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A Level H1 Chemistry Practice Paper 3

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TuitionGoWhere Practice Paper - Chemistry H1 A-Level

TuitionGoWhere Practice Paper (AI)

Subject: Chemistry H1 (8873) Level: A-Level Paper: Practice Paper – Version 3 Duration: 2 hours Total Marks: 80

Name: _________________________ Class: _________________________ Date: _________________________

Instructions to Candidates

This paper consists of three sections: Section A, Section B, and Section C.
Answer all questions in Section A and Section B.
Section C consists of two questions. Answer one question only.
Write your answers in the spaces provided.
Show all working clearly for calculation questions. Marks are awarded for method.
You may use a calculator. A Data Booklet is provided.
The number of marks is given in brackets [ ] at the end of each question or part question.

Section A: Multiple Choice (15 marks)

Answer all questions in this section. Circle the letter corresponding to the correct answer.

1. Which of the following statements best describes a Brønsted-Lowry base?

A. A substance that donates a proton. B. A substance that accepts a proton. C. A substance that produces OH⁻ ions in water. D. A substance that accepts an electron pair.

[1]

2. The pH of 0.10 mol dm⁻³ hydrochloric acid is 1.0. What is the pH of 0.10 mol dm⁻³ ethanoic acid?

A. 1.0 B. Between 1.0 and 7.0 C. Exactly 7.0 D. Greater than 7.0

[1]

3. Which pair of substances forms an acidic buffer solution when mixed in appropriate amounts?

A. HCl and NaCl B. CH₃COOH and CH₃COONa C. NH₃ and NH₄Cl D. NaOH and NaCl

[1]

4. The acid dissociation constant, Ka, of benzoic acid is 6.3 × 10⁻⁵ mol dm⁻³. What is the pKa of benzoic acid?

A. 3.20 B. 4.20 C. 5.20 D. 6.30

[1]

5. Which indicator is most suitable for a titration between a strong acid and a weak base?

A. Methyl orange (pH range 3.1–4.4) B. Bromothymol blue (pH range 6.0–7.6) C. Phenolphthalein (pH range 8.3–10.0) D. Litmus (pH range 5.0–8.0)

[1]

6. A student adds a few drops of universal indicator to a solution and observes a green colour. What is the approximate pH of the solution?

A. 1–2 B. 3–4 C. 7–8 D. 12–13

[1]

7. Which of the following is not a conjugate acid-base pair?

A. H₃O⁺ and H₂O B. NH₄⁺ and NH₃ C. HCl and Cl⁻ D. H₂SO₄ and SO₄²⁻

[1]

8. What is the ionic product of water, Kw, at 298 K?

A. 1.0 × 10⁻⁷ mol² dm⁻⁶ B. 1.0 × 10⁻¹⁴ mol² dm⁻⁶ C. 1.0 × 10⁻¹⁴ mol dm⁻³ D. 1.0 × 10⁻⁷ mol dm⁻³

[1]

9. A solution has a hydroxide ion concentration of 1.0 × 10⁻³ mol dm⁻³. What is the pH of the solution?

A. 3 B. 7 C. 11 D. 14

[1]

10. Which salt is produced when sulfuric acid is neutralised by potassium hydroxide?

A. KSO₄ B. K₂SO₄ C. K(SO₄)₂ D. KHSO₄

[1]

11. Which of the following is an example of an amphoteric oxide?

A. Na₂O B. MgO C. Al₂O₃ D. SO₂

[1]

12. A student dilutes 10.0 cm³ of 1.0 mol dm⁻³ HCl to 100.0 cm³ with distilled water. What is the new pH?

A. 0 B. 1 C. 2 D. 3

[1]

13. Which equation correctly represents the second dissociation of sulfuric acid in water?

A. H₂SO₄(aq) → 2H⁺(aq) + SO₄²⁻(aq) B. HSO₄⁻(aq) ⇌ H⁺(aq) + SO₄²⁻(aq) C. H₂SO₄(aq) ⇌ H⁺(aq) + HSO₄⁻(aq) D. HSO₄⁻(aq) → H⁺(aq) + SO₄²⁻(aq)

[1]

14. In the reaction NH₃ + H₂O ⇌ NH₄⁺ + OH⁻, which species acts as a Brønsted-Lowry acid?

A. NH₃ B. H₂O C. NH₄⁺ D. OH⁻

[1]

15. A buffer solution contains equal concentrations of a weak acid and its conjugate base. What is the relationship between pH and pKa?

