AQA A-Level Chemistry Paper 3, 2017: Question 26

1 mark · Medium difficulty · Multiple Choice

Calculate the concentration of chloride ions in a solution containing a given mass of lead(II) chloride.

Practise this question

Question

Question 26 asks: 'A solution of lead(II) chloride (Mr = 278.2) contains 1.08 g of PbCl2 in 100 cm3 of solution. In this solution, the lead(II) chloride is fully dissociated into ions. What is the concentration of chloride ions in this solution?' Four options are given: A: 3.88 × 10^-3 mol dm^-3, B: 7.76 × 10^-3 mol dm^-3, C: 3.88 × 10^-2 mol dm^-3, and D: 7.76 × 10^-2 mol dm^-3.
Question text

A solution of lead(ll) chloride (M = 278.2) contains 1.08 g of PbCl in 100 cm3 of

26 r 2

solution. In this solution, the lead(ll) chloride is fully dissociated into ions.

What is the concentration of chloride ions in this solution?

[1 mark]

A 3.88 × 10−3 mol dm−3

B 7.76 × 10−3 mol dm−3

C 3.88 × 10−2 mol dm−3

D 7.76 × 10−2 mol dm−3

Mark scheme

Show the mark scheme Mark scheme table row showing question number 26 with the correct answer D.

26 D

How to answer it

AQA A-Level Chemistry • Amount of Substance

Calculating Dissociated Ion Concentration

What this question tests

This multiple-choice question assesses your ability to calculate solution concentration from mass and volume, handle unit conversions (cm³ to dm³), and apply ionic dissociation stoichiometry (specifically recognising a 1 : 2 mole ratio of salt to constituent anion).

Question 26

Lead(II) chloride dissociation and chloride ion concentration

✅ Correct Answer

D — 7.76 × 10⁻² mol dm⁻³

Mark Scheme Breakdown:
• Selecting option D scores [1 mark].

💡 Key Knowledge

  • Mole equation: n = m / Mᵣ
  • Concentration equation: c = n / V (in dm³)
  • Volume conversion: 100 cm³ = 0.100 dm³ (divide by 1000)
  • Ionic dissociation:
    PbCl₂(s) → Pb²⁺(aq) + 2Cl⁻(aq)
    Each mole of PbCl₂ yields 2 moles of Cl⁻ ions.

📐 Step-by-Step Calculation

  1. Find moles of PbCl₂ dissolved:
    Moles = mass / Mᵣ = 1.08 / 278.2 = 3.882 × 10⁻³ mol
  2. Calculate the concentration of PbCl₂:
    Volume = 100 / 1000 = 0.100 dm³
    [PbCl₂] = 3.882 × 10⁻³ mol / 0.100 dm³ = 3.882 × 10⁻² mol dm⁻³
  3. Determine the concentration of Cl⁻ ions:
    Since PbCl₂ dissociates into 1 Pb²⁺ and 2 Cl⁻:
    [Cl⁻] = 2 × [PbCl₂]
    [Cl⁻] = 2 × (3.882 × 10⁻²) = 7.76 × 10⁻² mol dm⁻³

❌ Distractor Breakdown & Common Traps

  • A (3.88 × 10⁻³ mol dm⁻³): Candidate simply computed moles of PbCl₂ ( 1.08 / 278.2 ) and failed to divide by volume or multiply by 2.
  • B (7.76 × 10⁻³ mol dm⁻³): Candidate multiplied the moles by 2 to get moles of Cl⁻, but forgot to convert to concentration by dividing by 0.1 dm³.
  • C (3.88 × 10⁻² mol dm⁻³): The classic trap! Calculated the concentration of PbCl₂ (or Pb²⁺), but forgot the 1 : 2 stoichiometry for Cl⁻.

🧠 Exam Technique & Examiner Tips

  • Read the formula carefully: Whenever an ionic compound contains a polyatomic group or multiple halides (e.g., PbCl₂, CaCl₂, Na₂SO₄), write out the dissociation equation first.
  • Eliminate 50% immediately: You know 1 PbCl₂ gives 2 Cl⁻, so the answer must be a doubled value (7.76 × 10⁻ˣ), instantly narrowing your choice down to B or D.
  • Sanity check units: Always ensure volume is in dm³ before finalizing concentration.

Topics

Physical Chemistry · 3.1.2 Amount of Substance

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.