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TuitionGoWhere Practice Paper – Chemistry H1 A-Level (Version 4)

TuitionGoWhere Exam Practice (AI)

Field	Details
Subject:	Chemistry H1 (8873)
Level:	A-Level
Paper:	PRACTICE – Topic: Acids, Bases & Salts
Duration:	1 hour 15 minutes
Total Marks:	50
Name:	_________________________
Class:	_________________________
Date:	_________________________

Instructions to Candidates

This paper consists of 20 questions in three sections.
Answer all questions in the spaces provided.
Show all working clearly for calculation questions. Marks are awarded for method.
You may use a calculator and the A-Level Chemistry Data Booklet.
State symbols are required where appropriate.
The number of marks is given in brackets [ ] at the end of each question or part question.

Section A: Short Answer & Structured Response (15 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. Explain why the pH of 0.1 mol dm⁻³ ethanoic acid is higher than the pH of 0.1 mol dm⁻³ hydrochloric acid, even though both solutions have the same concentration. [2]

4. Write an expression for the acid dissociation constant, Kₐ, of ethanoic acid, CH₃COOH. State the units of Kₐ. [2]

5. A student measures the pH of a 0.050 mol dm⁻³ solution of a weak monoprotic acid, HA, and finds it to be 3.20. Calculate the Kₐ value of HA. [3]

6. State and explain the effect on the pH of a buffer solution containing CH₃COOH and CH₃COONa when a small amount of dilute hydrochloric acid is added. [2]

7. Identify the Period 3 element that forms an amphoteric oxide. Write equations to show the amphoteric nature of this oxide. [2]

Section B: Calculation & Data Interpretation (20 marks)

Answer all questions in this section.

8. A solution of sodium hydroxide was standardised by titrating it against 25.0 cm³ of 0.100 mol dm⁻³ hydrochloric acid. The volume of sodium hydroxide required was 23.40 cm³.

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

(b) Calculate the concentration of the sodium hydroxide solution in mol dm⁻³. [2]

9. Benzoic acid, C₆H₅COOH, is a weak monoprotic acid with Kₐ = 6.3 × 10⁻⁵ mol dm⁻³ at 298 K.

(a) Write an equation for the dissociation of benzoic acid in water. [1]

(b) Calculate the pH of a 0.020 mol dm⁻³ solution of benzoic acid at 298 K. [3]

10. A buffer solution is prepared by mixing 50.0 cm³ of 0.200 mol dm⁻³ ethanoic acid with 50.0 cm³ of 0.200 mol dm⁻³ sodium ethanoate. The Kₐ of ethanoic acid is 1.8 × 10⁻⁵ mol dm⁻³.

(a) Calculate the pH of this buffer solution. [2]

(b) Calculate the new pH after 5.0 cm³ of 0.100 mol dm⁻³ hydrochloric acid is added to 100 cm³ of the buffer solution. [4]

11. A student prepared a solution by dissolving 2.52 g of oxalic acid dihydrate, (COOH)₂·2H₂O, in distilled water and making the solution up to 250 cm³. Oxalic acid is a diprotic acid.

(a) Calculate the concentration of the oxalic acid solution in mol dm⁻³. [2] (Mᵣ of (COOH)₂·2H₂O = 126.0)

(b) 25.0 cm³ of this oxalic acid solution required 20.50 cm³ of sodium hydroxide solution for complete neutralisation. Calculate the concentration of the sodium hydroxide solution. [3]

12. The pH of a 0.100 mol dm⁻³ solution of a weak base, B, is 11.20 at 298 K.

(a) Calculate the pOH of the solution. [1]

(b) Calculate the concentration of hydroxide ions in the solution. [1]

(c) Calculate the base dissociation constant, Kբ, of B. [2]

Section C: Context-Based & Integrated Questions (15 marks)

Answer all questions in this section.

13. Carbonic acid, H₂CO₃, is formed when carbon dioxide dissolves in rainwater. It is a weak diprotic acid.

(a) Write an equation for the first dissociation of carbonic acid in water, including state symbols. [1]

(b) Write the expression for the first acid dissociation constant, Kₐ₁, of carbonic acid. [1]

(c) Explain why rainwater is naturally slightly acidic, even in unpolluted environments. [2]

14. In the production of yoghurt, bacteria convert lactose into lactic acid, CH₃CH(OH)COOH. Lactic acid is a weak monoprotic acid with Kₐ = 1.4 × 10⁻⁴ mol dm⁻³.

