Secondary 4 Combined Science Chemistry Preliminary Examination Paper 5

TuitionGoWhere Practice Paper — Combined Science Chemistry Secondary 4

Answer Key — Preliminary Examination Paper 2 (Version 5 of 5)

Section A: Multiple Choice Questions [10 marks]

1. C [1]

• Acids react with metals to produce hydrogen gas. (A and B are properties of alkalis; D is a property of alkalis.)

2. A [1]

• pH = –log₁₀[H⁺], so [H⁺] = 10^(–pH) = 10^(–3) = 0.001 mol/dm³.

3. B [1]

• H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O. The salt produced is sodium sulfate.

4. C [1]

• Calcium hydroxide (Ca(OH)₂) is a soluble base, hence an alkali. HCl is an acid; distilled water is neutral; CO₂ is an acidic oxide.

5. B [1]

• Universal indicator turns orange at approximately pH 5–6, indicating a weakly acidic solution.

6. C [1]

• An insoluble base reacts with acid. Excess base is used to ensure all acid is neutralised, then the excess (unreacted) base is removed by filtration since it is insoluble.

7. C [1]

• Universal indicator turns purple at pH 11–14, indicating a strong alkali.

8. B [1]

• HNO₃ + KOH → KNO₃ + H₂O. The salt is potassium nitrate (KNO₃). The nitrate ion is NO₃⁻, not NO₂⁻.

9. C [1]

• Barium sulfate is insoluble. All sodium, potassium, and ammonium salts are soluble (solubility rules).

10. B [1]

• HCl + NaOH → NaCl + H₂O (1:1 mole ratio). Since concentrations are equal (0.100 mol/dm³), the volume of NaOH required equals the volume of HCl = 25.0 cm³.

Section B: Structured Questions [30 marks]

11. [6 marks]

(a) Solution P is acidic [1] because it has a pH of 1, which is less than 7 [1].

• Marking note: Both the identification (P) and the explanation (pH < 7) are required for full marks.

(b) Solution Q is neutral [1].

(c) Any one of the following [1]:

• It turns red litmus paper blue.
• It reacts with acids to form a salt and water.
• It has a pH greater than 7.
• It feels slippery (if concentrated).
• It conducts electricity.

(d) Mg + 2H⁺ → Mg²⁺ + H₂ [2]

• Marking note: Award 1 mark for correct reactants and products, 1 mark for correct state symbols or correct balancing. Accept Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g) for full marks. Award 1 mark if the molecular equation Mg + 2HCl → MgCl₂ + H₂ is given instead.

12. [7 marks]

(a) Copper(II) oxide reacts with acids to form a salt and water [1].

• Marking note: A base is a substance that reacts with an acid to form a salt and water only (or a metal oxide/hydroxide that neutralises acid).

(b) CuO + H₂SO₄ → CuSO₄ + H₂O [2]

• Marking note: Award 1 mark for correct formulae of all reactants and products, 1 mark for balancing.

(c) Procedure [4]:

1. Add excess copper(II) oxide to warm dilute sulfuric acid in a beaker, stirring until no more reacts [1].
2. Filter the mixture using filter paper and funnel to remove the excess (unreacted) copper(II) oxide [1].
3. Heat the filtrate (copper(II) sulfate solution) to evaporate some water until the solution is saturated (or until a crystallisation point is reached — e.g., a drop placed on a glass rod forms crystals) [1].
4. Allow the saturated solution to cool so that copper(II) sulfate crystals form. Filter to collect the crystals, then dry the crystals between filter papers or in a warm oven [1].

• Explanation for using excess CuO: Excess copper(II) oxide is used to ensure that all the acid is completely neutralised [1]. Since CuO is insoluble, the excess can be removed by filtration, leaving only the copper(II) sulfate solution.
• Note: The explanation for excess CuO is embedded in the procedure marks above. Award the mark where the student clearly states that excess ensures all acid reacts and can be filtered off.

13. [7 marks]

(a) Precipitation (double decomposition) [1]. Lead(II) chloride is insoluble in water, so it cannot be prepared by titration or by reacting an acid with an excess insoluble base. It must be prepared by mixing two soluble solutions to form an insoluble salt [1].

