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AQA A Level Chemistry

Topic questions Chemical changes Weak acids and Ka

Weak acids and Ka

13 questions · 144 marks · about 3 hours

Weak acids and Ka questions

AQA 7405 Weak acids and Ka 13 questions

Paper 1 June 2025, Question 8 17 marks

This question is about acids and bases.

8.1

Nitric acid is a strong acid.

Calculate the pH of 0.150 mol dm–3 nitric acid.

Give your answer to 2 decimal places.

[2]
Reveal mark scheme Mark scheme for question 8.1
8.2

Propanoic acid is a weak acid.

Calculate the concentration of a solution of propanoic acid with pH = 2.89 at 25 Β°C

The acid dissociation constant (Ka) for propanoic acid at 25 Β°C

Question 8, shown as an image
[4]
Reveal mark scheme Mark scheme for question 8.2
8.3
Question 8, shown as an image

A student records the pH as an alkali is added to an acid.

Figure 3 shows the pH curve for this reaction.

Figure for question 8

Which of these combinations gives the pH curve shown in Figure 3 ?

Question 8, shown as an image
Figure for question 8
[1]
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8.4

Different indicators change colour across different pH ranges.

Question 8, shown as an image
Figure for question 8
[1]
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8.5

Buffer solution B is made by dissolving 0.656 g of sodium ethanoate (Mr = 82.0) in 50 cm3 of 0.120 mol dm–3 ethanoic acid.

This buffer solution has pH = 4.87

Explain, qualitatively, how this buffer solution resists pH change when a small amount of sodium hydroxide solution is added.

[2]
Reveal mark scheme Mark scheme for question 8.5
8.6

Calculate the pH change that occurs when 1.50 cm3 of 0.200 mol dm–3 HCl(aq) are added to buffer solution B.

Give your answer to 2 decimal places.

The acid dissociation constant (Ka) for ethanoic acid at 25 Β°C is 1.78 Γ— 10–5 mol dm–3

[7]
Reveal mark scheme Mark scheme for question 8.6

Paper 3 June 2024, Question 4 11 marks

Figure 4 shows how the pH changes as 0.100 mol dm–3 sodium hydroxide solution is added to 25.0 cm3 of 0.0800 mol dm–3 aqueous propanoic acid at 298 K

Figure for question 4
4.1

Propanoic acid is a weak acid.

State the meaning of weak in this context.

[1]
Reveal mark scheme Mark scheme for question 4.1
4.2

Suggest why a student doing an experiment to produce the curve in Figure 4 would add the sodium hydroxide solution dropwise around the equivalence point.

[1]
Reveal mark scheme Mark scheme for question 4.2
4.3

Give an expression for Ka for propanoic acid (CH3CH2COOH).

Use this expression to show that pH = pKa when half of the propanoic acid has reacted with sodium hydroxide.

[3]
Reveal mark scheme Mark scheme for question 4.3
4.4

Use the pH from Figure 4, when half of the propanoic acid has reacted, to calculate

Question 4, shown as an image
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4.5

When sodium hydroxide solution is added to aqueous propanoic acid, the solution formed acts as a buffer when between 5 cm3 and 15 cm3 have been added.

Explain why the pH stays approximately constant during this part of the experiment.

[2]
Reveal mark scheme Mark scheme for question 4.5
4.6

Methyl orange and universal indicator are not suitable indicators for the titration of solutions of propanoic acid with sodium hydroxide.

State the reason why each indicator is not suitable.

Universal indicator

[2]
Reveal mark scheme Mark scheme for question 4.6

Paper 1 June 2023, Question 10 13 marks

This question is about weak acids.

10.1

Table 5 shows the pH ranges of some indicators.

Figure for question 10

Identify the indicator that is most suitable for use in a titration between propanoic acid and sodium hydroxide.

[1]
Reveal mark scheme Mark scheme for question 10.1
10.2

Give the expression for the acid dissociation constant (Ka) for propanoic acid

Question 10, shown as an image
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10.3

Calculate the pH of a 0.100 mol dm–3 propanoic acid solution. Give your answer to 2 decimal places.

For propanoic acid, pKa = 4.87

[4]
Reveal mark scheme Mark scheme for question 10.3
10.4

For butanoic acid, Ka = 1.51 Γ— 10–5 mol dm–3

20.0 cm3 of 0.100 mol dm–3 sodium hydroxide solution are added to 25.0 cm3 of 0.100 mol dm–3 butanoic acid solution.

Calculate the pH of the solution formed.

