AQA A-Level Chemistry Paper 1, 2017: Question 2

9 marks · Medium difficulty · State/Explain/Numerical

Calculate the concentration of ethanoic acid in a buffer from Ka and pH, and calculate the pH of a different buffer after a given amount of NaOH is added.

Practise this question

Question

AQA A-Level Chemistry Paper 1, 2017: Question 2
Question text

02 This question is about acidic solutions.

02.1 The acid dissociation constant, Ka, for ethanoic acid is given by the expression

[CH COO− ] [H+ ]

K = 3

a

[CH3COOH]

The value of K for ethanoic acid is 1.74 × 10−5 mol dm−3 at 25 °C

a

A buffer solution with a pH of 3.87 was prepared using ethanoic acid and

sodium ethanoate. In the buffer solution, the concentration of ethanoate ions

was 0.136 mol dm−3

Calculate the concentration of the ethanoic acid in the buffer solution.

Give your answer to three significant figures.

[3 marks]

D

Concentration of acid5 mol dm−3

02.2 In a different buffer solution, the concentration of ethanoic acid was

0.260 mol dm−3 and the concentration of ethanoate ions was 0.121 mol dm−3

A 7.00 × 10−3 mol sample of sodium hydroxide was added to 500 cm3 of this

buffer solution.

Calculate the pH of the buffer solution after the sodium hydroxide was added.

Give your answer to two decimal places.

[6 marks]

pH of buffer solution

Mark scheme

Show the mark scheme Mark scheme for AQA A-Level Chemistry Paper 1, 2017: Question 2

Question Answers Mark Additional Comments/Guidance

[H+] = (10−3.87 =) 1.3489 x 10−4 1 Allow 1.35 x 10-4. If M1 wrong can only score M2.

[𝐇+][𝐂𝐇 𝐂𝐎𝐎–] [𝟏.𝟑𝟒𝟖𝟗 × 𝟏𝟎–𝟒][𝟎.𝟏𝟑𝟔]

𝟑

[CH3COOH] = [ ] = ( –𝟓 = 1.05436) 1

𝐾𝑎 [𝟏.𝟕𝟒 × 𝟏𝟎 ] Mark is for correctly rearranged equation.

1.05 – 1.06 (mol dm−3) 1 3 sf or more

02.1

– – –

If 0.007 moles in 500 cm3 seen follow Mark Scheme 1

16 of 35

Mark Scheme 1

moles ethanoic acid = 0.130

moles sodium ethanoate = 0.0605 1 Method 1

mol CH3COOH after addition = (0.130 - 0.007 ) = 0.123 1

mol CH COO- after addition = (0.0605+0.007) = 0.0675

31 For M3 allow M1 – 0.007

[𝟏.𝟕𝟒 × 𝟏𝟎–𝟓][𝟎.𝟏𝟐𝟑] 1 For M4 allow M2 + 0.007

+ [𝑲𝒂 ][𝐂𝐇𝟑𝐂𝐎𝐎𝐇] -5

[H ] =( – ) = [ ] (= 3.171 x 10 )

[𝐂𝐇𝟑𝐂𝐎𝐎 ] 𝟎.𝟎𝟔𝟕𝟓 1

pH = 4.50 (must be 2dp) Method 1 and 2

M5 = expression with their numbers

If 0.014 moles in 1 dm3 follow Mark Scheme 2 M6 = answer to 2 dp

Mark scheme 2

pH = 4.50 scores 6 marks

02.2 moles CH3COOH after addition = (0.260 - 0.014 ) = 0.246

(This scores 2 marks)

moles CH COO- after addition = (0.121+0.014) = 0.135 (This

3 If used in K expression, stop at M4

a

scores 2 marks) If divide by 2 after M5, lose M6

[𝟏.𝟕𝟒 × 𝟏𝟎–𝟓][𝟎.𝟐𝟒𝟔]

+ [𝑲𝒂 ][𝐂𝐇𝟑𝐂𝐎𝐎𝐇]

[H ] =( – ) = [𝟎.𝟏𝟑𝟓]

[𝐂𝐇𝟑𝐂𝐎𝐎 ]

pH = 4.50 (must be 2dp) Allow solutions which use Henderson-Hasselbach Equation

– – –

17 of 35

Total 9

How to answer it

Buffers with ethanoic acid / ethanoate (Kₐ & pH)

