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A Level H1 Chemistry Acids Bases Salts Quiz

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Questions

A-Level Chemistry H1 Quiz - Acids Bases Salts

Name: _________________________ Class: _________________________ Date: _________________________ Score: ______ / 50

Duration: 1 hour 15 minutes Total Marks: 50

Instructions:

Answer ALL questions in the spaces provided.
Show all working for calculation questions.
State units where appropriate.
A Data Booklet is provided for reference.
The number of marks is given in brackets [ ] at the end of each question or part question.

Section A: Short Answer and Structured Response (20 marks)

Answer all questions in this section.

1. What is meant by the term weak acid? Illustrate your answer with a suitable equation, including state symbols. [2]

2. Distinguish between a strong acid and a concentrated acid. [2]

3. The pH of a sample of rainwater was measured and found to be 5.6. Explain why rainwater is naturally slightly acidic, and write a balanced equation for the reaction that causes this acidity. [2]

4. Calcium hydroxide, Ca(OH)₂, is sometimes added to fermentation tanks to prevent the buildup of lactic acid. Explain why high acidity reduces the effectiveness of enzymes used in fermentation. [2]

5. Identify the Period 3 element that forms a sparingly soluble amphoteric oxide. Write equations to show the amphoteric nature of this oxide. [3]

6. A student prepared a buffer solution by mixing ethanoic acid and sodium ethanoate.

(a) Define a buffer solution. [1]

(b) Explain, using equations where appropriate, how this buffer solution resists changes in pH when a small amount of hydrochloric acid is added. [3]

7. The acid dissociation constant, Kₐ, for ethanoic acid is 1.8 × 10⁻⁵ mol dm⁻³ at 298 K.

(a) Write the expression for the acid dissociation constant of ethanoic acid. [1]

(b) Calculate the pH of a 0.10 mol dm⁻³ solution of ethanoic acid. State any assumption made. [4]

Section B: Calculation and Data Interpretation (18 marks)

Answer all questions in this section.

8. In a titration experiment, 25.0 cm³ of sodium hydroxide solution required 22.40 cm³ of 0.100 mol dm⁻³ hydrochloric acid for complete neutralisation.

(a) Write a balanced equation for the reaction. [1]

(b) Calculate the concentration of the sodium hydroxide solution in mol dm⁻³. [3]

9. A student titrated 25.0 cm³ of a weak monoprotic acid, HA, with 0.100 mol dm⁻³ sodium hydroxide solution. The pH was monitored throughout the titration. The equivalence point was reached when 20.00 cm³ of sodium hydroxide had been added, and the pH at this point was 8.7.

(a) Calculate the concentration of the acid, HA. [2]

(b) Explain why the pH at the equivalence point is greater than 7. [2]

(c) The pH at half-neutralisation (after adding 10.00 cm³ of NaOH) was found to be 4.8. Determine the Kₐ value for the acid, HA. [2]

10. A 0.0250 mol dm⁻³ solution of a weak base, B, has a pH of 10.50 at 298 K.

(a) Calculate the concentration of hydroxide ions, [OH⁻], in this solution. (K_w = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K) [2]

(b) Write the expression for the base dissociation constant, K_b, for B. [1]

11. A buffer solution is prepared by dissolving 0.0500 mol of ethanoic acid and 0.0500 mol of sodium ethanoate in water to make 1.00 dm³ of solution.

(a) Calculate the pH of this buffer solution. (Kₐ for ethanoic acid = 1.8 × 10⁻⁵ mol dm⁻³) [2]

(b) Calculate the new pH after 0.010 mol of sodium hydroxide is added to 1.00 dm³ of this buffer solution. Assume no change in volume. [3]

Section C: Application and Analysis (12 marks)

Answer all questions in this section.

