AQA A-Level Chemistry Paper 1, June 2025: Question 8

17 marks · Hard difficulty · State/Explain/Numerical

Calculate pH, weak acid concentration, analyze a pH curve, select indicators, explain buffer action, and calculate pH change of a buffer solution upon adding acid.

Practise this question

Question

A multi-part exam question about acids and bases. Part 08.1 asks to calculate the pH of 0.150 mol dm^-3 nitric acid. Part 08.2 asks to calculate the concentration of propanoic acid with pH 2.89 and Ka 1.35 x 10^-5. Figure 3 shows a pH titration curve starting at pH 3, rising gradually, then sharply from pH 6 to 12, and leveling off at pH 12.5. Part 08.3 asks to identify the acid-base combination for this curve. Part 08.4 asks to select indicators that change colour at the equivalence point from a table of six indicators with their pH ranges. Part 08.5 asks to explain qualitatively how buffer solution B resists pH change when NaOH is added. Part 08.6 asks to calculate the pH change when 1.50 cm^3 of 0.200 mol dm^-3 HCl is added to buffer B.
Question text

08 This question is about acids and bases.

08.1 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 marks]

pH

08.2 Propanoic acid is a weak acid.

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

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

is 1.35 × 10–5 mol dm–3

[4 marks]

28 –3

Concentration mol dm

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

Figure 3 shows the pH curve for this reaction.

Figure 3

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

[1 mark]

Tick ( ) one box.

Strong alkali added to strong acid

Strong alkali added to weak acid

Weak alkali added to strong acid

Weak alkali added to weak acid 29

08.4 Different indicators change colour across different pH ranges.

Tick ( ) the box of each indicator that would change colour at the equivalence point in

Figure 3.

[1 mark]

Name of indicator pH range

*28* Methyl orange 3.2–4.4

Bromocresol green 3.8–5.4

Cresol red 7.2–8.8

Naphtholphthalein 7.3–8.7

Cresolphthalein 8.2–9.8

Indigo carmine 30 11.5–14.0

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

08.5 Explain, qualitatively, how this buffer solution resists pH change when a small amount

of sodium hydroxide solution is added.

[2 marks]

08.6 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 (K ) for ethanoic acid at 25 °C is 1.78 × 10–5 mol dm–3

a

[7 marks]

pH change

Mark scheme

Show the mark scheme The mark scheme for question 08. 08.1 shows pH = -log10(0.150) = 0.82. 08.2 shows [H+] = 10^-2.89, Ka expression, and propanoic acid concentration calculation resulting in 0.123 mol dm^-3. 08.3 identifies 'strong alkali added to weak acid'. 08.4 lists cresol red, naphtholphthalein, and cresolphthalein as the three ticked indicators. 08.5 explains that OH- ions react with H+ (or HA) and the equilibrium shifts right to maintain [H+]. 08.6 shows a 7-step calculation: finding initial moles of HA (0.00600) and A- (0.00800), moles of H+ added (0.000300), new moles of HA (0.00630) and A- (0.00770), calculating new [H+] = 1.46 x 10^-5, new pH = 4.84, and pH change = 4.87 - 4.84 = 0.03.

Question Answers Additional comments/Guidelines Mark

+ Correct value to 2dp scores 2 marks

M1 pH = –log10[H ] or pH = –log10(0.150) 2

08.1 M2 pH = 0.82 (1 x AO1,

1 x AO2)

M1 [H+] = 10–pH = 10–2.89 (= 1.29 × 10–3)

[H+]2 [H+]2

M2 Ka = OR = M2 allow with numbers or symbols

[CH3CH2COOH] [𝐻𝐴]

+ 2 −3 2 4

08.2 [H ] (1.29 × 10 ) M3 allow rearrangement with numbers or

M3 [CH3CH2COOH] = = −5

Ka 1.35 × 10 symbols (4 x AO2)

M4 0.123 (mol dm–3) M4 0.12 to 0.13 to 2 significant figures or more

scores 4 marks

strong alkali added to weak acid 1

08.3

(1 x AO2)

– A-LEVEL CHEMISTRY – –

these 3 indicators all ticked

cresol red 1

08.4

naphtholphthalein (1 x AO3)

cresolphthalein

Alternative

M1 OH- ions react with H+ M1 OH– ions react with CH COOH/ethanoic

+ - + acid/HA

M2 The HA ⇌ H + A equilibrium shifts right to maintain [H ]

