S4 Chemistry Paper: Neutralisation, Titration And Salt Preparation (100 Marks)

S4 Chemistry Paper: Neutralisation, Titration and Salt Preparation (100 marks)

2026-11-11

S4 Chemistry – acids, bases and titration

S4 chemistry paper: 6 questions, 24 parts, 100 marks, 75 minutes. Acid-base strength, titration, salt preparation and yield, indicators and pH, with full marking scheme.

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Paper details　S4 chemistry acids bases titration practice paper with full marking scheme

Question 1 Definitions and strength of acids and bases (12 marks)

Two bottles of solution, both 0.10 mol dm^-3, contain hydrochloric acid (HCl) and ethanoic acid (CH3COOH). Ka of ethanoic acid is 1.8 x 10^-5 at 25 C. Answer the following.

#	Question	Answer and marking notes
1	State the Brønsted-Lowry definitions of an acid and a base.	Answer: an acid is a proton donor and a base is a proton acceptor. (Marking: 1 mark each.)
2	Find the pH of the 0.10 mol dm^-3 hydrochloric acid.	Answer: hydrochloric acid is a strong acid and ionises completely, so [H+] is 0.10 mol dm^-3 and pH is 1.00. (Marking: 1 mark for complete ionisation, 2 marks for the pH value.)
3	Find the pH of the 0.10 mol dm^-3 ethanoic acid (take [H+] as the square root of Ka x c).	Answer: [H+] is the square root of (1.8 x 10^-5 x 0.10), which is 1.34e-03 mol dm^-3, and pH is 2.87. (Marking: 2 marks for the square root, 1 for [H+], 1 for pH.)
4	Explain why the two solutions of equal concentration have different pH values.	Answer: hydrochloric acid ionises completely while ethanoic acid ionises only partially, so the ethanoic acid solution has a smaller [H+] and a higher pH. (Marking: 2 marks for comparing the extent of ionisation, 1 for the conclusion.)

Question 2 Molarity and dilution (18 marks)

A student prepares solutions of different concentrations using 2.00 mol dm^-3 hydrochloric acid, solid sodium hydroxide (M = 40.0 g mol^-1) and distilled water.

#	Question	Answer and marking notes
5	Find the number of moles of HCl in 35.0 cm3 of the acid.	Answer: 2.00 x 35.0 / 1000 gives 0.0700 mol. (Marking: 2 marks for converting to dm3, 2 for the value.)
6	The 35.0 cm3 of acid is diluted with water to a total volume of 250.0 cm3. Find the new concentration.	Answer: the number of moles is unchanged, so 0.0700 / 0.2500 gives 0.280 mol dm^-3. (Marking: 2 marks for unchanged moles, 2 for the value.)
7	How many cm3 of the 2.00 mol dm^-3 acid are needed to prepare 500.0 cm3 of 0.500 mol dm^-3 acid?	Answer: moles needed is 0.500 x 0.5000 = 0.2500 mol, so the volume is 0.2500 / 2.00 x 1000 = 125.0 cm3. (Marking: 2 marks for moles, 2 for the equation, 1 for the answer.)
8	What mass of solid sodium hydroxide is needed to prepare 250.0 cm3 of 0.200 mol dm^-3 solution?	Answer: moles is 0.200 x 0.2500 = 0.050 mol, so the mass is 0.050 x 40.0 = 2.00 g. (Marking: 2 marks for moles, 2 for multiplying by molar mass, 1 for the answer.)

Question 3 Titration calculations (20 marks)

25.0 cm3 of sodium hydroxide solution of unknown concentration is titrated with 0.150 mol dm^-3 hydrochloric acid. The mean titre of three runs is 21.40 cm3. Answer the following.

#	Question	Answer and marking notes
9	Write the equation for the neutralisation reaction.	Answer: HCl + NaOH gives NaCl + H2O. (Marking: 2 marks for reactants, 1 for products and balancing.)
10	Calculate the number of moles of hydrochloric acid used.	Answer: 0.150 x 21.40 / 1000 gives 3.21e-03 mol. (Marking: 2 marks for converting the volume, 2 for the value.)
11	Find the concentration of the sodium hydroxide solution.	Answer: the mole ratio is 1 to 1, so the moles of sodium hydroxide are also 3.21e-03 mol and the concentration is 3.21e-03 / 0.0250 = 0.128 mol dm^-3. (Marking: 2 marks for the ratio, 1 for the volume, 2 for the value.)
12	If 0.100 mol dm^-3 sulphuric acid replaced the hydrochloric acid, what volume in cm3 would reach the same end point?	Answer: H2SO4 + 2NaOH gives Na2SO4 + 2H2O, so the moles of acid needed are 3.21e-03 / 2 = 1.60e-03 mol and the volume is 1.60e-03 / 0.100 x 1000 = 16.05 cm3. (Marking: 3 marks for the equation and ratio, 3 for moles, 2 for the volume.)