A. pH = pKa B. pH = ½ pKa C. pH = 2 × pKa D. pH = pKa + 1

[1]

Section B: Structured Questions (45 marks)

Answer all questions in this section.

16. Methanoic acid, HCOOH, is a weak acid found in ant venom.

(a) Define the term weak acid. [1]

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

(d) The Ka of methanoic acid is 1.8 × 10⁻⁴ mol dm⁻³. Calculate the pH of a 0.050 mol dm⁻³ solution of methanoic acid. State any assumption you make. [3]

(e) Explain why the pH of 0.050 mol dm⁻³ methanoic acid is higher than that of 0.050 mol dm⁻³ hydrochloric acid. [2]

[Total: 8 marks]

17. A student performs a titration to determine the concentration of a sodium hydroxide solution. 25.0 cm³ of the NaOH solution is titrated against 0.100 mol dm⁻³ sulfuric acid. The student obtains the following results:

Titration	1	2	3	4
Final burette reading / cm³	24.50	47.80	23.60	46.90
Initial burette reading / cm³	0.00	23.50	0.00	23.60
Volume of H₂SO₄ used / cm³	24.50	24.30	23.60	23.30

(a) Complete the table by calculating the volume of H₂SO₄ used in each titration. [1]

(b) Identify which titrations should be used to calculate the average titre. Explain your choice. [2]

(d) The equation for the reaction is: 2NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2H₂O(l)

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

(e) The student rinses the conical flask with distilled water before the titration but does not dry it. Explain whether this affects the accuracy of the titre. [2]

[Total: 9 marks]

18. A buffer solution is prepared by dissolving 4.10 g of sodium ethanoate (CH₃COONa, Mr = 82.0) in 500 cm³ of 0.200 mol dm⁻³ ethanoic acid.

[Ka of ethanoic acid = 1.8 × 10⁻⁵ mol dm⁻³]

(a) Calculate the concentration of sodium ethanoate in the buffer solution. [2]

(b) Calculate the pH of this buffer solution. [3]

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

(d) A student claims that diluting the buffer solution with water will change its pH. Explain whether the student is correct. [2]

[Total: 10 marks]

19. The graph below shows the pH titration curve obtained when 25.0 cm³ of an unknown monoprotic acid is titrated against 0.100 mol dm⁻³ sodium hydroxide solution.

[Graph description: The curve starts at pH 2.9, rises gradually, has a steep rise between 22.0 cm³ and 26.0 cm³, with the equivalence point at 24.0 cm³ and pH 8.3. The curve ends at pH 12.5.]

(a) State the type of acid (strong or weak) being titrated. Use evidence from the graph to justify your answer. [2]

(b) Determine the concentration of the unknown acid. [2]

(d) Suggest a suitable indicator for this titration. Explain your choice by referring to the indicator's pH range and the titration curve. [2]

(e) On the axes provided, sketch the pH titration curve that would be obtained if the same acid were titrated against 0.100 mol dm⁻³ ammonia solution (a weak base). Label the equivalence point. [2]

[Blank axes provided for student sketch]

[Total: 10 marks]

20. Calcium oxide (CaO) reacts with water to form calcium hydroxide, which is used in agriculture to neutralise acidic soils.

(a) Write a balanced equation, including state symbols, for the reaction of calcium oxide with water. [1]

(b) Explain why calcium hydroxide is classified as a base. [1]

(c) A farmer adds 2.80 kg of calcium oxide to soil. Calculate the mass of calcium hydroxide produced, assuming complete reaction.

[Mr: CaO = 56.1; Ca(OH)₂ = 74.1] [3]

(d) The soil has a pH of 4.5. Calculate the hydrogen ion concentration in the soil water. [2]

(e) Explain why adding calcium hydroxide increases the pH of the soil. Include an ionic equation in your answer. [1]

[Total: 8 marks]

Section C: Free Response Questions (20 marks)

Answer one question only from this section. Indicate clearly which question you are answering.

EITHER

21. Acids and bases are fundamental to many chemical and biological processes.

(a) Distinguish between a strong acid and a concentrated acid, using hydrochloric acid as an example in your explanation. [3]

(b) The pH of blood is maintained at approximately 7.4 by several buffer systems, including the carbonic acid-hydrogencarbonate buffer.

