Secondary 4 Pure Chemistry Preliminary Examination Paper 1

TuitionGoWhere Practice Paper — Pure Chemistry Secondary 4

Preliminary Examination — Paper 2 (Version 1 of 5)

Answer Key and Marking Scheme

Section A — Multiple Choice Questions [10 marks]

Marking notes:

• Award 1 mark for each correct answer. No partial credit.
• Q1: Acids react with carbonates to produce CO₂. Acids have pH < 7, turn blue litmus red, and conduct electricity.
• Q2: pH > 7 indicates an alkaline solution. pH 12 is strongly alkaline.
• Q3: H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O. The salt is potassium sulfate.
• Q4: SO₂ is an acidic oxide (non-metal oxide). Na₂O, MgO, and CaO are basic oxides (metal oxides).
• Q5: CO₂ turns limewater milky. Only a carbonate produces CO₂ with acid.
• Q6: Ethanoic acid (CH₃COOH) is a weak acid; the others are strong acids.
• Q7: Insoluble salts are prepared by precipitation (mixing two soluble salt solutions).
• Q8: [OH⁻] = 0.01 M → pOH = 2 → pH = 14 − 2 = 12.
• Q9: SO₂ dissolves in rainwater to form H₂SO₃ (and further oxidises to H₂SO₄), causing acid rain.
• Q10: Moles of NaOH = moles of HCl at equivalence (1:1 ratio). M₁V₁ = M₂V₂ → 0.100 × 25.0 = 0.100 × V₂ → V₂ = 25.0 cm³.

Section B — Structured Questions [50 marks]

Question 11 [6 marks]

(a) [2 marks] A Brønsted–Lowry base is a proton (H⁺ ion) acceptor.

• Award 1 mark for "proton acceptor" or "H⁺ acceptor".
• Award 1 mark for stating it accepts (not donates) protons.
• Accept: "A substance that accepts a hydrogen ion (H⁺)."

(b) [2 marks] Any two of the following (1 mark each):

• Turns blue litmus paper red.
• Has a pH less than 7.
• Reacts with reactive metals (e.g., Mg, Zn) to produce hydrogen gas.
• Reacts with carbonates/hydrogen carbonates to produce carbon dioxide gas.
• Tastes sour (do not award if stated without qualification at this level — accept as a property but not a test method).
• Conducts electricity in aqueous solution.

(c) [2 marks]

• A solution of pH 1 has a higher concentration of H⁺ ions than a solution of pH 3. [1]
• Specifically, pH 1 has 100 times more H⁺ ions than pH 3 (each pH unit represents a tenfold difference). [1]
• Accept: "The H⁺ concentration at pH 1 is 0.1 mol/dm³, while at pH 3 it is 0.001 mol/dm³, so pH 1 is 100 times more concentrated in H⁺."

Question 12 [8 marks]

(a) [3 marks — 1 mark each] (i) Strongly acidic: Solution P (pH 1) (ii) Neutral: Solution Q (pH 7) (iii) Strongly alkaline: Solution R (pH 13)

(b) [3 marks — 1 mark each]

• P (pH 1): Red
• Q (pH 7): Green
• R (pH 13): Purple / Violet / Dark blue-purple

Marking notes: Accept "violet" or "dark blue" for pH 13. Do not accept "blue" alone (that corresponds to pH ~10–11).

(c) [2 marks] HCl(aq) → H⁺(aq) + Cl⁻(aq)

• Award 1 mark for correct reactants and products.
• Award 1 mark for correct state symbols (all aqueous).
• Accept: HCl(aq) → H⁺(aq) + Cl⁻(aq) or HCl + H₂O → H₃O⁺ + Cl⁻.
• Do not award if state symbols are missing or incorrect.

Question 13 [7 marks]

(a) [2 marks] CaCO₃(s) + 2HNO₃(aq) → Ca(NO₃)₂(aq) + H₂O(l) + CO₂(g)

• Award 1 mark for correct formulae of all reactants and products.
• Award 1 mark for correct balancing.
• State symbols not required but if given, must be correct.

(b) [2 marks]

• Test: Bubble the gas through limewater (calcium hydroxide solution). [1]
• Observation: Limewater turns milky / white precipitate forms. [1]
• Accept: "Pass the gas into calcium hydroxide solution" for the test.

(c) [3 marks]

• Ethanoic acid is a weak acid, so it is only partially ionised in solution. [1]
• This means the concentration of H⁺ ions is lower in ethanoic acid compared to nitric acid (a strong acid) of the same concentration. [1]
• With fewer H⁺ ions per unit volume, there are fewer effective collisions per unit time between H⁺ ions and CaCO₃ particles, so the initial rate of reaction is slower. [1]

Marking notes: The answer must reference: (i) weak vs. strong acid / partial ionisation, (ii) lower [H⁺], and (iii) collision theory (fewer effective collisions → slower rate). All three points needed for full marks.

