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O Level Chemistry Practice Paper 1

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TuitionGoWhere Practice Paper - Chemistry O-Level

ANSWER KEY AND MARKING SCHEME

Paper: PRACTICE - Acids, Bases & Salts
Version: 1 of 5
Total Marks: 60


Section A: Short Answer Questions [20 marks]


1. Define the term acid according to the Arrhenius theory. [1]

Answer: An acid is a substance that produces hydrogen ions (H⁺) when dissolved in water.
[Accept: An acid is a substance that ionises/dissociates in water to produce H⁺ ions.]

Marking note: Must mention H⁺ ions and aqueous solution/water. Award 1 mark for correct definition.


2. A student tests an unknown solution with Universal Indicator and observes a blue-violet colour.

(a) State the approximate pH of the solution. [1]

Answer: pH 10–12 (or any value in this range).
[Accept: pH 11, pH 10–11, strongly alkaline pH range.]

(b) Identify the type of substance present in the solution. [1]

Answer: An alkali / a base / an alkaline substance.
[Accept: Strong alkali, soluble base.]


3. Ethanoic acid (CH₃COOH) is described as a weak acid, while hydrochloric acid (HCl) is a strong acid. Explain the difference between a weak acid and a strong acid in terms of ionisation. [2]

Answer: A strong acid ionises/dissociates completely in water to produce H⁺ ions [1], whereas a weak acid ionises/dissociates only partially in water, so only a small fraction of molecules produce H⁺ ions [1].

Marking note: Award 1 mark for "completely ionised" (strong acid) and 1 mark for "partially ionised" (weak acid). Must reference ionisation/dissociation. Do not accept "concentrated/dilute" as this is a different concept.


4. Write a balanced chemical equation, including state symbols, for the reaction between dilute sulfuric acid and zinc metal. [2]

Answer: Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g)

Marking note: Award 1 mark for correct formulae, 1 mark for correct state symbols and balancing. Equation must be balanced. Deduct 1 mark if state symbols are missing or incorrect.


5. A student adds dilute nitric acid to a sample of copper(II) oxide powder and warms the mixture gently.

(a) State the colour of the solution formed. [1]

Answer: Blue.
[Accept: Blue solution, blue-green.]

(b) Write a balanced chemical equation for this reaction. [2]

Answer: CuO(s) + 2HNO₃(aq) → Cu(NO₃)₂(aq) + H₂O(l)

Marking note: Award 1 mark for correct formulae, 1 mark for correct balancing and state symbols. Accept CuO + 2HNO₃ → Cu(NO₃)₂ + H₂O with state symbols.


6. Explain why copper metal does not react with dilute hydrochloric acid, but magnesium metal reacts vigorously. [2]

Answer: Copper is less reactive than hydrogen / is below hydrogen in the reactivity series, so it cannot displace hydrogen ions from the acid [1]. Magnesium is more reactive than hydrogen / is above hydrogen in the reactivity series, so it can displace hydrogen ions from the acid [1].

Marking note: Must reference relative reactivity/position in reactivity series for both metals. Award 1 mark for copper explanation, 1 mark for magnesium explanation.


7. State the name and chemical formula of the salt formed when potassium hydroxide solution is neutralised by sulfuric acid. [2]

Answer:
Name: Potassium sulfate [1]
Formula: K₂SO₄ [1]

Marking note: Award 1 mark for correct name, 1 mark for correct formula. Formula must have correct subscripts.


8. A farmer finds that the soil in a field is too acidic for growing crops. Name a suitable substance that can be added to the soil to raise its pH, and explain why it is effective. [2]

Answer: Calcium hydroxide / slaked lime / quicklime (calcium oxide) / limestone (calcium carbonate) [1]. It is a base which neutralises the excess acid in the soil, raising the pH [1].

Marking note: Award 1 mark for naming a suitable basic substance, 1 mark for explaining neutralisation. Accept any suitable base used in agriculture.


9. Complete the following word equation for the reaction between an acid and a carbonate: [2]

Answer: Acid + Carbonate → Salt + Water + Carbon dioxide

Marking note: Award 1 mark for any two correct products, 2 marks for all three correct. Products can be in any order.