What this question tests
  • Using pH to find [H⁺] and substituting into the Kₐ expression.
  • Rearranging Kₐ = ([CH₃COO⁻][H⁺])/[CH₃COOH] to solve for an unknown concentration.
  • Buffer stoichiometry when adding strong base: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O.
  • Turning concentrations into moles (volume in dm³), updating amounts, then calculating pH (Kₐ or Henderson–Hasselbalch).
  • Marks for method, correct rearrangement, correct significant figures / decimal places.
Given: Kₐ (ethanoic acid) = 1.74 × 10⁻⁵ mol dm⁻³ at 25 °C.
Part 02.1 (3 marks)

Find [CH₃COOH] in a buffer from pH and [CH₃COO⁻]

pH = 3.87, [CH₃COO⁻] = 0.136 mol dm⁻³

💡 Key knowledge

  • pH = −log₁₀[H⁺] so [H⁺] = 10⁻ᵖᴴ
  • For ethanoic acid: Kₐ = ([CH₃COO⁻][H⁺])/[CH₃COOH]
  • Rearrange to: [CH₃COOH] = ([CH₃COO⁻][H⁺])/Kₐ

🧠 Exam technique (how to secure the marks)

  • Write [H⁺] = 10⁻ᵖᴴ explicitly: this is a standalone mark in the scheme.
  • Show the rearranged expression before substituting numbers (mark for rearrangement).
  • Round only at the end; give final answer to 3 significant figures.
Examiner note: “Allow 1.35 × 10⁻⁴” for [H⁺]. If [H⁺] is wrong, method marks later are limited.

📐 Calculations (step-by-step)

  1. Convert pH to [H⁺]:
    [H⁺] = 10⁻³⋅⁸⁷ = 1.3489 × 10⁻⁴ mol dm⁻³
    Mark (1): correct [H⁺]. Mark scheme allows 1.35 × 10⁻⁴.
  2. Rearrange Kₐ expression to make [CH₃COOH] the subject:
    [CH₃COOH] = ([H⁺][CH₃COO⁻]) / Kₐ
    Mark (1): correctly rearranged equation.
  3. Substitute values:
    [CH₃COOH] = (1.3489 × 10⁻⁴ × 0.136) / (1.74 × 10⁻⁵)
    = 1.05436 mol dm⁻³
    = 1.05 mol dm⁻³ (3 s.f.)
    Mark (1): answer 1.05–1.06 mol dm⁻³, ≥3 s.f.

✅ Correct answer (with units)

[CH₃COOH] = 1.05 mol dm⁻³ (to 3 s.f.)
Mark breakdown (3):
• [H⁺] from pH (1) • correct rearrangement (1) • correct value & s.f. (1)

❌ Common errors (seen by examiners)

  • Using [H⁺] = −log(pH) (wrong direction). Must be 10⁻ᵖᴴ .
  • Putting [CH₃COOH] in the numerator (rearrangement error).
  • Forgetting units or giving 2 s.f. (the scheme explicitly expects ≥3 s.f.).
Part 02.2 (6 marks)

Buffer pH after adding NaOH

[CH₃COOH] = 0.260 mol dm⁻³, [CH₃COO⁻] = 0.121 mol dm⁻³, volume = 500 cm³, moles NaOH added = 7.00 × 10⁻³ mol

💡 Key knowledge

  • Convert volume: 500 cm³ = 0.500 dm³
  • Neutralisation in a buffer (strong base added):
    CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O
  • Stoichiometry: 1 mol OH⁻ removes 1 mol CH₃COOH and forms 1 mol CH₃COO⁻.
  • After updating amounts, use:
    [H⁺] = (Kₐ × [CH₃COOH])/[CH₃COO⁻] then pH = −log₁₀[H⁺]
    (Henderson–Hasselbalch is also acceptable.)

