AQA A-Level Chemistry Paper 3, 2019: Question 2

✅ Correct Structure

Sulfur at the centre formed of 6 covalent bonds:

• Two double bonds to separate oxygen atoms: S=O and S=O
• Two single bonds to separate hydroxyl groups: S–O–H and S–O–H
• Every single bond must be shown explicitly, including both O–H bonds!

❌ Common Errors & Pitfalls

• Writing O–H as -OH: A displayed formula requires all bonds shown. Missing the bond between O and H loses the mark instantly.
• Lone pairs on Sulfur: Do NOT put a lone pair on the sulfur atom. Sulfur uses all 6 of its valence electrons to bond (expanding its octet to 12 electrons).
• Note: Lone pairs on oxygen atoms and bond angles are ignored by the examiner, but keeping them neat is good practice.

✅ Correct Equations

Equation 1 (Strong acid behaviour – full dissociation):

H₂SO₄ → HSO₄⁻ + H⁺

or: H₂SO₄ + H₂O → HSO₄⁻ + H₃O⁺

Equation 2 (Weak acid behaviour – partial dissociation):

HSO₄⁻ ⇌ SO₄²⁻ + H⁺

or: HSO₄⁻ + H₂O ⇌ SO₄²⁻ + H₃O⁺

🧠 Exam Technique: Watch the Arrows

• The prompt explicitly tells you: "Include appropriate arrows in your equations." This is your clue that arrows are strictly assessed!
• Equation 1: Must use a forward arrow (→). A reversible arrow (⇌) is strictly rejected because H₂SO₄ is a strong acid that dissociates fully.
• Equation 2: Must use a reversible equilibrium arrow (⇌). A single-headed arrow (→) is strictly rejected because HSO₄⁻ is a weak acid.
• State symbols are not required and can be safely ignored.

✅ Standard Practical Procedure (4 Marks)

1. M1 (Accurate weighing): Weigh the solid and transfer using a method that allows the exact mass to be known (e.g. weighing bottle by difference, or zeroing/taring the balance before transfer with washings).
2. M2 (Dissolving): Dissolve the solid in deionised/distilled water in a beaker or suitable container using less than 250 cm³ of water.
3. M3 (Transfer & Washings): Transfer the solution into a 250 cm³ volumetric flask (or graduated flask) with washings of the beaker, funnel, and stirring rod.
4. M4 (Make up & Invert): Make up to the graduation mark with deionised water (until the bottom of the meniscus is on the line), then stopper and invert/shake thoroughly to mix.

❌ Key Misconceptions & Lost Marks

• "Dissolve in 250 cm³ of water": Immediate loss of M2! If you add 250 cm³ of water to dissolve the solid, the final volume will exceed 250 cm³.
• Forgetting rinsing/washings: Without rinsing the beaker and stirring rod into the volumetric flask, trace solute remains behind.
• Using a beaker or conical flask for the final volume: Only a volumetric flask has the required precision.
• Forgetting to invert/shake: Adding water to the mark is not enough; the solution must be homogenised.

📐 Step-by-Step Calculation

Step 1: Calculate [H⁺] from pH (M1)

[H⁺] = 10^-pH = 10^-1.72 = 0.01905 mol dm⁻³

Step 2: Find moles of NaHSO₄ (M2)

Convert mg to g: 605 mg = 0.605 g

M_r(NaHSO₄) = 23.0 + 1.0 + 32.1 + 4(16.0) = 120.1

Moles = 0.605 / 120.1 = 5.0375 × 10⁻³ mol

Step 3: Calculate initial concentration in 100 cm³ (M3)

[HSO₄⁻]_initial = (5.0375 × 10⁻³ mol) × (1000 / 100) = 0.050375 mol dm⁻³

Step 4: Equilibrium concentrations & K_a expression (M4)

At equilibrium: [SO₄²⁻] = [H⁺] = 0.01905 mol dm⁻³

[HSO₄⁻]_eqm = [HSO₄⁻]_initial - [H⁺] = 0.050375 - 0.01905 = 0.031325 mol dm⁻³

K_a = ([H⁺][SO₄²⁻]) / [HSO₄⁻]_eqm = (0.01905)² / 0.031325

Step 5: Calculate K_a to 3 significant figures (M5)

K_a = 1.16 × 10⁻² mol dm⁻³ (Accept range: 1.15 × 10⁻² to 1.18 × 10⁻²)

Step 6: State the units (M6)

Units = (mol dm⁻³ × mol dm⁻³) / (mol dm⁻³) = mol dm⁻³

🧠 Examiner Insight & Common Traps

• The mg trap: Students forget to divide 605 mg by 1000 to convert to grams.
• Volume scaling: The solution volume is 100 cm³, NOT 1000 cm³ or 250 cm³. Moles must be multiplied by 10 to find concentration in mol dm⁻³.
• The Weak Acid Approximation: Usually for weak acids, we assume [HA]_eqm ≈ [HA]_initial . If a student uses this simplification:
  K_a = (0.01905)² / 0.0504 = 7.21 × 10⁻³ mol dm⁻³ .
  The mark scheme generously allows this for full credit across M1–M5 (range 7.15–7.26 × 10⁻³), but subtracting [H⁺] is the rigorous method!
• Significant figures: The question asks explicitly for three significant figures. Giving 2 sf or 4 sf forfeits mark M5.

✅ Model Explanation (2 Marks)

1. M1 (Equilibrium shift): Adding sodium sulfate increases the concentration of sulfate ions, [SO₄²⁻]. The equilibrium:
   HSO₄⁻ ⇌ SO₄²⁻ + H⁺
   shifts to the left (to oppose/counteract the increase in [SO₄²⁻]).
2. M2 (Change in [H⁺]): As the equilibrium shifts left, [H⁺] decreases (or H⁺ ions are consumed/react), which causes the pH to increase ( pH = -log₁₀[H⁺] ).

🧠 Examiner Notes

• Identify the correct ion: Specify that it is the added SO₄²⁻ (or sulfate) ion causing the shift. Do not refer vaguely to "sodium sulfate" shifting the equilibrium.
• Link [H⁺] directly to pH: To gain the second mark, explicitly connect the leftward shift to a decrease in hydrogen ion concentration, [H⁺]. Remember: lower [H⁺] means higher pH!
• Marks M1 and M2 are independent — you can still gain M2 if you correctly identify that [H⁺] decreases.