12. Carbonic acid, H₂CO₃, is a weak diprotic acid formed when carbon dioxide dissolves in water.

(a) Write equations for the two dissociation steps of carbonic acid in water, including state symbols. [2]

(b) Write the expressions for Kₐ₁ and Kₐ₂ for carbonic acid. [2]

(c) The Kₐ₁ of carbonic acid is 4.3 × 10⁻⁷ mol dm⁻³. Calculate the pH of a 0.050 mol dm⁻³ solution of carbonic acid, assuming only the first dissociation contributes significantly to [H⁺]. [3]

13. A student investigates the relative strengths of three acids: hydrochloric acid (HCl), ethanoic acid (CH₃COOH), and benzoic acid (C₆H₅COOH). Solutions of each acid, all at 0.10 mol dm⁻³, are tested.

(a) The pH values measured are: 1.0, 2.6, and 2.9. Assign each pH value to the correct acid, explaining your reasoning. [3]

(b) Suggest why the pH of 0.10 mol dm⁻³ hydrochloric acid is 1.0, while the pH of 0.10 mol dm⁻³ ethanoic acid is higher. [2]

14. A student claims: "Since ethanoic acid is a weak acid, a 0.10 mol dm⁻³ solution of ethanoic acid must have a higher pH than a 0.010 mol dm⁻³ solution of hydrochloric acid." Evaluate this claim with supporting calculations. [3]

15. Explain why a mixture of aqueous ammonia and ammonium chloride can act as a buffer solution. Include relevant equations in your answer. [3]

16. The table below shows the pH of equimolar solutions of four different acids, each at 0.10 mol dm⁻³.

Acid	Formula	pH
A	HCl	1.0
B	CH₃COOH	2.9
C	ClCH₂COOH	1.9
D	Cl₂CHCOOH	1.3

(a) Explain why acid C is a stronger acid than acid B. [2]

(b) Explain why acid D is a stronger acid than acid C. [2]

17. A student prepares a solution by mixing 50.0 cm³ of 0.20 mol dm⁻³ ethanoic acid with 50.0 cm³ of 0.10 mol dm⁻³ sodium hydroxide.

(a) Write an equation for the reaction that occurs. [1]

(b) Determine the pH of the resulting solution. (Kₐ for ethanoic acid = 1.8 × 10⁻⁵ mol dm⁻³) [4]

18. The acid dissociation constant, Kₐ, of benzoic acid is 6.3 × 10⁻⁵ mol dm⁻³ at 298 K. A solution of benzoic acid has a pH of 2.80. Calculate the concentration of the benzoic acid solution. State any assumption made. [4]

19. A buffer solution is required with a pH of 4.50. A student has access to ethanoic acid (Kₐ = 1.8 × 10⁻⁵ mol dm⁻³) and sodium ethanoate. Calculate the ratio of [CH₃COO⁻] to [CH₃COOH] required to prepare this buffer. [3]

20. Explain, using the Brønsted-Lowry theory, why the hydrogen sulfate ion, HSO₄⁻, can behave as both an acid and a base. Write equations to illustrate your answer. [3]

END OF QUIZ

Check your work carefully. Ensure all answers are in the spaces provided.

Answers

A-Level Chemistry H1 Quiz - Acids Bases Salts: ANSWER KEY

Total Marks: 50

Section A: Short Answer and Structured Response (20 marks)

1. What is meant by the term weak acid? Illustrate your answer with a suitable equation, including state symbols. [2]

Answer: A weak acid is an acid that only partially dissociates/ionises in aqueous solution. [1]

Equation (any valid weak acid accepted): CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq) [1] OR HCOOH(aq) ⇌ HCOO⁻(aq) + H⁺(aq) OR H₂CO₃(aq) ⇌ HCO₃⁻(aq) + H⁺(aq)

Marking notes:

[1] for correct definition (must include "partially dissociates/ionises")
[1] for correct equation with reversible arrow (⇌) and state symbols (aq)
Do NOT accept "dilute acid" or "does not dissociate"
Accept any valid weak acid equation