🧠 Exam technique (where the 6 marks come from)

  • Step 1: turn both buffer concentrations into moles in 0.500 dm³ (this is often 2 method marks).
  • Step 2: do the reaction change using moles of OH⁻ (subtract from acid, add to ethanoate).
  • Step 3: convert back to concentrations (or use mole ratio directly; volume cancels if it stays the same).
  • Final: calculate [H⁺], then pH to 2 d.p. (final accuracy mark).
Examiner guidance highlights:
• If you used Kₐ correctly, full marks possible.
• If you “divide by 2 after M5” (unnecessary halving), you lose the final pH mark.
• Henderson–Hasselbalch solutions are allowed.

📐 Calculations (step-by-step, method matching mark scheme)

  1. Convert initial concentrations to moles in 0.500 dm³:
    n(CH₃COOH) = 0.260 × 0.500 = 0.130 mol
    n(CH₃COO⁻) = 0.121 × 0.500 = 0.0605 mol
    These are the values used in “Mark Scheme 1”.
  2. Reaction with added base (7.00 × 10⁻³ mol OH⁻):
    CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O
    n(CH₃COOH) after = 0.130 − 0.00700 = 0.123 mol
    n(CH₃COO⁻) after = 0.0605 + 0.00700 = 0.0675 mol
    Mark scheme awards marks for the “−0.007” and “+0.007” updates.
  3. Convert back to concentrations (still 0.500 dm³):
    [CH₃COOH] after = 0.123 / 0.500 = 0.246 mol dm⁻³
    [CH₃COO⁻] after = 0.0675 / 0.500 = 0.135 mol dm⁻³
    This matches the mark scheme’s alternative “Mark scheme 2” concentrations directly. (Either moles route or concentration route can earn full credit.)
  4. Use Kₐ to find [H⁺]:
    [H⁺] = (Kₐ × [CH₃COOH])/[CH₃COO⁻]
    [H⁺] = (1.74 × 10⁻⁵ × 0.246) / 0.135
    [H⁺] = 3.171 × 10⁻⁵ mol dm⁻³
  5. Convert to pH:
    pH = −log₁₀(3.171 × 10⁻⁵) = 4.498…
    pH = 4.50 (2 d.p.)
    Final mark requires 2 d.p.

✅ Correct answer

pH after adding NaOH = 4.50 (to 2 d.p.)
Mark breakdown (6, aligned to scheme intent):
• Determine initial moles (or correct scaled amounts) of CH₃COOH and CH₃COO⁻ (≈2 marks)
• Correctly subtract 0.007 from acid (1) and add 0.007 to ethanoate (1)
• Correct [H⁺] expression with their numbers (1)
• pH to 2 d.p. (1)

❌ Common errors (and why they lose marks)

  • Not converting 500 cm³ to dm³ (or mixing cm³ and dm³). This breaks the mole calculations.
  • Updating the wrong way round: adding OH⁻ should decrease CH₃COOH and increase CH₃COO⁻.
  • Using Kₐ backwards: writing [H⁺] = (Kₐ × [CH₃COO⁻])/[CH₃COOH] gives the wrong pH trend.
  • Randomly “dividing by 2” after calculating [H⁺] or pH. Examiner note: this loses the final accuracy mark.
  • Wrong rounding: final pH must be to 2 d.p. (4.50, not 4.5).

🧠 What distinguished top responses

  • Clear stoichiometry table (before/after) showing the 1:1 change with OH⁻.
  • Consistent units (dm³, mol dm⁻³) and rounding only at the end.
  • Efficient method choice: many high scorers used updated moles, then converted to concentrations, then Kₐ → pH (exactly as the mark scheme shows).
Quick checklist

💡 Must-know formulas

  • pH = −log₁₀[H⁺]
  • [H⁺] = 10⁻ᵖᴴ
  • Kₐ = ([CH₃COO⁻][H⁺])/[CH₃COOH]
  • n = cV with V in dm³

🧠 Buffer “strong base added” routine

  • Convert concentrations → moles in the given volume.
  • Use CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O to update moles.
  • Convert back to concentrations (if needed).
  • Use Kₐ (or Henderson–Hasselbalch) to get pH.

❌ Fast traps to avoid

  • Forgetting 500 cm³ = 0.500 dm³.
  • Swapping acid and conjugate base in the Kₐ rearrangement.
  • Giving pH to the wrong precision (needs 2 d.p. here).

Topics

Physical Chemistry · 3.1.12 Acids and Bases

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 1, 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.