Secondary 3 Chemistry Semestral Assessment 2 (End of Year) Paper 3

TuitionGoWhere Practice Paper — Chemistry Secondary 3

SA2 (Version 3 of 5) — Answer Key and Marking Scheme

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Section A — Short Answer Questions

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Question 1 [3 marks]

Name of gas: Hydrogen [1]

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

• Award [1] for correct reactants and products (correct formulas).
• Award [1] for correct balancing and state symbols.

Common mistakes:

• Writing H instead of H₂.
• Forgetting state symbols (no mark deduction if not required by question, but full marks require them here).
• Writing the equation unbalanced (e.g., Zn + HCl → ZnCl₂ + H₂).

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Question 2 [3 marks]

(a) X [1] — pH 7 is neutral.

(b) W [1] — pH 1 is the lowest, hence most acidic.

(c) Y [1] — pH 9 is weakly alkaline (just above 7).

Common mistakes:

• Confusing "most acidic" with "most alkaline" — students may select Y instead of W.
• Selecting Z (pH 4) as weakly alkaline — Z is weakly acidic, not alkaline.

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Question 3 [2 marks]

1. Effervescence / bubbles of gas are produced. [1]
2. The solid dissolves / the solid disappears. [1]

Acceptable alternatives:

• "Fizzing" for effervescence.
• "Colourless gas is produced" for the first mark.

Common mistakes:

• Saying "gas is produced" without specifying effervescence/bubbles — may still be accepted at teacher's discretion.
• Describing the gas test (e.g., "limewater turns milky") — this is not an observable change of the reaction itself but a test for the product.

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Question 4 [2 marks]

Aqueous ammonia is a weak base because it only partially ionises / partially dissociates in water. [1]

Only a small proportion of ammonia molecules react with water to produce hydroxide ions (OH⁻). [1]

Acceptable alternative for [1]: "It does not fully ionise in water."

Common mistakes:

• Saying ammonia is a weak base because it has a low pH — incorrect; bases have pH > 7.
• Confusing "weak base" with "dilute base" — weakness refers to degree of ionisation, not concentration.

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Question 5 [3 marks]

(a) Calcium oxide / CaO OR calcium hydroxide / Ca(OH)₂ OR calcium carbonate / CaCO₃ [1]

Acceptable: Any suitable base or carbonate that is commonly used as a soil amendment.

Common mistakes:

• Naming an acid (e.g., sulfuric acid) — this would further decrease pH.
• Naming a neutral salt (e.g., NaCl) — does not affect pH.

(b) The compound reacts with / neutralises the hydrogen ions / H⁺(aq) in the acidic soil. [1]

This reduces the concentration of H⁺ ions, causing the pH to increase. [1]

Acceptable for [1]: "The base reacts with the acid in the soil" or "H⁺ ions react with OH⁻ ions to form water."

Common mistakes:

• Saying the compound "absorbs" the acid without explaining the ion reaction.
• Not mentioning H⁺ ions at all.

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Section B — Structured Questions

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Question 6 [6 marks]

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

• [1] for correct formulas of all reactants and products.
• [1] for correct balancing and state symbols.

(b) Observation: Limewater turns milky / turns cloudy white. [1]

Explanation: This confirms that the gas produced is carbon dioxide / CO₂. [1]

(c) Calcium sulfate / CaSO₄ is produced, which is insoluble / sparingly soluble. [1]

It forms a coating / layer on the surface of the calcium carbonate. [1]

This prevents further contact between the acid and the calcium carbonate, so the reaction slows down and stops. [1]

Common mistakes:

• Saying calcium sulfate is soluble — it is sparingly soluble and forms a protective layer.
• Not explaining the mechanism (coating prevents contact) — students may just say "the acid runs out" without explaining why.

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Question 7 [6 marks]

(a) [4 marks]

1. Add excess solid lead(II) carbonate (or oxide/hydroxide) to dilute hydrochloric acid. [1] (Alternative: Mix lead(II) nitrate solution with dilute hydrochloric acid / sodium chloride solution to precipitate lead(II) chloride.)

Using the precipitation method (more appropriate given the table):

1. Add dilute hydrochloric acid to lead(II) nitrate solution. [1] (Or mix lead(II) nitrate solution with a soluble chloride such as sodium chloride or dilute HCl.)

2. A white precipitate of lead(II) chloride forms. Filter the mixture to collect the precipitate. [1]

3. Wash the residue / precipitate with distilled water to remove impurities. [1]

4. Dry the crystals by pressing between filter paper / leaving in a warm place / in a desiccator. [1]

Marking notes:

• [1] for correct reactants that produce lead(II) chloride.
• [1] for filtration.
• [1] for washing with distilled water.
• [1] for a valid drying method.

(b) Lead is below hydrogen in the reactivity series / is not reactive enough. [1]

Therefore, lead does not react with dilute hydrochloric acid. [1]

Acceptable alternative: "Lead is unreactive with dilute acids."

Common mistakes:

• Saying lead reacts too slowly — the issue is thermodynamic (position in reactivity series), not kinetic.
• Not referencing the reactivity series.

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Question 8 [6 marks]

(a) Average volume = (24.80 + 24.70 + 24.90) ÷ 3 = 74.40 ÷ 3 = 24.80 cm³ [2]

• [1] for selecting the correct concordant titres (excluding the rough titre).
• [1] for correct calculation.

