AQA A-Level Chemistry Paper 1, 2019: Question 9

15 marks · Hard difficulty · State/Explain/Numerical

Explain water neutrality at 40 °C and find Kw, determine Ka from a pH curve, select an indicator, calculate the pH of a buffer after adding KOH, and explain why dilution has no effect on buffer pH.

Practise this question

Question

Question 9 consists of five parts: 9.1 asks to explain why pure water at 40 °C with pH 6.67 is not acidic and calculate Kw (4 marks). 9.2 presents a titration curve of 25 cm³ of 0.080 mol dm⁻³ propanoic acid titrated with sodium hydroxide, showing an equivalence point at approximately 19.5 cm³ and steep vertical region between pH 6.5 and 11.5, asking to determine Ka using the graph (3 marks). 9.3 asks to select the most appropriate indicator for this titration from a table of four indicators (methyl orange, bromothymol blue, cresolphthalein, indigo carmine) with given pH ranges (1 mark). 9.4 asks to calculate the pH to two decimal places of a buffer prepared from 0.0136 mol KX and 100 cm³ of 0.500 mol dm⁻³ HX after adding 3.00 × 10⁻⁴ mol of KOH, with Ka = 1.41 × 10⁻⁵ mol dm⁻³ (6 marks). 9.5 asks to use a mathematical expression to explain why buffer pH remains constant upon dilution (1 mark).
Question text

09 This question is about different pH values.

09.1 For pure water at 40 °C, pH = 6.67

A student thought that the water was acidic.

Explain why the student was incorrect.

Determine the value of Kw at this temperature.

[4 marks]

Explanation

22 2 –6

Kw mol dm

09.2 Sodium hydroxide solution was added gradually from a burette to 25 cm3 of

0.080 mol dm–3 propanoic acid at 25 °C

The pH was measured and recorded at regular intervals.

The results are shown in Figure 4.

Figure 4

Use Figure 4 to determine the value of Ka for propanoic acid at 25 °C

Show your working.

[3 marks]

K mol dm−3

a

09.3 Suggest which indicator is the most appropriate for the reaction in Question 09.2?

Tick ( ) one box.

[1 mark]

Indicator pH range Tick ( ) one box

methyl orange 3.1 – 4.4

bromothymol blue 6.0 – 7.6

cresolphthalein 8.2 – 9.8

indigo carmine 24 11.6 – 13.0

09.4 A student prepared a buffer solution by adding 0.0136 mol of a salt KX to

100 cm3 of a 0.500 mol dm–3 solution of a weak acid HX and mixing thoroughly.

The student then added 3.00 × 10–4 mol of potassium hydroxide to the buffer solution.

Calculate the pH of the buffer solution after adding the potassium hydroxide.

For the weak acid HX at 25 °C the value of the acid dissociation constant,

K = 1.41 × 10–5 mol dm–3.

a

Give your answer to two decimal places.

[6 marks]

25 pH

09.5 A buffer solution has a constant pH even when diluted.

Use a mathematical expression to explain this.

[1 mark]

Mark scheme

Show the mark scheme Mark scheme for question 9: 09.1 awards 4 marks for [H+] = [OH-], [H+] = 10^-6.67 = 2.138 × 10^-7, Kw = [H+]^2, giving Kw = 4.57 × 10^-14 mol^2 dm^-6. 09.2 awards 3 marks for determining volume at half-equivalence (9.75 cm³), reading pH = 4.80 to 4.95, and calculating Ka = 10^-pH = 1.26 × 10^-5 (allow 1.12 × 10^-5 to 1.58 × 10^-5). 09.3 awards 1 mark for cresolphthalein. 09.4 awards 6 marks: Ka expression, initial moles HX = 0.0500 mol, moles HX after reaction = 0.0497 mol, moles KX after reaction = 0.0139 mol, [H+] = 5.04 × 10^-5, pH = 4.30. 09.5 awards 1 mark for showing ratio [HX]/[X-] remains constant.

Question Answers Additional Comments/Guidelines Mark

+ - + -

M1: [H ] = [OH ] M1: accept equal number/amounts of H and OH 1

+ -pH -7 -7

M2: [H ] (= 10 ) = 2.138 x 10 M2: allow 2.14 x 10 1

09.1

+ 2 -7 2 M3: allow (M2)2

M3: Kw = [H ] or (2.138 x 10 ) 1

-14 -14

M4: Kw = 4.57 x 10 M4: allow 4.58 x 10 1

M4 is dependent on (an answer)2 in M3

View with Figure X (ie graph) as they may show working there. Ignore calculations of mols of salt or acid

19.5 3 3 M1: Allow reading on graph to be from 19.4 to 19.7 1

M1: Determines volume at half equivalence (= cm ) = 9.75 (cm )

