AQA A-Level Chemistry Paper 1, 2019: Question 5

✅ Correct Answers

Equation:

NaCl + H₂SO₄ → NaHSO₄ + HCl

(Also allowed: 2NaCl + H₂SO₄ → Na₂SO₄ + 2HCl)

Role of sulfuric acid:

Proton donor (or Brønsted-Lowry acid / acid)

💡 Key Knowledge

• Chloride ions (Cl⁻) are too weak a reducing agent to reduce the sulfur in H₂SO₄ (S remains at +6).
• This is strictly an acid-base reaction, not a redox reaction. Concentrated H₂SO₄ donates a proton (H⁺) to Cl⁻ to produce misty white fumes of HCl gas.

❌ Common Errors

• Stating the role is an "oxidising agent" or "catalyst" (0 marks). Cl⁻ cannot reduce concentrated sulfuric acid.
• Trying to balance the equation by including Cl₂ or SO₂.

Mark Breakdown:
• 1 mark for the balanced equation.
• 1 mark for stating "proton donor" or "acid".

✅ Correct Answers

Equation (any one of the following):

2NaBr + 2H₂SO₄ → Na₂SO₄ + SO₂ + Br₂ + 2H₂O

2NaBr + 3H₂SO₄ → 2NaHSO₄ + SO₂ + Br₂ + 2H₂O

2H⁺ + 2Br⁻ + H₂SO₄ → SO₂ + Br₂ + 2H₂O

Observation: Brown gas (or orange gas / brown fumes / orange fumes)

Role of sulfuric acid: Oxidising agent (or electron acceptor)

🧠 Exam Technique: Half-Equations

Derive the full redox equation easily using half-equations:

1. Oxidation: 2Br⁻ → Br₂ + 2e⁻

2. Reduction: H₂SO₄ + 2H⁺ + 2e⁻ → SO₂ + 2H₂O

Combining gives: 2H⁺ + 2Br⁻ + H₂SO₄ → SO₂ + Br₂ + 2H₂O

Then add spectator ions (Na⁺) if writing a full molecular equation.

❌ Common Errors & Examiner Notes

• Wrong observation: The question asks for the reaction producing SO₂. Do not write "misty fumes" or "steamy fumes" (those come from the acid-base step forming HBr, not the redox step forming SO₂).
• Do not write "yellow solid" (that is sulfur, S, which is formed with iodide, not bromide).
• Writing "acid" for the role of sulfuric acid scores 0 here—the question specifically refers to the redox step forming SO₂ where S goes from +6 to +4.

Mark Breakdown:
• 1 mark for balanced redox equation.
• 1 mark for "brown fumes/gas" or "orange fumes/gas".
• 1 mark for "oxidising agent" or "electron acceptor".

✅ Correct Answers

05.3 Oxidation States:

• In NaClO₃: +5 (or 5)
• In NaCl: -1

05.4 Redox Definition:

Chlorine is oxidised and reduced (or undergoes disproportionation / gains and loses electrons).

📐 Working Out Oxidation States

In NaClO₃:

• Na is Group 1: +1
• Oxygen is always -2 in compounds (except peroxides/OF₂): 3 × (-2) = -6
• Overall neutral: (+1) + Cl + (-6) = 0 ⇒ Cl = +5

In NaCl:

• Na = +1 ⇒ Cl = -1

❌ Common Errors

• Forgetting the sign: Always include the negative sign for -1. (Writing just "1" will be penalised).
• Incomplete statement in 05.4: Stating only that chlorine is "oxidised" or only that it is "reduced" gains 0 marks. You must state both, or use the term disproportionation.

Mark Breakdown:
• 05.3: 1 mark for both oxidation states correct (+5 and -1).
• 05.4: 1 mark for "oxidised and reduced" / "disproportionation".

✅ Correct Identities & Formulas

• Formula of D: AgBr [1 mark]
• Formula of E: Ag₂CO₃ [1 mark]
• Formula of F: CO₂ [1 mark]

✅ Required Equations

Ionic equation to form E:

2Ag⁺ + CO₃²⁻ → Ag₂CO₃

(1 mark: must be ionic and balanced)

Equation for conversion of D into G:

AgBr + 2NH₃ → [Ag(NH₃)₂]⁺ + Br⁻

(Also allowed: AgBr + 2NH₃ → Ag(NH₃)₂Br)

(2 marks: 1 mark for formula of [Ag(NH₃)₂]⁺, 1 mark for balanced equation)

❌ Common Traps & Where Marks Were Lost

• Incorrect charge on silver carbonate: Writing AgCO₃ instead of Ag₂CO₃ (Ag is +1, CO₃ is 2-).
• Not providing an ionic equation: For the formation of E, writing full spectator salts like 2AgNO₃ + Na₂CO₃ → Ag₂CO₃ + 2NaNO₃ will not score the mark. Only reacting ions should be shown: 2Ag⁺ + CO₃²⁻ → Ag₂CO₃ .
• Coordination number error: Ammonia acts as a monodentate ligand; silver(I) takes a coordination number of 2, giving [Ag(NH₃)₂]⁺, NOT [Ag(NH₃)₄]⁺ or [Ag(NH₃)]⁺.

Mark Breakdown (6 Marks Total):
• 1 mark for D (AgBr)
• 1 mark for E (Ag₂CO₃)
• 1 mark for F (CO₂)
• 1 mark for ionic equation forming E
• 2 marks for equation converting D to G (1 mark for [Ag(NH₃)₂]⁺, 1 mark for balanced equation).