Edexcel A-Level Chemistry Paper 1, June 2019: Question 9

✅ Correct Answer

D: resists changes in pH if small quantities of acid or base are added.

💡 Key Knowledge

• Buffers do not keep pH constant; they only minimize changes upon small additions.
• Buffers do not necessarily keep pH below 7 (alkaline buffers exist).
• Equimolar amounts are not required; as long as both weak acid and its conjugate base are present in significant quantities, a buffer works.

📐 Step-by-Step Calculation

1. Find [H⁺(aq)] from pH:
   [H⁺] = 10⁻³·⁹⁰ = 1.2589 × 10⁻⁴ mol dm⁻³
2. Rearrange Kₐ expression to find conjugate base concentration:
   Kₐ = [H⁺][CH₃COO⁻] / [CH₃COOH]
   [CH₃COO⁻] = (Kₐ × [CH₃COOH]) / [H⁺]
   [CH₃COO⁻] = (1.74 × 10⁻⁵ × 0.800) / (1.2589 × 10⁻⁴) = 0.11057 mol dm⁻³
3. Calculate moles of ethanoate needed in the buffer volume:
   Volume = 50.0 cm³ = 0.0500 dm³
   Moles = 0.11057 × 0.0500 = 5.5285 × 10⁻³ mol
4. Convert moles to mass:
   Mᵣ of CH₃COONa = 82.0
   Mass = 5.5285 × 10⁻³ × 82.0 = 0.45339 g
5. Apply Significant Figures:
   Give answer to 2 or 3 SF: 0.45 g or 0.453 g

❌ Common Errors & Exam Technique

• Unit conversion traps: Forgetting to convert cm³ to dm³ when calculating moles.
• Sig Fig penalties: Leaving answers to 1 SF or 5+ SF when data is given to 2 or 3 SF.
• TE (Error Carried Forward): The mark scheme allows consequential marking at each stage if an earlier step went wrong, provided working is clear.

💡 Key Knowledge & Explanation

H₂CO₃ ⇌ H⁺ + HCO₃⁻

• Point 1: Extra CO₂ dissolves in blood to form more carbonic acid (H₂CO₃), increasing its concentration.
• Point 2: Equilibrium shifts to the right to oppose the increase, producing more H⁺ ions.
• Point 3: The large pre-existing pool/reservoir of hydrogencarbonate ions (HCO₃⁻) suppresses the ionization of carbonic acid, or combines with excess H⁺ ions to control pH.

🧠 Exam Technique

• Always explicitly mention equilibrium shifting using Le Chatelier's principle.
• Avoid imprecise phrasing like "CO₂ reacts with H⁺ to form carbonic acid" — examiners penalize this as chemically inaccurate.

🧠 Curve Sketching Guidelines

• Starting Point: Curve must start at pH = 3.0 (matching the propanoic acid initial pH given in the stem).
• Initial rise: Rapid initial increase in pH before flattening out into the buffer region.
• Buffer Region: Clearly label the region of gradual pH change between roughly 5 cm³ and 35 cm³ of NaOH added.
• Equivalence Point: Steep vertical section centered around 40 cm³ (since 40.0 cm³ of 0.100 mol dm⁻³ acid reacts completely with 40.0 cm³ of 0.100 mol dm⁻³ NaOH).
• End Point: Levels off at high pH (~12–13) near 100 cm³.

❌ Common Errors

• Starting the curve at pH 0 or 1 instead of the specified starting pH of 3.0.
• Placing the steep vertical equivalence point at the wrong volume (failing to calculate that 40 cm³ of acid requires 40 cm³ of base for neutralization).

✅ Correct Answer & Method

• Step 1: Determine the pH at the point when half of the acid is neutralized (half-equivalence point, occurring at 20 cm³ of NaOH added).
• Step 2: Use the relationship Kₐ = 10⁻ᵖᴴ (or equivalent description stating that pH equals pKₐ at this point, so inverse log of minus pH gives Kₐ).

🧠 Exam Technique

• At the half-neutralization point, the concentration of the weak acid equals the concentration of its conjugate base ([HA] = [A⁻]), which simplifies the Kₐ expression to Kₐ = [H⁺] , meaning pH = pKₐ . Memorize this shortcut!