2. Distinguish between a strong acid and a concentrated acid. [2]

Answer: A strong acid is one that completely dissociates/ionises in aqueous solution, while a concentrated acid contains a large amount of acid dissolved per unit volume of solution. [1] Strength refers to the degree of dissociation, whereas concentration refers to the amount of acid present. [1]

Marking notes:

[1] for defining strong acid (complete dissociation) and concentrated acid (large amount per volume)
[1] for clearly distinguishing strength vs concentration
Accept: "Strength is about extent of ionisation; concentration is about how much acid is present"

3. The pH of a sample of rainwater was measured and found to be 5.6. Explain why rainwater is naturally slightly acidic, and write a balanced equation for the reaction that causes this acidity. [2]

Answer: Rainwater is naturally slightly acidic because carbon dioxide in the atmosphere dissolves in water to form carbonic acid, a weak acid. [1]

Equation: CO₂(g) + H₂O(l) ⇌ H₂CO₃(aq) [1] OR CO₂(g) + H₂O(l) ⇌ H⁺(aq) + HCO₃⁻(aq)

Marking notes:

[1] for explanation linking CO₂ dissolution to acidity
[1] for correct balanced equation with state symbols
Accept either equation; reversible arrow required

Answer: High acidity (low pH) denatures the enzymes. [1] The excess H⁺ ions disrupt the ionic and hydrogen bonds that maintain the enzyme's tertiary structure, changing the shape of the active site so the substrate can no longer bind, and the enzyme loses its catalytic activity. [1]

Marking notes:

[1] for stating enzymes are denatured at high acidity/low pH
[1] for explaining mechanism (disruption of bonds in tertiary structure → active site shape change → substrate cannot bind)
Accept reference to enzymes having an optimal pH range

5. Identify the Period 3 element that forms a sparingly soluble amphoteric oxide. Write equations to show the amphoteric nature of this oxide. [3]

Answer: Aluminium (Al). [1]

Reaction with acid: Al₂O₃(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂O(l) [1] OR Al₂O₃(s) + 6H⁺(aq) → 2Al³⁺(aq) + 3H₂O(l)

Reaction with base: Al₂O₃(s) + 2NaOH(aq) + 3H₂O(l) → 2Na[Al(OH)₄](aq) [1] OR Al₂O₃(s) + 2OH⁻(aq) + 3H₂O(l) → 2[Al(OH)₄]⁻(aq)

Marking notes:

[1] for identifying aluminium
[1] for correct equation showing reaction with acid
[1] for correct equation showing reaction with base
Accept balanced ionic equations
Do NOT accept Si (SiO₂ is acidic, not amphoteric)

6. A student prepared a buffer solution by mixing ethanoic acid and sodium ethanoate.

(a) Define a buffer solution. [1]

Answer: A buffer solution is one that resists changes in pH when small amounts of acid or base are added. [1]

Marking notes:

[1] for correct definition (must include "resists changes in pH" and reference to small additions of acid/base)

(b) Explain, using equations where appropriate, how this buffer solution resists changes in pH when a small amount of hydrochloric acid is added. [3]

Answer: The buffer contains CH₃COOH (weak acid) and CH₃COO⁻ (conjugate base from sodium ethanoate). [1]

When HCl (H⁺) is added: CH₃COO⁻(aq) + H⁺(aq) → CH₃COOH(aq) [1]

The added H⁺ ions react with the ethanoate ions to form undissociated ethanoic acid molecules. This removes most of the added H⁺ from solution, so the pH remains approximately constant. [1]

Marking notes:

[1] for identifying buffer components (weak acid + conjugate base)
[1] for correct equation showing H⁺ reacting with CH₃COO⁻
[1] for explanation that H⁺ is removed from solution, minimising pH change

7. The acid dissociation constant, Kₐ, for ethanoic acid is 1.8 × 10⁻⁵ mol dm⁻³ at 298 K.

(a) Write the expression for the acid dissociation constant of ethanoic acid. [1]

Answer: Kₐ = [CH₃COO⁻][H⁺] / [CH₃COOH] [1]

Marking notes:

