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A Level H2 Chemistry Practice Paper 3

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Questions

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TuitionGoWhere Practice Paper - Chemistry H2 A-Level

TuitionGoWhere Exam Practice (AI)

Subject: Chemistry H2
Level: A-Level
Paper: PRACTICE (Version 3 of 5)
Duration: 2 hours
Total Marks: 75

Name: _________________________
Class: _________________________
Date: _________________________

Instructions to Candidates

  1. This paper consists of three sections: Section A, Section B, and Section C.
  2. Answer all questions in the spaces provided.
  3. You are advised to spend no more than 40 minutes on Section A, 50 minutes on Section B, and 30 minutes on Section C.
  4. The use of an approved scientific calculator is expected, where appropriate.
  5. A Data Booklet is provided. You should make use of relevant data where necessary.
  6. Show all working clearly; marks will be awarded for correct method even if the final answer is wrong.
  7. State symbols should be included in all equations unless otherwise stated.

Section A: Structured Questions (30 marks)

Answer all questions in this section.

Question 1: Titration Analysis (6 marks)

A student carried out a titration to determine the concentration of a solution of ethanoic acid, CH₃COOH, using 0.100 mol dm⁻³ sodium hydroxide, NaOH(aq). The following burette readings were obtained:

TitrationRough123
Final burette reading / cm³24.5047.8524.1548.05
Initial burette reading / cm³0.0024.500.0024.15
Volume of NaOH used / cm³24.5023.3524.1523.90

(a) From the titration results, obtain a suitable volume of NaOH(aq) to be used in your calculations. Show clearly how you obtained this volume. (2 marks)

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(b) Calculate the number of moles of NaOH present in the volume obtained in (a). (1 mark)

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(c) The student pipetted 25.0 cm³ of ethanoic acid solution into a conical flask for each titration. The equation for the reaction is:

CH₃COOH(aq) + NaOH(aq) → CH₃COONa(aq) + H₂O(l)

Calculate the concentration, in mol dm⁻³, of the ethanoic acid solution. (2 marks)

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(d) Phenolphthalein was used as the indicator in this titration. State the colour change observed at the end-point and explain why this indicator is suitable for a weak acid–strong base titration. (1 mark)

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Question 2: Qualitative Analysis – Gas Tests (4 marks)

Complete the following table by stating the test and the expected observation for each gas.

GasTestObservation
Ammonia, NH₃(i)(ii)
Carbon dioxide, CO₂(iii)(iv)

(4 marks)

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Question 3: Reactions of Aqueous Cations (6 marks)

(a) Aqueous sodium hydroxide is added dropwise, then in excess, to separate solutions containing Al³⁺(aq) and Cu²⁺(aq) ions.

Complete the table below to describe what is observed and write the formula of any complex ion formed in excess NaOH.

CationObservation with NaOH(aq) added dropwiseObservation with excess NaOH(aq)Formula of species in excess NaOH
Al³⁺(aq)(i)(ii)(iii)
Cu²⁺(aq)(iv)(v)(vi)

(6 marks)

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Question 4: Acid–Base Equilibria (8 marks)

(a) Define the term Brønsted–Lowry acid. (1 mark)

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(b) Write an expression for the acid dissociation constant, Kₐ, of ethanoic acid, CH₃COOH. (1 mark)

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(c) The Kₐ of ethanoic acid at 298 K is 1.74 × 10⁻⁵ mol dm⁻³. Calculate the pH of a 0.100 mol dm⁻³ solution of ethanoic acid. State any assumption you make. (3 marks)

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(d) A buffer solution is prepared by mixing 50.0 cm³ of 0.200 mol dm⁻³ ethanoic acid with 50.0 cm³ of 0.100 mol dm⁻³ sodium ethanoate. Calculate the pH of this buffer solution. (3 marks)

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Question 5: Salt Hydrolysis (6 marks)

(a) Explain why an aqueous solution of sodium ethanoate, CH₃COONa, is alkaline. Write an ionic equation to support your answer. (3 marks)

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(b) Calculate the pH of a 0.0500 mol dm⁻³ solution of sodium ethanoate at 298 K.
[Kₐ of ethanoic acid = 1.74 × 10⁻⁵ mol dm⁻³; K_w = 1.00 × 10⁻¹⁴ mol² dm⁻⁶] (3 marks)

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Section B: Data Interpretation and Calculations (25 marks)

Answer all questions in this section.

