S6 Chemistry Paper: Acid-Base Equilibrium, Buffers And Titration Curves (100 Marks)

S6 Chemistry Paper: Acid-Base Equilibrium, Buffers and Titration Curves (100 marks)

2026-11-11

S6 Chemistry – acids, bases and buffers

S6 Chemistry paper: six questions on acid base equilibrium, buffers, titration curves, salt hydrolysis, buffer capacity and titration errors, with answers and marking notes.

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Paper details　S6 Chemistry practice paper with six questions on acid-base equilibrium: weak acid ionisation and pH, buffer preparation and pH change, titration curves, salt hydrolysis, buffer capacity and error analysis. Full answers with marking notes.

Question 1 Ionisation of a weak acid and pH (16 marks)

Ethanoic acid has Ka = 1.8 x 10^-5 at 25 degrees. Answer the following.

#	Question	Answer and marking notes
1	Find the hydrogen ion concentration and the pH of 0.10 M ethanoic acid.	Answer: taking the hydrogen ion concentration as x, Ka is about x squared over 0.10, so x is 1.34e-03 M and the pH is 2.87. (Marking: 1 mark for the Ka expression, 1 for the approximation, 1 for the concentration, 1 for the pH.)
2	Find the degree of ionisation of the same solution as a percentage.	Answer: the degree of ionisation is the hydrogen ion concentration divided by 0.10, giving about 1.34 per cent, so only a very small fraction is ionised. (Marking: 2 marks for the definition, 1 for substitution, 1 for the value.)
3	The solution is diluted ten times to 0.010 M. Find the new pH and the change in pH.	Answer: the hydrogen ion concentration becomes 4.24e-04 M, so the pH is 3.37, about 0.50 units higher. Dilution raises the degree of ionisation but the concentration still falls. (Marking: 2 marks for the concentration, 1 for the pH, 1 for the change.)
4	Enough hydrochloric acid is added to 0.10 M ethanoic acid to make the equilibrium hydrogen ion concentration 0.050 M. Find the ethanoate concentration and explain the change in ionisation.	Answer: the ethanoate concentration is Ka times 0.10 divided by 0.050, that is 3.6e-05 M. The common ion effect pushes the equilibrium to the left, so the degree of ionisation falls sharply. (Marking: 1 mark for the equilibrium, 2 for substitution, 1 for the explanation.)

Question 2 Preparing a buffer and its pH (18 marks)

A 1.0 L buffer is made from 0.20 M ethanoic acid and 0.30 M sodium ethanoate, Ka = 1.8 x 10^-5.

#	Question	Answer and marking notes
5	Find the pH of this buffer solution.	Answer: the pH is pKa plus the log of salt over acid, that is 4.74 plus 0.18, giving 4.92. (Marking: 1 mark for the equation, 1 for the ratio, 1 for the log value, 1 for the answer.)
6	0.010 mol of hydrochloric acid is added to 1.0 L of the buffer with no volume change. Find the new pH and the change.	Answer: the acid becomes 0.21 M and the salt 0.29 M, so the pH is 4.88, a fall of only -0.036 units, which shows the buffer action. (Marking: 1 mark for each new concentration, 2 for the pH, 1 for the change.)
7	For a buffer of the same system at pH 5.20, what concentration ratio of salt to acid is needed?	Answer: the log of the ratio is 0.46, so the ratio is about 2.85, that is roughly 2.9 times as much salt as acid. (Marking: 1 mark for rearranging, 1 for the antilog, 2 for the answer.)
8	What mass of anhydrous sodium ethanoate (M = 82.0) must be added to 0.25 L of 0.20 M ethanoic acid to give pH 5.00?	Answer: the ratio is 1.80, so the salt concentration is 0.360 M, giving 0.0900 mol in 0.25 L and a mass of 7.38 g. (Marking: 1 mark for the ratio, 1 for the concentration, 1 for the moles, 2 for the mass.)

Question 3 Titration curve of a weak acid with a strong base (16 marks)

25.0 cm3 of 0.100 M ethanoic acid is titrated with 0.100 M sodium hydroxide.

#	Question	Answer and marking notes
9	Find the pH at the start of the titration.	Answer: the hydrogen ion concentration is 1.34e-03 M, so the pH is 2.87, weakly acidic. (Marking: 2 marks for the calculation, 1 for the pH, 1 for the comment.)
10	After 12.5 cm3 of sodium hydroxide is added, find the pH and explain why it takes this value.	Answer: the acid and the salt are then present in equal amounts, so the pH equals pKa, that is 4.74, because the buffer ratio is one to one. (Marking: 2 marks for comparing the amounts, 1 for the pH, 1 for the explanation.)
11	Find the pH at the equivalence point and state a suitable indicator.	Answer: at the equivalence point the total volume is 50.0 cm3 and the ethanoate concentration is 0.050 M, so with Kb equal to 5.6 x 10^-10 the pH is 8.72. Phenolphthalein, which changes between 8.3 and 10.0, is suitable. (Marking: 1 mark for the concentration, 1 for Kb, 1.5 for the pH, 0.5 for the indicator.)
12	After 30.0 cm3 of sodium hydroxide has been added, find the pH.	Answer: the excess sodium hydroxide is 0.50 mmol in 55.0 cm3, so the hydroxide concentration is 0.0091 M, the pOH is 2.04 and the pH is 11.96. The excess strong base now controls the pH. (Marking: 2 marks for the excess amount, 1 for the concentration, 1 for the pH.)

