AQA A-Level Chemistry Paper 1, 2018: Question 5

15 marks · Hard difficulty · State/Explain/Numerical

Define a strong acid, calculate the pH of a mixture of hydrochloric acid and barium hydroxide, explain why water at 30 °C is neutral, identify an acidic Period 3 oxide, write the Ka expression for ethanoic acid, and calculate the pH of an ethanoic acid/sodium ethanoate buffer after addition of hydrochloric acid.

Practise this question

Question

Question 05 consists of six sub-questions about acids, bases, and buffer solutions. Part 05.1 asks to state the meaning of the term strong acid for 1 mark. Part 05.2 is a 6-mark calculation where 10.35 cm³ of 0.100 mol dm⁻³ HCl is added to 25.0 cm³ of 0.150 mol dm⁻³ Ba(OH)₂ at 30 °C, using Kw = 1.47 × 10⁻¹⁴ mol² dm⁻⁶, requesting pH to 2 decimal places. Part 05.3 asks why water at 30 °C with pH 6.92 is neutral (1 mark). Part 05.4 provides a tick-box question to identify the oxide (Al₂O₃, Na₂O, SiO₂, or SO₂) that forms a solution with pH = 2 (1 mark). Part 05.5 asks for the acid dissociation constant Ka expression for ethanoic acid (1 mark). Part 05.6 is a 5-mark buffer calculation where 5.00 cm³ of 2.00 mol dm⁻³ HCl is added to a buffer of 0.025 mol sodium ethanoate in 500 cm³ of 0.0700 mol dm⁻³ ethanoic acid, with Ka = 1.76 × 10⁻⁵ mol dm⁻³.
Question text

05 Hydrochloric acid is a strong acid and ethanoic acid is a weak acid.

05.1 State the meaning of the term strong acid.

[1 mark]

In an experiment, 10.35 cm3 of 0.100 mol dm–3 hydrochloric acid are added to

05.2

25.0 cm3 of 0.150 mol dm–3 barium hydroxide solution.

Calculate the pH of the solution that forms at 30°C

K = 1.47 x 10–14 mol2 dm–6 at 30 °C

w

Give your answer to 2 decimal places.

[6 marks]

14 pH

05.3 The pH of water at 30°C is 6.92

*13* Give the reason why water is neutral at this temperature.

[1 mark]

05.4 Identify the oxide that could react with water to form a solution with pH = 2

Tick ( ) one box.

[1 mark]

Al2O3

Na2O

SiO2

SO 15

05.5 Give the expression for the acid dissociation constant (Ka) for ethanoic acid

(CH3COOH).

[1 mark]

Ka

05.6 A buffer solution contains 0.025 mol of sodium ethanoate dissolved in

500 cm3 of 0.0700 mol dm–3 ethanoic acid at 25 °C

A sample of 5.00 cm3 of 2.00 mol dm–3 hydrochloric acid is added to this buffer

solution.

Calculate the pH of the solution formed.

For ethanoic acid, K = 1.76 × 10–5 mol dm–3 at 25 °C

a

[5 marks]

pH

Mark scheme

Show the mark scheme Mark scheme for Question 05 with a total of 15 marks. 05.1: completely dissociates/ionises to form H+ ions (1 mark). 05.2: 6 marks for moles of HCl, moles of OH⁻, excess OH⁻ moles, [OH⁻] in total volume 35.35 cm³, [H⁺] using Kw, and final pH = 13.09 to 13.10 to 2 d.p. 05.3: [H⁺] = [OH⁻] (1 mark). 05.4: SO₂ ticked (1 mark). 05.5: Ka = [CH₃COO⁻][H⁺] / [CH₃COOH] (1 mark). 05.6: 5 marks for moles of added HCl and initial ethanoic acid, moles of CH₃COO⁻ remaining, moles of CH₃COOH formed, [H⁺] calculation, and final pH = 4.28 (must be 2 d.p.).

Question Answers Additional Comments/Guidance Mark

05.1 completely dissociates/ionises (to form H+ ions) 1

M1 moles HCl = 1.035 x 10–3 and moles Ba(OH) = 3.75 x 10–3

2 If M1 incorrect, lose M1 and M6

M2 moles OH- = 2 x (3.75 x 10–3 = 7.50 x 10-3 ) If M2 not x2, lose M2 and M6

- 7.50 x 10-3 - 1.035 x 10–3 = –3 3 If no subtraction for M3, lose M3 and M6 1

M3 XS n OH = 6.465 x 10 mol in 35.35 cm

M4 [OH] = 6.465 x 10–3 / 35.35 x 10–3 = 0.182885 mol dm-3 = M3/35.35 x 10-3

If M4 not /35.35 x 10–3, lose M4 and M6

However, if divided by 35.35 only lose M4

and allow M6 (ecf = 10.09)