(a) Calculate the pH of a 0.010 mol dm⁻³ solution of lactic acid. [3]

(b) Explain why the pH of the yoghurt does not change significantly as more lactic acid is produced during fermentation, given that the yoghurt contains phosphate salts that act as a buffer. [2]

15. A student investigates the reaction between magnesium ribbon and two different acids of the same concentration: 0.50 mol dm⁻³ hydrochloric acid (a strong acid) and 0.50 mol dm⁻³ ethanoic acid (a weak acid).

(a) State and explain one similarity and one difference the student would observe in the reactions. [2]

(b) The student repeats the experiment using 0.50 mol dm⁻³ sulfuric acid instead of hydrochloric acid. Explain why the rate of reaction with sulfuric acid is faster, even though both are strong acids. [2]

16. Aluminium hydroxide, Al(OH)₃, is used in some antacid tablets to neutralise excess stomach acid (hydrochloric acid).

(a) Write a balanced equation for the reaction between aluminium hydroxide and hydrochloric acid. [1]

(b) An antacid tablet contains 250 mg of Al(OH)₃. Calculate the volume of 0.100 mol dm⁻³ hydrochloric acid that can be neutralised by one tablet. [3] (Mᵣ of Al(OH)₃ = 78.0)

17. The amino acid glycine, H₂NCH₂COOH, can act as both an acid and a base in aqueous solution.

(a) Write an equation to show glycine acting as an acid in water. [1]

(b) Write an equation to show glycine acting as a base in water. [1]

(c) Explain why aqueous solutions of glycine can act as buffer solutions. [2]

18. A student prepared a buffer solution by partially neutralising 25.0 cm³ of 0.100 mol dm⁻³ ethanoic acid with 10.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide solution.

(a) Calculate the number of moles of ethanoic acid remaining after the reaction. [2]

(b) Calculate the number of moles of sodium ethanoate formed. [1]

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

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

20. A 0.0250 mol dm⁻³ solution of a weak monoprotic acid has a pH of 3.80. A student claims that this acid is ethanoic acid (Kₐ = 1.8 × 10⁻⁵ mol dm⁻³). By calculation, determine whether the student's claim is valid. [3]

END OF PAPER

This paper was generated by TuitionGoWhere for practice purposes. It follows the structure and style of A-Level H1 Chemistry assessments.

Answers

TuitionGoWhere Practice Paper – Chemistry H1 A-Level (Version 4)

ANSWER KEY AND MARKING SCHEME

TuitionGoWhere Exam Practice (AI)

Field	Details
Subject:	Chemistry H1 (8873)
Level:	A-Level
Paper:	PRACTICE – Topic: Acids, Bases & Salts
Total Marks:	50

Section A: Short Answer & Structured Response (15 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 one that only partially dissociates/ionises in aqueous solution. [1]

Equation (any valid weak acid accepted): CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq) [1]

Marking notes:

Award [1] for correct definition including "partially dissociates/ionises"
Award [1] for correct equation with reversible arrow (⇌) and state symbols
Accept other weak acids e.g. HCOOH, H₂CO₃, HF
Do NOT accept single arrow (→) – this indicates complete dissociation
Do NOT accept "dilute acid" as definition of weak acid

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

Answer: A strong acid is one that completely dissociates/ionises in aqueous solution, whereas a concentrated acid contains a large amount of acid dissolved per unit volume of solution. [2]

Marking notes:

Award [1] for correct definition of strong acid (complete dissociation)
Award [1] for correct definition of concentrated acid (amount per volume)
Key distinction: strength refers to degree of dissociation; concentration refers to amount of solute
Accept: "Strength is about dissociation; concentration is about how much acid is present"

3. Explain why the pH of 0.1 mol dm⁻³ ethanoic acid is higher than the pH of 0.1 mol dm⁻³ hydrochloric acid, even though both solutions have the same concentration. [2]

