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IB · CHEMISTRY HL

Chemistry HL

Reactivity 2: acid-base equilibria (HL) — Reactivity 2 HL

Name: ____________________Date: October 10, 2026
  1. 1.

    At 298 K, calculate the pH of a 0.0200 mol dm⁻³ solution of the strong base NaOH, assuming complete dissociation.

    [3 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: States that [OH⁻] = 0.0200 mol dm⁻³ and calculates pOH = −log₁₀(0.0200) ≈ 1.70. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Uses pH + pOH = 14.00 at 298 K. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Develop this part of the answer: Obtains pH ≈ 12.30. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
    5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: For a strong base, calculate pOH first and then convert to pH using the stated temperature relationship.

    Marking points

    • States that [OH⁻] = 0.0200 mol dm⁻³ and calculates pOH = −log₁₀(0.0200) ≈ 1.70.
    • Uses pH + pOH = 14.00 at 298 K.
    • Obtains pH ≈ 12.30.

    Examiner tip: For a strong base, calculate pOH first and then convert to pH using the stated temperature relationship.

  2. 2.

    Marking analysis: A learner attempts the following task: “At 298 K, calculate the pH of a 0.0200 mol dm⁻³ solution of the strong base NaOH, assuming complete dissociation.” Their response addresses only this point: “States that [OH⁻] = 0.0200 mol dm⁻³ and calculates pOH = −log₁₀(0.0200) ≈ 1.70.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [3 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: States that [OH⁻] = 0.0200 mol dm⁻³ and calculates pOH = −log₁₀(0.0200) ≈ 1.70. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Uses pH + pOH = 14.00 at 298 K. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Obtains pH ≈ 12.30. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: States that [OH⁻] = 0.0200 mol dm⁻³ and calculates pOH = −log₁₀(0.0200) ≈ 1.70.
    • Identifies the missing requirement: Uses pH + pOH = 14.00 at 298 K.
    • Identifies the missing requirement: Obtains pH ≈ 12.30.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  3. 3.

    Ethanoic acid (CH₃COOH) has Ka = 1.8 × 10⁻⁵. Calculate the pH of a 0.100 mol dm⁻³ solution of ethanoic acid, stating one assumption made.

    [4 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: Sets up Ka = [H⁺][A⁻]/[HA] ≈ [H⁺]²/[HA]₀, assuming [H⁺] is small compared to the initial acid concentration so [HA] ≈ 0.100. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Rearranges: [H⁺] = √(Ka × [HA]₀) = √(1.8 × 10⁻⁵ × 0.100). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Calculates [H⁺] ≈ 1.34 × 10⁻³ mol dm⁻³. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Work through this mathematical step: Uses pH = −log₁₀[H⁺] to obtain pH ≈ 2.87. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: The simplifying assumption that only a small fraction of a weak acid dissociates is valid whenever Ka is small and the initial concentration isn't too dilute — always state it as part of the method.

    Marking points

    • Sets up Ka = [H⁺][A⁻]/[HA] ≈ [H⁺]²/[HA]₀, assuming [H⁺] is small compared to the initial acid concentration so [HA] ≈ 0.100.
    • Rearranges: [H⁺] = √(Ka × [HA]₀) = √(1.8 × 10⁻⁵ × 0.100).
    • Calculates [H⁺] ≈ 1.34 × 10⁻³ mol dm⁻³.
    • Uses pH = −log₁₀[H⁺] to obtain pH ≈ 2.87.

    Examiner tip: The simplifying assumption that only a small fraction of a weak acid dissociates is valid whenever Ka is small and the initial concentration isn't too dilute — always state it as part of the method.

  4. 4.

    Marking analysis: A learner attempts the following task: “Ethanoic acid (CH₃COOH) has Ka = 1.8 × 10⁻⁵. Calculate the pH of a 0.100 mol dm⁻³ solution of ethanoic acid, stating one assumption made.” Their response addresses only this point: “Sets up Ka = [H⁺][A⁻]/[HA] ≈ [H⁺]²/[HA]₀, assuming [H⁺] is small compared to the initial acid concentration so [HA] ≈ 0.100.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [4 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: Sets up Ka = [H⁺][A⁻]/[HA] ≈ [H⁺]²/[HA]₀, assuming [H⁺] is small compared to the initial acid concentration so [HA] ≈ 0.100. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Rearranges: [H⁺] = √(Ka × [HA]₀) = √(1.8 × 10⁻⁵ × 0.100). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Calculates [H⁺] ≈ 1.34 × 10⁻³ mol dm⁻³. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Uses pH = −log₁₀[H⁺] to obtain pH ≈ 2.87. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: Sets up Ka = [H⁺][A⁻]/[HA] ≈ [H⁺]²/[HA]₀, assuming [H⁺] is small compared to the initial acid concentration so [HA] ≈ 0.100.
    • Identifies the missing requirement: Rearranges: [H⁺] = √(Ka × [HA]₀) = √(1.8 × 10⁻⁵ × 0.100).
    • Identifies the missing requirement: Calculates [H⁺] ≈ 1.34 × 10⁻³ mol dm⁻³.
    • Identifies the missing requirement: Uses pH = −log₁₀[H⁺] to obtain pH ≈ 2.87.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  5. 5.

