AQA A-Level Chemistry Paper 1, 2017: Question 3

9 marks · Medium difficulty · State/Explain/Numerical

Use Kw at 10°C to choose the correct expression for Kw, calculate the pH of pure water and of a 0.0131 mol dm^-3 Ca(OH)2 solution at 10°C, explain why pure water at 10°C is not alkaline, and predict/justify whether a saturated Mg(OH)2 solution (from 0.0131 mol added) has a larger, smaller or same pH as the Ca(OH)2 solution.

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Question

Question text

03 The ionic product of water, K = 2.93 × 10−15 mol2 dm−6 at 10 °C

03.1 Which is the correct expression for Kw?

Tick ( ) one box.

[1 mark]

[H2O]

A Kw = + −

[H ][OH ]

B +

Kw = [H ][H2O]

C + −

Kw = [H ][OH ]

[H+ ][OH− ]

D Kw =

[H2O]

03.2 Calculate the pH of pure water at 10 °C

Give your answer to two decimal places.

[2 marks]

pH of water

03.3 Suggest why this pure water at 10 °C is not alkaline.

[1 mark]

03.4 Calculate the pH of a 0.0131 mol dm−3 solution of calcium hydroxide at 10 °C

Give your answer to two decimal places.

[3 marks]

pH of solution

03.5 The 0.0131 mol dm−3 calcium hydroxide solution at 10 °C was a saturated

solution.

A student added 0.0131 mol of magnesium hydroxide to 1.00 dm3 of water at

10 °C and stirred the mixture until no more solid dissolved.

Predict whether the pH of the magnesium hydroxide solution formed at 10 °C is

larger than, smaller than or the same as the pH of the calcium hydroxide

solution at 10 °C

Explain your answer.

[2 marks]

pH of magnesium hydroxide compared to calcium hydroxide

Explanation

Mark scheme

Show the mark scheme

Question Answers Mark Additional Comments/Guidance

03.1 Ans = C 1

[H+] = √K = √ 2.93 x 10−15 (= 5.41 x 10−8) 1

03.2 −8

pH = (− log (5.41 x 10 ) = 7.27 1 Must be 2dp 7.27 scores 2 marks

03.3 [H+] = [OH-] allow description in words

1 + -

equal moles/quantities/numbers/ratio of H and OH

[OH−] = 0.0131 x 2 = 0.0262 1 pH = 12.95 scores 3 marks

pH = 12.42 scores 2 marks (K = 1x10-14)

pH = 12.65 scores 1 mark (not multiplied by 2)

pH = 12.35 scores 1 mark (divided by 2)

pH = 12.12 scores 0 marks (no x2 and wrong Kw)

[H+] = (K / [OH−] ) = 2.93 x 10−15 / 0.0262 (= 1.118 x 10−13) 1

03 4 pH = (− log (1.118 x 10−13) = 12.9514 = 12.95 1

allow to 2dp or more

[OH−] = 0.0131 x 2 = 0.0262

pOH = (−log 0.0262) = 1.5817 – – –

pH −15 12.95

=(−log Kw− pOH = −log 2.93x10 −1.58 =14.53−1.58) =

18 of 35

smaller / lower pH / less alkaline / more acidic If not smaller CE = 0/2

Allow pH number between 8 and 12

03.5

M2 dependent on M1 but if blank mark on

(magnesium hydroxide) is less soluble / sparingly soluble/

1 Ignore concentration and dissociation

solubility of hydroxide increases down group II

Ignore incorrect formula

Do not allow Mg(OH)2 is insoluble

Total 9

How to answer it

Kw at 10 °C: pH, pOH & Group 2 Hydroxides

AQA A-Level Chemistry • Topic: acids, bases & ionic product of water (K W )

Total: 9 marks

What this question tests

Knowing the correct definition of K W and when [H₂O] is omitted.
Using K W = [H⁺][OH⁻] at a non‑standard temperature (10 °C), not assuming 1.0 × 10⁻¹⁴.
Calculating pH of pure water from [H⁺] = √K W and giving the answer to 2 d.p.
Recognising neutrality as [H⁺] = [OH⁻] (not “pH = 7”).
Strong base stoichiometry: Ca(OH)₂ gives 2 OH⁻ per formula unit; then use K W to find [H⁺] and pH.
Explaining pH comparison using solubility trends down Group 2 (Mg(OH)₂ less soluble ⇒ lower [OH⁻] ⇒ lower pH).

