HKDSE Chemistry Practice Paper: Acid-Base Titration And Electrochemistry

HKDSE Chemistry Practice Paper: Acid-Base Titration and Electrochemistry

2026-11-03

HKDSE Chemistry: Acid-Base Titration and Electrochemistry

HKDSE Chemistry practice paper: acid-base titration and electrochemistry, 12 questions, 100 marks, 75 minutes, with full answers and marking points for A4 printing.

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Paper details　Level: S6 | Subject: Chemistry | Time: 75 minutes | Marks: 100 | Questions: 12

Section A Basic acid-base calculations (Q1-Q4, 28 marks)

Section A has four compulsory questions.

#	Question	Answer and marking notes
1	A 25.0 cm³ sample of 0.100 mol dm⁻³ hydrochloric acid is titrated with 0.125 mol dm⁻³ sodium hydroxide solution. Calculate the volume of sodium hydroxide solution needed to reach the end point, in cm³, and write the balanced ionic equation. (6 marks)	Answer: n(HCl) = 0.100 × 0.0250 = 2.50 × 10⁻³ mol. Since H⁺ + OH⁻ → H₂O, n(NaOH) = 2.50 × 10⁻³ mol; V = 2.50 × 10⁻³ ÷ 0.125 = 0.0200 dm³ = 20.0 cm³. Marks: moles 2, 1:1 ratio 1, volume 2, correct unit 1.
2	Calculate the pH of 0.010 mol dm⁻³ hydrochloric acid and of 0.010 mol dm⁻³ ethanoic acid (Ka = 1.8 × 10⁻⁵ mol dm⁻³), and explain why the two values differ. (7 marks)	Answer: hydrochloric acid is strong and fully ionised, so [H⁺] = 0.010 mol dm⁻³ and pH = 2.00 (2 marks). Ethanoic acid is weak and only partly ionised: [H⁺] = √(1.8 × 10⁻⁵ × 0.010) = 4.24 × 10⁻⁴ mol dm⁻³, pH = 3.37 (3 marks). Only a small fraction of weak-acid molecules release hydrogen ions, so at equal molarity its hydrogen ion concentration is lower and its pH higher (2 marks).
3	10.0 cm³ of 0.500 mol dm⁻³ hydrochloric acid is diluted with water to a total volume of 250 cm³. Calculate the pH of the diluted solution and state the effect of a 25-fold dilution on the pH. (7 marks)	Answer: diluted concentration = 0.500 × 10.0 ÷ 250 = 0.0200 mol dm⁻³ (2 marks); pH = −log(0.0200) = 1.70 (3 marks). A 25-fold dilution lowers the hydrogen ion concentration, so the pH rises from 0.30 to about 1.70, an increase of 1.40 unit, while the solution stays strongly acidic (2 marks).
4	25.0 cm³ of 0.100 mol dm⁻³ sulphuric acid reacts with 0.200 mol dm⁻³ sodium hydroxide solution. Calculate the volume of alkali required and explain why sulphuric acid needs more alkali than hydrochloric acid of the same molarity. (8 marks)	Answer: n(H₂SO₄) = 0.100 × 0.0250 = 2.50 × 10⁻³ mol (1 mark); sulphuric acid is diprotic and fully ionised, giving two hydrogen ions per formula unit, so n(NaOH) = 5.00 × 10⁻³ mol (3 marks); V = 5.00 × 10⁻³ ÷ 0.200 = 0.0250 dm³ = 25.0 cm³ (2 marks). At equal molarity the diprotic acid supplies twice as many hydrogen ions, so twice the amount of alkali is needed (2 marks).

Section B Weak acids, buffers and titration curves (Q5-Q8, 32 marks)

Section B has four compulsory questions.

