OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2022: Question 19

✅ Correct Answers

• Completed Table (Titres): Trial = 25.25 | Titre 1 = 22.75 | Titre 2 = 22.70 | Titre 3 = 22.55 cm³
• Mean Titre: 22.50 cm³ (using Titres 2 and 3 that are within 0.10 cm³ of each other).

❌ Common Errors

• Including the trial titration in the mean calculation (the trial is always exploratory and discarded).
• Failing to write 22.50 to 2 decimal places (missing the trailing zero loses communication marks where 2 DP is required for burette readings).

📐 Step-by-Step Calculation

1. Moles of KOH in titre:
   0.0600 mol dm⁻³ × (22.50 / 1000) = 1.35 × 10⁻³ mol
2. Moles of acid in 10.0 cm³ portion:
   Since equation shows 1:2 ratio, divide by 2:
   1.35 × 10⁻³ / 2 = 6.75 × 10⁻⁴ mol
3. Moles of acid in 250 cm³ volumetric flask:
   6.75 × 10⁻⁴ × (250 / 10.0) = 0.016875 mol
4. Mass of acid in 4 tablets:
   Molar mass of HOOC(CH₂)₂COOH = 118.0 g mol⁻¹.
   Mass = 0.016875 × 118 = 1.99125 g
5. Mass in ONE tablet (in mg):
   (1.99125 / 4) × 1000 = 497.8... mg → Round to 3 SF: 498 mg

🧠 Exam Technique & Traps

• Reacting Ratios: Do not forget to account for the 1:2 stoichiometric ratio between the dibasic succinic acid and KOH.
• Significant Figures: The final answer must be given to exactly 3 significant figures as requested ( 498 mg ).
• Unit Conversions: Watch out for grams to milligrams conversion (× 1000) and scaling up from 10 cm³ to 250 cm³ (× 25).

💡 Key Knowledge

A buffer solution forms because glycolic acid (a weak acid) is present in excess when reacted with potassium hydroxide (a strong base). This results in a mixture containing unreacted weak acid and its conjugate base (glycolate ions, HOCH₂COO⁻ ).

✅ Acceptable Terminology

• Glycolic acid is in excess / partially neutralised.
• Glycolate / potassium glycolate ions are present or produced.

📐 Step-by-Step Calculation

1. Initial moles of acid (HA):
   0.750 mol dm⁻³ × (60.0 / 1000) = 0.0450 mol
2. Initial moles of base (OH⁻):
   0.625 mol dm⁻³ × (40.0 / 1000) = 0.0250 mol
3. Moles remaining in buffer solution (total volume = 100 cm³):
   Remaining HA = 0.0450 - 0.0250 = 0.0200 mol
   Produced A⁻ = 0.0250 mol
4. Find Ka from pKa:
   Ka = 10⁻³·⁸³ = 1.479 × 10⁻⁴ mol dm⁻³
5. Calculate [H⁺] using buffer expression:
   [H⁺] = Ka × (n(HA) / n(A⁻))
   [H⁺] = 1.479 × 10⁻⁴ × (0.0200 / 0.0250) = 1.183 × 10⁻⁴ mol dm⁻³
6. Calculate pH:
   pH = -log[H⁺] = -log(1.183 × 10⁻⁴) = 3.93 (to 2 decimal places)

❌ Common Errors & Examiner Notes

• Ratio mistake: Forgetting to subtract moles of reacted acid from initial moles to find n(HA) remaining .
• Volume division: Because concentrations are proportional to moles in the same total volume, dividing both moles by 0.100 dm³ is optional, but omitting it when using [HA]/[A⁻] directly without calculating moles first will cause errors.
• Precision: pH must be given to 2 decimal places to match the decimal places of the pKa value.

💡 Key Knowledge

Aqueous ammonia ( NH₃(aq) ) acts as a weak base, providing hydroxide ions ( OH⁻ ) or reacting with hydrogen ions.

• Neutralisation: Added OH⁻ reacts with the hydrogen ions ( H⁺ ) or unreacted glycolic acid ( HOCH₂COOH ).
• Equilibrium Shift: The weak acid dissociation equilibrium ( HOCH₂COOH ⇌ H⁺ + HOCH₂COO⁻ ) shifts to the right to replace the removed H⁺ ions, maintaining a nearly constant pH.

✅ Mark Scheme Requirements

• State that NH₃ / OH⁻ reacts with H⁺ or glycolic acid.
• Explicitly state that the equilibrium shifts to the right (or write the equilibrium equation showing the shift).