Part (a) • 03.1 • 1 mark

Choosing the correct expression for K W

✅ Correct answer (1/1)

C: K W = [H⁺][OH⁻]

Mark point: identify the correct expression for K W .

💡 Key knowledge

From: H₂O(l) ⇌ H⁺(aq) + OH⁻(aq)
Equilibrium: K = [H⁺][OH⁻] / [H₂O]
Because liquid water has (effectively) constant concentration, we define K W = [H⁺][OH⁻] .

❌ Common errors (examiner insight)

Including [H₂O] in the expression (options A or D). In K W , it is omitted.
Mixing up multiplication/division: K W is the product of ionic concentrations.

Part (b) • 03.2 • 2 marks

Calculate the pH of pure water at 10 °C

Given: K W = 2.93 × 10⁻¹⁵ mol² dm⁻⁶

📐 Calculation (step-by-step)

In pure water: [H⁺] = [OH⁻] so
K W = [H⁺]²
Find [H⁺]:
[H⁺] = √(2.93 × 10⁻¹⁵) = 5.41 × 10⁻⁸ mol dm⁻³
Convert to pH:
pH = −log(5.41 × 10⁻⁸) = 7.27

Mark breakdown (2 marks):
• 1 mark for [H⁺] = √K W (or equivalent).
• 1 mark for correct pH to 2 d.p. → 7.27.

✅ Correct final answer

pH = 7.27 (to 2 d.p.)

Examiner note: “Must be 2 d.p. — 7.27 scores 2 marks”.

❌ Common errors (where marks are lost)

Using K W = 1.0 × 10⁻¹⁴ out of habit (that is for 25 °C).
Forgetting the square root and using [H⁺] = K W .
Not rounding to 2 decimal places (the mark scheme is strict here).

🧠 Exam technique

Whenever K W is given at a temperature, use that value. The question is testing temperature dependence.
Write the key link early: [H⁺] = [OH⁻] in pure water. That justifies the √ step for full method credit.

Part (c) • 03.3 • 1 mark

Explain why pure water at 10 °C is not alkaline

✅ Correct answer (1/1)

[H⁺] = [OH⁻] (equal amounts / equal concentrations of H⁺ and OH⁻).

Mark point: accept in words: “equal moles/quantities/numbers/ratio of H⁺ and OH⁻”.

💡 Key knowledge

Alkaline means [OH⁻] > [H⁺] .
Neutral means [OH⁻] = [H⁺] .
At 10 °C, neutral pH is 7.27 (not 7), because K W changes with temperature.

❌ Common errors

Saying “it is not alkaline because pH is less than 7” (wrong here: pH is 7.27).
Saying “neutral means pH = 7” without referencing temperature dependence.

Part (d) • 03.4 • 3 marks

Calculate the pH of 0.0131 mol dm⁻³ Ca(OH)₂ at 10 °C

📐 Calculation (step-by-step, mark-winning method)

Use stoichiometry for hydroxide ions (strong base):
Ca(OH)₂(aq) → Ca²⁺(aq) + 2OH⁻(aq)

[OH⁻] = 2 × 0.0131 = 0.0262 mol dm⁻³

This “×2” is a key marking point.
Use K W at 10 °C to find [H⁺]:
[H⁺] = K W / [OH⁻] = (2.93 × 10⁻¹⁵) / 0.0262

[H⁺] = 1.118 × 10⁻¹³ mol dm⁻³
Convert to pH:
pH = −log(1.118 × 10⁻¹³) = 12.9514…

pH = 12.95 (to 2 d.p.)