[5]
Reveal mark scheme Mark scheme for question 10.4
10.5

A student plans to titrate butanoic acid solution with a solution of ethylamine.

Explain why this titration could not be done using an indicator.

[2]
Reveal mark scheme Mark scheme for question 10.5

Paper 2 June 2023, Question 1 7 marks

This question is about rates of reaction.

Figure 1 shows apparatus used to measure the rate of reaction when an acid reacts with an excess of solid sodium hydrogencarbonate, NaHCO3

When different monoprotic organic acids are used, the rates at which gas escapes can be used to compare the strengths of the acids.

A timer is started when the NaHCO3 is added to the acid and the mass of CO2 gas lost is recorded at regular intervals. (It is assumed that any change in mass is due to the loss of CO2)

Figure for question 1
1.1

Suggest a reason why using a conical flask instead of a beaker would give more accurate results in this experiment.

[1]
Reveal mark scheme Mark scheme for question 1.1
1.2

Figure 2 shows the results of this experiment when 25.0 cm3 of a 2.23 mol dm–3 solution of ethanoic acid reacts with an excess of NaHCO3

Figure for question 1

Use Figure 2 to calculate the rate of reaction at 2 minutes.

Deduce the units of your calculated rate.

[3]
Reveal mark scheme Mark scheme for question 1.2
1.3

Chloroethanoic acid is a stronger acid than ethanoic acid.

Sketch, on Figure 2, the curve you would expect when 25.0 cm3 of a 2.23 mol dm–3 solution of chloroethanoic acid reacts with an excess of NaHCO3

Suggest why chloroethanoic acid is a stronger acid than ethanoic acid.

[3]
Reveal mark scheme Mark scheme for question 1.3

Paper 3 June 2023, Question 13 1 mark

When HF is added to water at 298 K, this equilibrium is established.

Question 13, shown as an image

At equilibrium, [HF] = 7.70 Γ— 10–3 mol dm–3 and [F–] = 2.30 Γ— 10–3 mol dm–3

What is the value of the equilibrium constant, in mol dm–3, at 298 K?

Figure for question 13
[1]

Paper 1 June 2022, Question 4 16 marks

This question is about acids and bases.

4.1

Calculate the pH of a 0.150 mol dm–3 solution of ethanoic acid at 25oC Give your answer to 2 decimal places.

For ethanoic acid, Ka = 1.74 x 10–5 mol dm–3 at 25oC

[3]
Reveal mark scheme Mark scheme for question 4.1
4.2

Strontium is an element in Group 2.

Calculate the pH of a 0.0100 mol dm–3 solution of strontium hydroxide at 10oC You may assume that strontium hydroxide is completely dissociated in this solution.

At 10oC the ionic product of water, Kw = 2.93 x 10–15 mol2 dm–6

[3]
Reveal mark scheme Mark scheme for question 4.2
4.3

The pH of a barium hydroxide solution is lower at 50 Β°C than at 10 Β°C

At 50 Β°C a 25 cm3 sample of this barium hydroxide solution was neutralised by 22.45 cm3 of hydrochloric acid added from a burette.

Deduce the volume of this hydrochloric acid that should be added from a burette to neutralise another 25 cm3 sample of this barium hydroxide solution at 10 Β°C

Question 4, shown as an image

Explain your answer

[2]
Reveal mark scheme Mark scheme for question 4.3
4.4

State how a buffer solution can be made from solutions of potassium hydroxide and ethanoic acid.

Give an equation for the reaction between potassium hydroxide and ethanoic acid.

State how this buffer solution resists changes in pH when a small amount of acid is added.

How buffer solution is made

Equation

How buffer solution resists pH change

[3]
Reveal mark scheme Mark scheme for question 4.4
4.5

A buffer solution is made by adding 2.00 g of sodium hydroxide to 500 cm3 of 1.00 mol dm–3 ethanoic acid solution.

Calculate the pH of this buffer solution at 25 Β°C

Give your answer to 2 decimal places.

For ethanoic acid, Ka = 1.74 x 10–5 mol dm–3 at 25 Β°C

[5]
Reveal mark scheme Mark scheme for question 4.5

Paper 3 November 2021, Question 5 5 marks

A mixture of methanoic acid and sodium methanoate in aqueous solution acts as an acidic buffer solution.

The equation shows the dissociation of methanoic acid.