2 giving M1 = 9.7 to 9.85

M2: pH = 4.80 to 4.95 1

M2: Reads off pH at half equivalence

09.2

M3: K (= 10-pH) = 10-4.9 = 1.26 x 10–5

a M3: Allow 1.12 x 10–5 to 1.58 x 10-5

M3: Allow 2sf or more

Alternative method Alternative M1 if calculation incorrect:

M1: pH of pure acid = 3 Allow pH = pK or [H+] = K at half equivalence

–3 2 a a

M2: Ka = (10 ) / 0.080

M3: = 1.25 x 10–5

09.3 cresolphthalein 1

M1: K = [H+][X-] or [H+] = K x [HX] M1: allow [H+] = K x [acid] 1

a a a

[HX] [X-] [salt]

M2: amount of HX = 0.0500 mol 1

M3: amount of HX after addn of KOH = 0.05 - 3 x 10-4 = 0.0497 mol M3: = M2 – 3 x 10-4 1

M4: amount of KX after addn of KOH = 0.0136 + 3 x 10-4 = 0.0139 mol 1

09.4 M5: [H+] =( 1.41 x 10-5 x 0.0497 ) = 5.04(15) x 10-5

0.0139

M6: pH = -log 5.04(15) x 10-5 = 4.30 Answer to 2 decimal places

10 1

If no attempt at M3 and M4 max 2 marks

If M3 or M4 attempted using 3 x 10-4 max 4 (M1,

M2, M3 or M4 and M6)

Allow inverse expression

09.5 1

ratio remains (almost) constant

How to answer it

pH, Weak Acids, Titration Curves & Buffer Systems

📌 WHAT THIS QUESTION TESTS

Core A-Level Physical Chemistry (Topic: Acids, Bases, Buffers & pH):

  • The definition of neutrality in aqueous systems and temperature dependence of water auto-ionisation (Kw).
  • Reading weak acid–strong base titration curves and applying the half-equivalence point to find Ka and pKa.
  • Selecting appropriate acid–base indicators matched to the vertical pH inflection range.
  • Quantitative buffer calculations: calculating the modified pH following the addition of a strong base (KOH) to an acidic buffer system.
  • Explaining the resistance of buffer pH to dilution via equilibrium ratio expressions.
PART 09.1 · 4 MARKS

Pure Water at 40 °C and the Ionic Product of Water (Kw)

Explaining neutrality at non-standard temperatures and calculating Kw

✅ Mark Scheme Breakdown

  • M1: Pure water is neutral because [H⁺] = [OH⁻] (or equal amounts/moles of H⁺ and OH⁻).
  • M2: [H⁺] = 10−pH = 10−6.67 = 2.138 × 10−7 mol dm−3 (allow 2.14 × 10−7).
  • M3: Kw = [H⁺]² or (2.138 × 10−7)².
  • M4: Kw = 4.57 × 10−14 mol² dm−6 (allow 4.58 × 10−14).
Score: 4 marks total. M4 requires correct squaring of calculated [H⁺].

📐 Step-by-Step Calculation

1 Find [H⁺]:
[H⁺] = 10−6.67 = 2.138 × 10−7 mol dm−3

2 Apply Kw expression for pure water:
Because [H⁺] = [OH⁻], Kw = [H⁺][OH⁻] = [H⁺]²

3 Evaluate:
Kw = (2.138 × 10−7)² = 4.57 × 10−14 mol² dm−6

💡 Key Knowledge

  • Definition of Neutrality: Neutral does not fundamentally mean pH = 7. It strictly means [H⁺] = [OH⁻].
  • The dissociation of water ( H₂O ⇌ H⁺ + OH⁻ ) is endothermic. As temperature rises, equilibrium shifts right, increasing [H⁺] and [OH⁻] equally. Hence, pH decreases below 7, but the water remains neutral!

❌ Common Errors & Misconceptions

  • "Water is neutral because pH = 7": Direct loss of M1. pH is only 7 at exactly 25 °C (298 K).
  • Premature rounding: Rounding [H⁺] to 2.1 × 10−7 before squaring gives 4.41 × 10−14, losing M4. Keep calculator memory intact!
PART 09.2 · 3 MARKS

Determining Ka from a Weak Acid Titration Curve

Using the Half-Equivalence Point on Figure 4

✅ Mark Scheme Breakdown

  • M1: Volume at equivalence = 19.5 cm³ (allow 19.4 to 19.7 cm³). Volume at half-equivalence = 19.5 / 2 = 9.75 cm³ (allow 9.70 to 9.85 cm³).
  • M2: Reading pH at half-equivalence: pH = 4.80 to 4.95.
  • M3: Ka = 10−pH. Using pH = 4.90 gives Ka = 1.26 × 10−5 mol dm−3 (acceptable range: 1.12 × 10−5 to 1.58 × 10−5 mol dm−3, 2 s.f. or more).
Alternative: Pure acid starting pH = 3.0 gives [H⁺] = 10−3. Ka = [H⁺]² / [HA] = (10−3)² / 0.080 = 1.25 × 10−5 mol dm−3.