Answer: Ethanoic acid is a weak acid that only partially dissociates in water, producing a lower concentration of H⁺ ions. [1] Hydrochloric acid is a strong acid that completely dissociates, producing a higher concentration of H⁺ ions. Since pH = −log[H⁺], a lower [H⁺] gives a higher pH. [1]

Marking notes:

Award [1] for stating ethanoic acid is weak (partial dissociation) and HCl is strong (complete dissociation)
Award [1] for linking [H⁺] to pH
Accept: "Fewer H⁺ ions in ethanoic acid solution → higher pH"

4. Write an expression for the acid dissociation constant, Kₐ, of ethanoic acid, CH₃COOH. State the units of Kₐ. [2]

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

Units: mol dm⁻³ [1]

Marking notes:

Award [1] for correct expression with products in numerator, reactant in denominator
Award [1] for correct units
Do NOT include [H₂O] in expression
Accept: mol/dm³ or M

5. A student measures the pH of a 0.050 mol dm⁻³ solution of a weak monoprotic acid, HA, and finds it to be 3.20. Calculate the Kₐ value of HA. [3]

Answer: [H⁺] = 10⁻³·²⁰ = 6.31 × 10⁻⁴ mol dm⁻³ [1]

HA ⇌ H⁺ + A⁻ At equilibrium: [H⁺] = [A⁻] = 6.31 × 10⁻⁴ mol dm⁻³ [HA] ≈ 0.050 − 6.31 × 10⁻⁴ ≈ 0.0494 mol dm⁻³ [1]

Kₐ = [H⁺][A⁻] / [HA] = (6.31 × 10⁻⁴)² / 0.0494 = 8.06 × 10⁻⁶ mol dm⁻³ [1]

Marking notes:

Award [1] for correct [H⁺] from pH
Award [1] for correct equilibrium concentrations (allow approximation [HA] ≈ 0.050)
Award [1] for correct Kₐ value with units
Accept: 7.9 × 10⁻⁶ to 8.1 × 10⁻⁶ mol dm⁻³ (depending on rounding)

6. State and explain the effect on the pH of a buffer solution containing CH₃COOH and CH₃COONa when a small amount of dilute hydrochloric acid is added. [2]

Answer: The pH remains approximately constant / changes very little. [1]

The added H⁺ ions react with the conjugate base CH₃COO⁻ to form undissociated CH₃COOH: CH₃COO⁻ + H⁺ → CH₃COOH This removes the added H⁺ ions from solution, so [H⁺] and hence pH remain largely unchanged. [1]

Marking notes:

Award [1] for stating pH remains approximately constant
Award [1] for correct explanation with equation or description of H⁺ removal
Accept: "The ethanoate ions 'mop up' the added H⁺ ions"

7. Identify the Period 3 element that forms an amphoteric oxide. Write equations to show the amphoteric nature of this oxide. [2]

Answer: Aluminium (Al) [1]

Equations: Al₂O₃(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂O(l) (reaction with acid) Al₂O₃(s) + 2NaOH(aq) + 3H₂O(l) → 2Na[Al(OH)₄](aq) (reaction with base) [1]

Marking notes:

Award [1] for identifying aluminium
Award [1] for both equations showing reaction with acid AND base
Accept: Al₂O₃ + 6H⁺ → 2Al³⁺ + 3H₂O for acid reaction
Accept: Al₂O₃ + 2OH⁻ + 3H₂O → 2[Al(OH)₄]⁻ for base reaction
Do NOT accept silicon (SiO₂ is acidic, not amphoteric)

Section B: Calculation & Data Interpretation (20 marks)

8. A solution of sodium hydroxide was standardised by titrating it against 25.0 cm³ of 0.100 mol dm⁻³ hydrochloric acid. The volume of sodium hydroxide required was 23.40 cm³.

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

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

Marking notes:

Award [1] for correct balanced equation with state symbols
Accept: OH⁻ + H⁺ → H₂O

(b) Calculate the concentration of the sodium hydroxide solution in mol dm⁻³. [2]

Answer: n(HCl) = c × V = 0.100 × (25.0/1000) = 0.00250 mol [1]

From equation, 1:1 ratio: n(NaOH) = 0.00250 mol c(NaOH) = n/V = 0.00250 / (23.40/1000) = 0.107 mol dm⁻³ [1]

Marking notes:

Award [1] for correct moles of HCl
Award [1] for correct concentration with units
Accept: 0.1068 mol dm⁻³ (3 s.f.)

