O Level Chemistry Practice Paper 4

Answer Key - TuitionGoWhere Practice Paper - Chemistry O-Level

Topic: Acids, Bases & Salts
Version: 4 of 5

Section A: Multiple Choice & Short Answer

1. B [1]
Reasoning: Neutralisation is Acid + Base $\rightarrow$ Salt + Water. CuO is a base. A is Acid + Metal. C is Acid + Carbonate (produces CO2). D is Gas phase reaction forming salt only (no water).

2. B [1]
Reasoning: Ethanoic acid is a weak acid. Orange indicates pH 3-5. pH 1 is strong acid (Red). pH 7 is neutral (Green). pH 13 is strong alkali (Purple).

3. B [1]
Reasoning: Definition of weak acid is partial ionisation/dissociation in water. A is incorrect because a concentrated weak acid can have high [H+]. C is a consequence, not the definition. D is incorrect (acidic pH < 7).

4. A [1]
Reasoning: CuO is in excess and is insoluble. CuSO4 is soluble and passes through the filter. The residue is the unreacted solid CuO.

5. C [1]
Reasoning: Titration is used for Soluble Salt from Soluble Base (Alkali). KNO3 is soluble, made from KOH (alkali) and HNO3. BaSO4 is insoluble (precipitation). CuCl2 from insoluble base (filtration/excess). ZnCO3 is insoluble.

6. [2]
Reagent: Barium nitrate solution ($Ba(NO_3)_2$) OR Barium chloride solution ($BaCl_2$) AND dilute nitric acid ($HNO_3$) or dilute hydrochloric acid ($HCl$).
(Note: If using BaCl2, acidify with HNO3 to rule out carbonate/sulfite interference, though HCl is often accepted if carbonate is ruled out. Best practice: Ba(NO3)2 + HNO3)
Observation: White precipitate forms.

7. [2]
$H^+(aq) +$ $OH^-(aq)$ $\rightarrow$ $H_2O(l)$

8. [2]
It reacts with acids to form a salt and water [1] AND it reacts with alkalis/bases to form a salt and water [1].

9.
(a) To ensure all the sulfuric acid reacts / is neutralised. [1]
(b) To remove the excess/unreacted magnesium carbonate solid. [1]

10.
(a) To neutralise the acidity of the soil / raise the pH. [1]
(b) $Ca(OH)_2 + 2HNO_3 \rightarrow Ca(NO_3)_2 + 2H_2O$ [2]
(1 mark for correct formulas, 1 mark for balancing)

Section B: Structured Questions

11.
(a) $n = \frac{C \times V}{1000}$
$n = \frac{0.10 \times 25.0}{1000} = 0.0025 \text{ mol}$ [2]
(1 mark for substitution, 1 mark for answer)

(b) Mole ratio HCl : NaOH is 1 : 1.
Moles HCl needed = 0.0025 mol.
$V = \frac{n \times 1000}{C} = \frac{0.0025 \times 1000}{0.10} = 25.0 \text{ cm}^3$ [2]
(1 mark for mole logic, 1 mark for calculation)

(c)

1. Pour the solution into an evaporating basin.
2. Heat gently to evaporate some water until saturated (crystallisation point).
3. Leave to cool and crystallise.
4. Filter the crystals and wash with a little cold distilled water.
5. Dry between filter papers or in a warm oven.
   [3 marks: Any 3 distinct correct steps]

12.
(a) $Zn(s) + 2HCl(aq) \rightarrow ZnCl_2(aq) + H_2(g)$ [2]
(1 mark for formulas, 1 mark for balancing/state symbols)

(b)

• Hydrochloric acid is a strong acid and is fully ionised, producing a higher concentration of $H^+$ ions [1].
• Ethanoic acid is a weak acid and is partially ionised, producing a lower concentration of $H^+$ ions [1].
• Higher concentration of $H^+$ leads to more frequent effective collisions between $H^+$ and Zn atoms per unit time [1].

(c)

• The final volume is the same for both [1].
• Because the amount (moles) of acid and the amount of zinc (excess) are the same, so the total moles of hydrogen produced depends only on the limiting reactant (acid), which is equal in moles for both [1].

