AQA A-Level Chemistry Paper 1, 2019: Question 6

Step 1: Calculate moles of thiosulfate used in the titration
Moles of S₂O₃²⁻ = (volume × concentration) ÷ 1000
Moles of S₂O₃²⁻ = (9.00 × 0.0800) ÷ 1000 = 7.20 × 10⁻⁴ mol

Step 2: Determine the molar ratio between Cu²⁺ and S₂O₃²⁻
From Equation 1: 2Cu²⁺ produces 1 I₂
From Equation 2: 1 I₂ reacts with 2 S₂O₃²⁻
Therefore: 2 moles of Cu²⁺ ≡ 2 moles of S₂O₃²⁻, which simplifies to a 1 : 1 stoichiometric ratio.
Moles of Cu²⁺ in 25.0 cm³ sample = 7.20 × 10⁻⁴ mol

Step 3: Scale up to the total volumetric flask volume (250 cm³)
Scaling factor = 250 cm³ ÷ 25.0 cm³ = 10
Moles of Cu²⁺ in 250 cm³ = 7.20 × 10⁻⁴ × 10 = 7.20 × 10⁻³ mol

Step 4: Calculate mass of copper in the alloy
Atomic mass of Cu = 63.5 g mol⁻¹
Mass of Cu = 7.20 × 10⁻³ × 63.5 = 0.4572 g (0.457 g)

Step 5: Convert sample mass of alloy to matching units (mg to g)
Mass of alloy sample = 985 mg = 985 ÷ 1000 = 0.985 g

Step 6: Calculate percentage by mass to the correct significant figures
% Cu = (0.4572 ÷ 0.985) × 100 = 46.4%
Significant figures rule: The data given includes 985 mg (3 sf), 250 cm³ (3 sf), 25.0 cm³ (3 sf), 9.00 cm³ (3 sf), 0.0800 mol dm⁻³ (3 sf). The final answer must be quoted to 3 significant figures.

❌ Common Errors & Pitfalls

• Incorrect Ratio: Assuming Cu²⁺ to S₂O₃²⁻ is 2:1 or 1:2 instead of looking at the linking iodine (I₂) intermediary which gives a net 1:1 ratio.
• Unit Inconsistency: Dividing mass in grams (0.457 g) directly by mass in milligrams (985 mg) yielding an answer of 0.0464%.
• Forgetting the Aliquot: Forgetting to multiply by 10 to scale from the 25.0 cm³ conical flask portion to the 250 cm³ volumetric flask.
• Rounding Off Early: Rounding intermediate values can lead to final percentages outside the acceptable range.

🧠 Exam Technique

• Always highlight the phrase "appropriate number of significant figures". Trace each piece of data in the prompt: all quantities have 3 sf, so your final answer MUST have 3 sf.
• If you realise you made a unit slip (e.g. your percentage is over 100% or under 1%), check your unit conversions between mg and g first.

✅ Acceptable Answers (Any TWO)

• Use more of the alloy (a larger mass of alloy).
• Use a lower concentration of the sodium thiosulfate solution (or use a lower mass of Na₂S₂O₃ to make the solution).

💡 Key Knowledge

Percentage uncertainty is given by:

% uncertainty = (apparatus uncertainty ÷ titre volume) × 100

Because the prompt specifies "using the same apparatus", you cannot suggest switching to a more precise burette or micro-pipette. The only way to lower percentage uncertainty is to increase the titre volume.

❌ Common Errors

• Suggesting "repeat and calculate a mean" (repeats reduce random error and identify anomalies, but do not change apparatus uncertainty).
• Suggesting a "more precise burette / balance" (contradicts the constraint: "using the same apparatus").
• Suggesting to "use a higher concentration of thiosulfate" (this would make the titre even smaller, increasing uncertainty!).

🧠 Examiner Insight

The titre was only 9.00 cm³, which is relatively small for a 50 cm³ burette. Increasing the titre towards 25–30 cm³ significantly reduces percentage error. Always think: How do I make the burette volume larger? (More analyte or more dilute titrant).

✅ Correct Answer

Oxidising agent (or electron acceptor)

💡 Key Knowledge

Reaction: 2S₂O₃²⁻ + I₂ → 2I⁻ + S₄O₆²⁻

• Iodine changes oxidation state from 0 in I₂ to -1 in I⁻.
• Because iodine gains electrons and decreases in oxidation state, it is reduced.
• A species that is reduced acts as an oxidising agent.

✅ Correct Answer

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁹

❌ Common Errors

• Writing [Ar] 3d⁹ — the question explicitly asks for the full electron configuration, so noble gas shorthand is rejected.
• Writing ... 3d⁷ 4s² — electrons in the 4s subshell are always removed before 3d electrons when forming transition metal ions.
• Confusing Cu²⁺ (3d⁹) with Cu⁺ (3d¹⁰) or neutral Cu ([Ar] 3d¹⁰ 4s¹).

✅ Correct Answer & Marking Points

• M1: Copper(I) has a full (3)d subshell / full (3)d shell / has a 3d¹⁰ configuration.
• M2: No (d-to-d) electron transitions are possible / it cannot absorb visible light (or white light).

💡 Key Knowledge

For a transition metal complex/compound to be coloured:

• Ligands cause the d-orbitals to split into two different energy levels (ΔE).
• Electrons absorb frequencies of visible light corresponding to ΔE ( ΔE = hν ) to promote an electron from a lower d-orbital to a higher, partially filled d-orbital (d-d transition).
• In Cu⁺ ( 3d¹⁰ ), the d-orbitals are completely filled. There is no empty orbital for an excited electron to move into, so no visible light is absorbed, and the compound appears white/colourless.

Step 1: Calculate moles of iodine vapour (I₂)
Iodine vapour is diatomic (I₂), so Mᵣ = 2 × 126.9 = 253.8 g mol⁻¹
n = mass ÷ Mᵣ = 5.00 ÷ 253.8 = 0.01970 mol (allow 5.00 / 254 = 0.01969 mol)

Step 2: Convert standard units for temperature and pressure
Temperature: T = 185 °C + 273 = 458 K
Pressure: P = 100 kPa = 100 × 10³ Pa = 100 000 Pa

Step 3: Rearrange ideal gas equation and solve for volume in m³
pV = nRT ⇒ V = nRT ÷ p
V = (0.01970 × 8.31 × 458) ÷ 100 000 = 7.498 × 10⁻⁴ m³ (or 7.50 × 10⁻⁴ m³)

Step 4: Convert volume from m³ to cm³ and round to 3 significant figures
1 m³ = 10⁶ cm³ ⇒ Multiply by 1 000 000
V = 7.498 × 10⁻⁴ × 10⁶ = 750 cm³ (allow 749 cm³)

❌ Common Errors in Question 06.6

• Using Monatomic Iodine: Dividing by 126.9 instead of 253.8 (loses M1, though subsequent marks can be awarded by error carried forward).
• Unit Conversion Traps:
  • Leaving pressure as 100 instead of converting to Pa (100 000 Pa).
  • Multiplying m³ by 10³ instead of 10⁶ to get cm³. (Remember: 1 m³ = 1 000 dm³ = 1 000 000 cm³).
• Significant Figures: Writing "750" is 3 sf when written as 750 or 7.50 × 10², but omitting units or quoting to 2 sf (e.g. 750 without showing working or 0.75) loses marks.

🧠 Summary Table of SI Gas Equation Units

Term	Required Unit	Conversion
p	Pascals (Pa)	kPa × 10³
V	Cubic metres (m³)	cm³ ÷ 10⁶
T	Kelvin (K)	°C + 273