9. Benzoic acid, C₆H₅COOH, is a weak monoprotic acid with Kₐ = 6.3 × 10⁻⁵ mol dm⁻³ at 298 K.

(a) Write an equation for the dissociation of benzoic acid in water. [1]

Answer: C₆H₅COOH(aq) ⇌ C₆H₅COO⁻(aq) + H⁺(aq) [1]

Marking notes:

Award [1] for correct equation with reversible arrow and state symbols
Must use ⇌, not →

(b) Calculate the pH of a 0.020 mol dm⁻³ solution of benzoic acid at 298 K. [3]

Answer: Kₐ = [H⁺]² / [C₆H₅COOH] [H⁺]² = Kₐ × [C₆H₅COOH] = 6.3 × 10⁻⁵ × 0.020 = 1.26 × 10⁻⁶ [1] [H⁺] = √(1.26 × 10⁻⁶) = 1.12 × 10⁻³ mol dm⁻³ [1] pH = −log(1.12 × 10⁻³) = 2.95 [1]

Marking notes:

Award [1] for correct substitution into Kₐ expression
Award [1] for correct [H⁺]
Award [1] for correct pH (2.95 or 2.9)
Accept assumption that [C₆H₅COOH]eq ≈ 0.020 mol dm⁻³ (dissociation is small)

(a) Calculate the pH of this buffer solution. [2]

Answer: After mixing, total volume = 100 cm³ [CH₃COOH] = (0.200 × 50.0/1000) / (100/1000) = 0.100 mol dm⁻³ [CH₃COO⁻] = (0.200 × 50.0/1000) / (100/1000) = 0.100 mol dm⁻³ [1]

pH = pKₐ + log([CH₃COO⁻]/[CH₃COOH]) pKₐ = −log(1.8 × 10⁻⁵) = 4.74 pH = 4.74 + log(0.100/0.100) = 4.74 [1]

Marking notes:

Award [1] for correct concentrations after mixing (or recognising ratio = 1)
Award [1] for correct pH
Accept: pH = pKₐ when [acid] = [conjugate base]

(b) Calculate the new pH after 5.0 cm³ of 0.100 mol dm⁻³ hydrochloric acid is added to 100 cm³ of the buffer solution. [4]

Answer: Initial moles in 100 cm³ buffer: n(CH₃COOH) = 0.100 × (100/1000) = 0.0100 mol n(CH₃COO⁻) = 0.100 × (100/1000) = 0.0100 mol [1]

Moles of H⁺ added = 0.100 × (5.0/1000) = 5.0 × 10⁻⁴ mol [1]

Added H⁺ reacts with CH₃COO⁻: New n(CH₃COO⁻) = 0.0100 − 5.0 × 10⁻⁴ = 0.0095 mol New n(CH₃COOH) = 0.0100 + 5.0 × 10⁻⁴ = 0.0105 mol [1]

New total volume = 105 cm³ (but ratio used, so volumes cancel) pH = 4.74 + log(0.0095/0.0105) = 4.74 + log(0.905) = 4.74 − 0.043 = 4.70 [1]

Marking notes:

Award [1] for initial moles of acid and conjugate base
Award [1] for moles of H⁺ added
Award [1] for new moles after reaction
Award [1] for correct new pH
Accept: 4.69–4.70

11. A student prepared a solution by dissolving 2.52 g of oxalic acid dihydrate, (COOH)₂·2H₂O, in distilled water and making the solution up to 250 cm³. Oxalic acid is a diprotic acid.