Question 6: Titration Curve Analysis (8 marks)

The graph below shows the pH curve obtained when 25.0 cm³ of 0.100 mol dm⁻³ aqueous ammonia, NH₃(aq), is titrated with 0.100 mol dm⁻³ hydrochloric acid, HCl(aq).

[Assume a typical weak base–strong acid titration curve is provided, showing pH starting at ~11, decreasing gradually, with a steep drop between 20–30 cm³, and equivalence point at pH ~5.]

(a) Use the curve to determine the pH at the equivalence point. Explain why the pH at the equivalence point is less than 7. (2 marks)

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(b) Write an equation for the reaction occurring during the titration. (1 mark)

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(c) Suggest a suitable indicator for this titration. Justify your choice by referring to the pH curve. (2 marks)

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(d) At the half-equivalence point (12.5 cm³ of HCl added), the pH of the solution is 9.25. Use this information to calculate the K_b value for ammonia. (3 marks)

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Question 7: Buffer Calculations (9 marks)

A buffer solution is required to maintain a pH of 4.50. A student decides to use a mixture of ethanoic acid (CH₃COOH, Kₐ = 1.74 × 10⁻⁵ mol dm⁻³) and sodium ethanoate (CH₃COONa).

(a) Use the Henderson–Hasselbalch equation to calculate the ratio [CH₃COO⁻] / [CH₃COOH] required in the buffer solution. (2 marks)

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(b) The student prepares the buffer by mixing 100 cm³ of 0.200 mol dm⁻³ ethanoic acid with V cm³ of 0.150 mol dm⁻³ sodium ethanoate solution. Calculate the volume V required. (3 marks)

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(c) A small amount of 0.010 mol of HCl(g) is dissolved in 1.00 dm³ of the buffer solution prepared in (b). Assuming no change in volume, calculate the new pH of the buffer. (4 marks)

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Question 8: Solubility Product (8 marks)

(a) Write an expression for the solubility product, K_sp, of silver chloride, AgCl. (1 mark)

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(b) The solubility of AgCl in water at 298 K is 1.34 × 10⁻⁵ mol dm⁻³. Calculate the value of K_sp for AgCl. (2 marks)

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(c) Calculate the solubility of AgCl in 0.100 mol dm⁻³ NaCl(aq) at 298 K. State any assumption you make. (3 marks)

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(d) Explain, using Le Chatelier's principle, why the solubility of AgCl is lower in NaCl(aq) than in pure water. (2 marks)

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Section C: Extended Response (20 marks)

Answer all questions in this section.

Question 9: Acid Rain and Environmental Chemistry (10 marks)

Acid rain is primarily caused by the dissolution of sulfur dioxide, SO₂, and oxides of nitrogen, NOₓ, in atmospheric water.

(a) Write equations to show how SO₂ is converted to sulfuric acid, H₂SO₄, in the atmosphere. Include relevant state symbols. (3 marks)

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(b) Explain why acid rain has a more damaging effect on lakes in regions where the underlying rock is granite rather than limestone. Include relevant chemical equations in your answer. (4 marks)

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(c) Calcium oxide (quicklime) can be added to lakes to neutralise acid rain. Write an ionic equation for the reaction between calcium oxide and the acidic component of acid rain. Calculate the mass of calcium oxide required to neutralise a lake containing 1.00 × 10⁷ dm³ of water with a pH of 4.50. Assume the acid is entirely H₂SO₄ and that the lake water behaves as a dilute solution. (3 marks)

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Question 10: Complex Ions and Ligand Exchange (10 marks)

(a) When aqueous ammonia is added dropwise to a solution containing Cu²⁺(aq) ions, a pale blue precipitate forms initially, which dissolves in excess ammonia to give a deep blue solution.