Mark breakdown (3 marks):
• 1 mark: [OH⁻] = 0.0131 × 2 = 0.0262
• 1 mark: [H⁺] = K W /[OH⁻] with K W = 2.93 × 10⁻¹⁵
• 1 mark: pH = 12.95 (allow 2 d.p. or more)

✅ Correct final answer

pH = 12.95

Examiner guidance: “allow to 2 d.p. or more”.

Alternative valid method (also in mark scheme)

Find pOH from [OH⁻], then use pK W at 10 °C:
pOH = −log(0.0262) = 1.5817
pK W = −log(2.93 × 10⁻¹⁵) = 14.53
pH = pK W − pOH = 14.53 − 1.58 = 12.95

❌ Common calculation traps (examiner insight)

Forgetting the ×2 for Ca(OH)₂ → gives pH ≈ 12.65 (only limited credit).
Using the wrong K W (1.0 × 10⁻¹⁴) → gives pH ≈ 12.42 (mark scheme indicates this earns 2 marks, not 3).
Dividing by 2 instead of multiplying → pH ≈ 12.35 (low method credit).
Units: keep concentrations in mol dm⁻³ consistently.

Mark scheme “diagnostic” values show how accuracy affects marks:
• 12.95 → 3 marks
• 12.42 (used K W = 1×10⁻¹⁴) → 2 marks
• 12.65 (no ×2) → 1 mark

🧠 Exam technique

Underline ions produced per formula unit for hydroxides: Group 2 hydroxides always produce 2 OH⁻ when fully dissociated.
At non-25 °C, use: pH + pOH = pK W where pK W = −log K W .

Part (e) • 03.5 • 2 marks

Compare pH of saturated Mg(OH)₂ vs saturated Ca(OH)₂ at 10 °C

✅ Correct answer (2/2)

Prediction: pH of Mg(OH)₂ solution is smaller / lower (less alkaline / more acidic) than Ca(OH)₂.
Explanation: Mg(OH)₂ is less soluble (sparingly soluble), so it produces a lower [OH⁻]. Solubility of Group 2 hydroxides increases down the group (Mg(OH)₂ < Ca(OH)₂).

Mark breakdown (2 marks):
• 1 mark: “smaller/lower pH” (must be correct direction; otherwise CE = 0/2).
• 1 mark: “Mg(OH)₂ less soluble / solubility increases down Group 2”.

💡 Key knowledge

“Saturated solution” means maximum dissolved concentration at that temperature.
Lower solubility ⇒ fewer OH⁻ ions in solution ⇒ lower pH.
Group 2 hydroxide solubility trend: increases down the group.

Examiner note: they ignore detailed concentration/dissociation workings here—this is a solubility argument question.

❌ Common errors (examiner insight)

Stating “Mg(OH)₂ is insoluble” — not allowed. It is sparingly soluble.
Giving the correct explanation but the wrong comparison direction. The mark scheme applies consequential marking: if the pH direction is wrong, you lose both marks.
Talking about “stronger base” instead of solubility. The key is how much dissolves, not base strength.

Examiner guidance: an acceptable pH number could be “between 8 and 12” provided the direction is clearly lower than Ca(OH)₂.

🧠 Exam technique

For “predict and explain” comparisons, write a one-line prediction first (so you secure M1), then a one-line reason linked to a trend (to secure M2).
Use precise language: “less soluble” / “sparingly soluble” is safer than “insoluble”.

Quick checklist • full-mark habits

💡 Must-know relationships

K W = [H⁺][OH⁻]
Pure water: [H⁺] = [OH⁻] = √K W
pH = −log[H⁺] , pOH = −log[OH⁻]
At any temperature: pH + pOH = pK W , where pK W = −log K W

❌ Top 3 avoidable mistakes

Assuming neutral pH is always 7 (only at 25 °C).
Forgetting the “2 OH⁻” from Ca(OH)₂.
Using K W = 1.0 × 10⁻¹⁴ when a different K W is given.

🧠 Presentation for method marks

Show the key intermediate concentration ([H⁺] or [OH⁻]) before taking logs.
Keep to the required rounding (2 d.p. where asked).
Write units for concentrations: mol dm⁻³ .

Topics

Physical Chemistry · 3.1.12 Acids and Bases · 3.1.2 Amount of Substance

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