    Define a buffer solution, and explain how a mixture of ethanoic acid and sodium ethanoate resists changes in pH when a small amount of acid is added.

    [4 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
    2. Develop this part of the answer: Defines a buffer solution as one that resists changes in pH when small amounts of acid or base are added. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
    3. Develop this part of the answer: States that the mixture contains a significant reservoir of both the weak acid (CH₃COOH) and its conjugate base (CH₃COO⁻, from the salt). Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
    4. Develop this part of the answer: States that added H⁺ ions are consumed by reacting with the conjugate base: CH₃COO⁻ + H⁺ → CH₃COOH. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
    5. Develop this part of the answer: States that because the added H⁺ is largely removed from solution by this reaction, the pH changes only slightly. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: A buffer needs comparable amounts of both a weak acid and its conjugate base (or weak base and conjugate acid) so that it can neutralize additions of either H⁺ or OH⁻.

    Marking points

    • Defines a buffer solution as one that resists changes in pH when small amounts of acid or base are added.
    • States that the mixture contains a significant reservoir of both the weak acid (CH₃COOH) and its conjugate base (CH₃COO⁻, from the salt).
    • States that added H⁺ ions are consumed by reacting with the conjugate base: CH₃COO⁻ + H⁺ → CH₃COOH.
    • States that because the added H⁺ is largely removed from solution by this reaction, the pH changes only slightly.

    Examiner tip: A buffer needs comparable amounts of both a weak acid and its conjugate base (or weak base and conjugate acid) so that it can neutralize additions of either H⁺ or OH⁻.

  6. 6.

    Marking analysis: A learner attempts the following task: “Define a buffer solution, and explain how a mixture of ethanoic acid and sodium ethanoate resists changes in pH when a small amount of acid is added.” Their response addresses only this point: “Defines a buffer solution as one that resists changes in pH when small amounts of acid or base are added.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [4 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: Defines a buffer solution as one that resists changes in pH when small amounts of acid or base are added. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: States that the mixture contains a significant reservoir of both the weak acid (CH₃COOH) and its conjugate base (CH₃COO⁻, from the salt). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: States that added H⁺ ions are consumed by reacting with the conjugate base: CH₃COO⁻ + H⁺ → CH₃COOH. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: States that because the added H⁺ is largely removed from solution by this reaction, the pH changes only slightly. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: Defines a buffer solution as one that resists changes in pH when small amounts of acid or base are added.
    • Identifies the missing requirement: States that the mixture contains a significant reservoir of both the weak acid (CH₃COOH) and its conjugate base (CH₃COO⁻, from the salt).
    • Identifies the missing requirement: States that added H⁺ ions are consumed by reacting with the conjugate base: CH₃COO⁻ + H⁺ → CH₃COOH.
    • Identifies the missing requirement: States that because the added H⁺ is largely removed from solution by this reaction, the pH changes only slightly.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  7. 7.

    A buffer solution is prepared containing 0.200 mol dm⁻³ ethanoic acid (Ka = 1.8 × 10⁻⁵) and 0.150 mol dm⁻³ sodium ethanoate. Calculate the pH of this buffer using the Henderson-Hasselbalch equation.

    [4 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: States the Henderson-Hasselbalch equation: pH = pKa + log₁₀([A⁻]/[HA]). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Calculates pKa = −log₁₀(1.8 × 10⁻⁵) ≈ 4.74. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Substitutes: pH = 4.74 + log₁₀(0.150/0.200). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Develop this part of the answer: Obtains pH ≈ 4.62. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: The Henderson-Hasselbalch equation is only valid for buffer mixtures where both the weak acid and its conjugate base are present in comparable, known concentrations.

    Marking points

    • States the Henderson-Hasselbalch equation: pH = pKa + log₁₀([A⁻]/[HA]).
    • Calculates pKa = −log₁₀(1.8 × 10⁻⁵) ≈ 4.74.
    • Substitutes: pH = 4.74 + log₁₀(0.150/0.200).
    • Obtains pH ≈ 4.62.

    Examiner tip: The Henderson-Hasselbalch equation is only valid for buffer mixtures where both the weak acid and its conjugate base are present in comparable, known concentrations.

  8. 8.