(i) Identify the pale blue precipitate and write an ionic equation for its formation. (2 marks)

(ii) Write the formula of the deep blue complex ion formed in excess ammonia and state its shape. (2 marks)

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(b) Explain, in terms of d-orbital splitting, why the deep blue complex ion is coloured. (3 marks)

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(c) The K_stab (stability constant) for the formation of [Cu(NH₃)₄]²⁺ is 1.0 × 10¹³ dm⁹ mol⁻³. Write an expression for K_stab and use it to explain why the precipitate dissolves in excess ammonia. (3 marks)

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END OF PAPER

Check your work carefully. Ensure all equations are balanced and state symbols are included where required.

Answers

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TuitionGoWhere Practice Paper - Chemistry H2 A-Level

ANSWER KEY AND MARKING SCHEME

Version 3 of 5

Section A: Structured Questions (30 marks)

Question 1: Titration Analysis (6 marks)

(a) Obtain suitable volume of NaOH(aq). (2 marks)

Answer:

  • Rough titration (24.50 cm³) is excluded as it is a rough titre.
  • Titrations 1, 2, and 3 give volumes: 23.35, 24.15, 23.90 cm³.
  • Range check: 24.15 − 23.35 = 0.80 cm³ > 0.10 cm³, so Titration 2 (24.15 cm³) is an outlier and is excluded.
  • Concordant results: Titrations 1 and 3 (23.35 and 23.90 cm³).
  • Mean volume = (23.35 + 23.90) ÷ 2 = 23.63 cm³ (to 2 d.p., matching burette precision).

Marking:

  • 1 mark for correctly identifying concordant titres and excluding outliers with reasoning.
  • 1 mark for correct mean volume (23.63 cm³) with appropriate precision.

(b) Moles of NaOH in volume from (a). (1 mark)

Answer: n(NaOH) = c × V = 0.100 mol dm⁻³ × (23.63 ÷ 1000) dm³ = 2.363 × 10⁻³ mol (accept 2.36 × 10⁻³ mol)

Marking:

  • 1 mark for correct calculation with units.

(c) Concentration of ethanoic acid. (2 marks)

Answer: Mole ratio CH₃COOH : NaOH = 1 : 1
n(CH₃COOH) = n(NaOH) = 2.363 × 10⁻³ mol
Volume of CH₃COOH = 25.0 cm³ = 0.0250 dm³
c(CH₃COOH) = n ÷ V = 2.363 × 10⁻³ ÷ 0.0250 = 0.0945 mol dm⁻³ (3 s.f.)

Marking:

  • 1 mark for correct mole ratio and moles of acid.
  • 1 mark for correct concentration with appropriate significant figures.

(d) Indicator colour change and suitability. (1 mark)

Answer: Colour change: Colourless to pale pink (or pink).
Phenolphthalein is suitable because its pH range (8.3–10.0) falls within the steep vertical portion of the pH curve for a weak acid–strong base titration, where the equivalence point is >7.

Marking:

  • 1 mark for correct colour change and valid justification linking indicator range to equivalence point pH.

Question 2: Qualitative Analysis – Gas Tests (4 marks)

GasTestObservation
Ammonia, NH₃(i) Hold damp red litmus paper near the gas(ii) Damp red litmus paper turns blue
Carbon dioxide, CO₂(iii) Bubble gas through limewater (calcium hydroxide solution)(iv) Limewater turns milky/cloudy (white precipitate forms)

Marking:

  • 1 mark for each correct test (i, iii).
  • 1 mark for each correct observation (ii, iv).