[1] for correct expression with correct species
Must have [H⁺] not [H₃O⁺] (both acceptable in practice, but [H⁺] is standard)

(b) Calculate the pH of a 0.10 mol dm⁻³ solution of ethanoic acid. State any assumption made. [4]

Answer: CH₃COOH ⇌ CH₃COO⁻ + H⁺

Kₐ = [CH₃COO⁻][H⁺] / [CH₃COOH] = 1.8 × 10⁻⁵

Let [H⁺] = x Then [CH₃COO⁻] = x and [CH₃COOH] ≈ 0.10 (assumption: x << 0.10) [1]

1.8 × 10⁻⁵ = x² / 0.10 [1] x² = 1.8 × 10⁻⁶ x = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ mol dm⁻³ [1]

pH = −log₁₀(1.34 × 10⁻³) = 2.87 [1]

Assumption: The degree of dissociation is small, so [CH₃COOH] at equilibrium ≈ initial concentration (0.10 mol dm⁻³). [1 mark included above]

Marking notes:

[1] for correct setup with Kₐ expression and substitution
[1] for correct algebraic manipulation
[1] for correct [H⁺] calculation
[1] for correct pH and stating assumption
Check assumption: x/[HA]₀ = 1.34 × 10⁻³ / 0.10 = 1.34% < 5%, so assumption valid

Section B: Calculation and Data Interpretation (18 marks)

8. In a titration experiment, 25.0 cm³ of sodium hydroxide solution required 22.40 cm³ of 0.100 mol dm⁻³ hydrochloric acid for complete neutralisation.

(a) Write a balanced equation for the reaction. [1]

Answer: NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l) [1]

Marking notes:

[1] for correct balanced equation with state symbols

(b) Calculate the concentration of the sodium hydroxide solution in mol dm⁻³. [3]

Answer: n(HCl) = c × V = 0.100 × (22.40/1000) = 2.24 × 10⁻³ mol [1]

From equation, 1:1 ratio, so n(NaOH) = 2.24 × 10⁻³ mol [1]

c(NaOH) = n / V = 2.24 × 10⁻³ / (25.0/1000) = 0.0896 mol dm⁻³ [1]

Marking notes:

[1] for correct moles of HCl
[1] for using 1:1 mole ratio
[1] for correct concentration with units
Accept 0.090 mol dm⁻³ (2 sf) or 0.0896 mol dm⁻³ (3 sf)

9. A student titrated 25.0 cm³ of a weak monoprotic acid, HA, with 0.100 mol dm⁻³ sodium hydroxide solution. The equivalence point was reached when 20.00 cm³ of sodium hydroxide had been added, and the pH at this point was 8.7.

(a) Calculate the concentration of the acid, HA. [2]

Answer: n(NaOH) = 0.100 × (20.00/1000) = 2.00 × 10⁻³ mol [1]

HA + NaOH → NaA + H₂O (1:1 ratio) n(HA) = 2.00 × 10⁻³ mol

c(HA) = 2.00 × 10⁻³ / (25.0/1000) = 0.0800 mol dm⁻³ [1]

Marking notes:

[1] for correct moles of NaOH
[1] for correct concentration of HA with units

(b) Explain why the pH at the equivalence point is greater than 7. [2]

Answer: At the equivalence point, all the weak acid HA has been converted to its conjugate base A⁻. [1] The A⁻ ion undergoes hydrolysis, accepting a proton from water: A⁻(aq) + H₂O(l) ⇌ HA(aq) + OH⁻(aq) This produces OH⁻ ions, making the solution alkaline (pH > 7). [1]

Marking notes:

[1] for identifying that the salt formed (A⁻) is the conjugate base of a weak acid
[1] for explaining hydrolysis produces OH⁻, making solution alkaline
Accept equation showing hydrolysis

(c) The pH at half-neutralisation (after adding 10.00 cm³ of NaOH) was found to be 4.8. Determine the Kₐ value for the acid, HA. [2]

Answer: At half-neutralisation, [HA] = [A⁻]. [1]