    Marking analysis: A learner attempts the following task: “A buffer solution is prepared containing 0.200 mol dm⁻³ ethanoic acid (Ka = 1.8 × 10⁻⁵) and 0.150 mol dm⁻³ sodium ethanoate. Calculate the pH of this buffer using the Henderson-Hasselbalch equation.” Their response addresses only this point: “States the Henderson-Hasselbalch equation: pH = pKa + log₁₀([A⁻]/[HA]).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [4 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: States the Henderson-Hasselbalch equation: pH = pKa + log₁₀([A⁻]/[HA]). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Calculates pKa = −log₁₀(1.8 × 10⁻⁵) ≈ 4.74. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Substitutes: pH = 4.74 + log₁₀(0.150/0.200). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Obtains pH ≈ 4.62. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: States the Henderson-Hasselbalch equation: pH = pKa + log₁₀([A⁻]/[HA]).
    • Identifies the missing requirement: Calculates pKa = −log₁₀(1.8 × 10⁻⁵) ≈ 4.74.
    • Identifies the missing requirement: Substitutes: pH = 4.74 + log₁₀(0.150/0.200).
    • Identifies the missing requirement: Obtains pH ≈ 4.62.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  9. 9.

    Sketch, in words, the shape of the pH curve produced when a strong base is added gradually to a strong acid, and state the approximate pH at the equivalence point.

    [3 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
    2. Develop this part of the answer: States that the pH starts low and rises slowly at first as base is added. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
    3. Develop this part of the answer: States that there is a very steep, near-vertical rise in pH around the equivalence point. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
    4. Develop this part of the answer: States that the pH at the equivalence point is approximately 7 (neutral), since a strong acid and strong base produce a neutral salt solution. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
    5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: The identity of the equivalence-point pH (7 for strong-strong, above 7 for weak acid-strong base, below 7 for strong acid-weak base) depends on whether the resulting salt's ions hydrolyse.

    Marking points

    • States that the pH starts low and rises slowly at first as base is added.
    • States that there is a very steep, near-vertical rise in pH around the equivalence point.
    • States that the pH at the equivalence point is approximately 7 (neutral), since a strong acid and strong base produce a neutral salt solution.

    Examiner tip: The identity of the equivalence-point pH (7 for strong-strong, above 7 for weak acid-strong base, below 7 for strong acid-weak base) depends on whether the resulting salt's ions hydrolyse.

  10. 10.

    Marking analysis: A learner attempts the following task: “Sketch, in words, the shape of the pH curve produced when a strong base is added gradually to a strong acid, and state the approximate pH at the equivalence point.” Their response addresses only this point: “States that the pH starts low and rises slowly at first as base is added.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [3 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: States that the pH starts low and rises slowly at first as base is added. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: States that there is a very steep, near-vertical rise in pH around the equivalence point. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: States that the pH at the equivalence point is approximately 7 (neutral), since a strong acid and strong base produce a neutral salt solution. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: States that the pH starts low and rises slowly at first as base is added.
    • Identifies the missing requirement: States that there is a very steep, near-vertical rise in pH around the equivalence point.
    • Identifies the missing requirement: States that the pH at the equivalence point is approximately 7 (neutral), since a strong acid and strong base produce a neutral salt solution.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  11. 11.

    Explain why the equivalence point for the titration of a weak acid with a strong base occurs above pH 7.

    [3 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
    2. Develop this part of the answer: States that at the equivalence point, the weak acid has been fully converted to its conjugate base (a salt) in solution. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
    3. Develop this part of the answer: States that this conjugate base is a weak base and reacts with water (hydrolysis) to produce OH⁻ ions: A⁻ + H₂O ⇌ HA + OH⁻. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
    4. Develop this part of the answer: States that the resulting excess of OH⁻ over H⁺ makes the solution basic, so the pH is above 7. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
    5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: At the equivalence point of any weak-strong titration, the solution contains only the salt of the weak component, so its pH is governed by the hydrolysis of that salt's ion, not by 'leftover' acid or base.

    Marking points

    • States that at the equivalence point, the weak acid has been fully converted to its conjugate base (a salt) in solution.
    • States that this conjugate base is a weak base and reacts with water (hydrolysis) to produce OH⁻ ions: A⁻ + H₂O ⇌ HA + OH⁻.
    • States that the resulting excess of OH⁻ over H⁺ makes the solution basic, so the pH is above 7.

    Examiner tip: At the equivalence point of any weak-strong titration, the solution contains only the salt of the weak component, so its pH is governed by the hydrolysis of that salt's ion, not by 'leftover' acid or base.

  12. 12.

    Marking analysis: A learner attempts the following task: “Explain why the equivalence point for the titration of a weak acid with a strong base occurs above pH 7.” Their response addresses only this point: “States that at the equivalence point, the weak acid has been fully converted to its conjugate base (a salt) in solution.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [3 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: States that at the equivalence point, the weak acid has been fully converted to its conjugate base (a salt) in solution. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: States that this conjugate base is a weak base and reacts with water (hydrolysis) to produce OH⁻ ions: A⁻ + H₂O ⇌ HA + OH⁻. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: States that the resulting excess of OH⁻ over H⁺ makes the solution basic, so the pH is above 7. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: States that at the equivalence point, the weak acid has been fully converted to its conjugate base (a salt) in solution.
    • Identifies the missing requirement: States that this conjugate base is a weak base and reacts with water (hydrolysis) to produce OH⁻ ions: A⁻ + H₂O ⇌ HA + OH⁻.
    • Identifies the missing requirement: States that the resulting excess of OH⁻ over H⁺ makes the solution basic, so the pH is above 7.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  13. 13.

