AQA A-Level Chemistry Paper 1, 2022: Question 4

16 marks · Medium difficulty · State/Explain/Numerical

Calculate pH values of weak acid, strong base, and buffer solutions, explain neutralisation volumes at different temperatures, and describe buffer action.

Practise this question

Question

Question text

04 This question is about acids and bases.

04.1 Calculate the pH of a 0.150 mol dm–3 solution of ethanoic acid at 25 oC

Give your answer to 2 decimal places.

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

[3 marks]

04.2 Strontium is an element in Group 2.

Calculate the pH of a 0.0100 mol dm–3 solution of strontium hydroxide at 10 oC

You may assume that strontium hydroxide is completely dissociated in this solution.

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

[3 marks]

15 pH

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

At 50 °C a 25 cm3 sample of this barium hydroxide solution was neutralised by

22.45 cm3 of hydrochloric acid added from a burette.

*14* 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 °C

[2 marks]

Circle ( ) the correct answer.

˃ 22.45 cm3 = 22.45 cm3 ˂ 22.45 cm3

Explain your answer

04.4 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.

[3 marks]

How buffer solution is made

Equation

How buffer solution resists pH change

04.5 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.

*15* Calculate the pH of this buffer solution at 25 °C

Give your answer to 2 decimal places.

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

[5 marks]

Mark scheme

Show the mark scheme

Question Answers Additional Comments/Guidelines Mark

+ – + 2

[H ] [CH3COO ] [H ]

M1 Ka = = 1

[CH3COOH] [CH3COOH]

04.1 M2 [H+] = √1.74 × 10–5 × 0.150) = 1.62 × 10–3 (mol dm–3)

M3 pH = 2.79 M3 = – log M2 1

answer must be to 2 decimal places AO2

M1 [OH–] = 2 × 0.01 = 0.02 1

K 2.93 × 10-15 2.93 × 10-15

+ w –13

M2 [H ] = – = = 1.47 × 10 allow 1

[OH ] 0.02 M1

04.2

M3 pH = 12.83 allow 12.8 1

M3 = – log M2 AO2

if pH = 12.5(3) allow 2 marks (not used factor of 2

in M1)

– A-LEVEL CHEMISTRY – –

22.45 1

04.3 – –

same [OH ] or same amount/number of OH ions 1

AO3

add excess ethanoic acid to KOH 1

add enough KOH to the ethanoic acid so that the acid is partially

neutralised

add enough KOH so that the acid contains a mixture of ethanoic acid

and ethanoate ions

04.4

KOH + CH COOH CH COOK + H O allow KOH + CH COOH CH COO– K+ + H O 1

33 2 3 3 2

CH COO– (from salt) reacts with (added) acid/H+ ignore equilibrium shifts 1

AO3

– A-LEVEL CHEMISTRY – –

2.00

M1 (at start) n(NaOH) = = 0.05 (mol) 1

and

500 × 1.0

n(CH3COOH) = = 0.5 (mol)

1000

M2 (after adding NaOH) 1

n(CH3COOH) = (0.50 – 0.05) = 0.45 (mol)

M3 n(CH COO–) = (n NaOH) = 0.05 (mol) M3 = n(NaOH) from M1 1

04.5 3

K × [CH COOH] 1.74 × 10-5 × 0.9 1.74 × 10-5 × 0.45/V -5

a 3 1.74 × 10 × M2

M4 [H+] = or or M4 = 1

–

[CH3COO ] 0.1 0.05/V M3

= 1.57 × 10–4 (mol dm-3) V cancels out so not needed in this expression

M5 pH = 3.80 answer to 2 decimal places 1

M5 allow 3.81 AO2

allow pH = – log M4

– A-LEVEL CHEMISTRY – –

Henderson–Hasselbach method

2.00

M1 (at start) n(NaOH) = = 0.05 (mol) 1

and

500 × 1.0

n(CH3COOH) = = 0.5 (mol) 25

1000

M2 (after adding NaOH) 1

n(CH3COOH) = (0.50 – 0.05) = 0.45 (mol)

or [CH3COOH] = 0.9(0)

04.5

M3 n(CH COO–) = n(NaOH) = 0.05 (mol) M3 = n(NaOH) from M1 1

or [CH COO–] = 0.1(0)

–

[CH3COO ]

