OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), November 2020: Question 18

✅ Correct Answer

Equation: Mg + 2CH₃COOH → (CH₃COO)₂Mg + H₂

Oxidation: Mg changes from 0 to +2

Reduction: H (in acid) changes from +1 to 0

💡 Key Knowledge

• Acids react with reactive metals to form a salt and hydrogen gas.
• Oxidation is an increase in oxidation number (loss of electrons).
• Reduction is a decrease in oxidation number (gain of electrons).

🧠 Exam Technique

Make sure you explicitly link the change in oxidation number to the specific element. State both the initial and final oxidation states clearly to secure independent marks.

❌ Common Errors

Students often forget to balance the equation (missing the 2 in front of CH₃COOH ) or incorrectly state that the whole molecule is oxidized/reduced rather than the specific element.

✅ Correct Answer

HCOOH + CH₃COOH ⇌ HCOO⁻ + CH₃COOH₂⁺

Pair 1: HCOOH (A1) and HCOO⁻ (B1)

Pair 2: CH₃COOH₂⁺ (A2) and CH₃COOH (B2)
(Note: Labels A1/A2 and B1/B2 can be interchanged as long as matched correctly).

💡 Key Knowledge

The stronger acid (higher Kₐ , which is HCOOH ) donates a proton to the weaker acid ( CH₃COOH ), acting temporarily as a base. Charges must balance correctly on the products.

🧠 Exam Technique

Check the Kₐ values in Table 18.1 first! HCOOH has a higher Kₐ ( 1.82 × 10⁻⁴ ) than CH₃COOH ( 1.78 × 10⁻⁵ ), meaning methanoic acid acts as the acid here.

❌ Common Errors

Omitting or misplacing charges on ionic species (e.g., writing HCOO instead of HCOO⁻ ) will immediately lose the mark for the equilibrium products.

✅ Correct Answer

Concentration shown to be 0.204 mol dm⁻³

📐 Step-by-Step Calculation

1. Find [H⁺]:
   [H⁺] = 10⁻ᵖᴴ = 10⁻²·⁷² = 1.905 × 10⁻³ mol dm⁻³
2. Rearrange Kₐ expression for [HA]:
   Kₐ = [H⁺][CH₃COO⁻] / [CH₃COOH]
   Since [H⁺] = [CH₃COO⁻] :
   Kₐ = [H⁺]² / [CH₃COOH]
3. Calculate [CH₃COOH]:
   [CH₃COOH] = (1.905 × 10⁻³)² / (1.78 × 10⁻⁵) = 0.204 mol dm⁻³

🧠 Exam Technique

When a question says "Show that...", your working must be rigorous and clear. Do not round intermediate values too early on your calculator.

❌ Common Errors

Forgetting to square the [H⁺] term in the rearranged Kₐ expression is the most common pitfall.

✅ Correct Answer

Required concentration = 2.4 × 10⁻² mol dm⁻³ (or 0.024 mol dm⁻³ )

📐 Step-by-Step Calculation

1. Calculate [H⁺] in the buffer:
   [H⁺] = 10⁻⁴·⁰⁰ = 1.00 × 10⁻⁴ mol dm⁻³
2. <Calculate moles of weak acid (CH₃COOH) mixed:
   n = (0.204 × 400) / 1000 = 8.16 × 10⁻² mol
3. Rearrange Kₐ to find salt concentration in the buffer mixture:
   [CH₃COO⁻] = (Kₐ × [CH₃COOH]) / [H⁺]
   [CH₃COO⁻] = (1.78 × 10⁻⁵ × 8.16 × 10⁻²) / (1.00 × 10⁻⁴) = 0.0145 mol dm⁻³
4. Scale up to find original concentration in 600 cm³:
   Moles in buffer = 0.0145 × (1000/1000) = 0.0145 mol (in 1 dm³ total volume, or use dilution factor)
   Original concentration = (1.45248 × 10⁻² × 1000) / 600 = 2.4 × 10⁻² mol dm⁻³

🧠 Exam Technique

Total volume after mixing is 400 cm³ + 600 cm³ = 1000 cm³ (1 dm³) . Recognising this shortcut saves valuable time, though scaling via moles is completely foolproof.

❌ Common Errors

Inverting dilution factors (using 400 instead of 600 for the final salt solution step) or confusing acid and hydrogen ion concentrations in the Kₐ expression.