    State the criterion for choosing a suitable indicator for an acid-base titration, in terms of the indicator's pH range and the pH curve.

    [2 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
    2. Develop this part of the answer: States that a suitable indicator is one whose colour-change pH range falls entirely within the steep, near-vertical portion of the titration's pH curve. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
    3. Develop this part of the answer: States that this ensures the indicator changes colour at (or very close to) the equivalence point, at a volume that closely matches it. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
    4. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: The exact equivalence-point pH does not need to match the indicator's midpoint exactly — only the indicator's full colour-change range needs to fall within the vertical jump of the curve.

    Marking points

    • States that a suitable indicator is one whose colour-change pH range falls entirely within the steep, near-vertical portion of the titration's pH curve.
    • States that this ensures the indicator changes colour at (or very close to) the equivalence point, at a volume that closely matches it.

    Examiner tip: The exact equivalence-point pH does not need to match the indicator's midpoint exactly — only the indicator's full colour-change range needs to fall within the vertical jump of the curve.

  14. 14.

    Marking analysis: A learner attempts the following task: “State the criterion for choosing a suitable indicator for an acid-base titration, in terms of the indicator's pH range and the pH curve.” Their response addresses only this point: “States that a suitable indicator is one whose colour-change pH range falls entirely within the steep, near-vertical portion of the titration's pH curve.” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [2 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: States that a suitable indicator is one whose colour-change pH range falls entirely within the steep, near-vertical portion of the titration's pH curve. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: States that this ensures the indicator changes colour at (or very close to) the equivalence point, at a volume that closely matches it. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: States that a suitable indicator is one whose colour-change pH range falls entirely within the steep, near-vertical portion of the titration's pH curve.
    • Identifies the missing requirement: States that this ensures the indicator changes colour at (or very close to) the equivalence point, at a volume that closely matches it.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  15. 15.

    A buffer solution has a pH of 5.20. A small amount of concentrated HCl is added. Explain, referring to Le Chatelier's principle, why the pH of the buffer changes only slightly rather than dropping sharply.

    [3 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
    2. Develop this part of the answer: States that the added H⁺ increases the concentration of H⁺ in the equilibrium HA ⇌ H⁺ + A⁻. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
    3. Develop this part of the answer: States, by Le Chatelier's principle, that the equilibrium shifts to the left (towards HA) to partially counteract the increase in [H⁺], consuming most of the added H⁺ by reaction with A⁻. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
    4. Develop this part of the answer: States that because most of the added H⁺ is removed from free solution this way, the free [H⁺] and hence the pH change only slightly. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
    5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Buffer resistance is a direct application of Le Chatelier's principle to the weak acid dissociation equilibrium, combined with the large reserve of conjugate base available to react.

    Marking points

    • States that the added H⁺ increases the concentration of H⁺ in the equilibrium HA ⇌ H⁺ + A⁻.
    • States, by Le Chatelier's principle, that the equilibrium shifts to the left (towards HA) to partially counteract the increase in [H⁺], consuming most of the added H⁺ by reaction with A⁻.
    • States that because most of the added H⁺ is removed from free solution this way, the free [H⁺] and hence the pH change only slightly.

    Examiner tip: Buffer resistance is a direct application of Le Chatelier's principle to the weak acid dissociation equilibrium, combined with the large reserve of conjugate base available to react.

  16. 16.

    Marking analysis: A learner attempts the following task: “A buffer solution has a pH of 5.20. A small amount of concentrated HCl is added. Explain, referring to Le Chatelier's principle, why the pH of the buffer changes only slightly rather than dropping sharply.” Their response addresses only this point: “States that the added H⁺ increases the concentration of H⁺ in the equilibrium HA ⇌ H⁺ + A⁻.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [3 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: States that the added H⁺ increases the concentration of H⁺ in the equilibrium HA ⇌ H⁺ + A⁻. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: States, by Le Chatelier's principle, that the equilibrium shifts to the left (towards HA) to partially counteract the increase in [H⁺], consuming most of the added H⁺ by reaction with A⁻. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: States that because most of the added H⁺ is removed from free solution this way, the free [H⁺] and hence the pH change only slightly. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: States that the added H⁺ increases the concentration of H⁺ in the equilibrium HA ⇌ H⁺ + A⁻.
    • Identifies the missing requirement: States, by Le Chatelier's principle, that the equilibrium shifts to the left (towards HA) to partially counteract the increase in [H⁺], consuming most of the added H⁺ by reaction with A⁻.
    • Identifies the missing requirement: States that because most of the added H⁺ is removed from free solution this way, the free [H⁺] and hence the pH change only slightly.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  17. 17.

