes with temperature, the pH of neutral water also changes. Here, √(6.40 × 10⁻¹⁵) gives 8.00 × 10⁻⁸ mol dm⁻³, matching option A (and the mark scheme).