Question 3: Reactions of Aqueous Cations (6 marks)

CationObservation with NaOH(aq) added dropwiseObservation with excess NaOH(aq)Formula of species in excess NaOH
Al³⁺(aq)(i) White precipitate forms(ii) White precipitate dissolves, forming a colourless solution(iii) [Al(OH)₄]⁻ (accept Al(OH)₄⁻)
Cu²⁺(aq)(iv) Blue precipitate forms(v) Blue precipitate is insoluble; no further change(vi) Cu(OH)₂ (precipitate remains)

Marking:

  • 1 mark for each correct entry (i–vi).

Question 4: Acid–Base Equilibria (8 marks)

(a) Brønsted–Lowry acid definition. (1 mark)

Answer: A Brønsted–Lowry acid is a proton (H⁺) donor.

Marking: 1 mark for correct definition.

(b) Kₐ expression for ethanoic acid. (1 mark)

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

Marking: 1 mark for correct expression.

(c) pH of 0.100 mol dm⁻³ ethanoic acid. (3 marks)

Answer: Assumption: Degree of dissociation is small, so [CH₃COOH]ₑq ≈ 0.100 mol dm⁻³.
Kₐ = [H⁺]² / [CH₃COOH]
[H⁺] = √(Kₐ × [CH₃COOH]) = √(1.74 × 10⁻⁵ × 0.100) = √(1.74 × 10⁻⁶) = 1.319 × 10⁻³ mol dm⁻³
pH = −log₁₀(1.319 × 10⁻³) = 2.88 (2 d.p.)

Marking:

  • 1 mark for stating assumption.
  • 1 mark for correct [H⁺] calculation.
  • 1 mark for correct pH.

(d) pH of buffer solution. (3 marks)

Answer: After mixing:
[CH₃COOH] = (0.200 × 50.0) ÷ 100.0 = 0.100 mol dm⁻³
[CH₃COO⁻] = (0.100 × 50.0) ÷ 100.0 = 0.0500 mol dm⁻³
Kₐ = [H⁺][CH₃COO⁻] / [CH₃COOH]
[H⁺] = Kₐ × [CH₃COOH] / [CH₃COO⁻] = 1.74 × 10⁻⁵ × (0.100 / 0.0500) = 3.48 × 10⁻⁵ mol dm⁻³
pH = −log₁₀(3.48 × 10⁻⁵) = 4.46 (2 d.p.)

Marking:

  • 1 mark for correct diluted concentrations.
  • 1 mark for correct [H⁺] calculation.
  • 1 mark for correct pH.

Question 5: Salt Hydrolysis (6 marks)

(a) Why CH₃COONa(aq) is alkaline. (3 marks)

Answer: Sodium ethanoate is a salt of a weak acid (CH₃COOH) and a strong base (NaOH). The ethanoate ion, CH₃COO⁻, undergoes hydrolysis:
CH₃COO⁻(aq) + H₂O(l) ⇌ CH₃COOH(aq) + OH⁻(aq)
The production of OH⁻ ions makes the solution alkaline (pH > 7).

Marking:

  • 1 mark for identifying ethanoate ion as the conjugate base of a weak acid.
  • 1 mark for correct hydrolysis equation.
  • 1 mark for linking OH⁻ production to alkalinity.

(b) pH of 0.0500 mol dm⁻³ CH₃COONa. (3 marks)

Answer: K_b for CH₃COO⁻ = K_w / Kₐ = 1.00 × 10⁻¹⁴ / 1.74 × 10⁻⁵ = 5.747 × 10⁻¹⁰ mol dm⁻³
CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻
K_b = [CH₃COOH][OH⁻] / [CH₃COO⁻] ≈ [OH⁻]² / 0.0500
[OH⁻] = √(5.747 × 10⁻¹⁰ × 0.0500) = √(2.874 × 10⁻¹¹) = 5.361 × 10⁻⁶ mol dm⁻³
pOH = −log₁₀(5.361 × 10⁻⁶) = 5.27
pH = 14.00 − 5.27 = 8.73 (2 d.p.)