    Calculate the pOH and hence the pH of a 0.0250 mol dm⁻³ solution of the strong base NaOH at 298 K, given Kw = 1.0 × 10⁻¹⁴.

    [4 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: States that for a strong monobasic base, [OH⁻] equals the base concentration: [OH⁻] = 0.0250 mol dm⁻³. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Uses pOH = −log₁₀[OH⁻] to obtain pOH ≈ 1.60. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Uses the relation pH + pOH = 14 (from Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 298 K). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Develop this part of the answer: Obtains pH ≈ 12.40. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: The relation pH + pOH = 14 only holds exactly at 298 K, since Kw itself is temperature-dependent — at other temperatures Kw must be used directly.

    Marking points

    • States that for a strong monobasic base, [OH⁻] equals the base concentration: [OH⁻] = 0.0250 mol dm⁻³.
    • Uses pOH = −log₁₀[OH⁻] to obtain pOH ≈ 1.60.
    • Uses the relation pH + pOH = 14 (from Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 298 K).
    • Obtains pH ≈ 12.40.

    Examiner tip: The relation pH + pOH = 14 only holds exactly at 298 K, since Kw itself is temperature-dependent — at other temperatures Kw must be used directly.

  18. 18.

    Marking analysis: A learner attempts the following task: “Calculate the pOH and hence the pH of a 0.0250 mol dm⁻³ solution of the strong base NaOH at 298 K, given Kw = 1.0 × 10⁻¹⁴.” Their response addresses only this point: “States that for a strong monobasic base, [OH⁻] equals the base concentration: [OH⁻] = 0.0250 mol dm⁻³.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [4 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: States that for a strong monobasic base, [OH⁻] equals the base concentration: [OH⁻] = 0.0250 mol dm⁻³. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Uses pOH = −log₁₀[OH⁻] to obtain pOH ≈ 1.60. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Uses the relation pH + pOH = 14 (from Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 298 K). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Obtains pH ≈ 12.40. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: States that for a strong monobasic base, [OH⁻] equals the base concentration: [OH⁻] = 0.0250 mol dm⁻³.
    • Identifies the missing requirement: Uses pOH = −log₁₀[OH⁻] to obtain pOH ≈ 1.60.
    • Identifies the missing requirement: Uses the relation pH + pOH = 14 (from Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 298 K).
    • Identifies the missing requirement: Obtains pH ≈ 12.40.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  19. 19.

    Outline why a buffer solution has a limited buffering capacity, referring to what happens if a large excess of strong acid is added.

    [2 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
    2. Develop this part of the answer: States that the buffer's capacity to neutralize added acid depends on the finite amount of conjugate base (A⁻) present. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
    3. Develop this part of the answer: States that once a large excess of strong acid is added, the conjugate base is used up (fully converted to HA), so any further added H⁺ is no longer neutralized and the pH then falls sharply. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
    4. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: A buffer's capacity is set by the smaller of the two component concentrations (acid or conjugate base); once that component is exhausted, the solution behaves like an unbuffered one.

    Marking points

    • States that the buffer's capacity to neutralize added acid depends on the finite amount of conjugate base (A⁻) present.
    • States that once a large excess of strong acid is added, the conjugate base is used up (fully converted to HA), so any further added H⁺ is no longer neutralized and the pH then falls sharply.

    Examiner tip: A buffer's capacity is set by the smaller of the two component concentrations (acid or conjugate base); once that component is exhausted, the solution behaves like an unbuffered one.

  20. 20.

    Marking analysis: A learner attempts the following task: “Outline why a buffer solution has a limited buffering capacity, referring to what happens if a large excess of strong acid is added.” Their response addresses only this point: “States that the buffer's capacity to neutralize added acid depends on the finite amount of conjugate base (A⁻) present.” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [2 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: States that the buffer's capacity to neutralize added acid depends on the finite amount of conjugate base (A⁻) present. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: States that once a large excess of strong acid is added, the conjugate base is used up (fully converted to HA), so any further added H⁺ is no longer neutralized and the pH then falls sharply. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: States that the buffer's capacity to neutralize added acid depends on the finite amount of conjugate base (A⁻) present.
    • Identifies the missing requirement: States that once a large excess of strong acid is added, the conjugate base is used up (fully converted to HA), so any further added H⁺ is no longer neutralized and the pH then falls sharply.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  21. 21.

    Ammonia has Kb = 1.8 × 10⁻⁵. Calculate the pH of a 0.150 mol dm⁻³ solution of ammonia at 298 K, stating one assumption made.