M5 [H+] = K /[OH-] = 1.47 x 10-14 /0.182885 = 1.47 x 10-14 / M4

w

05.2 = 8.0378 x 10–14 mol dm-3

If incorrect rearrangement or wrong

equation, lose M5 and M6 1

If K = 1.00 x 10-14 used only lose M5 and

w

allow M6 (ecf = 13.26)

M6 pH = 13.09 to 13.10 Must be 2dp

Credit other methods eg limiting reagent

Alternative MS to get M2: method since HCl limiting reagent

Starting amounts are 1.035 x 10-3 mol HCl and 3.75 x 10–3 mol Ba(OH) M1

So the HCl will react with 5.175 x 10-4 mol of the Ba(OH) leaving an excess of – – –

3.2325 x 10-3 mol of Ba(OH) (as alternative M2)

So n OH- = 2 x 3.2325 x 10-3= 6.465 x 10–3 mol M3

05.3 [H+] = [OH–] 1

1805.4 SO 1

K [CH COO–][H+]

a = 3 1

05.5 [CH3COOH]

M1 moles HCl added = 0.01 mol and moles CH COOH = 0.035 mol If used 0.07 mol for CH3COOH can score

M2, M3 and M4 (pH = 4.03)

M2 mol CH COO– = 0.025 - 1

M2 moles CH COO– = 0.025 – 0.01 (= 0.015) 3

3 M1(HCl)

M3 moles CH3COOH = 0.035 + 0.01 (= 0.045)

M3 mol CH3COOH = M1 (CH3COOH) + 1

05.6 M1 (HCl)

M4 [H+] = 1.76 x 10–5 x M3 (/ V) (= 5.28 x 10–5 )

M2 (/ V)

M4 is conditional on an attempt at an

addition/subtraction in M2 OR M3

M5 pH = 4.28

M5 must be 2dp

Total 15

How to answer it

Acids, Bases, Buffers & Period 3 Oxides

📋 What This Question Tests

This 15-mark structured question evaluates key physical and inorganic concepts: defining strong acids, calculating excess-reagent pH involving a diprotic base and non-standard Kw, understanding neutrality in terms of ion concentrations, Period 3 oxide acid-base properties, and multistep calculations of acidic buffer systems upon addition of strong acid.

Part (a) / Question 05.1

Definition of a Strong Acid (1 Mark)

✅ Correct Answer

Completely dissociates (or completely ionises) to form H⁺ ions.

🧠 Exam Technique

Always state "completely" or "fully". Saying an acid "easily dissociates" or "dissociates a lot" scores 0 marks.

Mark scheme: 1 mark for stating completely dissociates/ionises (to form H⁺ ions).

Part (b) / Question 05.2

Calculating pH of a Mixture: Strong Acid + Diprotic Base (6 Marks)

📐 Step-by-Step Calculation

  1. 1 Calculate initial moles:
    Moles of HCl = (10.35 / 1000) × 0.100 = 1.035 × 10⁻³ mol
    Moles of Ba(OH)₂ = (25.0 / 1000) × 0.150 = 3.75 × 10⁻³ mol
  2. 2 Account for diprotic hydroxide stoichiometry:
    Each Ba(OH)₂ releases 2 OH⁻ ions.
    Total moles of OH⁻ = 2 × (3.75 × 10⁻³) = 7.50 × 10⁻³ mol
  3. 3 Determine excess moles of OH⁻:
    Reaction: H⁺ + OH⁻ → H₂O (1:1 ratio)
    Excess OH⁻ = 7.50 × 10⁻³ − 1.035 × 10⁻³ = 6.465 × 10⁻³ mol
  4. 4 Calculate [OH⁻] in total volume:
    Total volume = 10.35 + 25.0 = 35.35 cm³ = 35.35 × 10⁻³ dm³
    [OH⁻] = (6.465 × 10⁻³) / (35.35 × 10⁻³) = 0.1829 mol dm⁻³
  5. 5 Calculate [H⁺] using Kw at 30 °C:
    [H⁺] = Kw / [OH⁻] = (1.47 × 10⁻¹⁴) / 0.1829 = 8.038 × 10⁻¹⁴ mol dm⁻³
  6. 6 Calculate pH:
    pH = −log₁₀[H⁺] = −log₁₀(8.038 × 10⁻¹⁴) = 13.09 (or 13.10)

❌ Common Errors

  • Forgetting to multiply by 2: Missing that 1 mol of Ba(OH)₂ gives 2 mol of OH⁻ loses M2 and often locks out M6.
  • Using 1.00 × 10⁻¹⁴ for Kw: The question explicitly specifies 30 °C with Kw = 1.47 × 10⁻¹⁴ mol² dm⁻⁶.
  • Dividing by the wrong volume: Forgetting to sum both volumes (35.35 cm³) before finding concentration.

