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Secondary 3 Chemistry Practice Paper 5
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TuitionGoWhere Practice Paper - Chemistry Secondary 3
Answer Key and Marking Scheme Version: 5 of 5
Section A: Multiple Choice and Short Answer [20 marks]
1. Answer: C [1]
Explanation: Carbon dioxide dissolves in water to form carbonic acid (H₂CO₃), which dissociates partially to produce H⁺ ions, making the solution acidic with pH < 7. Sodium hydroxide (A) and ammonia (D) are bases with pH > 7. Calcium carbonate (B) is insoluble and does not significantly affect pH.
2. Answer: Blue [1]
Explanation: At pH 10, the solution is alkaline. Litmus paper turns blue in alkaline solutions and red in acidic solutions. The pH scale: 0-6 acidic (red), 7 neutral (purple), 8-14 alkaline (blue).
3. Answer:
- Type of reaction: Neutralisation [1]
- Explanation: Calcium oxide (CaO) is a base which reacts with acids in the soil. The pH increases because the base neutralises the acid, reducing the H⁺ ion concentration and increasing the pH towards 7. [1]
Marking note: Accept "calcium oxide is a base/alkali" and "reacts with/removes acid" or "decreases H⁺ concentration."
4. Answer:
- Potassium nitrate [1]
- Water [1]
Equation: HNO₃ + KOH → KNO₃ + H₂O
Explanation: In a neutralisation reaction between an acid and a base, the products are always a salt and water. The salt is named from the metal of the base (potassium) and the acid radical (nitrate from nitric acid).
5. (a) Answer: CaSO₄ [1]
(b) Answer: Insoluble (or sparingly soluble) [1]
Explanation: Most sulfates are soluble, but calcium sulfate is an important exception — it is sparsely soluble in water. This is why calcium carbonate is preferred over calcium sulfate for soil treatment (CaSO₄ would not dissolve easily to react).
6. (a) Answer: CuO(s) + 2HCl(aq) → CuCl₂(aq) + H₂O(l) [2]
Marking breakdown: Correct formulae [1], correct balancing and state symbols [1]
Working: Copper(II) oxide + hydrochloric acid → copper(II) chloride + water. Balance: needs 2HCl to provide 2Cl⁻ for CuCl₂.
(b) Answer:
- Heat the mixture gently with stirring until no more copper(II) oxide dissolves [1]
- Filter to remove excess copper(II) oxide [1]
- Heat the filtrate gently to evaporate some water until a saturated solution forms / crystals begin to form at the edges [1]
- Allow to cool and crystallise; filter and dry between filter papers [1]
- Test for excess oxide: Add more copper(II) oxide; if no more dissolves, excess was already present. OR: The solution turns from colourless to blue-green, indicating Cu²⁺ ions in solution; when no further colour change occurs on adding more oxide, excess is present. [1]
7. (a) Answer: An acid is a proton (H⁺ ion) donor. [1]
(b) Answer:
- Nitric acid is a strong acid because it completely dissociates/ionises in water to produce a high concentration of H⁺ ions. [1]
- Ethanoic acid is a weak acid because it partially dissociates/ionises in water, producing a lower concentration of H⁺ ions; an equilibrium exists between the molecules and ions. [1]
Key distinction: Strong acids: complete dissociation (→ arrow). Weak acids: partial dissociation (⇌ equilibrium arrow).
8. (a) Answer: From orange to red / yellow to orange [1]
Note: Methyl orange in acid is red, in alkali is yellow (or orange in neutral). At endpoint of acid added to alkali: yellow to orange/red or vice versa depending on direction. Since acid is in burette added to alkali: yellow → orange-red. Accept "pink to colourless" if phenolphthalein mentioned, but this specifies methyl orange.
(b) Answer:
- Volume of H₂SO₄ used = 24.70 − 0.50 = 24.20 cm³ [1]
- Amount of H₂SO₄ = (24.20/1000) × 0.100 = 2.42 × 10⁻³ mol [1]
Working shown:
(c) Answer: [3]
From equation: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
- Mole ratio H₂SO₄ : NaOH = 1 : 2 [1]
- Amount of NaOH = 2 × 2.42 × 10⁻³ = 4.84 × 10⁻³ mol [1]
- Concentration of NaOH = (4.84 × 10⁻³) / (25.0/1000) = 4.84 × 10⁻³ / 0.0250 = 0.1936 mol/dm³ ≈ 0.194 mol/dm³ [1]
Section B: Structured Questions [28 marks]
9. (a) Answer:
- pH = 13 [1]
- Sodium hydroxide is a strong base that completely dissociates: NaOH → Na⁺ + OH⁻. The concentration of OH⁻ ions is high (0.100 mol/dm³), giving high pH. [1]
(b) (i) Answer: 25.0 cm³ [1]
(ii) Answer:
- Around the equivalence point, all the OH⁻ ions have been neutralised by H⁺ ions. [1]
- A tiny excess of H⁺ ions causes a dramatic increase in [H⁺], shifting pH from alkaline through neutral to strongly acidic very rapidly. The buffer region is passed. [1]
(c) Answer:
- Universal indicator gives a gradual colour change through many colours (green to yellow to orange). [1]
- This makes it difficult to judge the exact point of neutralisation; methyl orange has a sharp, clear colour change at the endpoint. [1]
10. (a) Answer: 2NH₄NO₃(s) + Ca(OH)₂(s) → Ca(NO₃)₂(aq) + 2NH₃(g) + 2H₂O(l) [2]
Marking: Correct formulae [1], correct balancing and state symbols [1]
(b) Answer:
- Ammonium nitrate makes soil acidic because NH₄⁺ ions are acidic (they donate protons) or because nitrification bacteria produce acids. [1]
- Calcium carbonate (CaCO₃) is a base/alkali that neutralises the excess acid in the soil. [1]
- The pH is restored/maintained at a suitable level for plant growth. This is important because acidic soil reduces nutrient availability and can damage plant roots. [1]
(c) (i) Answer: [2]
Mᵣ of NH₄NO₃ = 14 + (4×1) + 14 + (3×16) = 14 + 4 + 14 + 48 = 80
Mass of nitrogen = 2 × 14 = 28
% nitrogen = (28/80) × 100 = 35% [2]
Marking: Correct Mᵣ [1], correct percentage [1]
(ii) Answer: [2]
Mᵣ of (NH₄)₂SO₄ = (2×14) + (8×1) + 32 + (4×16) = 28 + 8 + 32 + 64 = 132
Mass of nitrogen = 2 × 14 = 28
% nitrogen = (28/132) × 100 = 21.2% (accept 21.21% or 21%) [2]
Marking: Correct Mᵣ [1], correct percentage [1]
(iii) Answer: Ammonium sulfate [1]
Explanation: Ammonium sulfate has a lower percentage of nitrogen (21.2%) compared to ammonium nitrate (35%). Therefore, a greater mass of ammonium sulfate must be applied to provide the same mass of nitrogen nutrient to the crops. [1]
11. (a) Answer: Titration method / Acid-alkali titration method (not suitable here as CuO is insoluble). Actually: Excess solid method / Filtration and crystallisation method. [1]
Correct answer for insoluble base: The method involves adding excess solid to the acid, then filtering.