Marking:

  • 1 mark for correct K_b calculation.
  • 1 mark for correct [OH⁻] calculation.
  • 1 mark for correct pH.

Section B: Data Interpretation and Calculations (25 marks)

Question 6: Titration Curve Analysis (8 marks)

(a) pH at equivalence point and explanation. (2 marks)

Answer: pH at equivalence point ≈ 5 (accept 4.8–5.2 from typical curve).
The pH is less than 7 because the salt formed, NH₄Cl, contains NH₄⁺ ions which hydrolyse to produce H⁺:
NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq)
This makes the solution acidic.

Marking:

  • 1 mark for correct pH from curve.
  • 1 mark for explanation linking to hydrolysis of ammonium ions.

(b) Equation for titration reaction. (1 mark)

Answer: NH₃(aq) + HCl(aq) → NH₄Cl(aq)
or NH₃(aq) + H⁺(aq) → NH₄⁺(aq)

Marking: 1 mark for correct balanced equation.

(c) Suitable indicator with justification. (2 marks)

Answer: Methyl orange (or screened methyl orange) is suitable. Its pH range (3.1–4.4) falls within the steep vertical portion of the pH curve near the equivalence point (pH ~5). Phenolphthalein would not be suitable as its range (8.3–10.0) does not overlap with the steep part of this curve.

Marking:

  • 1 mark for correct indicator.
  • 1 mark for justification linking indicator range to the steep portion of the curve.

(d) K_b for ammonia. (3 marks)

Answer: At half-equivalence point, [NH₃] = [NH₄⁺], so pH = pKₐ(NH₄⁺).
pKₐ(NH₄⁺) = 9.25
Kₐ(NH₄⁺) = 10⁻⁹·²⁵ = 5.62 × 10⁻¹⁰ mol dm⁻³
K_b(NH₃) = K_w / Kₐ(NH₄⁺) = 1.00 × 10⁻¹⁴ / 5.62 × 10⁻¹⁰ = 1.78 × 10⁻⁵ mol dm⁻³

Marking:

  • 1 mark for recognising that pH = pKₐ at half-equivalence.
  • 1 mark for correct Kₐ calculation.
  • 1 mark for correct K_b.

Question 7: Buffer Calculations (9 marks)

(a) Ratio [CH₃COO⁻] / [CH₃COOH]. (2 marks)

Answer: pH = pKₐ + log₁₀([CH₃COO⁻] / [CH₃COOH])
pKₐ = −log₁₀(1.74 × 10⁻⁵) = 4.76
4.50 = 4.76 + log₁₀(ratio)
log₁₀(ratio) = −0.26
Ratio = 10⁻⁰·²⁶ = 0.55 (or 0.5495)

Marking:

  • 1 mark for correct pKₐ.
  • 1 mark for correct ratio.

(b) Volume V of sodium ethanoate required. (3 marks)

Answer: n(CH₃COOH) = 0.200 × 0.100 = 0.0200 mol
n(CH₃COO⁻) = 0.150 × (V/1000) mol
Ratio = n(CH₃COO⁻) / n(CH₃COOH) = 0.55
0.150 × (V/1000) / 0.0200 = 0.55
V = (0.55 × 0.0200 × 1000) / 0.150 = 73.3 cm³ (3 s.f.)

Marking:

  • 1 mark for correct moles of acid.
  • 1 mark for setting up ratio equation.
  • 1 mark for correct V.

(c) New pH after adding HCl. (4 marks)

Answer: Total volume = 100 + 73.3 = 173.3 cm³ = 0.1733 dm³
Initial moles: CH₃COOH = 0.0200 mol; CH₃COO⁻ = 0.150 × 0.0733 = 0.0110 mol
Added H⁺ = 0.010 mol reacts with CH₃COO⁻:
CH₃COO⁻ + H⁺ → CH₃COOH
New moles: CH₃COO⁻ = 0.0110 − 0.010 = 0.0010 mol
CH₃COOH = 0.0200 + 0.010 = 0.0300 mol
New concentrations: [CH₃COO⁻] = 0.0010 / 0.1733 = 0.00577 mol dm⁻³
[CH₃COOH] = 0.0300 / 0.1733 = 0.1731 mol dm⁻³
[H⁺] = Kₐ × [CH₃COOH] / [CH₃COO⁻] = 1.74 × 10⁻⁵ × (0.1731 / 0.00577) = 5.22 × 10⁻⁴ mol dm⁻³
pH = −log₁₀(5.22 × 10⁻⁴) = 3.28 (2 d.p.)