From Kₐ = [H⁺][A⁻]/[HA], when [HA] = [A⁻]: Kₐ = [H⁺]

pH = 4.8, so [H⁺] = 10⁻⁴·⁸ = 1.58 × 10⁻⁵ mol dm⁻³

Therefore, Kₐ = 1.58 × 10⁻⁵ mol dm⁻³ [1]

Marking notes:

[1] for recognising that at half-neutralisation, [HA] = [A⁻] and therefore Kₐ = [H⁺]
[1] for correct Kₐ value
Accept 1.6 × 10⁻⁵ mol dm⁻³

10. A 0.0250 mol dm⁻³ solution of a weak base, B, has a pH of 10.50 at 298 K.

(a) Calculate the concentration of hydroxide ions, [OH⁻], in this solution. (K_w = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K) [2]

Answer: pH = 10.50, so pOH = 14.00 − 10.50 = 3.50 [1]

[OH⁻] = 10⁻³·⁵⁰ = 3.16 × 10⁻⁴ mol dm⁻³ [1]

Marking notes:

[1] for correct pOH calculation
[1] for correct [OH⁻]
Accept 3.2 × 10⁻⁴ mol dm⁻³

(b) Write the expression for the base dissociation constant, K_b, for B. [1]

Answer: B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq) K_b = [BH⁺][OH⁻] / [B] [1]

Marking notes:

[1] for correct K_b expression
Water omitted from expression as its concentration is effectively constant

(c) Calculate the value of K_b for B. State any assumption made. [3]

Answer: From the equation: [BH⁺] = [OH⁻] = 3.16 × 10⁻⁴ mol dm⁻³ [1]

Assumption: Degree of dissociation is small, so [B] at equilibrium ≈ 0.0250 mol dm⁻³ [1]

K_b = (3.16 × 10⁻⁴)² / 0.0250 = 3.99 × 10⁻⁶ mol dm⁻³ [1]

Marking notes:

[1] for recognising [BH⁺] = [OH⁻]
[1] for stating assumption and correct substitution
[1] for correct K_b value with units
Accept 4.0 × 10⁻⁶ mol dm⁻³

11. A buffer solution is prepared by dissolving 0.0500 mol of ethanoic acid and 0.0500 mol of sodium ethanoate in water to make 1.00 dm³ of solution.

(a) Calculate the pH of this buffer solution. (Kₐ for ethanoic acid = 1.8 × 10⁻⁵ mol dm⁻³) [2]

Answer: [CH₃COOH] = 0.0500 mol dm⁻³ [CH₃COO⁻] = 0.0500 mol dm⁻³

Kₐ = [H⁺][CH₃COO⁻] / [CH₃COOH] 1.8 × 10⁻⁵ = [H⁺] × 0.0500 / 0.0500 [1] [H⁺] = 1.8 × 10⁻⁵ mol dm⁻³

pH = −log₁₀(1.8 × 10⁻⁵) = 4.74 [1]

Marking notes:

[1] for correct substitution into Kₐ expression
[1] for correct pH
Note: When [HA] = [A⁻], pH = pKₐ = 4.74

(b) Calculate the new pH after 0.010 mol of sodium hydroxide is added to 1.00 dm³ of this buffer solution. Assume no change in volume. [3]

Answer: Added OH⁻ reacts with CH₃COOH: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O

New amounts: n(CH₃COOH) = 0.0500 − 0.010 = 0.0400 mol [1] n(CH₃COO⁻) = 0.0500 + 0.010 = 0.0600 mol [1]

[H⁺] = Kₐ × [CH₃COOH] / [CH₃COO⁻] [H⁺] = 1.8 × 10⁻⁵ × (0.0400) / (0.0600) = 1.20 × 10⁻⁵ mol dm⁻³

pH = −log₁₀(1.20 × 10⁻⁵) = 4.92 [1]

Marking notes:

[1] for correct new amount of CH₃COOH
[1] for correct new amount of CH₃COO⁻
[1] for correct pH
Accept working in concentrations since volume unchanged

