AQA A-Level Chemistry Paper 1, 2023: Question 10

13 marks · Medium difficulty · State/Explain/Numerical

Answer questions on weak acids including indicator selection, writing a Ka expression, calculating the pH of propanoic acid, calculating the buffer pH from a partial neutralisation, and explaining why a weak acid–weak base titration cannot use an indicator.

Practise this question

Question

Question 10 consisting of five parts: 10.1 displays Table 5 with three indicators and their pH ranges (Bromocresol green 3.8-5.4, Bromothymol blue 6.0-7.6, Thymol blue 8.0-9.6) and asks to identify the suitable indicator for a propanoic acid and sodium hydroxide titration (1 mark); 10.2 asks for the Ka expression for propanoic acid (1 mark); 10.3 asks to calculate the pH of a 0.100 mol dm^-3 propanoic acid solution given pKa = 4.87 to 2 decimal places (4 marks); 10.4 asks to calculate the pH when 20.0 cm^3 of 0.100 mol dm^-3 NaOH is added to 25.0 cm^3 of 0.100 mol dm^-3 butanoic acid, Ka = 1.51 x 10^-5 mol dm^-3 (5 marks); 10.5 asks to explain why a titration of butanoic acid with ethylamine cannot be done using an indicator (2 marks).
Question text

10 This question is about weak acids.

10.1 Table 5 shows the pH ranges of some indicators.

Table 5

Indicator pH range

Bromocresol green 3.8 – 5.4

Bromothymol blue 6.0 – 7.6

Thymol blue 8.0 – 9.6

Identify the indicator that is most suitable for use in a titration between propanoic acid

and sodium hydroxide.

[1 mark]

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

(CH3CH2COOH).

[1 mark]

Ka

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 marks]

pH

10.4 For butanoic acid, K = 1.51 × 10–5 mol dm–3

a

20.0 cm3 of 0.100 mol dm–3 sodium hydroxide solution are added to

*24* 25.0 cm3 of 0.100mol dm–3 butanoic acid solution.

Calculate the pH of the solution formed.

[5 marks]

pH

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 marks]

Mark scheme

Show the mark scheme Mark scheme for Question 10: 10.1 gives Thymol blue (1 mark); 10.2 gives Ka = [H+][CH3CH2COO-] / [CH3CH2COOH] with square brackets essential (1 mark); 10.3 awards M1 for Ka = 10^-4.87 = 1.35 x 10^-5, M2 for Ka expression with square brackets or numbers, M3 for [H+] = sqrt(Ka x c) = 1.16 x 10^-3 mol dm^-3, M4 for pH = -log10[H+] = 2.94 (allow 2.93) (4 marks); 10.4 awards M1 for moles acid = 2.5 x 10^-3, M2 for moles NaOH = 2.0 x 10^-3, M3 for unreacted acid = 5.0 x 10^-4, M4 for [H+] = Ka x [HX]/[X-] = 3.775 x 10^-6, M5 for pH = 5.42 (5 marks); 10.5 awards M1 for stating weak acid and weak base titration, and M2 for pH change is too gradual / not sharp at the equivalence point / no vertical section on pH curve (2 marks).

Question Answers Additional comments/Guidelines Mark

Thymol blue 1

10.1

(AO3)

K = [H+][CH CH COO–] Square brackets essential 1

a 3 2

10.2

[CH3CH2COOH] (AO1)

M1 K = 10–pKa = 1.35 × 10–5

a

M2 K = [H+]2 OR [H+]2 = K [CH CH COOH] M2 Square brackets or with numbers

a a 3 2

[CH3CH2COOH]

10.3

M3 [H+] = √(1.35 × 10–5 x 0.1) = 1.16 x 10–3 mol dm–3 (4 x AO2)

M4 pH = –log (1.16 x 10–3) = 2.94

10 M4 = –log10 M3

Answer to 2 decimal places

Allow 2.93

M1 Initial amount of butanoic acid = 25 × 0.1 ×10–3 = 2.5 ×10–3 mol

M2 Initial amount of NaOH = 20 × 0.1 × 10–3 = 2.0 × 10–3 mol

M3 Final amount of acid = 2.5 × 10–3 – 2.0 × 10–3 = 5.0 × 10–4 mol M3 = M1-M2

+ M4 allow volumes cancelled out

M4 [H ] = Ka × [HX]

