Edexcel A-Level Chemistry Paper 1, June 2017: Question 8

✅ Correct Answer: C (3.4)

Diluting a weak acid by a factor of 100 increases its pH by 1 unit (since [H⁺] decreases by a factor of 10, following pH = -log[H⁺]). Wait—let's check carefully using the dissociation approximation: [H⁺] = √(Ka × [HA]) . If concentration drops by a factor of 100 (10⁻²), the square root means [H⁺] drops by a factor of 10 ( √10⁻² = 10⁻¹ ). Therefore, pH increases by 1 unit from 2.4 to 3.4.

❌ Common Errors

Selecting A (1.4)—mistaking dilution for an increase in concentration (100-fold increase). Selecting B (2.4)—assuming no change in pH upon dilution.

📐 Step-by-Step Calculation

1. Find [H⁺] from pH:
   [H⁺] = 10⁻ᵖᴴ = 10⁻²·⁰ = 0.010 mol dm⁻³ (1 mark)
2. State the Ka expression:
   Ka = [H⁺]² / [CH₂OHCOOH] (assuming [H⁺] = [A⁻] ) (1 mark)
3. Rearrange and calculate concentration:
   [CH₂OHCOOH] = (0.010)² / (1.5 × 10⁻⁴) = 0.667 mol dm⁻³ (1 mark)

🧠 Exam Technique & Mark Scheme Guidance

• Allow [HA] in place of the formula.
• Accept 0.67 mol dm⁻³ . Ignore units if omitted, but always include them for good practice.
• A correct final answer with no working scores all 3 marks, but showing steps guarantees partial credit if an arithmetic slip occurs.

💡 Key Knowledge: Indicator Requirements

• Mark 1: Name a suitable indicator (e.g., Phenolphthalein or Phenol red).
• Mark 2: State the exact matching colour change (e.g., phenolphthalein: pink/red to colourless, or phenol red: red to yellow). Do not allow "clear".
• Mark 3: Justify selection by stating that the indicator's pH range lies entirely within the vertical/steep region of the titration curve.

❌ Common Pitfalls

Choosing an indicator like methyl orange (pH range 3.1–4.4), which changes colour way too early in the acidic buffer region before the steep equivalence point vertical jump (~pH 7 to 11).

✅ Part (ii) Answer: C (0.125 mol dm⁻³)

Using c₁V₁ = c₂V₂ or moles relationship: At equivalence/inflection, read the volume of glycolic acid added from the x-axis (20.0 cm³) against 25.0 cm³ of 0.100 mol dm⁻³ NaOH.
Calculations: Moles NaOH = 25 × 0.100 / 1000 = 0.0025 mol .
Conc acid = 0.0025 / 0.0200 = 0.125 mol dm⁻³ .

✅ Part (iii) Answer: C (8.3)

At the equivalence point, the solution contains sodium glycolate (salt of a weak acid and strong base). Hydrolysis of the glycolate ion makes the solution alkaline. Looking at the steep vertical inflection midpoint or salt pH, 8.3 is the correct alkaline pH value.

💡 Key Knowledge: Inductive Effects

• Mark 1: The oxygen atom of the extra hydroxyl ( -OH ) group in the alpha position (-CH₂OH) is electronegative and withdraws/attracts electrons via the inductive effect.
• Mark 2: This negative inductive effect stabilises the resulting carboxylate anion ( CH₂OHCOO⁻ ) by spreading/delocalising the negative charge, weakening the O-H bond and making the proton easier to lose.

🧠 Top-Level Response Markers

To secure both marks, candidates must explicitly link the electronegative oxygen / electron-withdrawing group to the stabilisation of the anion, rather than just stating general differences.

✅ Correct Equation

CH₂OHCOOH + CH₃COOH → CH₂OHCOO⁻ + CH₃COOH₂⁺

🧠 Mark Scheme Details

• 1 mark for both correct species on the product side.
• Allow formulae in either order.
• Make sure charges are clearly written as superscripts on the right side of the ions ( CH₂OHCOO⁻ and CH₃COOH₂⁺ ).