Section C: Application and Analysis (12 marks)

12. Carbonic acid, H₂CO₃, is a weak diprotic acid formed when carbon dioxide dissolves in water.

(a) Write equations for the two dissociation steps of carbonic acid in water, including state symbols. [2]

Answer: First dissociation: H₂CO₃(aq) ⇌ H⁺(aq) + HCO₃⁻(aq) [1] Second dissociation: HCO₃⁻(aq) ⇌ H⁺(aq) + CO₃²⁻(aq) [1]

Marking notes:

[1] for each correct equation with reversible arrows and state symbols
Must show stepwise dissociation

(b) Write the expressions for Kₐ₁ and Kₐ₂ for carbonic acid. [2]

Answer: Kₐ₁ = [H⁺][HCO₃⁻] / [H₂CO₃] [1] Kₐ₂ = [H⁺][CO₃²⁻] / [HCO₃⁻] [1]

Marking notes:

[1] for each correct expression
Water not included in expressions

(c) The Kₐ₁ of carbonic acid is 4.3 × 10⁻⁷ mol dm⁻³. Calculate the pH of a 0.050 mol dm⁻³ solution of carbonic acid, assuming only the first dissociation contributes significantly to [H⁺]. [3]

Answer: H₂CO₃ ⇌ H⁺ + HCO₃⁻

Kₐ₁ = [H⁺][HCO₃⁻] / [H₂CO₃] = 4.3 × 10⁻⁷

Let [H⁺] = x, then [HCO₃⁻] = x Assume [H₂CO₃] ≈ 0.050 (x is small) [1]

4.3 × 10⁻⁷ = x² / 0.050 [1] x² = 2.15 × 10⁻⁸ x = √(2.15 × 10⁻⁸) = 1.47 × 10⁻⁴ mol dm⁻³

pH = −log₁₀(1.47 × 10⁻⁴) = 3.83 [1]

Marking notes:

[1] for correct setup and assumption
[1] for correct [H⁺] calculation
[1] for correct pH
Accept 3.8 or 3.83

(a) The pH values measured are: 1.0, 2.6, and 2.9. Assign each pH value to the correct acid, explaining your reasoning. [3]

Answer: pH 1.0 → HCl (strong acid, completely dissociates: [H⁺] = 0.10 mol dm⁻³, pH = 1.0) [1] pH 2.6 → C₆H₅COOH (benzoic acid, Kₐ = 6.3 × 10⁻⁵, weaker than HCl but stronger than ethanoic acid) [1] pH 2.9 → CH₃COOH (ethanoic acid, Kₐ = 1.8 × 10⁻⁵, weakest of the three) [1]

Marking notes:

[1] for correctly assigning HCl to pH 1.0 with reasoning (strong acid, complete dissociation)
[1] for correctly assigning benzoic acid to pH 2.6
[1] for correctly assigning ethanoic acid to pH 2.9
Accept reasoning based on relative Kₐ values

(b) Suggest why the pH of 0.10 mol dm⁻³ hydrochloric acid is 1.0, while the pH of 0.10 mol dm⁻³ ethanoic acid is higher. [2]

Answer: HCl is a strong acid that completely dissociates, so [H⁺] = 0.10 mol dm⁻³ and pH = 1.0. [1] Ethanoic acid is a weak acid that only partially dissociates, so [H⁺] is much less than 0.10 mol dm⁻³, resulting in a higher pH. [1]

Marking notes:

[1] for stating HCl is strong/completely dissociates
[1] for stating ethanoic acid is weak/partially dissociates
Must link degree of dissociation to [H⁺] and pH

Answer: For 0.10 mol dm⁻³ CH₃COOH: [H⁺] = √(Kₐ × c) = √(1.8 × 10⁻⁵ × 0.10) = 1.34 × 10⁻³ mol dm⁻³ pH = 2.87 [1]