(a) Calculate the concentration of the oxalic acid solution in mol dm⁻³. [2] (Mᵣ of (COOH)₂·2H₂O = 126.0)

Answer: n = m/Mᵣ = 2.52 / 126.0 = 0.0200 mol [1] c = n/V = 0.0200 / (250/1000) = 0.0800 mol dm⁻³ [1]

Marking notes:

Award [1] for correct moles
Award [1] for correct concentration with units

(b) 25.0 cm³ of this oxalic acid solution required 20.50 cm³ of sodium hydroxide solution for complete neutralisation. Calculate the concentration of the sodium hydroxide solution. [3]

Answer: Equation: (COOH)₂ + 2NaOH → (COONa)₂ + 2H₂O n(oxalic acid) = 0.0800 × (25.0/1000) = 0.00200 mol [1]

From equation, 1:2 ratio: n(NaOH) = 2 × 0.00200 = 0.00400 mol [1] c(NaOH) = n/V = 0.00400 / (20.50/1000) = 0.195 mol dm⁻³ [1]

Marking notes:

Award [1] for moles of oxalic acid
Award [1] for correct mole ratio (1:2) and moles of NaOH
Award [1] for correct concentration with units
Accept: 0.1951 mol dm⁻³

12. The pH of a 0.100 mol dm⁻³ solution of a weak base, B, is 11.20 at 298 K.

(a) Calculate the pOH of the solution. [1]

Answer: pOH = 14.00 − pH = 14.00 − 11.20 = 2.80 [1]

(b) Calculate the concentration of hydroxide ions in the solution. [1]

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

(c) Calculate the base dissociation constant, Kբ, of B. [2]

Answer: B + H₂O ⇌ BH⁺ + OH⁻ [OH⁻] = [BH⁺] = 1.58 × 10⁻³ mol dm⁻³ [B] ≈ 0.100 − 1.58 × 10⁻³ ≈ 0.0984 mol dm⁻³ [1]

Kբ = [BH⁺][OH⁻] / [B] = (1.58 × 10⁻³)² / 0.0984 = 2.54 × 10⁻⁵ mol dm⁻³ [1]

Marking notes:

Award [1] for correct equilibrium concentrations
Award [1] for correct Kբ value with units
Accept: 2.5 × 10⁻⁵ mol dm⁻³

Section C: Context-Based & Integrated Questions (15 marks)

13. Carbonic acid, H₂CO₃, is formed when carbon dioxide dissolves in rainwater. It is a weak diprotic acid.

(a) Write an equation for the first dissociation of carbonic acid in water, including state symbols. [1]

Answer: H₂CO₃(aq) ⇌ HCO₃⁻(aq) + H⁺(aq) [1]

Marking notes:

Award [1] for correct equation with ⇌ and state symbols
Must show first dissociation only

(b) Write the expression for the first acid dissociation constant, Kₐ₁, of carbonic acid. [1]

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

Marking notes:

Award [1] for correct expression
Do NOT include [H₂O]

(c) Explain why rainwater is naturally slightly acidic, even in unpolluted environments. [2]

Answer: Carbon dioxide in the atmosphere dissolves in rainwater to form carbonic acid: [1] CO₂(g) + H₂O(l) ⇌ H₂CO₃(aq) Carbonic acid is a weak acid that partially dissociates, producing H⁺ ions, making rainwater slightly acidic (pH ≈ 5.6). [1]

Marking notes:

Award [1] for mentioning CO₂ dissolving to form H₂CO₃
Award [1] for linking to H⁺ production and acidity
Accept: equation for CO₂ + H₂O ⇌ H₂CO₃

14. In the production of yoghurt, bacteria convert lactose into lactic acid, CH₃CH(OH)COOH. Lactic acid is a weak monoprotic acid with Kₐ = 1.4 × 10⁻⁴ mol dm⁻³.

(a) Calculate the pH of a 0.010 mol dm⁻³ solution of lactic acid. [3]

Answer: Kₐ = [H⁺]² / [HA] [H⁺]² = 1.4 × 10⁻⁴ × 0.010 = 1.4 × 10⁻⁶ [1] [H⁺] = √(1.4 × 10⁻⁶) = 1.18 × 10⁻³ mol dm⁻³ [1] pH = −log(1.18 × 10⁻³) = 2.93 [1]

Marking notes:

Award [1] for correct substitution
Award [1] for correct [H⁺]
Award [1] for correct pH
Accept: 2.92–2.93