(i) Write an equation to show how carbonic acid is formed when carbon dioxide dissolves in blood. [1]

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

(iii) Explain how the carbonic acid-hydrogencarbonate buffer system maintains blood pH when excess H⁺ ions enter the blood. Include an equation in your answer. [3]

(c) A student prepares a buffer solution by mixing 50.0 cm³ of 0.100 mol dm⁻³ propanoic acid (C₂H₅COOH) with 25.0 cm³ of 0.100 mol dm⁻³ sodium propanoate (C₂H₅COONa).

[Ka of propanoic acid = 1.3 × 10⁻⁵ mol dm⁻³]

(i) Calculate the pH of this buffer solution. [4]

(ii) Calculate the new pH after 5.0 cm³ of 0.100 mol dm⁻³ HCl is added to the buffer solution. [5]

(d) Explain why a mixture of hydrochloric acid and sodium chloride does not function as a buffer solution. [3]

[Total: 20 marks]

OR

22. The chemistry of acids and bases has important applications in industry and environmental science.

(a) Sulfuric acid is manufactured industrially by the Contact Process. One step involves the equilibrium:

2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH = –197 kJ mol⁻¹

(i) State Le Chatelier's principle. [1]

(ii) Explain why a low temperature favours the formation of SO₃, but in practice a temperature of 450°C is used. [3]

(iii) State and explain the effect of increasing pressure on the equilibrium yield of SO₃. [2]

(b) Acid rain is formed when sulfur dioxide and oxides of nitrogen dissolve in rainwater.

(i) Write an equation to show how sulfur dioxide forms an acidic solution in water. [1]

(ii) Explain why acid rain damages limestone buildings. Include an equation in your answer. [3]

(i) Calculate the hydrogen ion concentration in this rainwater sample. [2]

(ii) The main acid present in this rainwater is sulfuric acid, formed from the oxidation of SO₂. Assuming all the H⁺ ions come from sulfuric acid (a strong diprotic acid), calculate the concentration of sulfuric acid in the rainwater. [2]

(iii) Explain why the pH of pure water is not exactly 7.0 when measured in the laboratory. [2]

(d) Describe how you would prepare a pure, dry sample of copper(II) sulfate crystals (CuSO₄·5H₂O) starting from copper(II) oxide and dilute sulfuric acid. Include a balanced equation in your answer. [4]

[Total: 20 marks]

END OF PAPER

Answers

TuitionGoWhere Practice Paper - Chemistry H1 A-Level – Answer Key and Marking Scheme

Paper: Practice Paper – Version 3 Total Marks: 80

Section A: Multiple Choice (15 marks)

Question	Answer	Explanation
1	B	A Brønsted-Lowry base is a proton (H⁺) acceptor. A is the definition of an acid. C is the Arrhenius definition. D is a Lewis base.
2	B	Ethanoic acid is a weak acid, so it partially dissociates. [H⁺] < 0.10 mol dm⁻³, so pH > 1.0 but still acidic (< 7.0).
3	B	An acidic buffer requires a weak acid and its conjugate base (salt). CH₃COOH is a weak acid; CH₃COONa provides the conjugate base CH₃COO⁻. A and D are strong acid/base combinations. C is an alkaline buffer.
4	B	pKa = –log₁₀(Ka) = –log₁₀(6.3 × 10⁻⁵) = 4.20
5	A	Strong acid-weak base titration has equivalence point pH < 7. Methyl orange (3.1–4.4) changes colour in the acidic range and is suitable.
6	C	Green on universal indicator corresponds to neutral/very weakly acidic or alkaline, approximately pH 7–8.
7	D	A conjugate acid-base pair differs by one H⁺. H₂SO₄ and SO₄²⁻ differ by two H⁺, so they are not a conjugate pair. The conjugate base of H₂SO₄ is HSO₄⁻.
8	B	At 298 K, Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶.
9	C	pOH = –log₁₀(1.0 × 10⁻³) = 3. pH = 14 – pOH = 14 – 3 = 11.
10	B	H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O. The salt is potassium sulfate, K₂SO₄.
11	C	Al₂O₃ is amphoteric (reacts with both acids and bases). Na₂O and MgO are basic; SO₂ is acidic.
12	B	New concentration = (1.0 × 10.0) / 100.0 = 0.10 mol dm⁻³. pH = –log₁₀(0.10) = 1.
13	B	The second dissociation of H₂SO₄ is partial: HSO₄⁻ ⇌ H⁺ + SO₄²⁻. The first dissociation is complete (→).
14	B	H₂O donates a proton to NH₃, forming OH⁻. A Brønsted-Lowry acid is a proton donor.
15	A	When [HA] = [A⁻], the Henderson-Hasselbalch equation gives pH = pKa + log(1) = pKa.