For 0.010 mol dm⁻³ HCl (strong acid, complete dissociation): [H⁺] = 0.010 mol dm⁻³ pH = 2.00 [1]

The claim is incorrect. The 0.10 mol dm⁻³ ethanoic acid has pH 2.87, which is higher than the pH 2.00 of 0.010 mol dm⁻³ HCl. However, this is not simply because ethanoic acid is weak; the concentration difference also matters. A more dilute strong acid can have a lower pH than a more concentrated weak acid. [1]

Marking notes:

[1] for correct pH calculation of ethanoic acid
[1] for correct pH calculation of HCl
[1] for correct evaluation (claim is incorrect) with explanation
Accept: The claim is false; pH of 0.10 M CH₃COOH (2.87) > pH of 0.010 M HCl (2.00)

15. Explain why a mixture of aqueous ammonia and ammonium chloride can act as a buffer solution. Include relevant equations in your answer. [3]

Answer: The mixture contains a weak base (NH₃) and its conjugate acid (NH₄⁺ from NH₄Cl). [1]

When acid (H⁺) is added: NH₃(aq) + H⁺(aq) → NH₄⁺(aq) The ammonia neutralises the added H⁺. [1]

When base (OH⁻) is added: NH₄⁺(aq) + OH⁻(aq) → NH₃(aq) + H₂O(l) The ammonium ions neutralise the added OH⁻. [1]

Thus, the solution resists changes in pH.

Marking notes:

[1] for identifying buffer components (weak base + conjugate acid)
[1] for equation/explanation of resistance to added acid
[1] for equation/explanation of resistance to added base

16. The table below shows the pH of equimolar solutions of four different acids, each at 0.10 mol dm⁻³.

Acid	Formula	pH
A	HCl	1.0
B	CH₃COOH	2.9
C	ClCH₂COOH	1.9
D	Cl₂CHCOOH	1.3

(a) Explain why acid C is a stronger acid than acid B. [2]

Answer: Acid C (chloroethanoic acid) has an electronegative chlorine atom that withdraws electron density from the O−H bond through the inductive effect. [1] This weakens the O−H bond, making it easier for the acid to donate a proton, and also stabilises the conjugate base (ClCH₂COO⁻) by dispersing the negative charge. [1]

Marking notes:

[1] for identifying the electron-withdrawing/inductive effect of Cl
[1] for explaining how this increases acid strength (weakens O−H bond and/or stabilises conjugate base)

(b) Explain why acid D is a stronger acid than acid C. [2]

Answer: Acid D (dichloroethanoic acid) has two chlorine atoms, which exert a greater electron-withdrawing inductive effect than the single chlorine in acid C. [1] This further weakens the O−H bond and provides greater stabilisation of the conjugate base, making acid D a stronger acid. [1]

Marking notes:

[1] for recognising the greater inductive effect with two Cl atoms
[1] for linking to increased acid strength

17. A student prepares a solution by mixing 50.0 cm³ of 0.20 mol dm⁻³ ethanoic acid with 50.0 cm³ of 0.10 mol dm⁻³ sodium hydroxide.

(a) Write an equation for the reaction that occurs. [1]

Answer: CH₃COOH(aq) + NaOH(aq) → CH₃COONa(aq) + H₂O(l) [1] OR CH₃COOH(aq) + OH⁻(aq) → CH₃COO⁻(aq) + H₂O(l)

Marking notes:

[1] for correct balanced equation

(b) Determine the pH of the resulting solution. (Kₐ for ethanoic acid = 1.8 × 10⁻⁵ mol dm⁻³) [4]

Answer: Initial moles: n(CH₃COOH) = 0.20 × (50.0/1000) = 0.0100 mol n(NaOH) = 0.10 × (50.0/1000) = 0.00500 mol [1]

NaOH is the limiting reagent. After reaction: n(CH₃COOH) remaining = 0.0100 − 0.00500 = 0.00500 mol n(CH₃COO⁻) formed = 0.00500 mol [1]

