OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2022: Question 4

1 mark · Medium difficulty · Multiple Choice

Calculate the concentration of hydrogen peroxide in a disinfectant using titration data and a balanced redox equation, choosing from multiple-choice options A to D.

Practise this question

Question

Multiple-choice question 4 shows the oxidation of hydrogen peroxide by manganate(VII) ions with the equation 2MnO4-(aq) + 5H2O2(aq) + 6H+(aq) -> 2Mn2+(aq) + 5O2(g) + 8H2O(l). It states that 25.00 cm3 of disinfectant reacts with 22.00 cm3 of 0.125 mol dm-3 KMnO4(aq), and asks for the concentration of H2O2 in mol dm-3 from options A (0.0440), B (0.110), C (0.275), and D (0.550).
Question text

4 Hydrogen peroxide, H2O2, can be oxidised by manganate(VII) ions under acid conditions as

shown below.

2MnO –(aq) + 5H O (aq) + 6H+(aq) 2Mn2+(aq) + 5O (g) + 8H O(l)

42 2 2 2

In a titration, 25.00 cm3 of a disinfectant containing hydrogen peroxide reacts with 22.00 cm3 of

0.125 mol dm–3 KMnO (aq).

What is the concentration of H O , in mol dm–3, in the disinfectant?

Assume that KMnO4 only reacts with H2O2 in the disinfectant.

A 0.0440

B 0.110

C 0.275

D 0.550

Your answer [1]

Mark scheme

Show the mark scheme The mark scheme table indicates the correct answer for question 4 is C, with a mark value of 1.

4 C 1 2.8

How to answer it

Redox Titration Calculation

What this question tests

This question assesses your ability to use stoichiometry from a balanced redox equation to calculate the concentration of a solution from titration data. Key skills include calculating moles from volume and concentration, applying stoichiometric molar ratios (2:5), and rearranging concentration formulae.

Question 4 Analysis

Multiple-Choice Redox Titration

✅ Correct Answer: C (0.275)

Option C is the correct concentration of the hydrogen peroxide solution.

💡 Key Knowledge

  • Relating reacting volumes and concentrations using Moles = Concentration × Volume .
  • Using balanced ionic equations to determine reacting mole ratios ( 2 MnO₄⁻ : 5 H₂O₂ ).
  • Converting volumes from cm³ to dm³ by dividing by 1000.

🧠 Exam Technique

Never assume a 1:1 molar ratio just because numbers aren't written in front of reactants. Always inspect the stoichiometry of the provided balanced equation carefully before calculating moles.

❌ Common Errors

  • Forgetting to apply the 2:5 stoichiometric ratio, leading to option B (0.110).
  • Failing to convert cm³ to dm³, resulting in magnitude errors of 10⁻³.

📐 Step-by-Step Calculation Guide

  1. Calculate moles of KMnO₄ added:
    Moles = Concentration × Volume (in dm³)
    Moles of MnO₄⁻ = 0.125 × (22.00 / 1000) = 0.00275 mol
  2. Use the stoichiometric ratio to find moles of H₂O₂:
    From the equation, 2 moles of MnO₄⁻ react with 5 moles of H₂O₂.
    Ratio is 2 : 5 (or factor of 2.5).
    Moles of H₂O₂ = 0.00275 × (5 / 2) = 0.006875 mol
  3. Calculate the concentration of H₂O₂:
    Concentration = Moles / Volume (in dm³)
    Concentration = 0.006875 / (25.00 / 1000) = 0.275 mol dm⁻³
Examiner Insight: Distractor B (0.110) is a very common trap representing students who correctly calculated the moles of manganate(VII) but incorrectly treated the reaction as a 1:1 ratio instead of applying the 2:5 ratio from the balanced equation.

Topics

Module 2: Foundations in chemistry · Practical Activity Groups · 2.1 Atoms and reactions · PAG 2: Acid-base titration

Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.