HKDSE Chemistry Practice Paper: Acids, Bases, Titration And PH

HKDSE Chemistry Practice Paper: Acids, Bases, Titration and pH

2026-11-04

HKDSE Physics: Mechanics and Circular Motion

HKDSE Chemistry paper: 12 questions, 100 marks, 75 minutes on acids, bases, pH, titration mole ratios, buffers, indicator choice and sample purity, with full worked answers.

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Paper details　Level: S4-S6 | Subject: Physics | Time: 75 minutes | Marks: 100 | Questions: 12

Section A: Mole Calculations and pH (Questions 1-4, 32 marks)

Each question carries 8 marks; show the equation, the substitution and the unit, and give pH to two decimal places.

#	Question	Answer and marking notes
1	A 25.0 cm³ portion of 0.100 M hydrochloric acid is titrated with 0.0500 M sodium hydroxide solution. (a) Find the number of moles of HCl in 25.0 cm³ of the acid. (b) Find the pH of the acid. (c) Find the volume of 0.0500 M NaOH needed to neutralise all the acid. (d) Find the number of moles of water formed.	(a) n(HCl) = cV = 0.100 × 25.0/1000 = 2.50 × 10⁻³ mol. [2 marks] (b) HCl is a strong acid and is fully ionised, so [H⁺] = 0.100 M and pH = −log 0.100 = 1.00. [2 marks] (c) H⁺ and OH⁻ react in a 1:1 ratio, so V = n/c = 2.50 × 10⁻³/0.0500 = 0.0500 dm³ = 50.0 cm³. [3 marks] (d) 1 mol of HCl gives 1 mol of water, so n(H₂O) = 2.50 × 10⁻³ mol. [1 mark] Marking notes: a strong acid is fully ionised; pH depends only on [H⁺]; multiply dm³ by 1000 to obtain cm³.
2	A 0.200 M solution of ethanoic acid has Ka = 1.8 × 10⁻⁵ at 25 °C. (a) Find the hydrogen ion concentration. (b) Find the pH of the solution. (c) Find the degree of dissociation as a percentage. (d) Explain why diluting the solution ten times changes the pH by less than one unit.	(a) [H⁺] = √(Ka × c) = √(1.8 × 10⁻⁵ × 0.200) ≈ 1.90 × 10⁻³ M. [2 marks] (b) pH = −log(1.90 × 10⁻³) = 2.72. [2 marks] (c) Degree of dissociation = [H⁺]/c × 100% = 1.90 × 10⁻³/0.200 × 100% = 0.95%. [3 marks] (d) Dilution lowers the concentration, but Ka = [H⁺]²/c shows that [H⁺] falls only with the square root of c, so a tenfold dilution should raise the pH by about 0.5 unit; in addition the degree of dissociation rises on dilution, replacing part of the lost hydrogen ions, so the rise is less than one unit. [1 mark] Marking notes: a weak acid is only partly ionised; the approximation [H⁺] = √(Ka·c) needs a small degree of dissociation.
3	25.0 cm³ of 0.100 M sodium hydroxide is titrated with 0.125 M sulphuric acid. (a) Find the number of moles of hydroxide ions in the alkali. (b) Find the number of moles of sulphuric acid required. (c) Find the volume of the acid required. (d) Find the number of moles of hydrogen ions added at the equivalence point.	(a) n(OH⁻) = 0.100 × 25.0/1000 = 2.50 × 10⁻³ mol. [2 marks] (b) H₂SO₄ supplies two H⁺ ions, so n(H₂SO₄) = 2.50 × 10⁻³/2 = 1.25 × 10⁻³ mol. [2 marks] (c) V = n/c = 1.25 × 10⁻³/0.125 = 0.0100 dm³ = 10.0 cm³. [3 marks] (d) n(H⁺) added = 2 × n(H₂SO₄) = 2.50 × 10⁻³ mol, equal to n(OH⁻). [1 mark] Marking notes: sulphuric acid is diprotic, so the ratio is H₂SO₄ : OH⁻ = 1 : 2.
4	Equal volumes of 0.10 M ethanoic acid and 0.10 M sodium ethanoate are mixed to give 1.0 dm³ of buffer solution; Ka = 1.8 × 10⁻⁵. (a) Write the expression for the hydrogen ion concentration and evaluate it. (b) Find the pH of the buffer. (c) Find the new pH after 0.010 mol of solid sodium hydroxide is added with no change in volume. (d) Explain why the pH hardly changes on adding a small amount of strong alkali.	(a) [H⁺] = Ka × [acid]/[salt] = 1.8 × 10⁻⁵ × 0.10/0.10 = 1.8e-05 M. [1 mark] (b) pH = −log(1.8 × 10⁻⁵) = 4.74. [1 mark] (c) The added OH⁻ converts 0.010 mol of the acid into ethanoate: the acid becomes 0.090 M and the salt 0.110 M, so [H⁺] = 1.8 × 10⁻⁵ × 0.090/0.110 = 1.47e-05 M and pH = 4.83. [3 marks] (d) The added OH⁻ is mostly neutralised by the acid to give more ethanoate, so the ratio [acid]/[salt] changes only slightly; since the pH depends on this ratio (Henderson–Hasselbalch), the pH rises only a little. [3 marks] Marking notes: a buffer contains a weak acid and its conjugate base; the pH is set by the ratio, not by the absolute concentrations.