Total volume = 100.0 cm³ = 0.100 dm³ [CH₃COOH] = 0.00500 / 0.100 = 0.0500 mol dm⁻³ [CH₃COO⁻] = 0.00500 / 0.100 = 0.0500 mol dm⁻³ [1]

This is a buffer solution with [HA] = [A⁻]. [H⁺] = Kₐ × [CH₃COOH] / [CH₃COO⁻] = 1.8 × 10⁻⁵ × 0.0500/0.0500 = 1.8 × 10⁻⁵ mol dm⁻³ pH = −log₁₀(1.8 × 10⁻⁵) = 4.74 [1]

Marking notes:

[1] for correct initial moles
[1] for correct moles after reaction
[1] for correct concentrations in mixture
[1] for correct pH
Accept alternative method using Henderson-Hasselbalch equation

Answer: [H⁺] = 10⁻²·⁸⁰ = 1.58 × 10⁻³ mol dm⁻³ [1]

C₆H₅COOH ⇌ C₆H₅COO⁻ + H⁺ [C₆H₅COO⁻] = [H⁺] = 1.58 × 10⁻³ mol dm⁻³ [1]

Kₐ = [H⁺][C₆H₅COO⁻] / [C₆H₅COOH] 6.3 × 10⁻⁵ = (1.58 × 10⁻³)² / [C₆H₅COOH] [1]

[C₆H₅COOH] = (1.58 × 10⁻³)² / (6.3 × 10⁻⁵) = 0.0396 mol dm⁻³ [1]

Assumption: The acid is weak, so the equilibrium concentration of undissociated acid is approximately equal to the initial concentration. OR The degree of dissociation is small. [1 mark included in above allocation]

Marking notes:

[1] for correct [H⁺] from pH
[1] for recognising [C₆H₅COO⁻] = [H⁺]
[1] for correct substitution into Kₐ expression
[1] for correct concentration with units
Accept 0.040 mol dm⁻³ (2 sf)

Answer: [H⁺] = 10⁻⁴·⁵⁰ = 3.16 × 10⁻⁵ mol dm⁻³ [1]

Kₐ = [H⁺][CH₃COO⁻] / [CH₃COOH] 1.8 × 10⁻⁵ = (3.16 × 10⁻⁵) × [CH₃COO⁻] / [CH₃COOH] [1]

[CH₃COO⁻] / [CH₃COOH] = (1.8 × 10⁻⁵) / (3.16 × 10⁻⁵) = 0.57 [1]

Ratio [CH₃COO⁻] : [CH₃COOH] = 0.57 : 1 (or approximately 1 : 1.75)

Marking notes:

[1] for correct [H⁺] from target pH
[1] for correct substitution into Kₐ expression
[1] for correct ratio
Accept 0.57:1 or 1:1.75 or equivalent

20. Explain, using the Brønsted-Lowry theory, why the hydrogen sulfate ion, HSO₄⁻, can behave as both an acid and a base. Write equations to illustrate your answer. [3]

Answer: According to Brønsted-Lowry theory, an acid is a proton donor and a base is a proton acceptor. [1]

HSO₄⁻ acting as an acid (proton donor): HSO₄⁻(aq) + H₂O(l) ⇌ SO₄²⁻(aq) + H₃O⁺(aq) [1] OR HSO₄⁻(aq) ⇌ SO₄²⁻(aq) + H⁺(aq)

HSO₄⁻ acting as a base (proton acceptor): HSO₄⁻(aq) + H⁺(aq) ⇌ H₂SO₄(aq) [1] OR HSO₄⁻(aq) + H₃O⁺(aq) ⇌ H₂SO₄(aq) + H₂O(l)

Marking notes:

[1] for stating Brønsted-Lowry definitions (acid = proton donor, base = proton acceptor)
[1] for equation showing HSO₄⁻ as acid (donating H⁺)
[1] for equation showing HSO₄⁻ as base (accepting H⁺)
Accept any valid equations demonstrating amphiprotic behaviour

END OF ANSWER KEY