10. State two observations you would make when a piece of marble (calcium carbonate) is added to dilute hydrochloric acid. [2]

Answer: Any two from:

  • Effervescence / fizzing / bubbles of gas produced [1]
  • The marble / solid dissolves / gets smaller / disappears [1]
  • A colourless solution is formed [1]
  • The gas produced turns limewater milky (if tested) [1]

Marking note: Award 1 mark for each valid observation, up to 2 marks. Must be observable changes.


Section B: Structured Questions [20 marks]


11. Magnesium and acid reaction.

(a) Calculate the number of moles of magnesium used. [2]

Answer:
n = m / Mr
n = 0.60 / 24 [1]
n = 0.025 mol [1]

Marking note: Award 1 mark for correct formula/substitution, 1 mark for correct answer with units. Accept 0.0250 mol.

(b) Calculate the volume of hydrogen gas produced at r.t.p. [2]

Answer:
From equation: 1 mol Mg produces 1 mol H₂ [1]
Moles of H₂ = 0.025 mol
Volume = n × 24 = 0.025 × 24 = 0.60 dm³ [1]
[Accept: 600 cm³]

Marking note: Award 1 mark for correct mole ratio, 1 mark for correct volume calculation with units. ECF from part (a) allowed.

(c) Explain why the reaction is slower with ethanoic acid. [2]

Answer: Ethanoic acid is a weak acid, which only partially ionises in water [1], so the concentration of H⁺ ions in the solution is lower than in hydrochloric acid of the same concentration. Fewer H⁺ ions means a slower rate of reaction [1].

Marking note: Award 1 mark for identifying ethanoic acid as weak/partially ionised, 1 mark for linking lower H⁺ concentration to slower rate.


12. Unknown salt X.

(a) Identify the anion present in salt X. Explain your reasoning. [2]

Answer: Chloride ion (Cl⁻) [1]. The white precipitate formed with silver nitrate solution (acidified with nitric acid) is silver chloride (AgCl), which confirms the presence of chloride ions [1].

Marking note: Award 1 mark for chloride, 1 mark for correct reasoning referencing the test and precipitate.

(b) Identify the cation present in salt X. Explain your reasoning. [2]

Answer: Ammonium ion (NH₄⁺) [1]. When sodium hydroxide is added and the mixture is warmed, ammonia gas is produced, which turns damp red litmus paper blue. This is the test for ammonium ions [1].

Marking note: Award 1 mark for ammonium, 1 mark for correct reasoning referencing the test and litmus result.

(c) Suggest a name and chemical formula for salt X. [2]

Answer:
Name: Ammonium chloride [1]
Formula: NH₄Cl [1]

Marking note: Award 1 mark for correct name, 1 mark for correct formula. Both must be consistent with identified ions.


13. Preparation of copper(II) sulfate crystals.

(a) Describe the steps to obtain dry crystals. [4]

Answer:

  1. Filter the mixture to remove the excess/unreacted copper(II) oxide [1].
  2. Heat the filtrate (copper(II) sulfate solution) to evaporate some of the water / to concentrate the solution / until a saturated solution is obtained [1].
  3. Allow the saturated solution to cool slowly so that crystals form / crystallisation occurs [1].
  4. Filter the crystals and dry them between pieces of filter paper / leave them to dry in air [1].

Marking note: Award 1 mark for each step. Steps must be in logical order. Accept "heat until crystallisation point" or "heat until saturation".

(b) Explain why copper(II) oxide is added in excess. [2]

Answer: To ensure that all the sulfuric acid is completely reacted/neutralised [1], so that the resulting solution contains only copper(II) sulfate and water (no excess acid) [1].

Marking note: Award 1 mark for ensuring complete reaction, 1 mark for purity of product or removal of acid.

(c) State why this method is not suitable for preparing lead(II) sulfate. [2]

Answer: Lead(II) sulfate is insoluble in water [1]. If lead(II) oxide is reacted with sulfuric acid, the lead(II) sulfate formed would form an insoluble layer around the lead(II) oxide, preventing further reaction [1].

Marking note: Award 1 mark for identifying insolubility of lead(II) sulfate, 1 mark for explaining the consequence. Accept: lead(II) sulfate is insoluble, so precipitation is a more suitable method.


Section C: Data-Based and Application Questions [20 marks]


14. pH bar chart analysis.