OR 2

08.5 OR

M1 HA + OH– → H O + A– (2 x AO1)

M2 ratio of [CH COOH] / [CH COO–] remains roughly constant 2

M2 ratio of [CH COOH] / [CH COO–] remains

roughly constant

– A-LEVEL CHEMISTRY – –

50 M1 amount HA at start = 0.120 × = 0.00600

M1 amount HA at start = 0.120 × = 0.00600 1000

1000

– 0.656

– 0.656 M2 amount A at start = = 0.00800

M2 amount A at start = = 0.00800 82.0

82.0 + 1.50

+ 1.50 M3 amount of H added = 0.200 × = 0.000300

M3 amount of H added = 0.200 × = 0.000300 1000

1000

– M4 amount of HA after reaction = 0.0063 and

M4 amount of HA after reaction = M1 + M3 and amount of A –

amount of A after reaction = 0.0077

after reaction = M2 – M3

08.6

+ [HA answer from M4] –5 M1 + M3 M5 is dependent on an attempt at M4 (7 x AO2)

M5 [H ] = Ka [A− 𝑎𝑛𝑠𝑤𝑒𝑟 𝑓𝑟𝑜𝑚 𝑀4] = 1.78 × 10 × M2 – M3

+ [HA] –5 0.00630 –5

M5 [H ] = Ka [A−] = 1.78 × 10 × 0.00770 = 1.46 × 10

M6 pH = – log M5

M6 pH = –log 1.46 × 10–5 = 4.84

M7 = 4.87 – M6

M7 change = 4.87 – 4.84 = 0.03 (answer to 2

decimal places)

How to answer it

Acids, Bases and Buffer Solutions

What this question tests
  • Calculating the pH of a strong monoprotic acid.
  • Rearranging the acid dissociation constant ( Kₐ ) expression to find the concentration of a weak acid.
  • Interpreting pH titration curves to identify the strength of the acid and base used.
  • Selecting appropriate chemical indicators based on their pH transition ranges.
  • Explaining qualitatively how acidic buffer solutions resist changes in pH when alkali is added.
  • Performing complex, multi-step buffer calculations to determine the pH change when a strong acid is added.

Part 08.1

pH of a Strong Acid

Correct Answer

pH = -log₁₀[H⁺]

pH = -log₁₀(0.150) = 0.82

[2 Marks]
M1: Correct formula or working
M2: 0.82 (must be exactly 2 decimal places)

Key Knowledge

Nitric acid (HNO₃) is a strong monoprotic acid. It dissociates fully in aqueous solution:

HNO₃(aq) → H⁺(aq) + NO₃⁻(aq)

Therefore, the concentration of hydrogen ions is equal to the concentration of the acid: [H⁺] = [HNO₃] = 0.150 mol dm⁻³ .

Exam Technique

Always read the rounding instructions carefully! The question explicitly asks for 2 decimal places. Writing 0.8 or 0.824 will lose you the second mark.

Common Errors

Some students confuse decimal places with significant figures. For pH values, only the numbers after the decimal point are significant, but the exam specifically requests 2 decimal places here.

Part 08.2

Concentration of a Weak Acid

Step-by-Step Calculation

  1. Find [H⁺] from pH:
    [H⁺] = 10⁻ᵖᴴ = 10⁻².⁸⁹ = 1.29 × 10⁻³ mol dm⁻³
  2. Write the Kₐ expression:
    Kₐ = [H⁺]² / [CH₃CH₂COOH]
    (Assuming [H⁺] ≈ [A⁻] for a weak acid in water)
  3. Rearrange for weak acid concentration:
    [CH₃CH₂COOH] = [H⁺]² / Kₐ
  4. Substitute and solve:
    [CH₃CH₂COOH] = (1.29 × 10⁻³)² / (1.35 × 10⁻⁵) = 0.123 mol dm⁻³
[4 Marks]
M1: [H⁺] calculation
M2: Kₐ expression
M3: Rearrangement
M4: 0.123 (accepts 0.12 to 0.13)

Common Traps

  • Forgetting to square [H⁺]: A very common slip is calculating [H⁺] / Kₐ instead of [H⁺]² / Kₐ .
  • Incorrect rearrangement: Ensure you can confidently rearrange algebraic fractions before the exam.