Note: Titres 1, 2, and 3 are concordant (within 0.10 cm³ of each other). The rough titre is excluded.

(b) H₂SO₄(aq) + 2KOH(aq) → K₂SO₄(aq) + 2H₂O(l) [1]

Common mistakes:

• Writing an unbalanced equation (e.g., H₂SO₄ + KOH → K₂SO₄ + H₂O without balancing).
• Incorrect formula for potassium sulfate (e.g., KSO₄).

(c) [3 marks]

Step 1: Moles of H₂SO₄ used = concentration × volume = 0.100 × (24.80 ÷ 1000) = 0.100 × 0.02480 = 0.00248 mol [1]

Step 2: From the equation, mole ratio H₂SO₄ : KOH = 1 : 2 Moles of KOH = 2 × 0.00248 = 0.00496 mol [1]

Step 3: Concentration of KOH = moles ÷ volume in dm³ = 0.00496 ÷ (25.0 ÷ 1000) = 0.00496 ÷ 0.0250 = 0.198 mol/dm³ (or 0.1984 mol/dm³) [1]

Accept: 0.198 mol/dm³ or 0.1984 mol/dm³ (3 s.f.)

Common mistakes:

• Forgetting to convert cm³ to dm³.
• Using the wrong mole ratio (1:1 instead of 1:2).
• Using the rough titre in the average calculation.

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Question 9 [6 marks]

(a) [3 marks]

• Hydrochloric acid: Blue litmus paper turns red. Red litmus paper stays red. [1]
• Sodium hydroxide solution: Red litmus paper turns blue. Blue litmus paper stays blue. [1]
• Distilled water: Neither red nor blue litmus paper changes colour. [1]

Acceptable alternatives:

• Describing the use of universal indicator: acid → red/orange, neutral → green, alkali → blue/purple.
• Using a single indicator (e.g., phenolphthalein) with correct observations.

Marking note: Award [1] for each correctly identified solution with correct observation.

(b) pH = –log₁₀[H⁺], so [H⁺] = 10^(–pH) = 10^(–1) = 0.10 mol/dm³ [1]

(c) The pH would increase. [1]

Adding water dilutes the acid, reducing the concentration of H⁺ ions. Since pH is inversely related to [H⁺], the pH increases (moves closer to 7). [1]

Common mistakes:

• Saying the pH decreases — this would mean the solution becomes more acidic, which is incorrect.
• Saying the pH stays the same — dilution changes the concentration.
• Saying the pH becomes 7 — dilution alone cannot make a strong acid neutral.

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Question 10 [6 marks]

(a) [4 marks]

Oxide	Type of Oxide	Reason
Calcium oxide	Basic oxide	Reacts with acids to form salt and water [2]
Zinc oxide	Amphoteric oxide	Reacts with both acids and bases to form salt and water [2]

Marking: [1] for correct type, [1] for correct reason for each oxide.

(b) ZnO(s) + 2HCl(aq) → ZnCl₂(aq) + H₂O(l) [1]

(c) ZnO(s) + 2NaOH(aq) → Na₂ZnO₂(aq) + H₂O(l) [1]

Acceptable alternative for (c): ZnO + 2NaOH + H₂O → Na₂[Zn(OH)₄] (sodium zincate)

Common mistakes:

• Writing incorrect formulas (e.g., ZnOH instead of ZnO).
• Not balancing the equation.
• Not knowing that zinc oxide is amphoteric — students may classify it as basic.

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Section C — Source-Based / Data Interpretation

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Question 11 [10 marks]

(a) Brand Q was more effective. [1]

The pH after Brand Q was dissolved (5.8) is closer to 7 (neutral) than the pH after Brand P (3.2), meaning Brand Q neutralised more of the acid. [1]

Acceptable: "Brand Q raised the pH more than Brand P."

(b) [H⁺] = 10^(–pH) = 10^(–1.0) = 0.10 mol/dm³ [1]

(c) [H⁺] = 10^(–pH) = 10^(–5.8) = 1.58 × 10⁻⁶ mol/dm³ [1]

Accept: 1.6 × 10⁻⁶ mol/dm³ (2 s.f.) or 1.58 × 10⁻⁶ mol/dm³.

(d) The hydrogen ions / H⁺(aq) from the acid react with hydroxide ions / OH⁻(aq) from the antacid. [1]

They combine to form water / H₂O, which reduces the concentration of H⁺ ions in the solution. [1]

Acceptable: "H⁺ ions are neutralised by the base in the antacid" for [1].

(e) CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g) [2]

• [1] for correct formulas.
• [1] for correct balancing and state symbols.

(f) Mg(OH)₂(s) + 2HCl(aq) → MgCl₂(aq) + 2H₂O(l) [2]

• [1] for correct formulas.
• [1] for correct balancing and state symbols.

Common mistakes in (e) and (f):

• Unbalanced equations.
• Incorrect formulas (e.g., MgOH instead of Mg(OH)₂).
• Missing state symbols.

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Mark Summary

Question	Marks
1	3
2	3
3	2
4	2
5	3
6	6
7	6
8	6
9	6
10	6
11	10
Total	50

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End of Answer Key