(a) Which solution is the most alkaline? [1]

Answer: Solution S (pH ~13).

Marking note: Must identify S. Award 1 mark.

(b) What does pH 7 indicate about solution R? [1]

Answer: Solution R is neutral / neither acidic nor alkaline.

Marking note: Award 1 mark for "neutral".

(c) Explain why solutions P and Q have different pH values despite having the same concentration. [3]

Answer: Hydrochloric acid (P) is a strong acid which ionises completely in water, producing a high concentration of H⁺ ions [1]. Ethanoic acid (Q) is a weak acid which ionises only partially in water, producing a lower concentration of H⁺ ions [1]. Since pH is a measure of H⁺ ion concentration, the strong acid (P) has a lower pH than the weak acid (Q) at the same concentration [1].

Marking note: Award 1 mark for strong acid = complete ionisation, 1 mark for weak acid = partial ionisation, 1 mark for linking H⁺ concentration to pH difference.

(d) Explain what has occurred when equal volumes of P and S are mixed, and write the ionic equation. [3]

Answer: Neutralisation has occurred [1]. The H⁺ ions from the acid (P) have reacted with the OH⁻ ions from the alkali (S) to form water [1].
Ionic equation: H⁺(aq) + OH⁻(aq) → H₂O(l) [1]

Marking note: Award 1 mark for identifying neutralisation, 1 mark for explanation in terms of ions, 1 mark for correct ionic equation with state symbols.


15. Salt preparation methods.

(a) Sodium chloride [2]

Answer:
Method: A (Titration) [1]
Reason: Sodium chloride is a soluble salt formed from a soluble alkali (NaOH) and an acid (HCl). Both reactants are soluble, so titration is the most suitable method as it allows exact neutralisation without using excess reactants [1].

Marking note: Award 1 mark for correct method, 1 mark for valid reason referencing solubility of reactants.

(b) Lead(II) sulfate [2]

Answer:
Method: C (Precipitation) [1]
Reason: Lead(II) sulfate is insoluble in water. It can be prepared by mixing aqueous solutions of lead(II) nitrate and sodium sulfate (or any soluble lead salt and soluble sulfate), forming a precipitate of lead(II) sulfate [1].

Marking note: Award 1 mark for correct method, 1 mark for valid reason referencing insolubility.

(c) Zinc sulfate [2]

Answer:
Method: B (Reacting acid with excess insoluble base) [1]
Reason: Zinc oxide or zinc carbonate (insoluble bases) can be reacted with excess sulfuric acid. The excess solid can be removed by filtration, leaving a solution of zinc sulfate which can be crystallised [1].

Marking note: Award 1 mark for correct method, 1 mark for valid reason. Accept method A if justified (e.g., using zinc hydroxide which is soluble in excess alkali, but this is less common).


16. Ammonium nitrate fertiliser.

(a) Write a balanced chemical equation for the reaction between ammonia solution and nitric acid. [2]

Answer: NH₃(aq) + HNO₃(aq) → NH₄NO₃(aq)
[Accept: NH₄OH(aq) + HNO₃(aq) → NH₄NO₃(aq) + H₂O(l)]

Marking note: Award 1 mark for correct formulae, 1 mark for correct balancing and state symbols.

(b) Explain what is meant by neutralisation in terms of the ions involved. [2]

Answer: Neutralisation is the reaction between H⁺ ions (from the acid) and OH⁻ ions (from the alkali/base) to form water [1]. The ionic equation is: H⁺(aq) + OH⁻(aq) → H₂O(l) [1].

Marking note: Award 1 mark for identifying the reacting ions, 1 mark for stating water is formed. Accept reference to the ionic equation as explanation.

(c) Suggest why ammonium nitrate is an effective fertiliser. [2]

Answer: Ammonium nitrate contains nitrogen in two forms: ammonium ions (NH₄⁺) and nitrate ions (NO₃⁻) [1]. Nitrogen is an essential element for plant growth, needed for the synthesis of proteins and chlorophyll. Ammonium nitrate provides a readily available source of nitrogen for plants [1].

Marking note: Award 1 mark for identifying nitrogen content, 1 mark for explaining the importance of nitrogen to plants. Accept: contains nitrogen which promotes leaf growth / is a component of proteins.


END OF ANSWER KEY


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