Section B: Structured Questions (45 marks)

Question 16 (8 marks)

(a) A weak acid is an acid that partially dissociates/ionises in water, establishing an equilibrium between the undissociated acid molecules and its ions. [1]

(b) HCOOH(aq) ⇌ HCOO⁻(aq) + H⁺(aq) [1] Mark: Award [1] for correct equation with equilibrium arrow and state symbols.

(c) Ka = [HCOO⁻][H⁺] / [HCOOH] [1] Mark: Accept any correct format. Do not penalise if [H₂O] is omitted (correct).

(d)

Assumption: Degree of dissociation is small, so [HCOOH]eq ≈ 0.050 mol dm⁻³ [1]
Ka = [H⁺]² / [HCOOH]
[H⁺] = √(Ka × [HCOOH]) = √(1.8 × 10⁻⁴ × 0.050) = √(9.0 × 10⁻⁶) = 3.0 × 10⁻³ mol dm⁻³ [1]
pH = –log₁₀(3.0 × 10⁻³) = 2.52 [1] Mark: [1] for assumption, [1] for correct [H⁺] calculation, [1] for correct pH. Accept 2.5 or 2.52.

(e) HCl is a strong acid and dissociates completely, so [H⁺] = 0.050 mol dm⁻³, giving pH = 1.30. [1] Methanoic acid is a weak acid and only partially dissociates, so [H⁺] < 0.050 mol dm⁻³, giving a higher pH. [1] Mark: Must mention complete vs partial dissociation.

Question 17 (9 marks)

(a) Completed table:

Titration	1	2	3	4
Volume used / cm³	24.50	24.30	23.60	23.30
[1]

(b) Titrations 2, 3, and 4 should be used. [1] Titration 1 is a rough titration (or is inconsistent with the others). The other three titres are concordant (within 0.20 cm³ of each other). [1]

(c) Average titre = (24.30 + 23.60 + 23.30) / 3 = 23.73 cm³ [1] Accept 23.7 cm³.

(d)

n(H₂SO₄) = c × V = 0.100 × (23.73/1000) = 0.002373 mol [1]
From equation: 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
Mole ratio NaOH : H₂SO₄ = 2 : 1
n(NaOH) = 2 × 0.002373 = 0.004746 mol [1]
c(NaOH) = n / V = 0.004746 / (25.0/1000) = 0.190 mol dm⁻³ [1] Mark: [1] for moles of H₂SO₄, [1] for mole ratio, [1] for final concentration. Accept 0.19 mol dm⁻³.

(e) Rinsing with distilled water does not affect the accuracy of the titre. [1] The number of moles of NaOH in the conical flask is unchanged. Adding water does not change the amount of NaOH present, so the volume of acid required to neutralise it remains the same. [1]

Question 18 (10 marks)

(a)

n(CH₃COONa) = mass / Mr = 4.10 / 82.0 = 0.0500 mol [1]
c(CH₃COONa) = n / V = 0.0500 / (500/1000) = 0.100 mol dm⁻³ [1]

(b)

[CH₃COOH] = 0.200 mol dm⁻³; [CH₃COO⁻] = 0.100 mol dm⁻³ [1]
[H⁺] = Ka × [CH₃COOH] / [CH₃COO⁻] = (1.8 × 10⁻⁵) × (0.200 / 0.100) = 3.6 × 10⁻⁵ mol dm⁻³ [1]
pH = –log₁₀(3.6 × 10⁻⁵) = 4.44 [1] Mark: [1] for correct concentrations, [1] for correct [H⁺], [1] for correct pH.