(d) Measure the loss in mass of the reaction flask over time [1] OR Measure the change in pH over time [1].

13.
(a) Iron [1]

(b) The reaction can proceed in both forward and backward directions / The products can react to reform the reactants. [1]

(c)
(i) Nitric acid [1]
(ii) $NH_3 + HNO_3 \rightarrow NH_4NO_3$ [2]
(1 mark for formulas, 1 mark for balancing)

(d)
Test: Hold damp red litmus paper near the gas / mouth of the test tube. [1]
Result: The litmus paper turns blue. [1]

14.
(a) $Ba^{2+}(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s)$ [2]
(1 mark for ions, 1 mark for state symbols and balancing)

(b) To remove any soluble impurities / adhering solution (e.g., sodium chloride/nitrate) from the surface of the precipitate. [1]

(c)

• Nitric acid is added in halide tests to remove carbonate or sulfite ions which would also form white precipitates with silver ions ($Ag_2CO_3$) [1].
• For sulfate test, barium carbonate is also a white precipitate, so acid is technically needed to rule out carbonate. However, the question asks why it's emphasized for halides.
  Alternative acceptable answer: Dilute nitric acid is used to prevent the precipitation of other barium salts like barium carbonate or barium sulfite which are soluble in acid, ensuring the white precipitate is strictly barium sulfate. If the question implies why we don't use HCl for sulfate test: Because HCl introduces chloride ions, which doesn't interfere with BaSO4, but if we were testing for halides later it would.
  Standard Mark Scheme Logic: Acid is added to both to remove carbonate interference. If the question implies a difference, it might refer to the fact that $BaSO_4$ is insoluble in acid, whereas $AgCl$ is the target.
  Refined Answer: Acid is added to both to remove carbonate ions. However, for halide tests, we specifically use nitric acid to avoid introducing chloride ions (if using HCl) or bromide/iodide ions which would interfere with the silver nitrate test. For sulfate, barium chloride/nitrate is used; if BaCl2 is used, HCl is fine. If Ba(NO3)2 is used, HNO3 is fine. The key is removing interfering ions. [2]

Section C: Free Response

15. [10 Marks]

Step 1: Solubility Test

• Add distilled water to a small amount of each solid in separate test tubes.
• Observation:
  • Solid X (NaCl) dissolves to form a colourless solution.
  • Solid Y (Na2CO3) dissolves to form a colourless solution.
  • Solid Z (CaCO3) does not dissolve (remains as solid/residue).
• Conclusion: Solid Z is Calcium Carbonate (insoluble carbonate). Solids X and Y are soluble. [3 marks]

Step 2: Acid Test on Solids (or Solutions for X/Y)

• Add dilute hydrochloric acid to the remaining solids/solutions.
• For Solid Z (already identified): Effervescence/bubbles produced. Gas turns limewater milky. Confirms Carbonate.
• For Solution X (NaCl): No effervescence / No visible reaction.
• For Solution Y (Na2CO3): Effervescence/bubbles produced. Gas turns limewater milky.
• Conclusion: Solid Y is Sodium Carbonate. Solid X is likely Sodium Chloride. [4 marks]

Step 3: Confirmatory Test for Chloride (Solid X)

• To the solution of Solid X, add dilute nitric acid followed by silver nitrate solution.
• Observation: White precipitate forms.
• Conclusion: Presence of chloride ions confirms Solid X is Sodium Chloride. [3 marks]

(Alternative valid path: Flame tests. Na gives yellow flame for X and Y. Ca gives brick-red for Z. Then distinguish X and Y with acid. This is also valid.)

Marking Rubric:

• 2 marks for correct identification of Z via solubility.
• 2 marks for correct observation/reasoning for Z with acid (if done).
• 2 marks for distinguishing X and Y using acid (effervescence for Y, none for X).
• 2 marks for confirmatory test for Carbonate (limewater).
• 2 marks for confirmatory test for Chloride (AgNO3).
  (Total 10 marks available. Award marks for logical flow, correct reagents, and correct observations.)