Part 08.3

Identifying the Titration Curve

Correct Answer

Strong alkali added to weak acid

[1 Mark] For ticking the correct box.

Key Knowledge

Analyze the start and end points of the pH curve in Figure 3:

  • Starting pH: The curve starts at pH ≈ 3. This indicates a weak acid (a strong acid would start around pH 1-2).
  • Ending pH: The curve levels off at pH ≈ 12.5. This indicates a strong alkali (a weak alkali would level off around pH 10).

Part 08.4

Selecting Indicators

Correct Answer

Tick the boxes for:

  • Cresol red (pH range 7.2–8.8)
  • Naphtholphthalein (pH range 7.3–8.7)
  • Cresolphthalein (pH range 8.2–9.8)
[1 Mark] All three correct indicators must be ticked, and no others.

Exam Technique

An indicator is suitable if its entire pH transition range falls completely within the vertical section (the rapid pH change region) of the titration curve.

Looking at Figure 3, the vertical section spans from approximately pH 6.0 to 11.0. The three selected indicators fit perfectly inside this window.

Part 08.5

Qualitative Buffer Action

Model Answer

M1: The added OH⁻ ions react with H⁺ ions in the buffer:

H⁺ + OH⁻ → H₂O

(Alternatively: OH⁻ reacts directly with the weak acid: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O )

M2: The weak acid dissociation equilibrium shifts to the right to replace the lost H⁺ ions:

CH₃COOH ⇌ CH₃COO⁻ + H⁺

(Alternatively: The ratio of [CH₃COOH] / [CH₃COO⁻] remains virtually constant)

[2 Marks]

How to get full marks

To score both marks, you must show a clear two-step logical sequence:

  1. Identify what the added species reacts with.
  2. Explain how the equilibrium shifts to restore the balance, or state that the concentration ratio of the weak acid to its conjugate base remains almost constant.

Part 08.6

Quantitative Buffer Calculation

Highly Challenging - 7 Marks

Step-by-Step Calculation

Step 1: Calculate initial moles of weak acid (HA) and conjugate base (A⁻)

• Moles of ethanoic acid ( HA ):

n(HA) = conc × vol = 0.120 × (50 / 1000) = 0.00600 mol [M1]

• Moles of sodium ethanoate ( A⁻ ):

n(A⁻) = mass / Mᵣ = 0.656 / 82.0 = 0.00800 mol [M2]

Step 2: Calculate moles of H⁺ added from HCl

n(H⁺ added) = conc × vol = 0.200 × (1.50 / 1000) = 0.000300 mol [M3]

Step 3: Determine new moles of HA and A⁻ after reaction

The added H⁺ reacts with the conjugate base ( A⁻ ) to form more weak acid ( HA ):

A⁻ + H⁺ → HA

• New n(HA) = 0.00600 + 0.000300 = 0.00630 mol [M4]

• New n(A⁻) = 0.00800 - 0.000300 = 0.00770 mol [M4]

Step 4: Calculate new [H⁺] using Kₐ

Kₐ = [H⁺][A⁻] / [HA]   ➔   [H⁺] = Kₐ × [HA] / [A⁻]

[H⁺] = 1.78 × 10⁻⁵ × (0.00630 / 0.00770) = 1.456 × 10⁻⁵ mol dm⁻³ [M5]

Step 5: Calculate new pH

New pH = -log₁₀(1.456 × 10⁻⁵) = 4.84 [M6]

Step 6: Calculate the pH change

pH Change = Initial pH - New pH = 4.87 - 4.84 = 0.03 [M7]

[7 Marks] (1 mark per step successfully completed)

Calculation Traps

  • Sign Error: When acid is added, the amount of acid ( HA ) must increase (+ 0.000300) and the base ( A⁻ ) must decrease (- 0.000300). Getting these signs backwards is the most common way to lose 4 marks instantly.
  • Stopping too early: Many students calculated the new pH ( 4.84 ) but forgot to calculate the change in pH ( 0.03 ). Always re-read the final line of the question!

Examiner Commentary

This is a classic high-tariff A-Level question. Top-tier students structured their working clearly, labeling each step (e.g., "Moles before", "Moles after"). This ensures that even if an arithmetic error is made, consequential marking (ECF) can still secure most of the marks.

Topics

Physical Chemistry · 3.1.12 Acids and Bases

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.