Marking:

  • 1 mark for correct initial moles.
  • 1 mark for correct moles after reaction with H⁺.
  • 1 mark for correct new concentrations.
  • 1 mark for correct final pH.

Question 8: Solubility Product (8 marks)

(a) K_sp expression for AgCl. (1 mark)

Answer: K_sp = [Ag⁺][Cl⁻]

Marking: 1 mark for correct expression.

(b) K_sp value. (2 marks)

Answer: AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
Solubility, s = 1.34 × 10⁻⁵ mol dm⁻³, so [Ag⁺] = [Cl⁻] = s
K_sp = s² = (1.34 × 10⁻⁵)² = 1.80 × 10⁻¹⁰ mol² dm⁻⁶ (3 s.f.)

Marking:

  • 1 mark for correct relationship K_sp = s².
  • 1 mark for correct numerical value.

(c) Solubility in 0.100 mol dm⁻³ NaCl(aq). (3 marks)

Answer: In NaCl(aq), [Cl⁻] ≈ 0.100 mol dm⁻³ (from NaCl, contribution from AgCl is negligible).
Let solubility = s mol dm⁻³.
[Ag⁺] = s; [Cl⁻] = 0.100 + s ≈ 0.100 mol dm⁻³
K_sp = [Ag⁺][Cl⁻] = s × 0.100 = 1.80 × 10⁻¹⁰
s = 1.80 × 10⁻¹⁰ / 0.100 = 1.80 × 10⁻⁹ mol dm⁻³

Assumption: s << 0.100, so [Cl⁻] from AgCl is negligible.

Marking:

  • 1 mark for stating assumption.
  • 1 mark for correct substitution.
  • 1 mark for correct solubility.

(d) Explanation using Le Chatelier's principle. (2 marks)

Answer: The equilibrium is: AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq).
In NaCl(aq), there is a high concentration of Cl⁻ ions. By Le Chatelier's principle, the increased [Cl⁻] shifts the equilibrium position to the left, favouring the formation of solid AgCl. This reduces the solubility of AgCl compared to pure water (common ion effect).

Marking:

  • 1 mark for stating the equilibrium.
  • 1 mark for explaining the shift due to increased [Cl⁻] and linking to reduced solubility.

Section C: Extended Response (20 marks)

Question 9: Acid Rain and Environmental Chemistry (10 marks)

(a) Conversion of SO₂ to H₂SO₄. (3 marks)

Answer: SO₂(g) + H₂O(l) → H₂SO₃(aq) (sulfurous acid)
2H₂SO₃(aq) + O₂(g) → 2H₂SO₄(aq)
or SO₂(g) + ½O₂(g) + H₂O(l) → H₂SO₄(aq) (catalysed by NOₓ or metal ions in atmosphere)

Marking:

  • 1 mark for SO₂ dissolving in water.
  • 1 mark for oxidation step.
  • 1 mark for overall balanced equation with state symbols.

(b) Effect on granite vs. limestone regions. (4 marks)

Answer: Granite is primarily composed of silicates (e.g., SiO₂) which do not react with acids. Therefore, lakes in granite regions have no natural buffering capacity and become acidified rapidly.

Limestone is primarily CaCO₃, which reacts with acid:
CaCO₃(s) + 2H⁺(aq) → Ca²⁺(aq) + H₂O(l) + CO₂(g)
This reaction neutralises the acid, providing natural buffering. Lakes in limestone regions are therefore less affected by acid rain.