(c)

The buffer contains a reservoir of CH₃COO⁻ ions (conjugate base). [1]
When H⁺ is added: CH₃COO⁻(aq) + H⁺(aq) → CH₃COOH(aq) [1]
The added H⁺ ions are removed by reaction with the ethanoate ions, forming undissociated ethanoic acid. The [H⁺] and therefore pH remains approximately constant. [1]

(d) The student is incorrect. [1] Dilution changes both [CH₃COOH] and [CH₃COO⁻] by the same factor. The ratio [CH₃COOH]/[CH₃COO⁻] remains constant, so [H⁺] and pH remain unchanged. [1]

Question 19 (10 marks)

(a) The acid is a weak acid. [1] Evidence: The initial pH is 2.9 (not 1.0 as expected for a strong acid of similar concentration). The equivalence point pH is 8.3 (>7), characteristic of a weak acid-strong base titration. [1]

(b)

At equivalence point: n(NaOH) = 0.100 × (24.0/1000) = 0.00240 mol [1]
n(acid) = n(NaOH) = 0.00240 mol (monoprotic acid, 1:1 ratio)
c(acid) = n / V = 0.00240 / (25.0/1000) = 0.0960 mol dm⁻³ [1]

(c) At the equivalence point, all the weak acid has been converted to its conjugate base (salt). [1] The conjugate base undergoes hydrolysis: A⁻(aq) + H₂O(l) ⇌ HA(aq) + OH⁻(aq), producing OH⁻ ions and making the solution alkaline (pH > 7). [1]

(d) Phenolphthalein (pH range 8.3–10.0) is suitable. [1] The equivalence point pH (8.3) falls within the colour change range of phenolphthalein, so a sharp colour change will be observed at the end point. [1]

(e) Sketch should show:

Initial pH similar (~2.9) [½]
Buffer region before equivalence [½]
Equivalence point at pH < 7 (around 5–6) [½]
Curve less steep than strong base titration [½] Mark: Award marks for correct shape and labelled equivalence point.

Question 20 (8 marks)

(a) CaO(s) + H₂O(l) → Ca(OH)₂(aq/s) [1] Accept Ca(OH)₂(s) or (aq).

(b) Calcium hydroxide is a base because it accepts protons (H⁺) / neutralises acids / produces OH⁻ ions in water. [1]

(c)

n(CaO) = mass / Mr = 2800 / 56.1 = 49.91 mol [1]
From equation: 1 mol CaO produces 1 mol Ca(OH)₂
n(Ca(OH)₂) = 49.91 mol [1]
Mass Ca(OH)₂ = n × Mr = 49.91 × 74.1 = 3698 g ≈ 3.70 kg [1] Mark: [1] for moles of CaO, [1] for mole ratio, [1] for final mass with units. Accept 3.7 kg.

(d)

pH = –log₁₀[H⁺] [1]
[H⁺] = 10⁻⁴·⁵ = 3.16 × 10⁻⁵ mol dm⁻³ [1] Accept 3.2 × 10⁻⁵ mol dm⁻³.

(e) Calcium hydroxide neutralises H⁺ ions in the soil: OH⁻(aq) + H⁺(aq) → H₂O(l), reducing [H⁺] and increasing pH. [1]

Section C: Free Response Questions (20 marks)

Question 21 (20 marks)

(a) A strong acid is one that completely dissociates in water (e.g., HCl → H⁺ + Cl⁻). [1] A concentrated acid contains a large amount of acid dissolved per unit volume of water. [1] Example: 0.1 mol dm⁻³ HCl is a dilute solution of a strong acid; 10 mol dm⁻³ HCl is a concentrated solution of a strong acid. Strength refers to degree of dissociation; concentration refers to amount of solute. [1]

(b)(i) CO₂(aq) + H₂O(l) ⇌ H₂CO₃(aq) [1]

(b)(ii) H₂CO₃(aq) ⇌ H⁺(aq) + HCO₃⁻(aq) [1]

(b)(iii)

The buffer contains H₂CO₃ (weak acid) and HCO₃⁻ (conjugate base). [1]
When excess H⁺ enters the blood: HCO₃⁻(aq) + H⁺(aq) → H₂CO₃(aq) [1]
The added H⁺ ions are removed by reaction with hydrogencarbonate ions, forming carbonic acid. The [H⁺] remains approximately constant, maintaining pH around 7.4. [1]

(c)(i)

Total volume = 75.0 cm³
[C₂H₅COOH] = (0.100 × 50.0/1000) / (75.0/1000) = 0.0667 mol dm⁻³ [1]
[C₂H₅COO⁻] = (0.100 × 25.0/1000) / (75.0/1000) = 0.0333 mol dm⁻³ [1]
[H⁺] = Ka × [acid]/[salt] = (1.3 × 10⁻⁵) × (0.0667/0.0333) = 2.6 × 10⁻⁵ mol dm⁻³ [1]
pH = –log₁₀(2.6 × 10⁻⁵) = 4.59 [1]