Question 5, shown as an image

Calculate the mass, in g, of sodium methanoate (HCOONa) that must be added to 25.0 cm3 of 0.100 mol dmmaths symbol3 methanoic acid to produce a buffer solution with pH = 4.05 at 298 K

For methanoic acid, pKa = 3.75 at 298 K

Assume that the volume of the solution remains constant.

[5]
Reveal mark scheme Mark scheme for question 5

Paper 1 November 2020, Question 4 12 marks

Propanoic acid (C2H5COOH) is a weak acid.

The acid dissociation constant (Ka) for propanoic acid is 1.35 Γ— 10–5 mol dm–3

Question 4, shown as an image
4.1

State the meaning of the term weak acid.

[1]
Reveal mark scheme Mark scheme for question 4.1
4.2

Give an expression for the acid dissociation constant for propanoic acid.

[1]
Reveal mark scheme Mark scheme for question 4.2
4.3

A student dilutes 25.0 cm3 of 0.500 mol dm–3 propanoic acid by adding water until the

Question 4, shown as an image

Calculate the pH of this diluted solution of propanoic acid.

Give your answer to 2 decimal places.

[4]
Reveal mark scheme Mark scheme for question 4.3
4.4

A buffer solution with a pH of 4.50 is made by dissolving x g of sodium propanoate (C2H5COONa) in a solution of propanoic acid. The final volume of buffer solution is 500 cm3 and the final concentration of the propanoic acid is 0.250 mol dm–3

Calculate x in g For propanoic acid, Ka = 1.35 maths symbol 10–5 mol dm–3

[6]
Reveal mark scheme Mark scheme for question 4.4

Paper 1 June 2019, Question 9 15 marks

This question is about different pH values.

9.1

For pure water at 40 Β°C, pH = 6.67 A student thought that the water was acidic.

Explain why the student was incorrect.

Determine the value of Kw at this temperature.

Question 9, shown as an image
[4]
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9.2

Sodium hydroxide solution was added gradually from a burette to 25 cm3 of 0.080 mol dm–3 propanoic acid at 25 Β°C The pH was measured and recorded at regular intervals.

The results are shown in Figure 4.

Figure for question 9

Use Figure 4 to determine the value of Ka for propanoic acid at 25 Β°C

Show your working.

Question 9, shown as an image
[3]
Reveal mark scheme Mark scheme for question 9.2
9.3

Suggest which indicator is the most appropriate for the reaction in Question 09.2 ? Tick (βœ“) one box.

IndicatorpH rangeTick () one box
methyl orange3.1 – 4.4
bromothymol blue6.0 – 7.6
cresolphthalein8.2 – 9.8
indigo carmine11.6 – 13.0
[1]
Reveal mark scheme Mark scheme for question 9.3
9.4

A student prepared a buffer solution by adding 0.0136 mol of a salt KX to 100 cm3 of a 0.500 mol dm–3 solution of a weak acid HX and mixing thoroughly.

The student then added 3.00 Γ— 10–4 mol of potassium hydroxide to the buffer solution.

Calculate the pH of the buffer solution after adding the potassium hydroxide.

For the weak acid HX at 25 Β°C the value of the acid dissociation constant,

Question 9, shown as an image

Give your answer to two decimal places.

[6]
Reveal mark scheme Mark scheme for question 9.4
9.5

A buffer solution has a constant pH even when diluted.

Use a mathematical expression to explain this.

[1]
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Paper 3 June 2019, Question 2 15 marks

This question is about sulfuric acid and its salts.

2.1

Draw the displayed formula of a molecule of H2SO4

[1]
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2.2

In aqueous solution, sulfuric acid acts as a strong acid. The H2SO4 dissociates to form HSO4maths symbol ions and H+ ions.

The HSO4maths symbol ions act as a weak acid and dissociate to form SO42maths symbol ions and H+ ions.

Give an equation to show each stage in the dissociation of sulfuric acid in aqueous solution.

Include appropriate arrows in your equations.

[2]
Reveal mark scheme Mark scheme for question 2.2
2.3

A student is required to make 250 cm3 of an aqueous solution that contains an accurately measured mass of sodium hydrogensulfate (NaHSO4).

Describe the method that the student should use to make this solution.

[4]
Reveal mark scheme Mark scheme for question 2.3
2.4

A solution that contains 605 mg of NaHSO4 in 100 cm3 of solution has a pH of 1.72

Calculate the value of Ka for the hydrogensulfate ion (HSO4maths symbol) that is behaving as a weak acid. Give your answer to three significant figures.

Question 2, shown as an image
[6]
Reveal mark scheme Mark scheme for question 2.4
2.5

Some sodium sulfate is dissolved in a sample of the solution from question 02.4.