🧠 Exam Technique: Half-Equivalence

  • At half-equivalence: [HA] = [A⁻].
  • Substitute into Ka = ([H⁺][A⁻]) / [HA] → Ka = [H⁺], meaning pKa = pH.
  • Always draw construction lines on the graph:
    • Vertical equivalence line at ~19.5 cm³
    • Half volume line at 9.75 cm³ straight up to the curve and across to the pH axis.
PART 09.3 · 1 MARK

Selecting the Correct Acid–Base Indicator

Matching indicator pH range to the titration curve's vertical section

✅ Correct Answer

Tick: cresolphthalein (pH range 8.2 – 9.8)

1 Mark: Exactly one box ticked.

💡 How to Choose an Indicator

  • From Figure 4, the vertical equivalence section runs from approximately pH 6.5 to 11.5.
  • An indicator is suitable if its entire pH transition range falls completely within this steep vertical region:
    • Methyl orange (3.1 – 4.4): Too low (changes during buffer region).
    • Bromothymol blue (6.0 – 7.6): Starts before the steep vertical climb.
    • Cresolphthalein (8.2 – 9.8): Lies entirely within the vertical inflection.
    • Indigo carmine (11.6 – 13.0): Too high (changes after equivalence).
PART 09.4 · 6 MARKS

Buffer Solution Calculation with Added Strong Base

Calculating pH after KOH reacts with a weak acid / salt mixture

📐 Step-by-Step Calculation (Awarding 6 Full Marks)

1 State Ka expression:
Ka = [H⁺][X⁻] / [HX]  ⟹  [H⁺] = Ka × [HX] / [X⁻]  (M1)

2 Initial moles of weak acid (HX):
Moles of HX = (100 / 1000) × 0.500 = 0.0500 mol  (M2)
Initial moles of salt (X⁻) = 0.0136 mol

3 Reaction with added KOH:
Added strong base reacts: HX + OH⁻ → X⁻ + H₂O
Moles of added KOH = 3.00 × 10−4 mol
• New moles of HX = 0.0500 − (3.00 × 10−4) = 0.0497 mol  (M3)
• New moles of X⁻ = 0.0136 + (3.00 × 10−4) = 0.0139 mol  (M4)

4 Calculate [H⁺]:
(Note: Volumes cancel out, so ratio of moles = ratio of concentrations)
[H⁺] = (1.41 × 10−5 × 0.0497) / 0.0139 = 5.0415 × 10−5 mol dm−3  (M5)

5 Calculate final pH:
pH = −log₁₀(5.0415 × 10−5) = 4.2974... → 4.30 (must be 2 decimal places!)  (M6)

❌ Common Errors in Buffer Calculations

  • Sign mix-up: Adding KOH reduces acid [HX] and increases salt [X⁻]. Students often subtract from salt or add to acid.
  • Significant figures / Decimal places: The question explicitly specifies two decimal places. Writing 4.3 or 4.297 loses M6. Remember: pH values are recorded to 2 decimal places.
  • Ignoring the added base entirely: Attempting to calculate pH directly from initial quantities yields a maximum of 2 marks (M1 and M2).

🧠 Top Exam Technique Tip

Always do an intuitive check on your answer:

  • Initial buffer pH: −log₁₀(1.41 × 10−5 × 0.0500 / 0.0136) = 4.28.
  • Adding a tiny amount of strong base should cause the pH to rise slightly.
  • Calculated pH = 4.30 (an increase of +0.02) confirms logical correctness!
PART 09.5 · 1 MARK

Mathematical Explanation of Buffer Resistance to Dilution

Why dilution does not alter buffer pH

✅ Mark Scheme Answer

The ratio [HX] / [X⁻] (or [acid] / [salt]) remains constant / almost constant.

1 Mark: Allow the inverse ratio [X⁻] / [HX] remains constant.

💡 The Mathematical Proof

Rearranging the acid dissociation constant:

[H⁺] = Ka × ([HX] / [X⁻])

  • When distilled water is added, both [HX] and [X⁻] are diluted by the exact same factor (same new total volume V).
  • Because (moles HX / V) / (moles X⁻ / V) = moles HX / moles X⁻ , the volume terms cancel out.
  • Since Ka is constant at a fixed temperature, [H⁺] remains constant, maintaining a constant pH.

Topics

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

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 1, 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.