AQA A-Level Chemistry Paper 3, 2017: Question 10

1 mark · Easy difficulty · Multiple Choice

Calculate the pH of a 0.46 mol dm⁻³ potassium hydroxide solution at 298 K using the ionic product of water, Kw.

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Question

Multiple-choice question 10 asking: 'What is the pH of a 0.46 mol dm⁻³ solution of potassium hydroxide at 298 K? (Kw = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K)'. Four options are given: A 0.34, B 13.66, C 13.96, and D 14.34, each accompanied by an answer lozenge.
Question text

What is the pH of a 0.46 mol dm−3 solution of potassium hydroxide at 298 K?

(K = 1.0 × 10−14 mol2 dm−6 at 298 K)

w

[1 mark]

A 0.34

B 13.66

C 13.96

D 14.34

Mark scheme

Show the mark scheme Mark scheme table showing question number 10 corresponds to correct answer B.

10 B

How to answer it

Calculating the pH of a Strong Monobasic Base (KOH)

📋 What this question tests

This question assesses your ability to calculate the pH of a strong alkaline solution using the ionic product of water (Kw):

  • Recognising that potassium hydroxide (KOH) is a strong base that fully dissociates 1:1 into K⁺ and OH⁻.
  • Applying the expression for the ionic product of water: Kw = [H⁺][OH⁻].
  • Rearranging to find [H⁺] and calculating pH via pH = −log₁₀[H⁺] (or using pH = 14 − pOH).

Question 10 Walkthrough

Multiple Choice [1 Mark]

✅ Correct Answer

B: 13.66

Potassium hydroxide is a strong mono-acidic base with [OH⁻] = 0.46 mol dm⁻³. At 298 K, this yields a [H⁺] of 2.17 × 10⁻¹⁴ mol dm⁻³, which gives a pH of 13.66.

💡 Key Knowledge

  • KOH dissociation: KOH(aq) → K⁺(aq) + OH⁻(aq) (100% dissociated, so [OH⁻] = [KOH]).
  • Water equilibrium: Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K.
  • pH definition: pH = −log₁₀[H⁺].
  • pOH relationship: pH + pOH = 14.00 at 298 K.

📐 Step-by-Step Calculation

Method 1 (Standard Kw rearrangement):

  1. Find [OH⁻]:
    Because KOH fully dissociates:
    [OH⁻] = 0.46 mol dm⁻³
  2. Rearrange Kw to find [H⁺]:
    [H⁺] = Kw / [OH⁻]
    [H⁺] = (1.0 × 10⁻¹⁴) / 0.46 = 2.174 × 10⁻¹⁴ mol dm⁻³
  3. Calculate pH:
    pH = −log₁₀(2.174 × 10⁻¹⁴) = 13.66

Method 2 (Alternative pOH shortcut):

  • pOH = −log₁₀[OH⁻] = −log₁₀(0.46) = 0.337
  • pH = 14 − pOH = 14 − 0.337 = 13.66

❌ Common Errors & Distractor Analysis

  • Option A (0.34): Calculated −log₁₀(0.46) directly and mistakenly called it pH. This is actually the pOH, representing a strongly acidic value rather than an alkaline one!
  • Option C (13.96): Arises from arithmetic slips or misuse of powers of ten on scientific calculators.
  • Option D (14.34): Incorrectly calculated 14 + pOH (14 + 0.34) instead of subtracting it from 14.

🧠 Exam Technique & Sanity Check

  • Sense-check the number: KOH is a strong alkali at a substantial concentration (0.46 mol dm⁻³). The pH must be well above 7, immediately eliminating A (0.34) without needing a calculator.
  • Watch brackets on your calculator: When dividing by scientific notation, ensure you enter 1.0 × 10⁻¹⁴ / 0.46 correctly using the EXP or ×10ˣ button to avoid order-of-operation errors.

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.