Explain why this increases the pH of the solution.

[2]
Reveal mark scheme Mark scheme for question 2.5

Paper 1 June 2018, Question 5 15 marks

Also involves Strong and weak acids

Hydrochloric acid is a strong acid and ethanoic acid is a weak acid.

5.1

State the meaning of the term strong acid.

[1]
Reveal mark scheme Mark scheme for question 5.1
5.2

In an experiment, 10.35 cm3 of 0.100 mol dm–3 hydrochloric acid are added to 25.0 cm3 of 0.150 mol dm–3 barium hydroxide solution.

Calculate the pH of the solution that forms at 30 Β°C

Question 5, shown as an image

Give your answer to 2 decimal places.

[6]
Reveal mark scheme Mark scheme for question 5.2
5.3

The pH of water at 30 Β°C is 6.92

Give the reason why water is neutral at this temperature.

[1]
Reveal mark scheme Mark scheme for question 5.3
5.4

Identify the oxide that could react with water to form a solution with pH = 2

Tick (βœ“) one box.

Figure for question 5
[1]
Reveal mark scheme Mark scheme for question 5.4
5.5

Give the expression for the acid dissociation constant (Ka) for ethanoic acid

Question 5, shown as an image
[1]
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5.6

A buffer solution contains 0.025 mol of sodium ethanoate dissolved in 500 cm3 of 0.0700 mol dm–3 ethanoic acid at 25 Β°C A sample of 5.00 cm3 of 2.00 mol dm–3 hydrochloric acid is added to this buffer solution.

Calculate the pH of the solution formed.

For ethanoic acid, Ka = 1.76 Γ— 10–5 mol dm–3 at 25 Β°C

[5]
Reveal mark scheme Mark scheme for question 5.6

Paper 3 June 2018, Question 4 16 marks

Also involves Chemical measurements and uncertainty

A student carried out an experiment to find the temperature rise for a reaction between hydrochloric acid and sodium hydroxide solution.

β€’ The student used a measuring cylinder to place 50 cm3 of 0.400 mol dm–3 hydrochloric acid into a glass beaker.

β€’ The student recorded the temperature at one-minute intervals for three minutes.

β€’ At the fourth minute the student added 50 cm3 of 0.400 mol dm–3 sodium hydroxide solution and stirred to mix the solutions, but did not record the temperature.

maths symbol The student recorded the temperature at one-minute intervals for a further eight minutes.

The results are shown in Table 3.

Figure for question 4
4.1

Plot a graph of temperature against time on the grid opposite. Use your graph to find the temperature rise, βˆ† T, at the fourth minute. Show your working on the graph by drawing suitable lines of best fit.

[5]
Reveal mark scheme Mark scheme for question 4.1
4.2

The uncertainty in each of the temperature readings from the thermometer used in this experiment was Β±0.1Β° C

Calculate the percentage uncertainty in the value for the temperature rise.

Question 4, shown as an image
[1]
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4.3

Suggest a change to the experiment that would minimise heat loss.

[1]
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4.4

Suggest and explain another change to the experiment that would decrease the percentage uncertainty in the use of the same thermometer.

[2]
Reveal mark scheme Mark scheme for question 4.4
4.5

A second student completed an experiment to determine the enthalpy of neutralisation for the reaction between ethanedioic acid solution (HOOCCOOH) and potassium hydroxide solution.

The student added 25 cm3 of 0.80 mol dm–3 ethanedioic acid solution to 75 cm3 of 0.60 mol dm–3 potassium hydroxide solution. The temperature increased by 3.2 Β° C

Give an equation for the reaction between ethanedioic acid solution and potassium hydroxide solution. Calculate the enthalpy change (βˆ† H) per mole of water formed in this reaction. Assume that the specific heat capacity of the reaction mixture is 4.2 J K–1g–1 Assume that the density of the reaction mixture is 1.00 g cm–3

Question 4, shown as an image
[5]
Reveal mark scheme Mark scheme for question 4.5
4.6

In a similar experiment to that in Question 04.5, the enthalpy of neutralisation for the reaction between sulfuric acid and potassium hydroxide solution was found to be –57.0 kJ mol–1 per mole of water formed.

Suggest an explanation for the difference between this value and your answer to Question 04.5. (If you were unable to obtain an answer to Question 04.5 you should assume a value of –28.5 kJ mol–1. This is not the correct answer.)

Turn over for Section B

[2]
Reveal mark scheme Mark scheme for question 4.6

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