Question 14 [8 marks]

(a) [2 marks]

• Sodium oxide is a metal oxide. [1]
• It reacts with water to form an alkaline solution (sodium hydroxide) / it neutralises acids to form salt and water. [1]
• Accept: "It dissolves in water to form NaOH, which is an alkali."

(b) [2 marks] SO₂(g) + H₂O(l) → H₂SO₃(aq)

• Award 1 mark for correct reactants and products.
• Award 1 mark for balancing (already balanced as written).
• Accept state symbols if correct; not required.

(c) [2 marks] Al₂O₃(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂O(l)

• Award 1 mark for correct products (AlCl₃ and H₂O).
• Award 1 mark for correct balancing.
• Accept: Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O (state symbols optional).

(d) [2 marks]

• Carbon monoxide does not react with water to form an acid or a base. [1]
• It is neither acidic nor basic / it does not neutralise acids or bases / it does not form an acid or alkali in water. [1]
• Accept: "It does not dissolve in water to form H⁺ or OH⁻ ions."

Question 15 [6 marks]

(a) [4 marks]

1. Add copper(II) oxide (CuO) to a beaker containing dilute sulfuric acid. [1]
2. Warm the mixture gently and stir to speed up the reaction. [1]
3. Continue adding CuO until no more reacts (excess CuO remains) — this ensures all the acid is used up. [1]
4. Filter the mixture to remove the excess (unreacted) CuO. Collect the filtrate (copper(II) sulfate solution). [1]
5. Heat the filtrate to evaporate some water until the solution is saturated, then allow it to cool to crystallise. [1]
6. Filter off the crystals and dry them between filter papers or in a warm oven. [1]

Award 4 marks for any four distinct, correct procedural steps with purposes. Key points: warming, excess CuO, filtration of excess, crystallisation/evaporation.

(b) [2 marks] CuO(s) + H₂SO₄(aq) → CuSO₄(aq) + H₂O(l)

• Award 1 mark for correct formulae.
• Award 1 mark for balancing.

Question 16 [5 marks]

(a) [4 marks]

• Reagent: Sodium hydroxide solution (NaOH), added dropwise and then in excess. [1]

• Observation with aluminium chloride: A white precipitate forms initially, which dissolves in excess NaOH to give a colourless solution. [1]

  • Accept: "White precipitate soluble in excess NaOH."

• Observation with magnesium chloride: A white precipitate forms, which is insoluble in excess NaOH. [1]

  • Accept: "White precipitate insoluble in excess NaOH."

• Award 1 mark for correctly identifying the difference (i.e., Al(OH)₃ is amphoteric and dissolves in excess; Mg(OH)₂ is not). [1]

Marking notes: The key distinguishing feature is the behaviour in excess NaOH. If the student only states "white precipitate forms" for both without mentioning excess, award a maximum of 2 marks for this part.

(b) [1 mark] Neutralisation (or acid–base neutralisation).

Question 17 [5 marks]

(a) [2 marks] S(s) + O₂(g) → SO₂(g)

• Award 1 mark for correct reactants and products.
• Award 1 mark for balancing.
• Accept state symbols if correct.

(b) [1 mark] Any one of the following:

• Corrodes limestone buildings and statues.
• Acidifies lakes/rivers, harming aquatic life.
• Damages forests / leaches nutrients from soil.
• Corrodes metal structures (e.g., bridges).

(c) [2 marks]

• Sulfur dioxide dissolves in rainwater to form sulfurous acid (H₂SO₃), which makes the rain acidic. [1]
• Equation: SO₂(g) + H₂O(l) → H₂SO₃(aq) [1]
• Accept: Further oxidation to H₂SO₄ (sulfuric acid) with appropriate equation: 2SO₂ + O₂ + 2H₂O → 2H₂SO₄.

Question 18 [5 marks]

(a) [2 marks] HCl(aq) + KOH(aq) → KCl(aq) + H₂O(l)

• Award 1 mark for correct formulae of all species.
• Award 1 mark for balancing.