Marking:

  • 1 mark for identifying granite as unreactive (no buffering).
  • 1 mark for identifying limestone as reactive (buffering).
  • 1 mark for correct equation for limestone reaction.
  • 1 mark for clear explanation linking rock type to lake pH.

(c) Mass of CaO required. (3 marks)

Answer: pH = 4.50, so [H⁺] = 10⁻⁴·⁵⁰ = 3.16 × 10⁻⁵ mol dm⁻³
H₂SO₄ → 2H⁺ + SO₄²⁻, so [H₂SO₄] = ½ × [H⁺] = 1.58 × 10⁻⁵ mol dm⁻³
Moles of H₂SO₄ = 1.58 × 10⁻⁵ × 1.00 × 10⁷ = 158 mol
CaO(s) + 2H⁺(aq) → Ca²⁺(aq) + H₂O(l)
or CaO(s) + H₂SO₄(aq) → CaSO₄(aq) + H₂O(l)
Mole ratio CaO : H₂SO₄ = 1 : 1, so n(CaO) = 158 mol
Mᵣ(CaO) = 40.1 + 16.0 = 56.1 g mol⁻¹
Mass = 158 × 56.1 = 8.86 × 10³ g = 8.86 kg (3 s.f.)

Marking:

  • 1 mark for correct [H⁺] and moles of H₂SO₄.
  • 1 mark for correct mole ratio and moles of CaO.
  • 1 mark for correct mass with units.

Question 10: Complex Ions and Ligand Exchange (10 marks)

(a)(i) Pale blue precipitate and ionic equation. (2 marks)

Answer: Pale blue precipitate: Copper(II) hydroxide, Cu(OH)₂
Ionic equation: Cu²⁺(aq) + 2OH⁻(aq) → Cu(OH)₂(s)

Marking:

  • 1 mark for correct identity.
  • 1 mark for correct balanced ionic equation with state symbols.

(a)(ii) Deep blue complex ion and shape. (2 marks)

Answer: Formula: [Cu(NH₃)₄]²⁺ (accept [Cu(NH₃)₄(H₂O)₂]²⁺)
Shape: Square planar (or distorted octahedral if water ligands included).

Marking:

  • 1 mark for correct formula.
  • 1 mark for correct shape.

(b) Explanation of colour using d-orbital splitting. (3 marks)

Answer: In [Cu(NH₃)₄]²⁺, the Cu²⁺ ion (d⁹ configuration) is surrounded by four NH₃ ligands in a square planar arrangement. The ligands cause the d-orbitals to split into two energy levels. The energy gap, ΔE, between the higher and lower d-orbitals corresponds to the energy of visible light (orange/red region). An electron in a lower d-orbital absorbs a photon of this energy and is promoted to a higher d-orbital (d–d transition). The complementary colour (blue) is transmitted, giving the complex its deep blue colour.

Marking:

  • 1 mark for d-orbital splitting due to ligands.
  • 1 mark for absorption of visible light corresponding to ΔE.
  • 1 mark for linking absorbed colour to observed complementary colour.

(c) K_stab expression and explanation. (3 marks)

Answer: K_stab = [[Cu(NH₃)₄]²⁺] / ([Cu²⁺][NH₃]⁴)
The large K_stab value (1.0 × 10¹³) indicates that the equilibrium:
Cu²⁺(aq) + 4NH₃(aq) ⇌ [Cu(NH₃)₄]²⁺(aq)
lies far to the right. In excess ammonia, the high [NH₃] drives the equilibrium further right, dissolving the Cu(OH)₂ precipitate as Cu²⁺ ions are removed from solution to form the stable complex ion.

Marking:

  • 1 mark for correct K_stab expression.
  • 1 mark for linking large K_stab to equilibrium position.
  • 1 mark for explaining how excess NH₃ shifts equilibrium to dissolve precipitate.

END OF ANSWER KEY

Total marks: 75