OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2023: Question 20

✅ Correct Answers

• (i) pH = 6.77
• (ii) In pure water, [H⁺] = [OH⁻] always, regardless of temperature shifts.

💡 Key Knowledge

• Autoionization of water is endothermic. As temperature increases, K_w increases, meaning [H⁺] increases and pH drops.
• Despite a lower pH at 40 °C, water remains strictly neutral because [H⁺] = [OH⁻] .

🧠 Exam Technique

• Step 1: State formula K_w = [H⁺][OH⁻] = [H⁺]² .
• Step 2: Take square root of K_w to find [H⁺] .
• Step 3: Apply pH = -log[H⁺] and format correctly to 2 decimal places.

❌ Common Errors

• Using room temperature K_w (1.00 × 10⁻¹⁴) instead of the 40 °C table value (2.92 × 10⁻¹⁴) .
• Failing to provide answers to the requested 2 decimal places.
• Stating water is acidic at 40 °C because pH < 7.

✅ Correct Answers

• Equation: Sr + 2H₂O → Sr(OH)₂ + H₂ (State symbols optional, multiples allowed).
• Calculation pH: 11.51

💡 Key Knowledge

• Group 2 metals react with water to form metal hydroxides and hydrogen gas.
• Sr(OH)₂ is a strong base that fully dissociates, releasing two moles of OH⁻ per mole of compound.

📐 Step-by-Step Calculation

1. Moles of Sr(OH)₂: 0.145 g / 121.6 g mol⁻¹ = 1.1924 × 10⁻³ mol
2. Concentration of OH⁻: Account for 250 cm³ volume (0.250 dm³) and 2:1 stoichiometry:
   [OH⁻] = (1.1924 × 10⁻³ / 0.250) × 2 = 9.539 × 10⁻³ mol dm⁻³
3. Find [H⁺] at 40 °C: Use updated K_w (2.92 × 10⁻¹⁴) :
   [H⁺] = K_w / [OH⁻] = (2.92 × 10⁻¹⁴) / (9.539 × 10⁻³) = 3.061 × 10⁻¹² mol dm⁻³
4. Calculate pH: pH = -log(3.061 × 10⁻¹²) = 11.51

❌ Common Errors

• Forgetting to multiply the hydroxide concentration by 2 (missing the stoichiometry of Sr(OH)₂ ).
• Using standard K_w (1.00 × 10⁻¹⁴) instead of the specified 40 °C value, leading to pH = 11.98 .

✅ Correct Answers

• (i) Equation: SrCO₃ + 2HNO₃ → Sr(NO₃)₂ + H₂O + CO₂
• (ii) Explanation: M_r of SrCO₃ is greater than M_r of CaCO₃ . Therefore, equal masses mean fewer moles of SrCO₃ are present, producing fewer moles (and a smaller volume) of CO₂ gas.

🧠 Exam Technique

• When comparing gas volumes from equal masses of reagents, always discuss molar mass ( M_r ) differences and relative initial moles.
• Remember stoichiometry is 1:1 for metal carbonates to CO₂ .

❌ Common Errors

• Writing H₂CO₃ as a product instead of decomposing into H₂O + CO₂ .
• Vague answers referencing "reactivity" instead of citing molar masses and mole calculations.

✅ Correct Answers

• (i) Ionic Equation: Mg + 2H⁺ → Mg²⁺ + H₂
• (ii) Explanation: HCl is a strong acid and fully dissociates, whereas CH₃COOH is a weak acid and partially dissociates. Consequently, HCl has a higher initial [H⁺] , causing a faster initial rate of reaction/more frequent collisions. Both produce the same total volume of gas because both acids are monobasic and have identical initial concentrations and volumes, releasing equal total moles of H⁺ .

💡 Key Knowledge

• Strong acids dissociate completely; weak acids establish an equilibrium position far to the left.
• Reaction rate depends on initial [H⁺] , whereas total gas volume depends on total moles of acid available to react.

❌ Common Errors

• Stating that ethanoic acid produces less gas (forgetting that weak acids still release all their protons eventually as the equilibrium shifts during the reaction).
• Confusing strong/weak terminology with concentrated/dilute concepts.

✅ Correct Answers

• (i) Definition: One mole of acid donates/dissociates to form one mole of protons ( H⁺ ).
• (ii) Ka Calculation: 1.5 × 10⁻⁵ mol dm⁻³ (accept 1.53 × 10⁻⁵ ).

📐 Step-by-Step Ka Calculation

1. Find [H⁺] from pH: [H⁺] = 10⁻ᵖᴴ = 10⁻⁵·⁰⁷ = 8.51 × 10⁻⁶ mol dm⁻³
2. Calculate initial moles and buffer components: Moles of KOH added = 3.39 g / 56.1 g mol⁻¹ = 0.0604 mol . This converts an equivalent amount of HA into salt A⁻ .
3. Find equilibrium concentrations in buffer volume (0.250 dm³):
   • [A⁻] = 0.0604 / 0.250 = 0.242 mol dm⁻³
   • Remaining [HA] = (0.376 × 0.250 - 0.0604) / 0.250 = 0.134 mol dm⁻³
4. Rearrange Ka expression: K_a = ([H⁺][A⁻]) / [HA]
   K_a = (8.51 × 10⁻⁶ × 0.242) / 0.134 = 1.53 × 10⁻⁵ mol dm⁻³

❌ Common Errors

• Forgetting to subtract moles of KOH added from the initial moles of weak acid when finding remaining [HA] .
• Using incorrect unit formatting or improper rounding mid-calculation.

✅ Correct Answers

• The ratio/proportion of [HA] / [A⁻] remains unchanged upon dilution with a small volume of water.

💡 Key Knowledge

• In the buffer equation [H⁺] = K_a × ([HA] / [A⁻]) , adding water scales both [HA] and [A⁻] down by the exact same dilution factor. Their ratio stays constant, keeping [H⁺] and therefore pH stable.