AQA A-Level Chemistry Paper 1, 2017: Question 11

Skills & knowledge

• Using two titrations to separate contributions from a salt vs an acid (subtraction strategy).
• Redox stoichiometry with MnO₄⁻ and C₂O₄²⁻ in acidic conditions (heated).
• Acid–base stoichiometry for H₂C₂O₄ with NaOH (2:1 ratio).
• Scaling from 25.0 cm³ aliquot to 250 cm³ and converting moles → mass → %.
• Ligand substitution in complexes; why ΔH is close to 0; geometry & isomerism.

Exam focus

• Track what each titration measures: total ethanedioate vs ethanedioate from the acid only.
• Use the given balanced equations to get mole ratios.
• Write units every line (cm³→dm³) and use correct significant figures.

💡 Key knowledge (what each titration measures)

• KMnO₄ redox titration measures total C₂O₄²⁻ present after acidifying: this includes C₂O₄²⁻ that comes from both Na₂C₂O₄ and H₂C₂O₄ (once deprotonated in acidic medium, it is still the same ethanedioate skeleton being oxidised).
• NaOH titration measures only H₂C₂O₄ present (it’s an acid–base neutralisation). From that you can calculate the moles of C₂O₄²⁻ that originated from the acid.
• Then: (total C₂O₄²⁻) − (C₂O₄²⁻ from acid) = C₂O₄²⁻ from Na₂C₂O₄.

🧠 Exam technique (how to score full marks)

• Do the redox moles first: n = cV with V in dm³ .
• Use the ratio from the equation: 2 MnO₄⁻ : 5 C₂O₄²⁻ so multiply by 5/2 .
• For NaOH: H₂C₂O₄ + 2OH⁻ → C₂O₄²⁻ + 2H₂O , so n(C₂O₄²⁻ from acid) = n(OH⁻)/2 .
• Only at the end: scale ×10 (25.0 cm³ → 250 cm³), then convert to mass, then %.

✅ Correct answer (with accepted range)

Percentage by mass of sodium ethanedioate = 56.6% (3 s.f.)
Mark scheme allows 56.5%–56.8%.

❌ Common errors (that lose marks fast)

• Forgetting cm³ → dm³ (divide volumes by 1000).
• Using the wrong redox ratio: it is 2 MnO₄⁻ to 5 C₂O₄²⁻, so multiply moles of MnO₄⁻ by 5/2.
• Not halving the NaOH moles: H₂C₂O₄ needs 2OH⁻ per 1 acid molecule.
• Subtracting the wrong way round: you must do (total ethanedioate) − (ethanedioate from acid). A negative value signals a logic error.
• Scaling error: 25.0 cm³ to 250 cm³ is ×10 (not ×1000).
• Significant figures: final % should reflect given data (typically 3 s.f.).

🧠 Examiner-style commentary (what distinguishes top answers)

• Top answers make it explicit that the KMnO₄ titration gives total C₂O₄²⁻, while NaOH isolates the contribution from H₂C₂O₄ only, then clearly show the subtraction step.
• Many students lose marks by treating the two titrations as independent and never combining them (or combining without clear reasoning).
• Clear, line-by-line units and ratios helps prevent “carry-forward” mistakes.

✅ Correct equation (1 mark)

[Fe(H₂O)₆]³⁺ + 3C₂O₄²⁻ → [Fe(C₂O₄)₃]³⁻ + 6H₂O

Marking: must show ligand substitution on the hexaaqua iron(III) ion and correct stoichiometry (3 ethanedioate ligands replace 6 waters).

💡 Why ΔH is close to zero (1 mark)

• Approximately the same number and same type of bonds are broken and made (Fe–O bonds to water are replaced by Fe–O bonds to ethanedioate).
• So energy required ≈ energy released → ΔH ≈ 0.

✅ Displayed formula to draw (1 mark)

The complex is [Fe(C₂O₄)₃]³⁻ .

✅ O—Fe—O bond angle (1 mark)

90° or 180°

Reason: octahedral complex → adjacent ligands 90°, opposite 180°.

✅ Type of isomerism (1 mark)

Optical isomerism

Tris-bidentate octahedral complexes like [Fe(C₂O₄)₃]³⁻ can be chiral (non-superimposable mirror images).

💡 Must-know ratios

• 2 MnO₄⁻ : 5 C₂O₄²⁻ (acidic redox)
• H₂C₂O₄ : OH⁻ = 1 : 2 (neutralisation)
• 25.0 cm³ → 250 cm³ is ×10

🧠 High-score structure

• Compute totals → compute acid contribution → subtract → scale up → mass → %.
• Keep each line to one idea, with units.

✅ Final numeric answer to remember

56.6% Na₂C₂O₄ by mass (allowed ~56.5–56.8%).