[X–] 1.51 × 10–5 × 5.0 × 10–4

2.0 × 10–3 5

10.4 Or [H+] = 1.51 × 10–5 × 0.0111

+ –6 –3 (5 x AO2)

0.0444 M4 Allow [H ] = 3.775 × 10 mol dm

Alternative method for M4

[𝑋𝑋−] 0.0444

pH = pKa + log = 4.82 + log ( )

[𝐻𝐻𝑋𝑋] 0.0111

M5 pH = 5.42 +

M5 = dependent on a correct expression for [H ]

in M4

M1 This is a weak acid and weak base/alkali titration

10.5 M2 pH change is too gradual/not sharp (at the equivalence point M2 Allow no vertical/steep section on pH curve

so colour change of indicator is difficult to judge) (2 x AO3)

How to answer it

Weak Acids, Buffers & Titration pH Curves

📋 What This Question Tests

This question evaluates your core competency across AQA Year 2 Physical Chemistry (Acids, Bases and Buffers):

  • Indicator Selection: Matching indicator transition ranges to titration equivalence points.
  • Equilibrium Expressions: Writing strict, formal expressions for the acid dissociation constant ( Ka ).
  • Weak Acid pH: Solving for [H⁺] and pH using pKa and initial weak acid concentration.
  • Acidic Buffer Calculations: Handling partial neutralisation to find remaining acid and conjugate base concentrations.
  • Titration Curves: Explaining why weak acid–weak base combinations cannot produce an accurate indicator endpoint.
Question 10.1 • 1 Mark

Indicator Selection for Weak Acid–Strong Base Titration

Identifying the correct indicator from pH transition intervals

✅ Correct Answer

Thymol blue

💡 Key Knowledge

Propanoic acid is a weak acid and sodium hydroxide is a strong base. At the equivalence point, the conjugate base (propanoate ion, CH₃CH₂COO⁻) hydrolyses water, yielding an alkaline equivalence point (typically pH 8–10). The indicator range must fall entirely within the steep vertical section of the pH curve.

Mark Scheme: 1 mark for Thymol blue (AO3).
Question 10.2 • 1 Mark

Acid Dissociation Constant Expression

Writing the equilibrium law for propanoic acid

✅ Correct Answer

Ka = [H⁺][CH₃CH₂COO⁻] / [CH₃CH₂COOH]

Also accepted: [H₃O⁺] in place of [H⁺] .

🧠 Exam Technique

  • Square brackets are strictly essential: Writing round brackets ( ) loses the mark because square brackets denote molar concentration in mol dm⁻³.
  • Do not omit charges on ions ( H⁺ , CH₃CH₂COO⁻ ).
  • Do not write [H⁺]² here; that simplification is only valid during mathematical calculation, not for the formal equilibrium expression.
Mark Scheme: 1 mark (AO1). Square brackets essential.
Question 10.3 • 4 Marks

pH Calculation of a Monoprotic Weak Acid

Converting pKa to Ka and determining solution pH

📐 Step-by-Step Calculation

1 Find Ka from pKa:
Ka = 10-pKa = 10-4.87 = 1.349 × 10⁻⁵ mol dm⁻³

2 State the simplified expression:
For a pure weak acid in water, assuming [H⁺] ≈ [CH₃CH₂COO⁻] and dissociation is negligible:
Ka = [H⁺]² / [CH₃CH₂COOH] → [H⁺]² = Ka × [CH₃CH₂COOH]

3 Calculate [H⁺]:
[H⁺] = √(1.349 × 10⁻⁵ × 0.100) = √(1.349 × 10⁻⁶) = 1.161 × 10⁻³ mol dm⁻³

4 Calculate pH:
pH = -log₁₀(1.161 × 10⁻³) = 2.94 (must be given to 2 decimal places)

🧠 Exam Technique

AQA questions specifying "give your answer to 2 decimal places" apply this rigorously to pH values. Even if an answer were 3.00, writing 3 or 3.0 forfeits the final mark.