    [5 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: Sets up Kb = [NH₄⁺][OH⁻]/[NH₃] ≈ [OH⁻]²/[NH₃]₀, assuming [OH⁻] is small compared to the initial base concentration. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Rearranges: [OH⁻] = √(Kb × [NH₃]₀) = √(1.8 × 10⁻⁵ × 0.150). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Calculates [OH⁻] ≈ 1.64 × 10⁻³ mol dm⁻³. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Work through this mathematical step: Uses pOH = −log₁₀[OH⁻] ≈ 2.78. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    6. Work through this mathematical step: Uses pH = 14.00 − pOH to obtain pH ≈ 11.22. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Weak base pH calculations mirror weak acid calculations exactly, but solve for [OH⁻] via Kb first, then convert through pOH to pH — keep track of which quantity (H⁺ or OH⁻) you have found at each stage.

    Marking points

    • Sets up Kb = [NH₄⁺][OH⁻]/[NH₃] ≈ [OH⁻]²/[NH₃]₀, assuming [OH⁻] is small compared to the initial base concentration.
    • Rearranges: [OH⁻] = √(Kb × [NH₃]₀) = √(1.8 × 10⁻⁵ × 0.150).
    • Calculates [OH⁻] ≈ 1.64 × 10⁻³ mol dm⁻³.
    • Uses pOH = −log₁₀[OH⁻] ≈ 2.78.
    • Uses pH = 14.00 − pOH to obtain pH ≈ 11.22.

    Examiner tip: Weak base pH calculations mirror weak acid calculations exactly, but solve for [OH⁻] via Kb first, then convert through pOH to pH — keep track of which quantity (H⁺ or OH⁻) you have found at each stage.

  22. 22.

    Marking analysis: A learner attempts the following task: “Ammonia has Kb = 1.8 × 10⁻⁵. Calculate the pH of a 0.150 mol dm⁻³ solution of ammonia at 298 K, stating one assumption made.” Their response addresses only this point: “Sets up Kb = [NH₄⁺][OH⁻]/[NH₃] ≈ [OH⁻]²/[NH₃]₀, assuming [OH⁻] is small compared to the initial base concentration.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [5 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: Sets up Kb = [NH₄⁺][OH⁻]/[NH₃] ≈ [OH⁻]²/[NH₃]₀, assuming [OH⁻] is small compared to the initial base concentration. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Rearranges: [OH⁻] = √(Kb × [NH₃]₀) = √(1.8 × 10⁻⁵ × 0.150). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Calculates [OH⁻] ≈ 1.64 × 10⁻³ mol dm⁻³. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Uses pOH = −log₁₀[OH⁻] ≈ 2.78. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Requirement 5: Identifies the missing requirement: Uses pH = 14.00 − pOH to obtain pH ≈ 11.22. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: Sets up Kb = [NH₄⁺][OH⁻]/[NH₃] ≈ [OH⁻]²/[NH₃]₀, assuming [OH⁻] is small compared to the initial base concentration.
    • Identifies the missing requirement: Rearranges: [OH⁻] = √(Kb × [NH₃]₀) = √(1.8 × 10⁻⁵ × 0.150).
    • Identifies the missing requirement: Calculates [OH⁻] ≈ 1.64 × 10⁻³ mol dm⁻³.
    • Identifies the missing requirement: Uses pOH = −log₁₀[OH⁻] ≈ 2.78.
    • Identifies the missing requirement: Uses pH = 14.00 − pOH to obtain pH ≈ 11.22.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  23. 23.

    State the relationship between Ka of a weak acid and Kb of its conjugate base, and use this relationship to calculate Kb for the ethanoate ion, CH₃COO⁻, given Ka(CH₃COOH) = 1.8 × 10⁻⁵ and Kw = 1.0 × 10⁻¹⁴.

    [3 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Work through this mathematical step: States that Ka × Kb = Kw for a conjugate acid-base pair. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    3. Work through this mathematical step: Rearranges: Kb = Kw / Ka = (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Develop this part of the answer: Obtains Kb ≈ 5.6 × 10⁻¹⁰. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
    5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: This relationship shows that the stronger a weak acid is (the larger its Ka), the weaker its conjugate base is (the smaller its Kb), and vice versa — the two are always inversely linked through Kw.

    Marking points

    • States that Ka × Kb = Kw for a conjugate acid-base pair.
    • Rearranges: Kb = Kw / Ka = (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵).
    • Obtains Kb ≈ 5.6 × 10⁻¹⁰.

    Examiner tip: This relationship shows that the stronger a weak acid is (the larger its Ka), the weaker its conjugate base is (the smaller its Kb), and vice versa — the two are always inversely linked through Kw.

  24. 24.

