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Secondary 4 Pure Chemistry Preliminary Examination Paper 3

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TuitionGoWhere Practice Paper – Pure Chemistry Secondary 4

PRELIMINARY EXAMINATION – Version 3 of 5

ANSWER KEY AND MARKING SCHEME

Total Marks: 75

Section A: Multiple Choice (15 marks)

QuestionAnswerMark
1C[1]
2B[1]
3D[1]
4D[1]
5C[1]
6C[1]
7C[1]
8B[1]
9B[1]
10B[1]
11C[1]
12B[1]
13C[1]
14A[1]
15C[1]

Section B: Structured Questions (40 marks)

Question 16 (8 marks)

(a) Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g) [2]

Marking:

  • Correct formulae: 1 mark
  • Correct state symbols: 1 mark
  • Accept: Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g) for ionic equation

(b) Any two from: [2]

  • Effervescence / bubbles of gas produced (1)
  • Magnesium ribbon dissolves / disappears (1)
  • Solution becomes warm / temperature increases (1)
  • Colourless solution formed (1)

(c) Magnesium powder has a larger surface area than magnesium ribbon [1]. This increases the frequency of effective collisions between magnesium atoms and hydrogen ions, increasing the rate of reaction [1].

(d) No change / limewater remains colourless [1]. Hydrogen gas is produced, not carbon dioxide. Hydrogen does not react with limewater [1].

Question 17 (8 marks)

(a) Sodium sulfate [1]

(b) Yellow to orange / peach [1]
Accept: yellow to pink

(c) Titration is suitable for preparing soluble salts of Group I metals and ammonium salts [1]. Sodium sulfate is soluble in water, so titration can be used [1]. Lead(II) sulfate is insoluble in water, so precipitation must be used instead. Titration would not produce a visible end-point because lead(II) sulfate would precipitate during the titration [1].

(d) Moles of NaOH = 0.40 × (25.0/1000) = 0.0100 mol [1]
Moles of H₂SO₄ required = 0.0100 ÷ 2 = 0.00500 mol [1]
Volume of H₂SO₄ = 0.00500 ÷ 0.25 = 0.0200 dm³ = 20.0 cm³ [1]

Question 18 (9 marks)

(a) Al³⁺ (aluminium ion) [1]. A white precipitate forms with NaOH(aq) which dissolves in excess NaOH(aq), characteristic of Al³⁺ forming Al(OH)₃ which is amphoteric and dissolves to form [Al(OH)₄]⁻ [1].

(b) Cu²⁺ (copper(II) ion) [1]. A blue precipitate forms with NaOH(aq) which is insoluble in excess. With NH₃(aq), the blue precipitate dissolves in excess to form a deep blue solution (formation of [Cu(NH₃)₄]²⁺ complex) [1].

(c) Fe²⁺ (iron(II) ion) [1]. A green precipitate forms with NaOH(aq) which turns brown on standing due to oxidation of Fe(OH)₂ to Fe(OH)₃ by air [1].

(d) Cl⁻ (chloride ion) [1]. White precipitate of AgCl forms with AgNO₃(aq) [1].
Ag⁺(aq) + Cl⁻(aq) → AgCl(s) [1]

Question 19 (10 marks)

(a) Nitrogen: from fractional distillation of liquid air [1]
Hydrogen: from cracking of hydrocarbons / from natural gas (methane) reacting with steam [1]

(b) At lower temperatures, the rate of reaction would be too slow to be economical [1], even though the equilibrium yield of ammonia would be higher [1]. 450 °C is a compromise temperature that gives a reasonably fast rate of reaction while still producing an acceptable yield [1].

(c) The forward reaction produces fewer moles of gas (4 moles → 2 moles) [1]. According to Le Chatelier's principle, increasing pressure shifts the equilibrium to the right, increasing the yield of ammonia [1].

(d) The iron catalyst increases the rate of reaction [1] without being consumed in the process [1]. A catalyst works by providing an alternative reaction pathway with a lower activation energy, so more particles have energy greater than or equal to the activation energy, increasing the frequency of effective collisions [1].