🧠 Exam Technique: 2 Decimal Places

The question states: "Give your answer to 2 decimal places." Any pH value given to 1 or 3 decimal places immediately forfeits the final mark (M6), even with perfect working!

Part (c) / Question 05.3

Water Neutrality at 30 °C (1 Mark)

✅ Correct Answer

[H⁺] = [OH⁻] (or concentration of hydrogen ions equals concentration of hydroxide ions).

❌ Common Error

Stating "pH = 7". At 30 °C, pure water has a pH of 6.92 and is still neutral because neutrality is strictly defined by equal concentrations of H⁺ and OH⁻ ions, not by pH being 7.00.

Mark scheme: 1 mark for [H⁺] = [OH⁻].

Part (d) / Question 05.4

Period 3 Oxides (1 Mark)

✅ Correct Answer

Tick box 4: SO₂

💡 Key Knowledge

  • Na₂O: Basic oxide, reacts violently with water to give NaOH (pH ≈ 13–14).
  • Al₂O₃: Insoluble amphoteric oxide (pH ≈ 7).
  • SiO₂: Giant covalent, insoluble in water (pH ≈ 7).
  • SO₂: Acidic non-metal oxide, dissolves to form sulfurous acid (H₂SO₃), giving a solution around pH 2–3.
Mark scheme: 1 mark for SO₂ only.

Part (e) / Question 05.5

Ka Expression for Ethanoic Acid (1 Mark)

✅ Correct Answer

Ka = [CH₃COO⁻][H⁺] / [CH₃COOH]

(Square brackets denoting concentrations are essential)

🧠 Exam Technique

Do not write [H⁺]² on the top for a general Ka expression. That simplification is only valid for a pure weak acid solution, not for the true equilibrium expression.

Mark scheme: 1 mark for [CH₃COO⁻][H⁺] / [CH₃COOH]. Square brackets required.

Part (f) / Question 05.6

Buffer Solution Calculation with Added Acid (5 Marks)

📐 Step-by-Step Calculation

  1. 1 Initial moles in the mixture:
    Moles of CH₃COO⁻ (from salt) = 0.025 mol (given)
    Moles of CH₃COOH = (500 / 1000) × 0.0700 = 0.035 mol
    Moles of HCl added = (5.00 / 1000) × 2.00 = 0.010 mol
  2. 2 Buffer reaction on addition of strong acid:
    The added H⁺ reacts with ethanoate ions: CH₃COO⁻ + H⁺ → CH₃COOH
    New moles of CH₃COO⁻ = 0.025 − 0.010 = 0.015 mol (M2)
    New moles of CH₃COOH = 0.035 + 0.010 = 0.045 mol (M3)
  3. 3 Rearrange Ka to find [H⁺]:
    [H⁺] = Ka × ([CH₃COOH] / [CH₃COO⁻])
    Note: Because both species share the same total volume, the volume terms cancel, so ratio of moles = ratio of concentrations.
    [H⁺] = (1.76 × 10⁻⁵) × (0.045 / 0.015) = (1.76 × 10⁻⁵) × 3 = 5.28 × 10⁻⁵ mol dm⁻³ (M4)
  4. 4 Calculate final pH:
    pH = −log₁₀(5.28 × 10⁻⁵) = 4.28 (M5)

❌ Common Errors

  • Mixing up addition and subtraction: Adding H⁺ decreases the salt ([CH₃COO⁻]) and increases the acid ([CH₃COOH]). Doing the opposite gives an inverted ratio and is heavily penalised.
  • Treating 0.0700 as moles: 0.0700 is a concentration in mol dm⁻³ in 500 cm³; you must multiply by 0.500 dm³ to get 0.035 mol.

🧠 Top-Student Tip: Volume Cancellation

In buffer ratio calculations, dividing moles by total volume is valid but unnecessary: (nacid/V) / (nbase/V) = nacid / nbase . Working directly with moles saves crucial exam time!

Mark scheme: M1 (moles HCl & CH₃COOH), M2 (new CH₃COO⁻), M3 (new CH₃COOH), M4 ([H⁺] calculation), M5 (pH = 4.28 to 2 d.p.).

Topics

Physical Chemistry · Inorganic Chemistry · 3.1.12 Acids and Bases · 3.2.4 Properties of Period 3 Elements and Their Oxides

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 1, 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.