Ten original SPM-style multiple-choice questions (A–D) with answers and one-line reasons, plus three structured questions with full model answers, covering properties, the pH scale, molarity, dilution, titration, salt preparation and qualitative analysis.
Attempt every question with your answer covered, then check the reason. These are original questions written in SPM style on the same content standards, not past-year papers. Relative atomic masses are given where needed. For the multiple-choice section choose one option A–D.
Paper 1 style, multiple choice
1. Which substance produces hydrogen ions when dissolved in water?
- A. Sodium chloride
- B. Hydrogen chloride
- C. Copper
- D. Calcium carbonate
Answer: B. An acid such as hydrogen chloride ionises in water to release H⁺ ions.
2. Which of the following does not show acidic properties?
- A. Dilute hydrochloric acid
- B. Dry hydrogen chloride gas
- C. Aqueous sulfuric acid
- D. Aqueous ethanoic acid
Answer: B. Without water there are no free H⁺ ions, so dry hydrogen chloride shows no acidic properties.
3. Solution X has pH 3 and solution Y has pH 6. Which statement is correct?
- A. Y is more acidic than X
- B. X and Y have the same H⁺ concentration
- C. X has a higher concentration of H⁺ ions than Y
- D. Y is alkaline
Answer: C. A lower pH means a higher H⁺ concentration, so X (pH 3) is more acidic than Y.
4. 4.0 g of sodium hydroxide, NaOH, is dissolved to make 250 cm³ of solution. What is its molarity? [H = 1, O = 16, Na = 23]
- A. 0.10 mol dm⁻³
- B. 0.40 mol dm⁻³
- C. 1.0 mol dm⁻³
- D. 16 mol dm⁻³
Answer: B. Moles = 4.0 ÷ 40 = 0.10 mol; molarity = 0.10 ÷ 0.250 = 0.40 mol dm⁻³.
5. What is the concentration of 0.10 mol dm⁻³ sulfuric acid in g dm⁻³? [H = 1, O = 16, S = 32]
- A. 0.98 g dm⁻³
- B. 4.9 g dm⁻³
- C. 9.8 g dm⁻³
- D. 98 g dm⁻³
Answer: C. Molar mass of H₂SO₄ = 98; 0.10 × 98 = 9.8 g dm⁻³.
6. How many moles of hydrochloric acid are in 20.0 cm³ of 0.50 mol dm⁻³ HCl?
- A. 0.010 mol
- B. 0.025 mol
- C. 0.10 mol
- D. 10 mol
Answer: A. Moles = 0.50 × 20.0 ÷ 1000 = 0.010 mol.
7. 25.0 cm³ of 0.80 mol dm⁻³ solution is diluted with water to 100 cm³. What is the new molarity?
- A. 0.20 mol dm⁻³
- B. 0.32 mol dm⁻³
- C. 0.80 mol dm⁻³
- D. 3.2 mol dm⁻³
Answer: A. M₂ = M₁V₁ ÷ V₂ = (0.80 × 25.0) ÷ 100 = 0.20 mol dm⁻³.
8. What volume of 0.10 mol dm⁻³ sodium hydroxide is needed to neutralise 25.0 cm³ of 0.10 mol dm⁻³ sulfuric acid? The equation is H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.
- A. 12.5 cm³
- B. 25.0 cm³
- C. 50.0 cm³
- D. 100 cm³
Answer: C. Moles of acid = 0.0025; moles of NaOH = 2 × 0.0025 = 0.005; volume = 0.005 ÷ 0.10 = 0.050 dm³ = 50.0 cm³.
9. Which salt is insoluble in water?
- A. Sodium nitrate
- B. Potassium chloride
- C. Barium sulfate
- D. Ammonium sulfate
Answer: C. Most sulfates are soluble, but barium sulfate is one of the exceptions, so it must be prepared by precipitation.
10. When silver nitrate solution is added to a salt solution, a white precipitate forms that dissolves in ammonia solution. Which ion is present?
- A. Carbonate
- B. Sulfate
- C. Nitrate
- D. Chloride
Answer: D. A white precipitate with silver nitrate that dissolves in ammonia solution confirms the chloride ion.
Paper 2 style, structured questions
Question 1 (titration). In a titration, 25.0 cm³ of sodium hydroxide solution is exactly neutralised by 22.5 cm³ of 0.20 mol dm⁻³ hydrochloric acid.
- (a) Name the apparatus used to measure the 25.0 cm³ of sodium hydroxide.
- (b) Write the balanced equation for the reaction.
- (c) Calculate the molarity of the sodium hydroxide.
- (d) Express this concentration in g dm⁻³. [H = 1, O = 16, Na = 23]
Model answer.
- (a) A pipette.
- (b) HCl + NaOH → NaCl + H₂O.
- (c) Moles of HCl = 0.20 × 22.5 ÷ 1000 = 4.5 × 10⁻³ mol. The ratio is 1 : 1, so moles of NaOH = 4.5 × 10⁻³ mol. Molarity = 4.5 × 10⁻³ ÷ (25.0 ÷ 1000) = 4.5 × 10⁻³ ÷ 0.0250 = 0.18 mol dm⁻³.
- (d) Molar mass of NaOH = 40 g mol⁻¹; concentration = 0.18 × 40 = 7.2 g dm⁻³.
Question 2 (standard solution and dilution). A student prepares 250 cm³ of a 0.10 mol dm⁻³ standard solution of anhydrous sodium carbonate, Na₂CO₃. [C = 12, O = 16, Na = 23]
- (a) Calculate the mass of sodium carbonate needed.
- (b) Name the apparatus used to make the volume up accurately to 250 cm³.
- (c) The student takes 25.0 cm³ of this standard solution and dilutes it to 100 cm³. Calculate the molarity of the diluted solution.
Model answer.
- (a) Molar mass of Na₂CO₃ = 2(23) + 12 + 3(16) = 106 g mol⁻¹. Moles = 0.10 × 0.250 = 0.025 mol. Mass = 0.025 × 106 = 2.65 g.
- (b) A volumetric flask.
- (c) M₂ = M₁V₁ ÷ V₂ = (0.10 × 25.0) ÷ 100 = 0.025 mol dm⁻³.
Question 3 (salt preparation and qualitative analysis). Zinc sulfate is a soluble salt.
- (a) Describe how to prepare a pure, dry sample of zinc sulfate crystals from zinc oxide and dilute sulfuric acid.
- (b) State the test and observation that confirm the sulfate ion.
- (c) State the test and observation that confirm the zinc ion using sodium hydroxide solution.
Model answer.
- (a) Add zinc oxide in excess to warm dilute sulfuric acid and stir until no more dissolves, so that all the acid reacts. Filter to remove the unreacted zinc oxide. Heat the filtrate to the point of crystallisation, then let it cool so crystals form. Filter the crystals, wash with a little cold distilled water and dry between filter papers.
- (b) Add dilute hydrochloric acid, then barium chloride solution; a white precipitate confirms the sulfate ion.
- (c) Add sodium hydroxide solution: a white precipitate forms that dissolves in excess sodium hydroxide, confirming the zinc ion.
Marking yourself
For every calculation, check that you converted cm³ to dm³, used the mole ratio from the balanced equation, and wrote the unit on the final line. A one-to-one teacher can go through your titration layout to make sure each step earns its mark. When you are confident, revisit the worked examples and then the common mistakes for this chapter.
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