(b) Explain why the pH of the yoghurt does not change significantly as more lactic acid is produced during fermentation, given that the yoghurt contains phosphate salts that act as a buffer. [2]

Answer: The phosphate salts act as a buffer, resisting changes in pH. [1] The added H⁺ ions from lactic acid react with the conjugate base (HPO₄²⁻ or PO₄³⁻) in the phosphate buffer, forming the weak acid component. This removes excess H⁺ from solution, so the pH remains relatively constant. [1]

Marking notes:

Award [1] for stating buffer resists pH change
Award [1] for explanation of how buffer works (H⁺ reacts with conjugate base)
Accept: reference to equilibrium shifting to remove added H⁺

(a) State and explain one similarity and one difference the student would observe in the reactions. [2]

Answer: Similarity: Both reactions produce hydrogen gas / both reactions produce the same salt (magnesium chloride/ethanoate). [1]

Difference: The reaction with hydrochloric acid is faster / more vigorous than with ethanoic acid. This is because HCl is a strong acid that completely dissociates, giving a higher [H⁺], whereas ethanoic acid is weak and partially dissociates, giving a lower [H⁺]. [1]

Marking notes:

Award [1] for valid similarity
Award [1] for valid difference with explanation linking to [H⁺]
Accept: "Same volume of H₂ produced eventually" as similarity
Accept: "HCl reaction produces more effervescence initially" as difference

Answer: Sulfuric acid is diprotic (H₂SO₄), producing 2 moles of H⁺ per mole of acid, whereas HCl is monoprotic, producing only 1 mole of H⁺ per mole of acid. [1] Therefore, 0.50 mol dm⁻³ H₂SO₄ has [H⁺] = 1.0 mol dm⁻³, which is twice the [H⁺] of 0.50 mol dm⁻³ HCl (0.50 mol dm⁻³). The higher [H⁺] leads to a faster reaction rate. [1]

Marking notes:

Award [1] for stating H₂SO₄ is diprotic
Award [1] for linking higher [H⁺] to faster rate
Accept: calculation showing [H⁺] in H₂SO₄ = 1.0 mol dm⁻³

16. Aluminium hydroxide, Al(OH)₃, is used in some antacid tablets to neutralise excess stomach acid (hydrochloric acid).

(a) Write a balanced equation for the reaction between aluminium hydroxide and hydrochloric acid. [1]

Answer: Al(OH)₃(s) + 3HCl(aq) → AlCl₃(aq) + 3H₂O(l) [1]

Marking notes:

Award [1] for correct balanced equation with state symbols

(b) An antacid tablet contains 250 mg of Al(OH)₃. Calculate the volume of 0.100 mol dm⁻³ hydrochloric acid that can be neutralised by one tablet. [3] (Mᵣ of Al(OH)₃ = 78.0)

Answer: n(Al(OH)₃) = m/Mᵣ = 0.250 / 78.0 = 3.205 × 10⁻³ mol [1]

From equation, 1:3 ratio: n(HCl) = 3 × 3.205 × 10⁻³ = 9.615 × 10⁻³ mol [1]

V(HCl) = n/c = 9.615 × 10⁻³ / 0.100 = 0.0962 dm³ = 96.2 cm³ [1]

Marking notes:

Award [1] for correct moles of Al(OH)₃ (must convert mg to g)
Award [1] for correct mole ratio and moles of HCl
Award [1] for correct volume with units
Accept: 96.0–96.5 cm³

17. The amino acid glycine, H₂NCH₂COOH, can act as both an acid and a base in aqueous solution.

(a) Write an equation to show glycine acting as an acid in water. [1]

Answer: H₂NCH₂COOH(aq) + H₂O(l) ⇌ H₂NCH₂COO⁻(aq) + H₃O⁺(aq) [1]

Marking notes:

Award [1] for equation showing glycine donating H⁺ (from COOH group)
Accept: H₂NCH₂COOH ⇌ H₂NCH₂COO⁻ + H⁺

(b) Write an equation to show glycine acting as a base in water. [1]

Answer: H₂NCH₂COOH(aq) + H₂O(l) ⇌ ⁺H₃NCH₂COOH(aq) + OH⁻(aq) [1]

Marking notes:

Award [1] for equation showing glycine accepting H⁺ (on NH₂ group)
Accept: H₂NCH₂COOH + H⁺ ⇌ ⁺H₃NCH₂COOH

(c) Explain why aqueous solutions of glycine can act as buffer solutions. [2]

Answer: Glycine is amphoteric – it can act as both a weak acid (donating H⁺ from COOH) and a weak base (accepting H⁺ on NH₂). [1] In solution, glycine exists as a zwitterion (⁺H₃NCH₂COO⁻) which can react with both added acid and added base, resisting changes in pH. [1]

Marking notes:

Award [1] for identifying amphoteric nature
Award [1] for explaining buffer action (reacts with both H⁺ and OH⁻)
Accept: "The COO⁻ group neutralises added acid; the NH₃⁺ group neutralises added base"

18. A student prepared a buffer solution by partially neutralising 25.0 cm³ of 0.100 mol dm⁻³ ethanoic acid with 10.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide solution.

(a) Calculate the number of moles of ethanoic acid remaining after the reaction. [2]

Answer: Initial n(CH₃COOH) = 0.100 × (25.0/1000) = 2.50 × 10⁻³ mol [1] n(NaOH) added = 0.100 × (10.0/1000) = 1.00 × 10⁻³ mol

CH₃COOH + NaOH → CH₃COONa + H₂O n(CH₃COOH) remaining = 2.50 × 10⁻³ − 1.00 × 10⁻³ = 1.50 × 10⁻³ mol [1]

Marking notes:

Award [1] for initial moles of CH₃COOH
Award [1] for correct remaining moles

(b) Calculate the number of moles of sodium ethanoate formed. [1]

Answer: n(CH₃COONa) = n(NaOH) = 1.00 × 10⁻³ mol [1]

Marking notes:

Award [1] for correct moles (1:1 ratio with NaOH)

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

Answer: Total volume = 25.0 + 10.0 = 35.0 cm³ (ratio used, volumes cancel) pH = pKₐ + log([CH₃COO⁻]/[CH₃COOH]) pKₐ = −log(1.8 × 10⁻⁵) = 4.74 [1] pH = 4.74 + log(1.00 × 10⁻³ / 1.50 × 10⁻³) = 4.74 + log(0.667) = 4.74 − 0.176 = 4.56 [1]

Marking notes:

Award [1] for pKₐ
Award [1] for correct pH
Accept: 4.56–4.57

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

Answer: Aqueous ammonia is a weak base: NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq) [1] Ammonium chloride provides the conjugate acid, NH₄⁺: NH₄Cl(aq) → NH₄⁺(aq) + Cl⁻(aq) [1]

When a small amount of acid (H⁺) is added, it reacts with NH₃: NH₃ + H⁺ → NH₄⁺ When a small amount of base (OH⁻) is added, it reacts with NH₄⁺: NH₄⁺ + OH⁻ → NH₃ + H₂O This removes the added H⁺ or OH⁻, so the pH remains approximately constant. [1]

Marking notes:

Award [1] for identifying NH₃ as weak base with equilibrium equation
Award [1] for identifying NH₄⁺ as conjugate acid
Award [1] for explaining buffer action with both equations
Accept: reference to equilibrium shifting to counteract changes

Answer: [H⁺] = 10⁻³·⁸⁰ = 1.58 × 10⁻⁴ mol dm⁻³ [1]

Kₐ = [H⁺]² / [HA] = (1.58 × 10⁻⁴)² / 0.0250 = 2.50 × 10⁻⁸ / 0.0250 = 1.00 × 10⁻⁶ mol dm⁻³ [1]

The calculated Kₐ (1.00 × 10⁻⁶) is significantly different from the Kₐ of ethanoic acid (1.8 × 10⁻⁵). Therefore, the student's claim is NOT valid – the acid is much weaker than ethanoic acid. [1]

Marking notes:

Award [1] for correct [H⁺] from pH
Award [1] for correct calculation of Kₐ
Award [1] for correct conclusion with justification (comparison of values)
Accept: "No, because calculated Kₐ ≠ 1.8 × 10⁻⁵"

END OF ANSWER KEY

Total marks: 50