DSE Chemistry Practice Paper: Acids, Bases And Titration

DSE Chemistry Practice Paper: Acids, Bases and Titration

2026-11-03

DSE Chemistry – Acids and Bases Practice

DSE Chemistry practice paper: 12 questions, 100 marks, 75 minutes. Covers pH, titration, dilution, Ka, neutralisation enthalpy, salt hydrolysis, buffers and indicators.

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Paper details　DSE Chemistry, titration, pH, Ka, salt hydrolysis, indicator, 12 questions, 100 marks

Section A (Questions 1 to 4, 26 marks)

Read each question and write the correct answer with an explanation. Show working and units for any calculation.

#	Question	Answer and marking notes
1	What is the pH of 0.100 mol dm-3 hydrochloric acid (a strong acid that ionises completely) at 25 degrees Celsius? Show your working and give the answer to three significant figures.	Answer: pH = 1.00. Hydrochloric acid is a strong acid and ionises completely, so the hydrogen ion concentration equals the acid concentration, 0.100 mol dm-3; pH = -log(0.100) = 1.00. Marking: complete ionisation stated (2); pH formula applied (2); three significant figures (1); link between Kw and neutral pH at 25 degrees Celsius (1).
2	What is the pH of a 0.0100 mol dm-3 sodium hydroxide solution at 25 degrees Celsius?	Answer: pH = 12.00. First pOH = -log(0.0100) = 2.00, then pH = 14.00 – 2.00 = 12.00. Marking: pOH calculated (2); Kw relationship used (2); value and significant figures (2).
3	25.0 cm3 of 0.100 mol dm-3 sodium hydroxide is titrated with 0.125 mol dm-3 hydrochloric acid. What volume of acid is needed to reach the equivalence point?	Answer: 20.0 cm3. n(NaOH) = 0.0250 x 0.100 = 0.00250 mol; the acid and base react in a 1:1 ratio, so the same amount of acid is needed; V = 0.00250 / 0.125 = 0.0200 dm3 = 20.0 cm3. Marking: amount of base (2); 1:1 ratio (1); volume calculation (2); unit conversion (2).
4	How many moles of sodium hydroxide are present in 25.0 cm3 of 0.100 mol dm-3 solution, and what is that mass? (Mr = 40.0)	Answer: 0.00250 mol and 0.100 g. n = 0.0250 x 0.100 = 0.00250 mol; m = n x Mr = 0.00250 x 40.0 = 0.100 g. Marking: volume converted to dm3 (2); amount in moles (2); mass with unit (2); significant figures (1).

Section B (Questions 5 to 8, 32 marks)

Answer the following questions with working, units and evidence drawn from the data given.

#	Question	Answer and marking notes
5	10.0 cm3 of 2.00 mol dm-3 hydrochloric acid is diluted with water to 250.0 cm3. Calculate the pH of the diluted solution.	Answer: pH = 1.10. Dilution does not change the amount of acid, so c1V1 = c2V2 and c2 = (2.00 x 10.0) / 250.0 = 0.0800 mol dm-3; pH = -log(0.0800) = 1.10. Marking: dilution relationship (3); concentration (2); pH and significant figures (2).
6	A 0.100 mol dm-3 solution of a weak acid HA has a pH of 2.90 at 25 degrees Celsius. Calculate Ka and state the assumptions you make.	Answer: Ka is about 1.6 x 10-5 mol dm-3. From pH 2.90 the hydrogen ion concentration is 1.26 x 10-3 mol dm-3. Assuming the acid ionises only slightly, the concentrations of hydrogen ion and conjugate base are equal and the equilibrium concentration of HA is about 0.100 – 1.26 x 10-3. So Ka = (1.26 x 10-3)^2 / 0.0987 = 1.6 x 10-5. Marking: hydrogen ion concentration from pH (2); charge balance assumption (2); equilibrium concentration (2); value with unit (2).
7	50.0 cm3 of 1.00 mol dm-3 hydrochloric acid is mixed with 50.0 cm3 of 1.00 mol dm-3 sodium hydroxide and the temperature rises by 6.7 degrees Celsius. Taking the density of the solution as 1.00 g cm-3 and its specific heat capacity as 4.18 J g-1 K-1, calculate the enthalpy change of neutralisation.	Answer: about -56.0 kJ mol-1. The total mass is 100.0 g and the heat released is q = mc dT = 100.0 x 4.18 x 6.7 = 2.80 x 10^3 J. The amount of water formed is 0.0500 mol, so the enthalpy change is -2.80 x 10^3 / 0.0500 = -5.60 x 10^4 J mol-1, that is -56.0 kJ mol-1. Marking: mass of solution (2); heat released (2); amount in moles (2); sign and unit (2).
8	25.0 cm3 of a vinegar sample is titrated with 0.100 mol dm-3 sodium hydroxide and 22.5 cm3 is required. Assuming ethanoic acid is monobasic with Mr = 60.0, calculate the mass of acid in the sample.	Answer: 0.135 g. n(NaOH) = 0.0225 x 0.100 = 0.00225 mol. Ethanoic acid is monobasic, so the same amount of acid is present and m = 0.00225 x 60.0 = 0.135 g. Marking: amount of base (3); 1:1 ratio (2); mass with unit (2); significant figures (2).

