AQA A-Level Chemistry Paper 3, 2024: Question 4

Skills & knowledge

• Meaning of a weak acid in water (partial dissociation)
• Reading a titration curve: recognising the steep region at equivalence
• Writing the Kₐ expression for a carboxylic acid
• Using the half-neutralisation point: pH = pKₐ
• Calculating Kₐ from pH (and good significant figures)
• Explaining buffer action during weak acid/strong base titration
• Choosing suitable/unsuitable indicators based on pH transition ranges

How marks are won here

• Use the exact language AQA wants: “partially dissociates/ionises”
• At equivalence: mention rapid/large pH change for a small volume
• For pH = pKₐ proof: show the key step [HA] = [A⁻]
• For Kₐ: use pH at half-equivalence (not initial pH)
• Indicators: link choice to steep part of curve and transition range

✅ Correct answer (1)

Propanoic acid is weak because it only partially dissociates/ionises in water (only a small fraction of molecules form ions in aqueous solution).

❌ Common errors (why marks are lost)

• Writing “it’s reversible” (AQA: not credited).
• Talking about concentration (weak ≠ dilute).
• Stating incorrect ions (e.g. vague “hydrogen ions don’t dissociate”).

💡 Key knowledge

Weak acid equilibrium in water:

CH₃CH₂COOH(aq) ⇌ H⁺(aq) + CH₃CH₂COO⁻(aq)

“Weak” means equilibrium lies mostly to the left.

✅ Correct answer (1)

Because around the equivalence point there is a very steep/rapid change in pH, so a small volume of NaOH causes a large pH change.

🧠 Exam technique

• Use curve language: “steep gradient” / “large pH change for small volume”.
• If you add “so the equivalence/end-point can be located accurately”, keep the steep-change idea as your main sentence.

💡 Link to the graph

On the titration curve, the near-vertical section is where pH changes dramatically. That is why dropwise addition is used there.

✅ Correct answers (with the logic that earns marks)

M1:

Kₐ = [H⁺][CH₃CH₂COO⁻] / [CH₃CH₂COOH]

M2: When half of the propanoic acid has reacted with NaOH, the amount (and therefore concentration) of acid remaining equals the amount (concentration) of conjugate base formed:

[CH₃CH₂COOH] = [CH₃CH₂COO⁻]

M3: Substitute into Kₐ:

Kₐ = [H⁺] × ([CH₃CH₂COO⁻]/[CH₃CH₂COOH]) = [H⁺] × 1 = [H⁺]

So pKₐ = −log Kₐ = −log[H⁺] = pH. Therefore pH = pKₐ.

🧠 Exam technique (what top scripts do)

• Explicitly state the “half reacted” reasoning: moles of HA left = moles of A⁻ formed.
• Keep it tight: the key is making the ratio [A⁻]/[HA] = 1 so log 1 = 0 .
• AQA allows [H₃O⁺] instead of [H⁺] .

❌ Common errors

• Using round brackets instead of concentration brackets, or missing them entirely.
• Writing Kₐ = [H⁺]² / [HA] (wrong here because [A⁻] is not equal to [H⁺] in a buffer mixture).
• Forgetting to state [HA] = [A⁻] at half-neutralisation, so you can’t justify pH = pKₐ.

📐 Calculations (step-by-step)

1. Find equivalence volume (from data in question):
   Moles propanoic acid initially = 0.0800 mol dm⁻³ × 25.0 cm³ × (1 dm³ / 1000 cm³)
   = 0.0800 × 0.0250 = 0.00200 mol
   Volume of 0.100 mol dm⁻³ NaOH to neutralise = 0.00200 / 0.100 = 0.0200 dm³ = 20.0 cm³.
2. Half-neutralisation volume = 20.0 cm³ / 2 = 10.0 cm³.
3. Read pH at 10.0 cm³ from Figure 4: typically around 4.5–4.6 (mark scheme allows 4.4–4.7).
4. At half-neutralisation, pH = pKₐ, so Kₐ = 10⁻ᵖᴴ.
5. Example using pH = 4.60:
   Kₐ = 10⁻⁴⋅⁶⁰ = 2.51 × 10⁻⁵ mol dm⁻³ (3 s.f.)

❌ Common calculation traps

• Using the starting pH (≈2.80). The question says when half has reacted.
• Reading the pH at the wrong volume (must be half of equivalence volume, not “halfway along the x-axis”).
• Forgetting cm³ ↔ dm³ conversion when finding equivalence volume.
• Giving Kₐ with too few/too many significant figures; aim for ≥ 2 s.f..

🧠 Exam technique

• Always calculate (or identify from curve) the equivalence volume first, then halve it.
• When reading from the graph, quote a sensible value and then show Kₐ clearly as 10⁻ᵖᴴ .

✅ Correct explanation (2)

M1: Added OH⁻ reacts with propanoic acid, removing OH⁻:

OH⁻ + CH₃CH₂COOH → CH₃CH₂COO⁻ + H₂O

M2: Because both CH₃CH₂COOH and CH₃CH₂COO⁻ are present, the ratio [CH₃CH₂COOH]/[CH₃CH₂COO⁻] changes only slightly, so [H⁺] (and therefore pH) stays nearly constant. Equivalently, the dissociation equilibrium shifts to maintain/replace H⁺.

💡 Key knowledge

• A buffer contains a weak acid and its conjugate base.
• During this titration, the mixture becomes a buffer after some NaOH has converted some HA to A⁻.
• Buffer region on the curve is the gently rising section (here stated between 5 cm³ and 15 cm³).

❌ Common errors

• Only saying “it’s a buffer” with no chemistry of OH⁻ being removed.
• Claiming pH is constant because the solution is “neutralising” (too vague).
• Forgetting to mention the presence of both acid and salt (conjugate base).

✅ Correct answers (1 + 1)

Methyl orange: Its pH transition range does not match the steep pH change at the equivalence point (so it would not change colour at the equivalence/end-point).

Universal indicator: It shows a range of colours during the titration, so there is no distinct/sharp colour change at the equivalence/end-point.

🧠 Exam technique (indicator questions)

• Always link the indicator to the curve: “indicator must change colour within the vertical/steep part”.
• For universal indicator, the key phrase is “no distinct end-point”, not “it’s inaccurate”.

❌ Common errors

• Just stating “methyl orange is for strong acids” without referencing transition range vs equivalence pH (not what the mark scheme rewards).
• Saying universal indicator is “not precise” without explaining the multiple colours / no sharp change.
• Bringing in “weak acid–strong base” as the only reason (mark scheme says ignore this on its own).

🧠 Do this in the exam

• Define weak acid as partial dissociation in water.
• At equivalence: write “steep gradient / large pH change for small volume”.
• Write Kₐ = [H⁺][A⁻]/[HA] using correct species.
• State [HA] = [A⁻] at half-neutralisation → pH = pKₐ.
• Half-neutralisation volume here is 10.0 cm³; read pH there, then Kₐ = 10⁻ᵖᴴ.
• Buffer explanation must include: OH⁻ removed + ratio stays ~constant / equilibrium replaces H⁺.
• Indicators: talk about transition range and sharp colour change.

💡 One-line takeaways

• Half-equivalence point is where pH = pKₐ.
• Buffers work because added acid/base is mopped up by a conjugate pair.
• Good indicators change colour over the steep section of the curve.