❌ Common Errors

  • Forgetting to square root Ka × [HA] before taking the negative logarithm.
  • Using premature rounding (e.g. keeping only 1 sig fig for Ka) which leads to an inaccurate final pH.
Mark Breakdown:
• M1: Ka = 1.35 × 10⁻⁵
• M2: Ka = [H⁺]² / [CH₃CH₂COOH] or substituted numbers
• M3: [H⁺] = 1.16 × 10⁻³ mol dm⁻³
• M4: pH = 2.94 (allow 2.93 depending on rounding)
Question 10.4 • 5 Marks

pH of an Acidic Buffer Solution

Partial neutralisation of butanoic acid with sodium hydroxide

📐 Step-by-Step Calculation

1 Calculate initial moles of acid:
Moles HX = (25.0 / 1000) × 0.100 = 2.50 × 10⁻³ mol

2 Calculate initial moles of NaOH added:
Moles OH⁻ = (20.0 / 1000) × 0.100 = 2.00 × 10⁻³ mol

3 Determine composition after neutralisation:
Neutralisation reaction: CH₃CH₂CH₂COOH + OH⁻ → CH₃CH₂CH₂COO⁻ + H₂O
• Unreacted acid remaining: 2.50 × 10⁻³ - 2.00 × 10⁻³ = 5.0 × 10⁻⁴ mol
• Salt/conjugate base formed: 2.00 × 10⁻³ mol

4 Calculate [H⁺]:
Rearranging buffer equilibrium Ka = [H⁺][A⁻] / [HA] :
[H⁺] = Ka × [HA] / [A⁻]
Note: Because both species share the same total volume (45.0 cm³), the volume cancels out, allowing direct substitution of mole values:
[H⁺] = (1.51 × 10⁻⁵ × 5.0 × 10⁻⁴) / (2.00 × 10⁻³) = 3.775 × 10⁻⁶ mol dm⁻³

5 Determine final pH:
pH = -log₁₀(3.775 × 10⁻⁶) = 5.42

🧠 Alternative Method (Henderson-Hasselbalch)

pKa = -log₁₀(1.51 × 10⁻⁵) = 4.821
pH = pKa + log₁₀([A⁻] / [HA])
pH = 4.821 + log₁₀(2.00 × 10⁻³ / 5.0 × 10⁻⁴) = 4.821 + log₁₀(4) = 4.821 + 0.602 = 5.42

❌ Critical Buffer Traps

  • Mixing up remaining acid vs initial acid: You must subtract the moles of NaOH added from initial acid moles to find the unreacted acid.
  • Inverting the fraction: Always check: there is more conjugate base ( 2.0 × 10⁻³ ) than remaining acid ( 0.5 × 10⁻³ ), so the resulting pH must be higher than the pKa (4.82).
Mark Breakdown:
• M1: Moles of acid = 2.5 × 10⁻³ mol
• M2: Moles of NaOH = 2.0 × 10⁻³ mol
• M3: Remaining acid = 5.0 × 10⁻⁴ mol (M1 - M2)
• M4: Correct calculation of [H⁺] = 3.775 × 10⁻⁶ mol dm⁻³ or valid Henderson-Hasselbalch substitution
• M5: pH = 5.42 (dependent on correct [H⁺] expression in M4)
Question 10.5 • 2 Marks

Titration Feasibility: Weak Acid vs Weak Base

Explaining why indicators fail in weak–weak titrations

✅ Model Answer

  • This is a reaction between a weak acid and a weak base.
  • The pH change at the equivalence point is too gradual / not sharp (there is no vertical / steep section on the titration curve), making it impossible to observe a sharp indicator colour change.

💡 Visualising the Curve

In a strong acid–strong base titration, the curve possesses a rapid 4 to 6 pH unit vertical drop over a fraction of a drop of titrant. For a weak acid–weak base (e.g. butanoic acid and ethylamine), the curve has an inflection point but no vertical region. An indicator would change colour very gradually over several cubic centimetres of added titrant, giving no precise end-point.

Mark Scheme:
• M1: Weak acid and weak base/alkali titration (1 mark)
• M2: pH change is too gradual / not sharp at equivalence point OR no vertical/steep section on pH curve (1 mark)

Topics

Physical Chemistry · 3.1.12 Acids and Bases

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 1, 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.