M4 pH = 4.759 + log 1

[CH3COOH]

0.1 0.05/V V cancels out so not needed in this expression

or 4.759 + log or 4.759 + log

0.9 0.45/V

M5 pH = 3.80 answer to 2 decimal places 1

M5 allow 3.81 AO2

How to answer it

Acids, Bases, Buffer Action & pH Calculations

Syllabus Focus: Physical Chemistry (3.1.12)

What This Question Tests

Weak acid dissociation: Setting up the equilibrium expression for K_a and calculating the pH of a weak monoprotic acid.
Strong alkaline solutions (Group 2): Working with diprotic hydroxides, stoichiometry ( 2 OH⁻ per formula unit), and non-standard temperatures using K_w .
Conceptual neutralisation: Differentiating between solution pH (which varies with temperature and K_w ) and the absolute amount (moles) of OH⁻ requiring neutralisation.
Acidic buffer formation & action: Identifying reagents to form a buffer (partial neutralisation), writing molecular equations, and explaining resistance to pH changes when acid is added.
Multi-step buffer calculations: Calculating moles before and after partial neutralisation, determining equilibrium concentrations, and calculating final pH to 2 decimal places.

Question 04.1 • 3 Marks

pH of a Weak Acid Solution (Ethanoic Acid)

Calculate the pH of 0.150 mol dm⁻³ ethanoic acid at 25 °C (K_a = 1.74 × 10⁻⁵ mol dm⁻³)

📐 Step-by-Step Calculation

1 State expression for K_a:
K_a = [H⁺][CH₃COO⁻] / [CH₃COOH] = [H⁺]² / [CH₃COOH] [1 mark]

2 Rearrange and solve for [H⁺]:
[H⁺] = √(K_a × [CH₃COOH])
[H⁺] = √(1.74 × 10⁻⁵ × 0.150) = 1.6155 × 10⁻³ mol dm⁻³ [1 mark]

3 Calculate pH:
pH = -log₁₀(1.6155 × 10⁻³) = 2.7918... → 2.79 [1 mark]

❌ Common Errors & Pitfalls

Incorrect Decimal Places: Giving pH as 2.8 (1 d.p.) or 2.792 (3 d.p.). pH values in AQA must strictly be quoted to 2 decimal places.
Forgetting to square root: Calculating [H⁺]² and taking -log([H⁺]²) , giving an impossibly high acidic pH (around 5.58).
Writing equilibrium without simplification: Forgetting to state [H⁺]² / [CH₃COOH] or failing to clearly indicate the assumption that [H⁺] ≈ [CH₃COO⁻] .

Mark Scheme Breakdown:
• M1: Expression K_a = [H⁺]² / [CH₃COOH] or full expression equating to [H⁺]² .
• M2: [H⁺] = √(1.74 × 10⁻⁵ × 0.150) = 1.62 × 10⁻³ mol dm⁻³ .
• M3: pH = 2.79 (must be 2 d.p., derived from -log M2).

Question 04.2 • 3 Marks

pH of a Group 2 Hydroxide at 10 °C

Calculate pH of 0.0100 mol dm⁻³ Sr(OH)₂ at 10 °C (K_w = 2.93 × 10⁻¹⁵ mol² dm⁻⁶)

📐 Step-by-Step Calculation

1 Determine [OH⁻] from stoichiometry:
Sr(OH)₂ dissolves completely: Sr(OH)₂ → Sr²⁺ + 2 OH⁻
[OH⁻] = 2 × 0.0100 = 0.0200 mol dm⁻³ [1 mark]

2 Calculate [H⁺] using K_w at 10 °C:
[H⁺] = K_w / [OH⁻] = (2.93 × 10⁻¹⁵) / 0.0200 = 1.465 × 10⁻¹³ mol dm⁻³ [1 mark]

3 Calculate final pH:
pH = -log₁₀(1.465 × 10⁻¹³) = 12.834... → 12.83 [1 mark]

🧠 Exam Technique & Examiner Insight

Factor of 2 Trap: Group 2 hydroxides M(OH)₂ yield two moles of hydroxide ions per mole of base. Forgetting the factor of 2 gives [OH⁻] = 0.0100 and pH = 12.53 (marks capped at 2/3).
Check your K_w value: Do not instinctively use 1.00 × 10⁻¹⁴ ! The question provides K_w at 10 °C ( 2.93 × 10⁻¹⁵ ).
Alternative method using pK_w : pK_w = -log(2.93 × 10⁻¹⁵) = 14.533 ; pOH = -log(0.02) = 1.699 ; pH = 14.533 - 1.699 = 12.83 .

