AQA A-Level Chemistry Paper 3, 2019: Question 6

1 mark · Easy difficulty · Multiple Choice

Determine which volume and concentration of sodium hydroxide reacts completely with 7.5 g of a diprotic acid.

Practise this question

Question

Multiple choice question asking which amount of sodium hydroxide would react exactly with 7.5 g of a diprotic acid, H2A (Mr = 150). Four options are provided: A (50 cm3 of 0.05 mol dm-3 NaOH), B (100 cm3 of 0.50 mol dm-3 NaOH), C (100 cm3 of 1.0 mol dm-3 NaOH), and D (100 cm3 of 2.0 mol dm-3 NaOH), each with a selection oval next to it.
Question text

06 Which amount of sodium hydroxide would react exactly with 7.5 g of a diprotic acid,

H2A (Mr = 150)?

[1 mark]

A 50 cm3 of 0.05 mol dm–3 NaOH(aq)

B 100 cm3 of 0.50 mol dm–3 NaOH(aq)

C 100 cm3 of 1.0 mol dm–3 NaOH(aq)

D 100 cm3 of 2.0 mol dm–3 NaOH(aq)

Mark scheme

Show the mark scheme Mark scheme table row showing question number 6 with the correct answer C and 1 mark.

6 C 1

How to answer it

Titration Calculations: Neutralisation of a Diprotic Acid

📋 What this question tests

This multiple-choice question assesses your ability to calculate reacting amounts in solution stoichiometry:

  • Converting mass to moles using molar mass: n = m / Mr
  • Recognising the stoichiometry of a diprotic acid reacting with a monoprotic base (1 : 2 mole ratio)
  • Calculating amounts of substance in solutions: n = c × V

Question 06 Breakdown

Identifying the correct reacting volume and concentration [1 mark]

✅ Correct Answer

C — 100 cm³ of 1.0 mol dm⁻³ NaOH(aq)

Mark Scheme: Award 1 mark for option C.

💡 Key Knowledge

  • Diprotic acid (H₂A): Each mole of H₂A releases 2 moles of H⁺ ions in aqueous solution.
  • Stoichiometric Equation:
    H₂A + 2NaOH → Na₂A + 2H₂O
  • Molar Ratio: 1 mole of diprotic acid requires 2 moles of NaOH for complete neutralisation.

📐 Step-by-Step Calculation

  1. Calculate the moles of diprotic acid (H₂A):
    Moles = mass / Mr
    Moles of H₂A = 7.5 g / 150 g mol⁻¹ = 0.050 mol
  2. Determine the moles of NaOH required:
    Because the acid is diprotic (H₂A), the mole ratio is 1 H₂A : 2 NaOH .
    Moles of NaOH = 0.050 mol × 2 = 0.10 mol
  3. Evaluate each given option to find which provides 0.10 mol of NaOH:
    Option Calculation (n = c × V in dm³) Moles of NaOH Conclusion
    A (50 / 1000) × 0.05 0.0025 mol Incorrect
    B (100 / 1000) × 0.50 0.050 mol Incorrect (Trap!)
    C (100 / 1000) × 1.0 0.10 mol Correct
    D (100 / 1000) × 2.0 0.20 mol Incorrect

❌ Common Traps & Errors

  • The 1:1 Ratio Trap (Option B): Students frequently calculate 0.05 mol of acid and immediately look for 0.05 mol of NaOH. This leads directly to choosing B. Always underline the word "diprotic"!
  • Volume Conversion: Forgetting to divide cm³ by 1000 when using n = c × V .
  • Doubling the Wrong Species (Option D): Some students incorrectly divide the required moles of NaOH by 2 or double the base twice, landing on 0.20 mol.

🧠 Exam Technique

  • Active Reading: Annotate the question prompt immediately. Circle "diprotic" and write "× 2" above it.
  • Pre-compute your target value: Find that you need exactly 0.10 mol NaOH before scanning the options. This stops distractor options from swaying your thinking.
  • Mental Math Shortcut: Notice options B, C, and D all use 100 cm³ (0.1 dm³). To get 0.10 mol from 0.1 dm³, the concentration must simply be 0.10 / 0.1 = 1.0 mol dm⁻³ .

Topics

Physical Chemistry · 3.1.2 Amount of Substance

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.