(c)(ii)

Moles of HCl added = 0.100 × (5.0/1000) = 0.00050 mol [1]
HCl reacts with C₂H₅COO⁻: C₂H₅COO⁻ + H⁺ → C₂H₅COOH
New moles: C₂H₅COOH = 0.00500 + 0.00050 = 0.00550 mol [1]
New moles: C₂H₅COO⁻ = 0.00250 – 0.00050 = 0.00200 mol [1]
New total volume = 80.0 cm³
[H⁺] = Ka × (n_acid/n_salt) = (1.3 × 10⁻⁵) × (0.00550/0.00200) = 3.575 × 10⁻⁵ mol dm⁻³ [1]
pH = –log₁₀(3.575 × 10⁻⁵) = 4.45 [1] Note: Since ratio of moles is used, volume cancels out.

(d) A buffer requires a weak acid and its conjugate base in significant concentrations. [1] HCl is a strong acid (completely dissociated); Cl⁻ is a very weak conjugate base with negligible tendency to accept protons. [1] When H⁺ is added, there is no species present to remove it. When OH⁻ is added, there is insufficient weak acid to neutralise it. The mixture cannot resist pH changes. [1]

Question 22 (20 marks)

(a)(i) Le Chatelier's principle states that if a system at dynamic equilibrium is subjected to a change in conditions, the position of equilibrium shifts to counteract (oppose) the change. [1]

(a)(ii)

The forward reaction is exothermic (ΔH = –197 kJ mol⁻¹). A low temperature favours the exothermic forward reaction, increasing the equilibrium yield of SO₃. [1]
However, at low temperatures, the rate of reaction is too slow to be economically viable. [1]
A compromise temperature of 450°C is used, which gives a reasonable yield at an acceptable rate. A catalyst (V₂O₅) is also used to increase the rate. [1]

(a)(iii)

Increasing pressure shifts the equilibrium to the right (favours SO₃ production). [1]
There are 3 moles of gas on the left (2SO₂ + O₂) and 2 moles on the right (2SO₃). Increasing pressure favours the side with fewer gas molecules, increasing the yield of SO₃. [1]

(b)(i) SO₂(g) + H₂O(l) → H₂SO₃(aq) OR SO₂(g) + H₂O(l) ⇌ H⁺(aq) + HSO₃⁻(aq) [1]

(b)(ii)

Limestone is calcium carbonate, CaCO₃. [1]
Acid rain contains H⁺ ions which react with CaCO₃: CaCO₃(s) + 2H⁺(aq) → Ca²⁺(aq) + CO₂(g) + H₂O(l) [1]
The solid calcium carbonate dissolves, causing erosion and damage to the building. [1]

(c)(i)

[H⁺] = 10⁻pH [1]
[H⁺] = 10⁻⁴·² = 6.31 × 10⁻⁵ mol dm⁻³ [1] Accept 6.3 × 10⁻⁵ mol dm⁻³.

(c)(ii)

H₂SO₄ is diprotic: H₂SO₄ → 2H⁺ + SO₄²⁻
[H₂SO₄] = [H⁺] / 2 = (6.31 × 10⁻⁵) / 2 = 3.16 × 10⁻⁵ mol dm⁻³ [1] Mark: [1] for correct division by 2, [1] for correct answer with units.

(c)(iii) Pure water absorbs carbon dioxide from the atmosphere. [1] CO₂ dissolves and reacts: CO₂ + H₂O ⇌ H₂CO₃ ⇌ H⁺ + HCO₃⁻, producing H⁺ ions and lowering the pH below 7.0 (typically around 5.6). [1]

(d)

Add excess copper(II) oxide (black solid) to warm dilute sulfuric acid and stir. [1]
CuO(s) + H₂SO₄(aq) → CuSO₄(aq) + H₂O(l) [1]
Filter the mixture to remove unreacted CuO. [½]
Heat the filtrate (copper(II) sulfate solution) to evaporate some water until crystallisation point is reached. [½]
Allow the solution to cool slowly; blue CuSO₄·5H₂O crystals will form. [½]
Filter the crystals, wash with a little cold distilled water, and dry between filter papers. [½]

END OF ANSWER KEY