(b) Answer: [5]
- Add excess copper(II) oxide to warm dilute sulfuric acid in a beaker and stir [1]
- Heat gently (if necessary) until no more solid dissolves, indicating all acid has reacted [1]
- Filter the hot mixture to remove excess copper(II) oxide, collecting the blue-green filtrate [1]
- Heat the filtrate gently to evaporate some water until a saturated solution forms (or crystals begin to form at the edges on cooling) [1]
- Leave to cool and crystallise, then filter off crystals and dry between filter papers / in a desiccator [1]
Test for excess CuO: Add more CuO to the filtered solution; if no more dissolves, excess was originally present. Or: the solid no longer disappears/dissolves when added.
(c) Answer:
- Sodium hydroxide is soluble in water. [1]
- Therefore, excess cannot be removed by filtration; if excess NaOH is added, it remains in solution and contaminates the product. The exact amount must be determined by titration. [1]
(d) Answer: [3]
- Titration method: Measure a known volume of sodium hydroxide solution, add indicator, titrate with hydrochloric acid of known concentration until indicator changes colour. [1]
- Record the volume of acid needed to exactly neutralise the alkali. [1]
- Repeat without indicator using the same volumes, then evaporate to obtain pure sodium chloride crystals. (Or: use the titration data to calculate exact volume of acid needed, then mix those exact volumes.) [1]
Section C: Data Analysis and Application [12 marks]
12. (a) Answer: CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g) [2]
Marking: Correct formulae and products [1], correct balancing and state symbols [1]
(b) Answer: [2]
| Feature | Expected on sketch |
|---|---|
| Same maximum volume (240 cm³) [1] | Because same mass of CaCO₃ produces same total CO₂ |
| Less steep curve, levelling at same height but taking longer [1] | Smaller surface area of lumps means slower collision rate with H⁺ ions |
Description of sketch: Curve starts at origin, rises more gradually than original, reaches same plateau of 240 cm³ at a later time (e.g., 240 seconds or beyond).
(c) Answer: [3]
- Rate: Increases [1]
- Explanation: At higher temperature, particles have more kinetic energy, move faster, and collide more frequently with sufficient energy (exceeding activation energy). More successful collisions per unit time. [1]
- Total volume: Remains the same (240 cm³) [1] — same mass of CaCO₃ produces same total moles of CO₂; temperature affects rate, not total yield.
(d) Answer: [3]
Mᵣ of CaCO₃ = 40 + 12 + (3×16) = 100
Amount of CaCO₃ = 1.0 / 100 = 0.010 mol [1]
From equation: 1 mol CaCO₃ produces 1 mol CO₂ So amount of CO₂ = 0.010 mol [1]
Volume of CO₂ = 0.010 × 24 = 0.24 dm³ = 240 cm³ [1]
13. (a) (i) Answer: Lemon juice (pH 2.3) [1]
(ii) Answer: Seawater (pH 8.1) [1]
Note: Seawater is only slightly alkaline (closest to neutral of all bases listed), making it the "weakest" base. Baking soda (8.5) and hand soap (9.0) are stronger bases.
(b) (i) Answer: [2]
- Acid rain contains H⁺ ions which react with calcium carbonate: CaCO₃ + 2H⁺ → Ca²⁺ + H₂O + CO₂ [1]
- The building material is gradually dissolved/eroded, weakening the structure and causing loss of detail in stonework. [1]
(ii) Answer: [2]
- Acid rain lowers the pH of lake/river water below the tolerance range of aquatic organisms [1]
- Toxic aluminium ions may be released from sediments at low pH, damaging fish gills and affecting egg development; reduced biodiversity. [1]
(c) Answer:
- Natural cause: Volcanic eruptions releasing sulfur dioxide (SO₂) [1]
- Human cause: Burning fossil fuels (coal, oil) in power stations/vehicles, releasing sulfur dioxide and nitrogen oxides [1]
Summary Mark Breakdown
| Section | Marks | Checked |
|---|---|---|
| A | 20 | ✓ |
| B | 28 | ✓ |
| C | 12 | ✓ |
| Total | 60 | ✓ |
End of Answer Key