Mark Scheme Breakdown:
• M1: [OH⁻] = 2 × 0.01 = 0.02 mol dm⁻³ .
• M2: [H⁺] = (2.93 × 10⁻¹⁵) / 0.02 = 1.47 × 10⁻¹³ mol dm⁻³ (allow K_w / M1).
• M3: pH = 12.83 (allow 12.8; if factor of 2 missed, allow pH = 12.53 for 2 marks).

Question 04.3 • 2 Marks

Temperature Effect on Neutralisation Titration

Deduce the volume of HCl needed to neutralise 25 cm³ of barium hydroxide at 10 °C compared to 50 °C

✅ Correct Answer

Circled option: = 22.45 cm³ [1 mark]

Explanation: Both samples have the same concentration of hydroxide ions ( [OH⁻] ) / the same number/amount of OH⁻ ions. [1 mark]

💡 Key Knowledge

Why did pH change with temperature? The self-ionisation of water is endothermic, so as temperature decreases from 50 °C to 10 °C, K_w decreases. Because [H⁺] = K_w / [OH⁻] , a lower K_w means a lower [H⁺] for the same [OH⁻] , hence a higher pH at lower temperature (or vice versa: lower pH at 50 °C).
Neutralisation stoichiometry: Titration neutralisation is purely stoichiometric: H⁺ + OH⁻ → H₂O . Changing temperature does not change the amount of dissolved Ba(OH)₂ or OH⁻ ions in a fixed 25 cm³ volume.

Mark Scheme Breakdown:
• M1: Selects = 22.45 cm³ .
• M2: Reason: same [OH⁻] OR same amount/number of OH⁻ ions.

Question 04.4 • 3 Marks

Buffer Solution Preparation & Action

Formation of a buffer from KOH and CH₃COOH, and its mechanism of resisting acid additions

✅ Model Answers

How buffer solution is made:
Add excess ethanoic acid to potassium hydroxide (KOH)
(OR: add enough KOH to partially neutralise the ethanoic acid so a mixture of CH₃COOH and CH₃COO⁻ ions remains). [1 mark]

Equation:
KOH + CH₃COOH → CH₃COOK + H₂O
(also accepted: KOH + CH₃COOH → CH₃COO⁻K⁺ + H₂O ) [1 mark]

How buffer resists pH change on adding acid:
The ethanoate ions ( CH₃COO⁻ ) react with the added acid / added H⁺ ions:
CH₃COO⁻ + H⁺ → CH₃COOH [1 mark]

❌ Common Misconceptions

Equimolar quantities: Stating "mix equal volumes/moles of KOH and ethanoic acid". This completely neutralises the acid into potassium ethanoate—no buffer forms! Acid must be in excess.
Vague resistance descriptions: Saying "the equilibrium shifts left" without specifying which species reacts with the added H⁺ . Top marks require naming the CH₃COO⁻ ion reacting with added H⁺ .
Wrong salt formula: Writing incorrect formulas like KCH₃COO or omitting water in the neutralisation equation.

Mark Scheme Breakdown:
• M1: Add excess ethanoic acid to KOH (or partial neutralisation to produce acid-salt mixture).
• M2: Balanced equation: KOH + CH₃COOH → CH₃COOK + H₂O .
• M3: CH₃COO⁻ (from salt) reacts with (added) acid/H⁺. (Mark scheme note: ignore equilibrium shifts).