• Marking note: Award 1 mark for naming the method, 1 mark for explaining that the salt is insoluble.

(b)(i) A white precipitate is formed [1].

(b)(ii) Pb(NO₃)₂ + 2NaCl → PbCl₂ + 2NaNO₃ [2]

• Marking note: Award 1 mark for correct formulae, 1 mark for balancing.

(b)(iii) Pb²⁺ + 2Cl⁻ → PbCl₂ [2]

• Marking note: Award 1 mark for correct ions, 1 mark for balancing. State symbols are not required but if included, they must be correct.

14. [6 marks]

(a) The rough titration gives an approximate volume of sodium hydroxide needed, so that subsequent titrations can be carried out more quickly and accurately by running the burette rapidly to near the expected end-point [1].

(b) Titration 2 (25.30 cm³) and Titration 3 (25.40 cm³) are concordant (within 0.20 cm³ of each other). Titration 1 (25.30 cm³) is also concordant with these.

• Average volume = (25.30 + 25.30 + 25.40) ÷ 3 = 25.33 cm³ [2]
• Marking note: Award 1 mark for selecting the correct concordant titres, 1 mark for the correct average. Accept 25.33 cm³ or 25.3 cm³ (to 1 d.p.). Do NOT include the rough titration (26.50 cm³) in the average.

(c) Working [3]:

Step 1: Moles of NaOH used = concentration × volume = 0.200 × (25.33 ÷ 1000) = 0.005066 mol [1]

Step 2: From the equation H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, the mole ratio of H₂SO₄ : NaOH = 1 : 2.

• Moles of H₂SO₄ = 0.005066 ÷ 2 = 0.002533 mol [1]

Step 3: Concentration of H₂SO₄ = moles ÷ volume = 0.002533 ÷ (25.0 ÷ 1000) = 0.101 mol/dm³ [1]

• Final answer: 0.101 mol/dm³ (to 3 s.f.)
• Marking note: Award 1 mark for each correct step. If the student uses 25.33 cm³ and gets 0.101 mol/dm³, award full marks. Accept answers in the range 0.100–0.102 mol/dm³ depending on rounding.

15. [6 marks]

(a) A strong alkali dissociates (ionises) completely in water to produce hydroxide ions (OH⁻) [1]. A weak alkali dissociates (ionises) only partially in water, so fewer hydroxide ions are produced [1].

• Marking note: The key distinction is "completely" vs. "partially". Both marks require this contrast.

(b)(i) NH₃ + HCl → NH₄Cl [1]

• Accept NH₄OH + HCl → NH₄Cl + H₂O. Award 1 mark for correct equation.

(b)(ii) Procedure [3]:

1. Use a pipette to measure a known volume of hydrochloric acid into a conical flask and add a few drops of methyl orange indicator [1].
2. Add ammonia solution from a burette until the indicator just changes colour (from red to orange/yellow), noting the volume used [1].
3. Repeat the titration without indicator using the exact volumes determined, then evaporate the solution to crystallisation point and allow to cool to obtain ammonium chloride crystals. Filter and dry the crystals [1].

• Alternative acceptable method: Simply mix the two solutions in the correct proportions (using titration data), then evaporate to concentrate and crystallise.
• Marking note: Award 1 mark for using titration to determine the correct proportions, 1 mark for evaporating/crystallising, 1 mark for filtering and drying.

16. [5 marks]

(a) Dilute nitric acid is added to remove any carbonate ions (or other interfering ions) that might also form a precipitate with silver nitrate, which would give a false positive result [1]. Carbonate ions would form a white precipitate (Ag₂CO₃) with AgNO₃, but this dissolves in dilute HNO₃ [1].

• Marking note: Award 1 mark for stating the purpose (to remove interfering ions/carbonates), 1 mark for explaining why (to avoid false positives / carbonates also give precipitates).

(b) Bromide ion (Br⁻) [1]. The precipitate is silver bromide, AgBr [1].