Section B: Titration and Buffers (Questions 5-8, 36 marks)

Each question carries 9 marks; state the mole ratio in titration calculations and use the total volume for excess problems.

#	Question	Answer and marking notes
5	A 20.0 cm³ portion of 0.0500 M sulphuric acid. (a) Find the number of moles of the acid. (b) Find the hydrogen ion concentration, assuming complete ionisation. (c) Find the pH. (d) Find the mass of sodium hydroxide needed to neutralise the acid completely.	(a) n(H₂SO₄) = 0.0500 × 20.0/1000 = 1.00 × 10⁻³ mol. [2 marks] (b) Sulphuric acid is a diprotic strong acid, so [H⁺] = 2 × 0.0500 = 0.100 M. [2 marks] (c) pH = −log 0.100 = 1.00. [2 marks] (d) n(NaOH) needed = 2 × 1.00 × 10⁻³ = 2.00 × 10⁻³ mol, mass = 2.00 × 10⁻³ × 40.0 = 0.0800 g. [3 marks] Marking notes: mass = moles × molar mass; remember the factor of two for a diprotic acid.
6	30.0 cm³ of 0.150 M potassium hydroxide is titrated with 0.100 M hydrochloric acid. (a) Find the number of moles of KOH. (b) Find the volume of acid needed to reach the equivalence point. (c) If only 25.0 cm³ of acid is added, find the pH of the mixture. (d) State whether the mixture is acidic, alkaline or neutral, and explain.	(a) n(KOH) = 0.150 × 30.0/1000 = 4.50 × 10⁻³ mol. [3 marks] (b) V(HCl) = 4.50 × 10⁻³/0.100 = 0.0450 dm³ = 45.0 cm³. [2 marks] (c) n(H⁺) added = 0.100 × 25.0/1000 = 2.50 × 10⁻³ mol, so OH⁻ is in excess by 4.50 × 10⁻³ − 2.50 × 10⁻³ = 2.00 × 10⁻³ mol; total volume = 55.0 cm³, [OH⁻] = 2.00 × 10⁻³/0.0550 = 0.036 M, pOH = 1.44, pH = 12.56. [3 marks] (d) The mixture is alkaline because hydroxide ions are in excess and some KOH remains unneutralised, so the pH is above 7. [1 mark] Marking notes: subtract the reacted amount before dividing by the total volume; pH + pOH = 14.
7	At 25 °C the ionic product of water is Kw = 1.0 × 10⁻¹⁴. (a) Find the pH of 0.010 M nitric acid. (b) For a solution of pH 2.00, find its pOH and hydroxide ion concentration. (c) Find the pH of 0.0010 M barium hydroxide solution. (d) Write the relationship between pH and pOH at 25 °C.	(a) Nitric acid is a strong acid, so [H⁺] = 0.010 M and pH = −log 0.010 = 2.00. [2 marks] (b) pOH = 14.00 − 2.00 = 12.00, so [OH⁻] = 10⁻¹² M = 1.0 × 10⁻¹² M. [3 marks] (c) Ba(OH)₂ gives two OH⁻ ions, so [OH⁻] = 2 × 0.0010 = 0.002 M, pOH = 2.70, pH = 11.30. [2 marks] (d) Kw = [H⁺][OH⁻] = 1e-14 at 25 °C, hence pH + pOH = 14.00. [2 marks] Marking notes: barium hydroxide is a diacidic base; Kw equals 1.0 × 10⁻¹⁴ only at 25 °C.
8	25.0 cm³ of sodium hydroxide solution is titrated with 0.100 M hydrochloric acid. The titres are 24.80, 24.90 and 24.85 cm³. (a) Find the mean titre. (b) Find the number of moles of HCl used. (c) Find the concentration of the sodium hydroxide. (d) Explain why only concordant titres are averaged.	(a) Mean titre = (24.80 + 24.90 + 24.85)/3 = 24.85 cm³. [2 marks] (b) n(HCl) = 0.100 × 24.85/1000 = 2.485 × 10⁻³ mol. [2 marks] (c) NaOH and HCl react in a 1:1 ratio, so [NaOH] = 2.485 × 10⁻³/0.0250 = 0.0994 M. [3 marks] (d) Concordant titres agree to within 0.20 cm³ and so reflect a small random error; including a widely different reading would introduce error and reduce the accuracy of the result. [2 marks] Marking notes: select the concordant titres first, then average; quote an appropriate number of significant figures.