Question 04.5 • 5 Marks

Buffer pH Calculation After Adding Strong Base

Calculate pH when 2.00 g NaOH is added to 500 cm³ of 1.00 mol dm⁻³ ethanoic acid (K_a = 1.74 × 10⁻⁵ mol dm⁻³)

📐 Step-by-Step Calculation

1 Calculate initial moles:
M_r(NaOH) = 23.0 + 16.0 + 1.0 = 40.0 g mol⁻¹
n(NaOH) = 2.00 / 40.0 = 0.050 mol
n(CH₃COOH)_initial = (500 × 1.00) / 1000 = 0.500 mol [1 mark]

2 Calculate equilibrium moles of unreacted acid:
n(CH₃COOH)_remaining = 0.500 - 0.050 = 0.450 mol [1 mark]

3 Calculate equilibrium moles of salt formed:
n(CH₃COO⁻) = n(NaOH) = 0.050 mol [1 mark]

4 Calculate [H⁺] (volumes cancel):
[H⁺] = K_a × [CH₃COOH] / [CH₃COO⁻]
[H⁺] = (1.74 × 10⁻⁵) × (0.450 / 0.050) = 1.566 × 10⁻⁴ mol dm⁻³ [1 mark]

5 Calculate final pH:
pH = -log₁₀(1.566 × 10⁻⁴) = 3.805 → 3.80 (or 3.81) [1 mark]

🧠 Alternative Method: Henderson-Hasselbalch

Once moles are found:

pK_a = -log₁₀(1.74 × 10⁻⁵) = 4.759
pH = pK_a + log₁₀([A⁻] / [HA])
pH = 4.759 + log₁₀(0.050 / 0.450)
pH = 4.759 + log₁₀(0.1111)
pH = 4.759 - 0.954 = 3.80

Key Tip: Notice that total volume (500 cm³) cancels out in the ratio [CH₃COOH] / [CH₃COO⁻] , so you can work directly with moles!

Mark Scheme Breakdown:
• M1: Initial moles: n(NaOH) = 0.05 mol AND n(CH₃COOH) = 0.50 mol .
• M2: Moles of acid remaining: 0.50 - 0.05 = 0.45 mol (or conc = 0.90 mol dm⁻³).
• M3: Moles of ethanoate formed: 0.05 mol (or conc = 0.10 mol dm⁻³).
• M4: Rearranging for [H⁺] = 1.57 × 10⁻⁴ mol dm⁻³ (or correct log ratio in Henderson-Hasselbalch).
• M5: pH = 3.80 (allow 3.81; answer must be to 2 decimal places).

Topics

Physical Chemistry · 3.1.12 Acids and Bases · 3.1.2 Amount of Substance

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 1, 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.

AQA A-Level Chemistry Paper 1, 2022: Question 4

Question

Mark scheme

How to answer it

Acids, Bases, Buffer Action & pH Calculations

What This Question Tests

pH of a Weak Acid Solution (Ethanoic Acid)

Calculate the pH of 0.150 mol dm⁻³ ethanoic acid at 25 °C (Ka = 1.74 × 10⁻⁵ mol dm⁻³)

📐 Step-by-Step Calculation

❌ Common Errors & Pitfalls

pH of a Group 2 Hydroxide at 10 °C

Calculate pH of 0.0100 mol dm⁻³ Sr(OH)₂ at 10 °C (Kw = 2.93 × 10⁻¹⁵ mol² dm⁻⁶)

📐 Step-by-Step Calculation

🧠 Exam Technique & Examiner Insight

Temperature Effect on Neutralisation Titration

Deduce the volume of HCl needed to neutralise 25 cm³ of barium hydroxide at 10 °C compared to 50 °C

✅ Correct Answer

💡 Key Knowledge

Buffer Solution Preparation & Action

Formation of a buffer from KOH and CH₃COOH, and its mechanism of resisting acid additions

✅ Model Answers

❌ Common Misconceptions

Buffer pH Calculation After Adding Strong Base

Calculate pH when 2.00 g NaOH is added to 500 cm³ of 1.00 mol dm⁻³ ethanoic acid (Ka = 1.74 × 10⁻⁵ mol dm⁻³)

📐 Step-by-Step Calculation

🧠 Alternative Method: Henderson-Hasselbalch

Topics

Calculate the pH of 0.150 mol dm⁻³ ethanoic acid at 25 °C (K_a = 1.74 × 10⁻⁵ mol dm⁻³)

Calculate pH of 0.0100 mol dm⁻³ Sr(OH)₂ at 10 °C (K_w = 2.93 × 10⁻¹⁵ mol² dm⁻⁶)

Calculate pH when 2.00 g NaOH is added to 500 cm³ of 1.00 mol dm⁻³ ethanoic acid (K_a = 1.74 × 10⁻⁵ mol dm⁻³)