Question 4 Salt preparation and yield (18 marks)

25.0 cm3 of 1.00 mol dm^-3 sulphuric acid is reacted with excess copper(II) oxide (M = 79.5 g mol^-1) to prepare copper(II) sulphate crystals (CuSO4.5H2O, M = 249.5 g mol^-1). 2.20 g of the oxide is weighed out and 5.30 g of pure crystals is obtained.

#	Question	Answer and marking notes
13	Explain why excess copper(II) oxide is used.	Answer: to ensure all the acid reacts so that no acid remains; the excess solid can be filtered off. (Marking: 2 marks for complete reaction, 1 for filtration.)
14	Find the minimum mass of copper(II) oxide required in theory.	Answer: moles of acid is 1.00 x 0.0250 = 0.0250 mol, the ratio is 1 to 1, so the mass of oxide is 0.0250 x 79.5 = 1.99 g. (Marking: 3 marks for moles, 2 for multiplying by molar mass.)
15	Find the theoretical yield of copper(II) sulphate crystals.	Answer: moles of crystals is 0.0250 mol, so the theoretical yield is 0.0250 x 249.5 = 6.24 g. (Marking: 2 marks for the ratio, 1 for the molar mass, 2 for the value.)
16	Calculate the percentage yield of this preparation.	Answer: percentage yield is 5.30 / 6.24 x 100%, which is 85.0%. (Marking: 3 marks for the expression, 2 for the value.)

Question 5 Titration curves and indicators (16 marks)

The following questions concern titration apparatus, titration curves and the choice of indicator. Phenolphthalein changes over pH 8.2 to 10.0 and methyl orange over pH 3.1 to 4.4.

#	Question	Answer and marking notes
17	Explain why a burette rather than a measuring cylinder is used in a titration.	Answer: a burette delivers the solution drop by drop and can be read to 0.05 cm3, while a measuring cylinder is too coarse to control the end point. (Marking: 2 marks for dropwise delivery, 1 for accuracy.)
18	Compare the pH at the equivalence point for a strong acid with a strong base and for a weak acid with a strong base.	Answer: the equivalence point is about pH 7 for a strong acid with a strong base, and above 7 for a weak acid with a strong base because the weak acid salt formed hydrolyses to give an alkaline solution. (Marking: 2 marks for the strong-strong case, 2 for the weak-strong case and its reason.)
19	A student uses phenolphthalein for the titration of ethanoic acid with sodium hydroxide. State whether this is suitable, with a reason.	Answer: it is suitable, since the equivalence point lies in the alkaline region above pH 7 and the phenolphthalein range of 8.2 to 10.0 covers it. (Marking: 2 marks for the judgement, 2 for the position of the end point, 1 for comparing the range.)
20	If the true volume is 21.10 cm3 but the student reads 21.40 cm3, find the relative error caused.	Answer: the error is (21.40 – 21.10) / 21.10 x 100%, which is 1.42%, so the result is too high. (Marking: 1 mark each for the difference, dividing by the true value, the value and the direction.)

Question 6 pH and hydrogen ion concentration (16 marks)

At 25 C, Kw is 1.0 x 10^-14. Calculate the pH and ion concentrations below and answer the last part.

#	Question	Answer and marking notes
21	For a solution of pH 3, find [H+] and [OH-].	Answer: [H+] is 10^-3 = 1e-03 mol dm^-3 and [OH-] is 1.0 x 10^-14 / 10^-3 = 1e-11 mol dm^-3. (Marking: 2 marks each.)
22	What is the pH after this solution is diluted 100 times?	Answer: [H+] falls to one hundredth of its value, so pH is 3 + 2 = 5. (Marking: 2 marks for the new concentration, 2 for the pH.)
23	Gastric juice has a pH of about 1.5. Find its [H+].	Answer: [H+] is 10^-1.5 = 0.0316 mol dm^-3, about 3.16e-02 mol dm^-3. (Marking: 2 marks for the expression, 2 for the value.)
24	Explain why a solution of pH 2 must not be used to clean a marble (calcium carbonate) surface.	Answer: the acid reacts with calcium carbonate, releasing carbon dioxide and dissolving the surface, so the marble is etched and loses its shine. (Marking: 2 marks for reacting with calcium carbonate, 1 for carbon dioxide, 1 for surface damage.)

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