OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2025: Question 4

1 mark · Medium difficulty · Multiple Choice

Calculate the volume in cm³ of 0.250 mol dm⁻³ barium hydroxide required to neutralise 25.0 cm³ of 0.115 mol dm⁻³ hydrochloric acid.

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Question

Multiple choice question asking: 'What volume, in cm³, of 0.250 mol dm⁻³ barium hydroxide solution is required to exactly neutralise 25.0 cm³ of 0.115 mol dm⁻³ hydrochloric acid?' followed by the balanced equation: Ba(OH)2(aq) + 2HCl(aq) → BaCl2(aq) + 2H2O(l). Four options are given: A 5.75, B 11.50, C 23.00, D 54.35.

Mark scheme

Show the mark scheme Mark scheme row showing question number 4 with the correct answer 'A' and 1 mark.

How to answer it

Titration Neutralisation: Ba(OH)₂ and HCl

📋 What this question tests

This question tests your ability to carry out quantitative solution stoichiometry calculations: calculating amounts in moles from concentration and volume ( n = c × V ), applying chemical equation stoichiometry (1 : 2 reacting ratio), and rearranging the concentration formula to solve for an unknown volume in cm³.

Question 4 Analysis

Multiple Choice Neutralisation Calculation

✅ Correct Answer: A (5.75 cm³)

The required volume of 0.250 mol dm⁻³ Ba(OH)₂ solution is 5.75 cm³.

📐 Step-by-Step Calculation

  1. Identify given data:
    Hydrochloric acid: V = 25.0 cm³ = 0.0250 dm³ , c = 0.115 mol dm⁻³
    Barium hydroxide: c = 0.250 mol dm⁻³ , V = ?
  2. Calculate moles of HCl:
    n(HCl) = c × V = 0.115 mol dm⁻³ × (25.0 / 1000) dm³ = 2.875 × 10⁻³ mol
  3. Determine reacting mole ratio:
    From the balanced equation: Ba(OH)₂(aq) + 2HCl(aq) → BaCl₂(aq) + 2H₂O(l)
    Ba(OH)₂ : HCl = 1 : 2
    n(Ba(OH)₂) = n(HCl) ÷ 2 = 2.875 × 10⁻³ ÷ 2 = 1.4375 × 10⁻³ mol
  4. Calculate volume of Ba(OH)₂ solution:
    V(dm³) = n / c = 1.4375 × 10⁻³ mol ÷ 0.250 mol dm⁻³ = 5.75 × 10⁻³ dm³
    V(cm³) = 5.75 × 10⁻³ × 1000 = 5.75 cm³

💡 Key Knowledge

  • Concentration formula: n = c × V (where V is in dm³).
  • Unit conversion: Convert cm³ to dm³ by dividing by 1000 ( 1 dm³ = 1000 cm³ ).
  • Diprotic/Dibasic nature: Barium hydroxide supplies two hydroxide ions ( 2OH⁻ ), reacting with two hydrogen ions ( 2H⁺ ) from two molecules of monoprotic HCl.

❌ Distractor Breakdown & Traps

  • B (11.50 cm³): Forgetting the stoichiometric ratio (1 : 1 assumed) gives 2.875 × 10⁻³ / 0.250 × 1000 = 11.50 cm³ .
  • C (23.00 cm³): Multiplying by 2 instead of dividing by 2 for the acid-base ratio gives (2 × 2.875 × 10⁻³) / 0.250 × 1000 = 23.00 cm³ .
  • D (54.35 cm³): Inverting the calculation formula completely ( 25.0 × 0.250 ÷ 0.115 ).

🧠 Exam Technique & Examiner Insight

  • Look at the equation coefficients first: In multiple-choice questions, examiners intentionally build distractors around mole ratio errors. Always circle the stoichiometric numbers in the equation before calculating.
  • Quick sanity check: The acid concentration is roughly half the base concentration, and it takes 2 moles of acid per mole of base. Therefore, the base volume must be roughly one-quarter of the acid volume ( 25.0 ÷ 4 ≈ 6.25 cm³ ). Only 5.75 is in this ballpark!
Mark Scheme: Option A [1 mark total]

Topics

Module 2: Foundations in chemistry · 2.1 Atoms and reactions

Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.