#	Question	Answer and marking notes
5	A 0.100 mol dm⁻³ solution of the monobasic weak acid HA has a pH of 2.90 at 25 °C. Calculate Ka to three significant figures and state whether HA is a strong or a weak acid. (8 marks)	Answer: [H⁺] = 10⁻²·⁹⁰ = 1.26 × 10⁻³ mol dm⁻³ (2 marks); Ka = (1.26 × 10⁻³)² ÷ 0.100 = 1.58 × 10⁻⁵ mol dm⁻³ (3 marks). Ka is far smaller than 1 and of the same order as that of ethanoic acid (1.8 × 10⁻⁵), so HA is a weak acid that is only partly ionised (3 marks).
6	A buffer is made from 0.10 mol dm⁻³ ethanoic acid and 0.10 mol dm⁻³ sodium ethanoate; Ka of ethanoic acid is 1.8 × 10⁻⁵ mol dm⁻³. Calculate the pH of the buffer and the concentration ratio of ethanoate to ethanoic acid needed for a pH of 5.00. (8 marks)	Answer: pH = pKa + log([A⁻]/[HA]) = 4.74 + log 1 = 4.74 (4 marks). For pH 5.00: log([A⁻]/[HA]) = 5.00 − 4.74 = 0.26, so [A⁻]/[HA] = 1.8 (4 marks), i.e. an ethanoate to ethanoic acid ratio of about 1.8 to 1.
7	20.0 cm³ of 0.100 mol dm⁻³ aqueous ammonia is titrated with 0.100 mol dm⁻³ hydrochloric acid. Calculate the volume of acid needed to reach the equivalence point and the pH of the solution at that point (Kb of NH₃ = 1.8 × 10⁻⁵, Kw = 1.0 × 10⁻¹⁴). (8 marks)	Answer: n(NH₃) = 0.100 × 0.0200 = 2.00 × 10⁻³ mol; the reaction is 1:1, so the acid volume = 2.00 × 10⁻³ ÷ 0.100 = 0.0200 dm³ = 20.0 cm³ (3 marks). Total volume at the equivalence point is 40.0 cm³, c(NH₄Cl) = 2.00 × 10⁻³ ÷ 0.0400 = 0.0500 mol dm⁻³ (2 marks). Ka(NH₄⁺) = Kw ÷ Kb = 5.6 × 10⁻¹⁰; [H⁺] = √(5.6 × 10⁻¹⁰ × 0.0500) = 5.3 × 10⁻⁶ mol dm⁻³, pH = 5.28 (3 marks).
8	25.0 cm³ of 0.100 mol dm⁻³ iron(II) sulphate solution is titrated in acidic medium with 0.0200 mol dm⁻³ potassium permanganate solution, MnO₄⁻ being reduced to Mn²⁺. Calculate the volume of permanganate solution required. (8 marks)	Answer: n(Fe²⁺) = 0.100 × 0.0250 = 2.50 × 10⁻³ mol (2 marks). Balanced equation: MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O, ratio 1 to 5 (3 marks); n(MnO₄⁻) = 5.00 × 10⁻⁴ mol; V = 5.00 × 10⁻⁴ ÷ 0.0200 = 0.0250 dm³ = 25.0 cm³ (3 marks).

Section C Integrated calculations (Q9-Q12, 40 marks)

Section C has four compulsory questions.

#	Question	Answer and marking notes
9	A 1.20 g sample containing sodium carbonate is dissolved and titrated with 0.900 mol dm⁻³ hydrochloric acid; 22.5 cm³ is used to reach the end point. Assuming the impurities do not react with acid, calculate the mass and the percentage by mass of sodium carbonate in the sample. (12 marks)	Answer: n(HCl) = 0.900 × 0.0225 = 2.025 × 10⁻² mol (2 marks); Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂, so n(Na₂CO₃) = 1.0125 × 10⁻² mol (3 marks); mass = 1.0125 × 10⁻² × 106 = 1.073 g (3 marks); percentage = 1.073 ÷ 1.20 × 100 = 89.4 per cent (4 marks).
10	For a zinc-copper cell, the standard electrode potentials are −0.76 V for Zn²⁺/Zn and +0.34 V for Cu²⁺/Cu. Calculate the standard cell e.m.f.; hence calculate the cell e.m.f. when [Zn²⁺] = 0.010 mol dm⁻³ and [Cu²⁺] = 1.0 mol dm⁻³ at 25 °C. (10 marks)	Answer: E°cell = 0.34 − (−0.76) = 1.10 V (3 marks). By the Nernst equation E = E° − (0.0592 ÷ 2) × log([Zn²⁺]/[Cu²⁺]) (3 marks); log(0.010 ÷ 1.0) = −2, so E = 1.10 + 0.0592 = 1.16 V (4 marks). The lower zinc ion concentration makes the cell e.m.f. higher than the standard value.
11	A steady current of 2.00 A is passed through copper(II) sulphate solution for 60.0 minutes. Calculate the mass of copper deposited at the cathode (Faraday constant 96500 C mol⁻¹, relative atomic mass of copper 63.5). (10 marks)	Answer: charge Q = 2.00 × 3600 = 7200 C (2 marks); amount of electrons = 7200 ÷ 96500 = 7.461 × 10⁻² mol (3 marks); Cu²⁺ + 2e⁻ → Cu gives n(Cu) = 3.731 × 10⁻² mol (3 marks); mass = 3.731 × 10⁻² × 63.5 = 2.37 g (2 marks).
12	At room temperature and pressure the molar volume is 24.0 dm³ mol⁻¹. 2.24 dm³ of carbon dioxide reacts completely with excess calcium hydroxide solution. Calculate the amount of carbon dioxide and the mass of calcium carbonate formed (relative formula mass of CaCO₃ = 100). (8 marks)	Answer: n(CO₂) = 2.24 ÷ 24.0 = 9.33 × 10⁻² mol (3 marks); CO₂ + Ca(OH)₂ → CaCO₃ + H₂O is 1:1, so n(CaCO₃) = 9.33 × 10⁻² mol (2 marks); mass = 9.33 × 10⁻² × 100 = 9.33 g (3 marks).

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