Section C (Questions 9 to 12, 42 marks)

Write fuller answers using correct subject terms and support them with examples from Hong Kong or daily life.

#	Question	Answer and marking notes
9	At the same concentration, strong and weak acids differ greatly in electrical conductivity and pH. Explain why.	Answer: A strong acid ionises completely in water, giving many mobile ions, so conductivity is high; the hydrogen ion concentration is large and the pH is low. A weak acid ionises only partially, so most of the acid remains as molecules, the ion concentration is low, conductivity is weaker and the pH is higher. On dilution the degree of ionisation of a weak acid rises, so its pH changes less than that of a strong acid. Marking: complete ionisation of strong acid (2); ions and conductivity (3); equilibrium of weak acid (3); pH comparison and dilution effect (2).
10	A 0.100 mol dm-3 ammonia solution and a 0.100 mol dm-3 sodium hydroxide solution have the same concentration but very different pH values. Explain why.	Answer: Sodium hydroxide is a strong base and ionises completely, so the hydroxide concentration equals 0.100 mol dm-3, pOH is 1.00 and pH is 13.00. Ammonia is a weak base and only a small fraction reacts with water: NH3 + H2O ⇌ NH4+ + OH-. The hydroxide concentration is far below 0.100 mol dm-3, so the pH is much lower, about 11. Marking: complete ionisation and pH 13.00 (3); partial ionisation of ammonia (3); equilibrium equation (2); comparison (2).
11	Explain why ammonium chloride solution is acidic, sodium ethanoate solution is alkaline, and sodium chloride solution is neutral.	Answer: Ammonium chloride dissolves to give ammonium ions, which hydrolyse to form ammonia and hydroxonium ions: NH4+ + H2O ⇌ NH3 + H3O+, so the solution is acidic. Sodium ethanoate gives ethanoate ions, which hydrolyse to form ethanoic acid and hydroxide ions: CH3COO- + H2O ⇌ CH3COOH + OH-, so the solution is alkaline. Sodium chloride gives ions that do not hydrolyse, so the solution stays neutral. Marking: hydrolysis of ammonium ion with equation (4); hydrolysis of ethanoate ion with equation (4); sodium chloride not hydrolysed (2); conclusion (1).
12	Which indicator should be used for the titration of ammonia solution with hydrochloric acid? Give reasons and describe the shape of the titration curve near the equivalence point.	Answer: Methyl orange should be used. In a strong acid with a weak base the equivalence point is acidic, at about pH 5 to 6; methyl orange changes colour between about pH 3.1 and 4.4, which is close to that value, whereas phenolphthalein changes between about pH 8.3 and 10.0 and would change colour before the equivalence point. The curve shows only a short steep section near the equivalence point, with pH falling from about 6 to about 4, so the indicator range must lie inside that section. Marking: choice of methyl orange (3); acidic equivalence point (3); comparison of ranges (3); steep section described (2).

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