    Marking analysis: A learner attempts the following task: “State the relationship between Ka of a weak acid and Kb of its conjugate base, and use this relationship to calculate Kb for the ethanoate ion, CH₃COO⁻, given Ka(CH₃COOH) = 1.8 × 10⁻⁵ and Kw = 1.0 × 10⁻¹⁴.” Their response addresses only this point: “States that Ka × Kb = Kw for a conjugate acid-base pair.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [3 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: States that Ka × Kb = Kw for a conjugate acid-base pair. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Rearranges: Kb = Kw / Ka = (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Obtains Kb ≈ 5.6 × 10⁻¹⁰. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: States that Ka × Kb = Kw for a conjugate acid-base pair.
    • Identifies the missing requirement: Rearranges: Kb = Kw / Ka = (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵).
    • Identifies the missing requirement: Obtains Kb ≈ 5.6 × 10⁻¹⁰.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  25. 25.

    Phosphoric acid, H₃PO₄, is triprotic and has three successive acid dissociation constants, Ka1, Ka2 and Ka3. State how the relative sizes of Ka1, Ka2 and Ka3 compare, and explain why.

    [3 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
    2. Develop this part of the answer: States that Ka1 > Ka2 > Ka3, i.e. each successive dissociation constant is smaller than the last. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
    3. Develop this part of the answer: Explains that it becomes progressively harder to remove a positively charged H⁺ ion from an increasingly negatively charged ion. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
    4. Develop this part of the answer: States that this is because the increasing negative charge on the ion after each dissociation step more strongly attracts (holds onto) the remaining H⁺ ions, resisting further dissociation. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
    5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: This pattern (Ka1 > Ka2 > Ka3) holds for essentially every polyprotic acid, since the underlying electrostatic reasoning — a more negative species holds onto H⁺ more tightly — is universal.

    Marking points

    • States that Ka1 > Ka2 > Ka3, i.e. each successive dissociation constant is smaller than the last.
    • Explains that it becomes progressively harder to remove a positively charged H⁺ ion from an increasingly negatively charged ion.
    • States that this is because the increasing negative charge on the ion after each dissociation step more strongly attracts (holds onto) the remaining H⁺ ions, resisting further dissociation.

    Examiner tip: This pattern (Ka1 > Ka2 > Ka3) holds for essentially every polyprotic acid, since the underlying electrostatic reasoning — a more negative species holds onto H⁺ more tightly — is universal.

  26. 26.

    Marking analysis: A learner attempts the following task: “Phosphoric acid, H₃PO₄, is triprotic and has three successive acid dissociation constants, Ka1, Ka2 and Ka3. State how the relative sizes of Ka1, Ka2 and Ka3 compare, and explain why.” Their response addresses only this point: “States that Ka1 > Ka2 > Ka3, i.e. each successive dissociation constant is smaller than the last.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [3 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: States that Ka1 > Ka2 > Ka3, i.e. each successive dissociation constant is smaller than the last. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Explains that it becomes progressively harder to remove a positively charged H⁺ ion from an increasingly negatively charged ion. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: States that this is because the increasing negative charge on the ion after each dissociation step more strongly attracts (holds onto) the remaining H⁺ ions, resisting further dissociation. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: States that Ka1 > Ka2 > Ka3, i.e. each successive dissociation constant is smaller than the last.
    • Identifies the missing requirement: Explains that it becomes progressively harder to remove a positively charged H⁺ ion from an increasingly negatively charged ion.
    • Identifies the missing requirement: States that this is because the increasing negative charge on the ion after each dissociation step more strongly attracts (holds onto) the remaining H⁺ ions, resisting further dissociation.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  27. 27.

    State two different methods that could be used to prepare a buffer solution containing ethanoic acid and ethanoate ions.

    [2 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
    2. Develop this part of the answer: States that a solution of a weak acid (e.g. ethanoic acid) can be mixed directly with a solution of a soluble salt of its conjugate base (e.g. sodium ethanoate). Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
    3. Develop this part of the answer: States that a solution of the weak acid can be partially neutralised with a solution of a strong base (e.g. NaOH), converting only some of the acid to its conjugate base and leaving a mixture of both. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
    4. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Both methods produce chemically identical buffer solutions, containing a comparable mixture of a weak acid and its conjugate base — the choice between them is simply one of practical convenience in the laboratory.

    Marking points

    • States that a solution of a weak acid (e.g. ethanoic acid) can be mixed directly with a solution of a soluble salt of its conjugate base (e.g. sodium ethanoate).
    • States that a solution of the weak acid can be partially neutralised with a solution of a strong base (e.g. NaOH), converting only some of the acid to its conjugate base and leaving a mixture of both.

    Examiner tip: Both methods produce chemically identical buffer solutions, containing a comparable mixture of a weak acid and its conjugate base — the choice between them is simply one of practical convenience in the laboratory.

  28. 28.