(b) [3 marks]

Step 1: Calculate moles of KOH. Moles of KOH = concentration × volume = 0.100 mol/dm³ × (25.0 / 1000) dm³ = 0.00250 mol [1]

Step 2: Use stoichiometry. From the equation, the mole ratio of HCl : KOH = 1 : 1. ∴ Moles of HCl required = 0.00250 mol [1]

Step 3: Calculate volume of HCl. Volume of HCl = moles ÷ concentration = 0.00250 ÷ 0.200 = 0.0125 dm³ = 12.5 cm³ [1]

Marking notes:

• Award 1 mark for correct moles of KOH.
• Award 1 mark for correct mole ratio application.
• Award 1 mark for correct final answer with unit.
• If the student uses the formula M₁V₁ = M₂V₂ (where 1 = acid, 2 = base): 0.200 × V₁ = 0.100 × 25.0 → V₁ = 12.5 cm³. Award full marks if working is shown.
• Common error: Forgetting to convert cm³ to dm³ — penalise once only if the method is otherwise correct.

Section C — Free Response Questions [20 marks]

Question 19 [10 marks]

(a) [6 marks]

Procedure:

1. Add a few drops of Universal Indicator to a small sample of each liquid in separate test tubes. [1]
   • The liquid that turns the indicator red/orange (pH ≈ 1–3) is hydrochloric acid. [½]
   • The liquid that turns the indicator purple/blue (pH ≈ 11–14) is sodium hydroxide solution. [½]
   • The liquid that turns the indicator green (pH ≈ 7) is distilled water. [½]

Alternative / confirmatory test (award marks for either approach):

2. Add zinc granules to a sample of each liquid. [1]

   • The liquid that produces bubbles of gas (hydrogen) is hydrochloric acid. [½]
   • The liquids that show no reaction are NaOH and water. [½]

3. Add copper(II) carbonate to the remaining two liquids. [1]

   • No reaction with either (both are not acids), so this does not distinguish them. Instead, use Universal Indicator or conductivity.

Conductivity test (alternative):

• Test the electrical conductivity of each liquid using a conductivity apparatus. [1]
  • The liquid that conducts electricity well (bulb glows brightly) is NaOH (strong electrolyte). [½]
  • The liquid that does not conduct (bulb does not glow) is distilled water. [½]

Marking notes for (a):

• Award up to 2 marks for identifying a valid reagent/test method.
• Award up to 2 marks for correct observations for all three liquids.
• Award up to 2 marks for correct conclusions (identifying each liquid).
• The student does not need to use all reagents listed — any valid combination that correctly identifies all three liquids scores full marks.
• Accept any logically sound procedure. The key is that the student distinguishes all three liquids with correct observations.

(b) [4 marks]

Equation: Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g) [2]

• Award 1 mark for correct formulae; 1 mark for balancing.

Observations (1 mark each, any two):

• Bubbles / effervescence of gas are produced. [1]
• The zinc granules dissolve / get smaller / disappear. [1]
• The test tube feels warm (exothermic reaction). [1]
• A pop sound is heard when a lighted splint is held to the gas (confirms H₂). [1]

Question 20 [10 marks]

(a) [4 marks — 1 mark each]

Marking notes:

• For NaOH (soluble base): Titration is required because there is no visible sign of neutralisation; an indicator is needed.
• For CuO (insoluble base): React with acid, filter off excess CuO. Accept: "Acid + insoluble base" or "add excess insoluble base to acid, then filter."
• For Zn (reactive metal): React with acid, filter off excess Zn. Accept: "Acid + metal" or "react metal with acid, filter excess."
• For Ag (unreactive metal): Silver does not react readily with dilute acid. The salt is prepared via precipitation (e.g., mixing AgNO₃ from another route, or using silver carbonate). Accept: "Precipitation" or "react with silver carbonate then filter."

(b) [3 marks]

• Sodium hydroxide and nitric acid are both colourless solutions, and the reaction has no visible sign (no precipitate, no colour change, no gas). [1]
• Without titration and an indicator, it is impossible to know when exact neutralisation (equivalence point) has been reached. [1]
• If the solutions are simply mixed, one reagent may be in excess, contaminating the salt product. Titration ensures the exact stoichiometric amounts react, giving a pure salt without excess acid or alkali. [1]

Marking notes: Key points: (1) no visible change, (2) need to detect equivalence point, (3) avoid contamination from excess reagent.

(c) [3 marks]

• Excess silver carbonate is used to ensure that all the nitric acid is completely reacted / used up, so that no acid remains in the solution to contaminate the salt. [1]
• The excess (unreacted) silver carbonate is insoluble, so it can be removed by filtration. [1]
• The filtrate contains only silver nitrate solution. The pure salt is obtained by evaporating the water to crystallisation (heat to saturate, then cool to crystallise; filter and dry the crystals). [1]

Marking notes: Key points: (1) excess ensures complete reaction of acid, (2) excess carbonate is removed by filtration (insoluble), (3) crystallisation/evaporation to obtain pure salt crystals.

Summary of Marks

END OF ANSWER KEY