Question 20 (10 marks)

(a) BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq) [2]
Marking: Correct formulae (1), correct state symbols (1)

(b) Precipitation [1]

(c)

  1. Filter the mixture to separate the solid barium sulfate (residue) from the solution (filtrate) [1].
  2. Wash the residue with distilled water to remove any soluble impurities (NaCl) [1].
  3. Dry the residue between sheets of filter paper or in a warm oven [1].

(d) Moles of BaCl₂ = 0.20 × (50.0/1000) = 0.0100 mol [1]
Moles of BaSO₄ = 0.0100 mol (1:1 ratio) [1]
Mass of BaSO₄ = 0.0100 × 233 [1]
= 2.33 g [1]

Section C: Free-Response Questions (20 marks)

Question 21

(a) An acid is a substance that produces hydrogen ions (H⁺) in aqueous solution [1]. A base is a substance that produces hydroxide ions (OH⁻) in aqueous solution [1].

(b) A strong acid ionises completely in aqueous solution to produce H⁺ ions [1].
Example: HCl(aq) → H⁺(aq) + Cl⁻(aq) [1]
A weak acid ionises partially in aqueous solution to produce H⁺ ions [1].
Example: CH₃COOH(aq) ⇌ H⁺(aq) + CH₃COO⁻(aq) [1]
Accept: use of reversible arrow for weak acid essential

(c) Diagram (2 marks): Labelled diagram showing beaker with CuO and H₂SO₄, Bunsen burner/tripod/gauze for heating, filter funnel and filter paper, evaporating dish for crystallisation.

Procedure (6 marks):

  1. Add excess copper(II) oxide to warm dilute sulfuric acid and stir [1]. Excess CuO ensures all acid is neutralised [1].
  2. Filter the mixture to remove unreacted CuO [1]. This separates the insoluble excess CuO from the copper(II) sulfate solution [1].
  3. Heat the filtrate to evaporate some water until a saturated solution is obtained (crystals form on cooling rod / first signs of crystallisation) [1].
  4. Allow the saturated solution to cool slowly. Blue crystals of CuSO₄·5H₂O will form [1].
  5. Filter the crystals and wash with a little cold distilled water to remove impurities [1].
  6. Dry the crystals between sheets of filter paper [1].

Note: Award marks for clear, logical sequence with explanations.

(d)(i) CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l) [2]
Marking: Correct formulae (1), correct state symbols (1)

(d)(ii) Moles of HCl = 0.50 × (50.0/1000) = 0.0250 mol [1]
Moles of CO₂ = 0.0250 ÷ 2 = 0.0125 mol [1]
Volume of CO₂ = 0.0125 × 24 [1]
= 0.30 dm³ (or 300 cm³) [1]

Question 22

(a)(i) Acids react with reactive metals to produce a salt and hydrogen gas [1].
Example: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g) [2]
Marking: Correct observation (1), correct balanced equation with state symbols (2)

(a)(ii) Acids react with metal carbonates to produce a salt, carbon dioxide, and water [1].
Example: CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l) [2]
Marking: Correct products (1), correct balanced equation with state symbols (2)

(b) Add aqueous sodium hydroxide to the fertiliser sample and warm gently [1]. Test any gas produced with damp red litmus paper [1]. If ammonium ions are present, ammonia gas is produced which turns damp red litmus paper blue [1].
NH₄⁺(aq) + OH⁻(aq) → NH₃(g) + H₂O(l) [1]

(c) Soil pH affects the availability of nutrients to plants [1]. If soil is too acidic, essential nutrients may become unavailable or toxic metals may dissolve [1]. The farmer can add calcium oxide (quicklime) or calcium carbonate (limestone) to neutralise the acidic soil [1].
CaO(s) + 2H⁺(aq) → Ca²⁺(aq) + H₂O(l) [1]
Accept: CaCO₃ + 2H⁺ → Ca²⁺ + CO₂ + H₂O

(d)(i) Moles of H₂SO₄ = 0.10 × (20.0/1000) = 0.00200 mol [2]
Marking: Correct substitution (1), correct answer (1)

(d)(ii) Moles of KOH = 2 × 0.00200 = 0.00400 mol [1]
Concentration of KOH = 0.00400 ÷ (25.0/1000) = 0.160 mol/dm³ [1]

(d)(iii) Concentration in g/dm³ = 0.160 × 56 [1]
= 8.96 g/dm³ [1]

END OF ANSWER KEY