    Marking analysis: A learner attempts the following task: “State two different methods that could be used to prepare a buffer solution containing ethanoic acid and ethanoate ions.” Their response addresses only this point: “States that a solution of a weak acid (e.g. ethanoic acid) can be mixed directly with a solution of a soluble salt of its conjugate base (e.g. sodium ethanoate).” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [2 marks] · no calculator

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: States that a solution of a weak acid (e.g. ethanoic acid) can be mixed directly with a solution of a soluble salt of its conjugate base (e.g. sodium ethanoate). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: States that a solution of the weak acid can be partially neutralised with a solution of a strong base (e.g. NaOH), converting only some of the acid to its conjugate base and leaving a mixture of both. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: States that a solution of a weak acid (e.g. ethanoic acid) can be mixed directly with a solution of a soluble salt of its conjugate base (e.g. sodium ethanoate).
    • Identifies the missing requirement: States that a solution of the weak acid can be partially neutralised with a solution of a strong base (e.g. NaOH), converting only some of the acid to its conjugate base and leaving a mixture of both.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

  29. 29.

    A 0.200 dm³ buffer initially contains 0.0300 mol CH₃COOH and 0.0200 mol CH₃COO⁻. A student adds 0.00500 mol HCl and assumes the volume change is negligible. The pKa of ethanoic acid is 4.76. (a) Write the reaction of added H⁺ with the buffer base. (b) Calculate the moles of acid and base after reaction. (c) Calculate the new pH. (d) Explain why the pH change is limited.

    [5 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
    2. Develop this part of the answer: Writes CH₃COO⁻ + H⁺ → CH₃COOH. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
    3. Work through this mathematical step: Calculates n(CH₃COO⁻) = 0.0150 mol after reaction. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    4. Work through this mathematical step: Calculates n(CH₃COOH) = 0.0350 mol after reaction. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    5. Work through this mathematical step: Uses pH = pKa + log(nbase/nacid) to obtain pH = 4.76 + log(0.0150/0.0350) ≈ 4.39. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
    6. Develop this part of the answer: Explains that the conjugate base consumes most added H⁺, converting it to weak acid, so free [H⁺] changes much less than in an unbuffered solution. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
    7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Complete the stoichiometric neutralisation before using the Henderson-Hasselbalch relationship.

    Marking points

    • Writes CH₃COO⁻ + H⁺ → CH₃COOH.
    • Calculates n(CH₃COO⁻) = 0.0150 mol after reaction.
    • Calculates n(CH₃COOH) = 0.0350 mol after reaction.
    • Uses pH = pKa + log(nbase/nacid) to obtain pH = 4.76 + log(0.0150/0.0350) ≈ 4.39.
    • Explains that the conjugate base consumes most added H⁺, converting it to weak acid, so free [H⁺] changes much less than in an unbuffered solution.

    Examiner tip: Complete the stoichiometric neutralisation before using the Henderson-Hasselbalch relationship.

  30. 30.

    Marking analysis: A learner attempts the following task: “A 0.200 dm³ buffer initially contains 0.0300 mol CH₃COOH and 0.0200 mol CH₃COO⁻. A student adds 0.00500 mol HCl and assumes the volume change is negligible. The pKa of ethanoic acid is 4.76. (a) Write the reaction of added H⁺ with the buffer base. (b) Calculate the moles of acid and base after reaction. (c) Calculate the new pH. (d) Explain why the pH change is limited.” Their response addresses only this point: “Writes CH₃COO⁻ + H⁺ → CH₃COOH.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.

    [5 marks]

    Answer explanation

    Draft walkthroughs are based on marking guidance, not independently verified derivations.

    1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
    2. Requirement 1: Recognises credit for the stated point: Writes CH₃COO⁻ + H⁺ → CH₃COOH. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    3. Requirement 2: Identifies the missing requirement: Calculates n(CH₃COO⁻) = 0.0150 mol after reaction. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    4. Requirement 3: Identifies the missing requirement: Calculates n(CH₃COOH) = 0.0350 mol after reaction. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    5. Requirement 4: Identifies the missing requirement: Uses pH = pKa + log(nbase/nacid) to obtain pH = 4.76 + log(0.0150/0.0350) ≈ 4.39. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    6. Requirement 5: Identifies the missing requirement: Explains that the conjugate base consumes most added H⁺, converting it to weak acid, so free [H⁺] changes much less than in an unbuffered solution. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
    7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

    Marking points

    • Recognises credit for the stated point: Writes CH₃COO⁻ + H⁺ → CH₃COOH.
    • Identifies the missing requirement: Calculates n(CH₃COO⁻) = 0.0150 mol after reaction.
    • Identifies the missing requirement: Calculates n(CH₃COOH) = 0.0350 mol after reaction.
    • Identifies the missing requirement: Uses pH = pKa + log(nbase/nacid) to obtain pH = 4.76 + log(0.0150/0.0350) ≈ 4.39.
    • Identifies the missing requirement: Explains that the conjugate base consumes most added H⁺, converting it to weak acid, so free [H⁺] changes much less than in an unbuffered solution.

    Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.

Original, unofficial practice questions from SubjectScout — not official past-paper questions.