• Marking note: Both the ion and the formula must be correct for 2 marks. Award 1 mark if only one is correct.

(c) NaI + AgNO₃ → AgI + NaNO₃ [2]

• Marking note: Award 1 mark for correct formulae, 1 mark for balancing. Accept the ionic equation Ag⁺ + I⁻ → AgI for full marks.

17. [5 marks]

(a) Acid Y is the weak acid [1]. At the same concentration (0.1 mol/dm³), a strong acid would have a pH of 1.0 (fully dissociated), but acid Y has a pH of 2.5, indicating it is only partially dissociated [1].

• Marking note: Award 1 mark for identifying Y, 1 mark for the explanation linking higher pH to partial dissociation.

(b) Acids X and Z could both be strong acids (or both be the same acid) at the same concentration [1]. Since both have pH 1.0 at 0.1 mol/dm³, they are both fully dissociated, meaning they are both strong acids (e.g., HCl and HNO₃).

• Accept: They are both strong monoprotic acids at the same concentration.

(c) Acid X would react faster [1]. Acid X is a strong acid (pH 1.0), so it has a higher concentration of H⁺ ions than acid Y (pH 2.5, weak acid). A higher concentration of H⁺ ions means more frequent effective collisions with magnesium, resulting in a faster rate of reaction [1].

• Marking note: Award 1 mark for identifying X, 1 mark for the explanation involving H⁺ concentration and rate of reaction.

Section C: Free Response Question [10 marks]

18. [10 marks]

(a) Dip a glass rod into the river water and touch it to a piece of universal indicator paper (or add a few drops of universal indicator solution to a sample of the river water) [1]. If the water is acidic, the universal indicator will turn red or orange (pH below 7) [1].

• Accept: Use a pH meter — it will give a reading below 7.

(b) Any two of the following [1 each, total 2]:

• Acid rain (due to dissolved sulfur dioxide or nitrogen oxides from industrial emissions / vehicle exhaust).
• Discharge from factories releasing acidic waste.
• Runoff from mines containing acidic compounds.
• Dissolution of carbon dioxide from the atmosphere forming weak carbonic acid (natural acidity).
• Leaching of acidic soils into the river.

(c)(i) Effervescence / bubbles of gas are produced [1]. The solid calcium carbonate gradually dissolves (decreases in amount) [1].

• Marking note: Award 1 mark for bubbles/effervescence, 1 mark for solid dissolving.

(c)(ii) Excess calcium carbonate is used to ensure that all the acid in the river water is completely neutralised [1].

(c)(iii) CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ [2]

• Marking note: Award 1 mark for correct formulae, 1 mark for balancing. Accept state symbols if correct.

(d) The residue is excess (unreacted) calcium carbonate [1].

• Marking note: Accept "calcium carbonate" or "CaCO₃". The residue is the excess solid that did not react, which is filtered off.

Mark Summary

Common Mistakes to Watch For

• Q1: Confusing acid and alkali properties (e.g., choosing A or D which describe alkalis).
• Q2: Incorrectly calculating [H⁺] from pH (e.g., choosing C, 0.1 mol/dm³, by forgetting the negative sign in the logarithm).
• Q8: Writing the wrong formula for the salt (e.g., KNO₂ instead of KNO₃).
• Q11(d): Forgetting to write the ionic equation in terms of H⁺ ions; writing the molecular equation instead (award partial credit).
• Q12(c): Forgetting to explain why excess CuO is used (to ensure all acid reacts) or forgetting to filter off the excess.
• Q13(b)(iii): Including spectator ions (Na⁺ and NO₃⁻) in the ionic equation.
• Q14(b): Including the rough titration in the average volume calculation.
• Q14(c): Forgetting the 1:2 mole ratio between H₂SO₄ and NaOH.
• Q15(a): Not using the terms "completely" and "partially" when comparing strong and weak alkalis.
• Q16(a): Not explaining that carbonates would interfere by also forming precipitates.
• Q17(c): Not linking the rate of reaction to the concentration of H⁺ ions.
• Q18(c)(iii): Forgetting to balance the equation or writing CO instead of CO₂.