Question 4 Salt hydrolysis and the acidity of solutions (16 marks)

At 25 degrees Kw = 1.0 x 10^-14, Kb of ammonia = 1.8 x 10^-5 and Ka of ethanoic acid = 1.8 x 10^-5.

#	Question	Answer and marking notes
13	Find the pH of 0.10 M ammonium chloride and write the hydrolysis equation.	Answer: the ammonium ion is a weak acid with Ka equal to Kw over Kb, that is 5.56e-10, so the hydrogen ion concentration is 7.45e-06 M and the pH is 5.13. (Marking: 1 mark for the equation, 1 for Ka, 1 for the concentration, 1 for the pH.)
14	Find the pH of 0.10 M sodium ethanoate solution.	Answer: the ethanoate ion hydrolyses to give hydroxide, with Kb equal to 5.56e-10, so the hydroxide concentration is 7.45e-06 M and the pH is 8.87, which is basic. (Marking: 1 mark for Kb, 1 for the concentration, 2 for the pH.)
15	Find the pH of 0.10 M sodium chloride solution and explain your answer.	Answer: the pH is 7.00 because neither ion hydrolyses, so the hydrogen and hydroxide concentrations are both 1.0 x 10^-7 M and the solution is neutral. (Marking: 2 marks for the pH, 2 for the explanation.)
16	Arrange 0.10 M ammonium chloride, sodium ethanoate and ammonium ethanoate in order of increasing pH, with reasons.	Answer: ammonium chloride at 5.13, then ammonium ethanoate at 7.00, then sodium ethanoate at 8.87. In ammonium ethanoate the acid and base constants are equal, so the two hydrolyses cancel and the pH is close to neutral. (Marking: 2 marks for the order, 1.5 for the ammonium ethanoate reason, 0.5 for the rest.)

Question 5 Buffer capacity and choosing a buffer system (18 marks)

A 1.0 L solution contains 0.10 M weak acid HA and 0.10 M its sodium salt NaA, pKa = 4.76.

#	Question	Answer and marking notes
17	After 0.020 mol of sodium hydroxide is added with no volume change, find the pH.	Answer: HA falls to 0.08 M and the ion rises to 0.12 M, so the pH is 4.94. (Marking: 1 mark for each new concentration, 1 for the working, 1 for the answer.)
18	If the same amount of sodium hydroxide is added to 1.0 L of pure water, find the pH and compare.	Answer: the hydroxide concentration is 0.020 M, so the pH is 12.30, which is 7.36 units higher than in the buffer. The buffer resists a large pH change. (Marking: 2 marks for the pH, 1 for the comparison, 1 for the explanation.)
19	How many moles of hydrochloric acid must be added to lower the pH of the buffer by 1.00 unit, assuming no volume change?	Answer: the required ratio is 0.10, so solving the ratio equation gives x equal to 0.0818 mol. (Marking: 1.5 marks for the target ratio, 2 for the equation, 1.5 for the solution.)
20	To prepare a buffer of pH 9.0, which pair should be chosen: NH4+/NH3 with pKa 9.25, CH3COOH/CH3COO- with pKa 4.76 or H2CO3/HCO3- with pKa 6.35? Find the concentration ratio needed.	Answer: the ammonium pair should be chosen because its pKa is closest to the target pH, differing by only 0.25. The ratio of ammonia to ammonium ion is 0.56, close to one, so the capacity is good. (Marking: 2 marks for the choice, 1 for the reason, 2 for the ratio.)

Question 6 Titration data and error analysis (16 marks)

25.0 cm3 of sodium hydroxide is titrated with 0.100 M ethanoic acid using phenolphthalein. The four titres are 24.80, 25.00, 24.85 and 24.90 cm3.

#	Question	Answer and marking notes
21	Select the acceptable titres, find the mean and state which reading is rejected.	Answer: reject 25.00 cm3, which differs from the others by 0.15 cm3, beyond the 0.10 cm3 limit. The mean of the other three is 24.85 cm3. (Marking: 1 mark for the rejection, 1 for the reason, 2 for the mean.)
22	Find the concentration of the sodium hydroxide solution.	Answer: the acid used is 0.0025 mol and it reacts one to one with the base, so the sodium hydroxide concentration is 0.1006 M. (Marking: 2 marks for the moles, 1 for the ratio, 1 for the concentration.)
23	Find the mass of solute in 250.0 cm3 of the sodium hydroxide solution, M = 40.0.	Answer: there are 0.02515 mol in 250.0 cm3, so the mass is 1.01 g. (Marking: 2 marks for the moles, 2 for the mass.)
24	If the burette is rinsed with distilled water but not with the acid, will the calculated sodium hydroxide concentration be too high or too low? Explain.	Answer: too high. The residual water dilutes the acid, so more of it is needed to neutralise the same amount of base; using 0.100 M overestimates the moles of acid and hence the result. (Marking: 1 mark for the conclusion, 1 for the dilution, 1 for the larger volume, 1 for the deduction.)

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