AQA A-Level Chemistry Paper 3, June 2022: Question 7

✅ Correct Answer: C (HOOCCOOH)

The correct option is C — ethanedioic acid (oxalic acid).

• Carbon count: 1 mole of acid gives 2 moles of CO₂, meaning each molecule contains exactly 2 carbon atoms.
• Acidic protons: 1 mole of acid reacts with 2 moles of NaOH, meaning each molecule contains 2 –COOH groups (it is diprotic).
• HOOCCOOH contains 2 carbons and 2 carboxylic groups, satisfying both pieces of evidence.

📐 Step-by-Step Deduction

1 Find number of carbon atoms per molecule:
Moles of acid = 0.0100 mol
Moles of CO₂ = 0.0200 mol
Ratio of C : Acid = 0.0200 / 0.0100 = 2 carbons per molecule.
Eliminates A (1 carbon) and D (3 carbons).

2 Find moles of NaOH used:
Volume = 20 cm³ = 20 / 1000 = 0.020 dm³
Concentration = 1.00 mol dm⁻³
Moles of NaOH = 0.020 dm³ × 1.00 mol dm⁻³ = 0.0200 mol

3 Determine proticity (number of –COOH groups):
Moles of NaOH : Moles of Acid = 0.0200 : 0.0100 = 2 : 1
Reaction equation: H₂A + 2NaOH → Na₂A + 2H₂O
The acid must be a dicarboxylic acid. Between B (monoprotic) and C (diprotic), only C is correct.

💡 Key Knowledge

• Combustion ratio: In complete combustion, all carbon in the organic compound converts to CO₂:
  C_nH_mO_x + ... → n CO₂ . Therefore, n = moles of CO₂ / moles of compound .
• Carboxylic acid neutralisation: Each –COOH group supplies 1 H⁺ ion that reacts with 1 OH⁻ ion:
  R–COOH + OH⁻ → R–COO⁻ + H₂O .
• Monocarboxylic acids react in a 1 : 1 ratio with NaOH.
• Dicarboxylic acids react in a 1 : 2 ratio with NaOH.

🧠 Exam Technique & Strategy

• Two-step elimination: Use each piece of information sequentially to eliminate half the options:
  • Step 1 (Combustion): Eliminates A (1 carbon) and D (3 carbons). Remaining: B and C.
  • Step 2 (Titration): Eliminates B (monoprotic). Remaining: C.
• Quick mental math: In multiple-choice questions, spot the 1 : 2 ratio immediately: 0.0100 : 0.0200 means double the moles of both CO₂ and NaOH!

❌ Common Errors & Pitfalls

• Selecting B (CH₃COOH): Forgetting to test the titration data. Ethanoic acid has 2 carbons, but it is monoprotic and would only require 0.0100 mol of NaOH (10 cm³), not 20 cm³.
• Selecting D (HOOCCH₂COOH): Correctly recognising that a dicarboxylic acid is required, but failing to check carbon stoichiometry (propandioic acid has 3 carbons and would produce 0.0300 mol CO₂).
• Volume conversion omission: Forgetting to divide 20 cm³ by 1000 when calculating moles from concentration ( n = c × V ).