OCR A-Level Chemistry Unified chemistry (03), June 2019: Question 2

✅ Correct Answer

CO₃²⁻ + H₂O → OH⁻ + HCO₃⁻

Also accepted:
CO₃²⁻ + 2H₂O → 2OH⁻ + H₂CO₃
CO₃²⁻ + H₂O → 2OH⁻ + CO₂

💡 Key Knowledge

Carbonate ions ( CO₃²⁻ ) act as a Brønsted–Lowry base by accepting a proton ( H⁺ ) from water molecules, generating hydroxide ions ( OH⁻ ), which makes the solution alkaline.

❌ Common Errors

• Writing a reaction with acid (e.g. CO₃²⁻ + 2H⁺ → H₂O + CO₂ ) instead of reacting directly with water.
• Writing a full un-ionised equation like Na₂CO₃ + H₂O → 2NaOH + CO₂ when an ionic equation is explicitly requested.

🧠 Exam Technique

• Check atom and charge balances carefully: Left side = (2−) charge; Right side = (1−) + (1−) = (2−).
• State symbols are not explicitly penalised if omitted, but writing them ( aq , l ) prevents balance mistakes.

✅ Correct Answer

• Acid / H⁺ / HCl reacts with OR protonates benzoate ions (or C₆H₅COO⁻ / sodium benzoate) to form benzoic acid.
• Alternatively, give the ionic equation:
  C₆H₅COO⁻ + H⁺ → C₆H₅COOH

💡 Key Knowledge

In alkaline conditions (Step 1), oxidation yields water-soluble sodium benzoate, not free benzoic acid. Adding concentrated HCl protonates benzoate ions. Benzoic acid has low cold-water solubility due to its bulky benzene ring, so it precipitates as crystals.

❌ Common Errors

• Stating only that the acid "neutralises the excess sodium carbonate / alkali". While true, this fails to explain why crystals of benzoic acid appear.
• Vaguely stating "it acts as a catalyst" or "it lowers the pH" without mentioning the benzoate ion or protonation.

🧠 Exam Technique

Whenever an organic salt is converted to an organic acid, use the precise vocabulary: "protonates" or "reacts with benzoate ions to produce benzoic acid".

✅ Correct Answer

C₆H₅CH₂OH + 2[O] → C₆H₅COOH + H₂O

Molecular formula version also allowed:
C₇H₈O + 2[O] → C₇H₆O₂ + H₂O

💡 Key Knowledge

• Refluxing a primary alcohol with excess oxidising agent produces a carboxylic acid + H₂O.
• This requires 2[O] equivalents (one removes 2 hydrogen atoms to form an aldehyde intermediate, the second inserts an oxygen atom).

❌ Common Errors

• Forgetting the by-product H₂O on the right-hand side.
• Writing only 1[O] (which would only oxidise it to benzaldehyde, C₆H₅CHO ).

🧠 Exam Technique

Count hydrogens on both sides:
Left: 5 (ring) + 2 (CH₂) + 1 (OH) = 8.
Right: 5 (ring) + 1 (COOH) + 2 (H₂O) = 8. Balanced!

Step 1: Calculate mass of reactant (phenylmethanol)
mass = volume × density = 4.00 cm³ × 1.04 g cm⁻³ = 4.16 g

Step 2: Calculate theoretical moles of phenylmethanol
Molar mass ( Mᵣ ) of C₆H₅CH₂OH = (7 × 12.0) + (8 × 1.0) + 16.0 = 108.0 g mol⁻¹
theoretical moles = 4.16 / 108.0 = 0.038519... mol (Mark 1)

Step 3: Calculate actual moles of product (benzoic acid)
Molar mass ( Mᵣ ) of C₆H₅COOH = (7 × 12.0) + (6 × 1.0) + 32.0 = 122.0 g mol⁻¹
actual moles = 1.59 g / 122.0 g mol⁻¹ = 0.013033... mol (Mark 2)

Step 4: Calculate percentage yield (3 significant figures)
Reacting ratio is 1:1, so theoretical moles of benzoic acid = 0.038519... mol
% yield = (actual moles / theoretical moles) × 100
% yield = (0.013033... / 0.038519...) × 100 = 33.8% (or 33.9% depending on intermediate rounding) (Mark 3)

❌ Common Calculation Traps

• Ignoring density: Using 4.00 g instead of 4.16 g gives 35.2% (maximum 2 marks awarded).
• Dividing by density: Calculating 4.00 / 1.04 = 3.846 g gives 36.5% or 36.6% (maximum 2 marks awarded).
• Incorrect sig figs: Writing 34% or 33.83% instead of the requested 3 significant figures loses the final mark.

🧠 Alternative Method (Mass-Based)

• Theoretical mass of benzoic acid = 0.038519 mol × 122.0 g mol⁻¹ = 4.70 g
• % yield = (1.59 / 4.70) × 100 = 33.8%
• Keep unrounded numbers in calculator memory to avoid rounding discrepancy.

✅ Correct Answer

Mark 1: Dissolve the impure crystals in the minimum volume of hot solvent (or hot water).

Mark 2: Cool the solution, filter the crystals (under reduced pressure / Buchner funnel), and leave to dry (e.g. in a desiccator or warm oven). (All three steps required for Mark 2).

💡 Why Each Step Is Necessary

• Minimum volume of hot solvent: Ensures solution is saturated at high temperature, maximising crystal yield on cooling.
• Cooling: Benzoic acid is far less soluble in cold water; pure crystals reform leaving soluble impurities in the filtrate.
• Filter & Dry: Separates solid crystals from solvent; drying removes residual moisture.

❌ Common Errors & Examiner Warnings

• Do NOT suggest chemical drying agents: Anhydrous salts like MgSO₄ or CaCl₂ are used to dry organic liquids, not solid crystals! This is an immediate disqualifier.
• Missing out either "minimum" or "hot" when describing dissolving forfeits Mark 1.
• Forgetting to state dry at the end loses Mark 2.

🧠 Exam Technique: The 4-Word Mantra

Remember the recrystallisation sequence with this checklist:

1. Dissolve (minimum volume of hot solvent)
2. Cool (ice bath / room temperature)
3. Filter (Buchner funnel / vacuum)
4. Dry (pat with filter paper / desiccator)