Section C: Enthalpy, Mixing and Sample Analysis (Questions 9-12, 32 marks)

Each question carries 8 marks; enthalpy changes must carry a sign and purity must be given as a percentage.

#	Question	Answer and marking notes
9	50.0 cm³ of 1.00 M hydrochloric acid and 50.0 cm³ of 1.00 M sodium hydroxide are mixed in a calorimeter and the temperature rises by 6.8 °C. Assume the density of the solution is 1.0 g cm⁻³ and its specific heat capacity is 4.2 J g⁻¹ K⁻¹. (a) Find the heat released in J. (b) Find the number of moles of water formed. (c) Find the enthalpy change of neutralisation in kJ mol⁻¹. (d) Explain why the experimental value is usually smaller than the textbook value.	(a) Heat released = mcΔT = 100.0 × 4.2 × 6.8 = 2856 J. [2 marks] (b) n(H₂O) = 1.00 × 0.0500 = 0.0500 mol. [2 marks] (c) ΔH = −Q/n = −2856/1000/0.0500 = -57.1 kJ mol⁻¹. [2 marks] (d) Some heat is lost to the surroundings and absorbed by the calorimeter, thermometer and walls, so the measured temperature rise is too small and the experimental heat released is less than the textbook value. [1 mark] Marking notes: neutralisation is exothermic, so ΔH carries a negative sign; divide by 1000 to convert J to kJ.
10	100.0 cm³ of 0.100 M hydrochloric acid is mixed with 50.0 cm³ of 0.100 M sodium hydroxide. (a) Find the number of moles of hydrogen ions in the acid. (b) Find the number of moles of hydroxide ions in the alkali. (c) Find the pH of the mixture. (d) State whether the mixture is acidic or alkaline.	(a) n(H⁺) = 0.100 × 100.0/1000 = 0.0100 mol. [2 marks] (b) n(OH⁻) = 0.100 × 50.0/1000 = 0.00500 mol. [2 marks] (c) Excess H⁺ = 0.0100 − 0.00500 = 0.0050 mol in a total volume of 150.0 cm³, so [H⁺] = 0.0050/0.150 = 0.0333 M and pH = 1.48. [3 marks] (d) The mixture is acidic because the strong acid is in excess and hydrogen ions remain (pH below 7). [1 mark] Marking notes: the volumes are additive (150.0 cm³); base the pH on the excess H⁺.
11	25.0 cm³ of 0.100 M ethanoic acid (Ka = 1.8 × 10⁻⁵) is titrated with 0.100 M sodium hydroxide. (a) Find the pH when half the equivalence volume of alkali has been added. (b) Use the answer to (a) to find Ka. (c) State the pH range of the buffer region of this titration curve. (d) Explain why phenolphthalein, rather than methyl orange, is a suitable indicator here.	(a) At the half-equivalence point [acid] = [salt], so pH = pKa = −log(1.8 × 10⁻⁵) = 4.74. [2 marks] (b) Ka = 10^(−4.74) = 1.8 × 10⁻⁵. [2 marks] (c) The buffer region is roughly pH 3.74 to 5.74, i.e. pKa ± 1. [3 marks] (d) The equivalence point of a weak acid–strong base titration lies above pH 7 (about 8–9), which falls within the colour-change range of phenolphthalein (8.3–10.0); methyl orange changes at pH 3.1–4.4 and would signal the end point too early. [1 mark] Marking notes: pH = pKa at the half-equivalence point is the key feature of a weak acid titration.
12	1.20 g of impure sodium hydroxide pellets is dissolved and made up to 250.0 cm³. A 25.0 cm³ portion requires 22.50 cm³ of 0.100 M hydrochloric acid for complete neutralisation. (a) Find the number of moles of HCl used. (b) Find the concentration of the sample solution. (c) Find the number of moles of NaOH in 250.0 cm³ of the solution. (d) Find the percentage purity of the sample.	(a) n(HCl) = 0.100 × 22.50/1000 = 2.250 × 10⁻³ mol. [2 marks] (b) [NaOH] = 2.250 × 10⁻³/0.0250 = 0.0900 M. [2 marks] (c) n(NaOH) = 0.0900 × 0.250 = 0.0225 mol. [2 marks] (d) Mass of pure NaOH = 0.0225 × 40.0 = 0.900 g, percentage purity = 0.900/1.20 × 100% = 75.0%. [2 marks] Marking notes: the titration measures only one 25.0 cm³ portion, so scale up by 250/25 = 10.

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