AQA A-Level Chemistry Paper 1, June 2024: Question 7

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Question Answers Additional comments/Guidelines Mark

M1 K = [H+] [OH–]

w

07.1 M2 [H+] = √2.92 × 10–14

(1 × AO1,

2 × AO2)

M3 = –log10 M2

M3 pH = 6.77

answer to 2 decimal places

M1 n OH– = 5.25 × 10–3 and n H+ = 2.00 × 10–3 mol

– –3 M2 = n(OH–) – n(H+) in M1

M2 excess OH = 3.25 × 10 mol

–3 𝑀𝑀2

– 3.25×10 –3 M3 =

M3 [OH ] = = 0.0591 mol dm 55 𝑥𝑥 10−3

55.0×10–3

07.2

(5 x AO2)

2.92 𝑥𝑥 10−14

2.92×10–14 M4 =

M4 [H+] = = 4.94 × 10–13 mol dm–3

𝑀𝑀3

0.0591

M5 12.31 M5 = –log10(M4)

How to answer it

Water Dissociation & Acid-Base Mixture pH

📋 What This Question Tests

This question assesses quantitative acid-base equilibrium concepts from Physical Chemistry:

Writing the equilibrium expression for the ionic product of water ( K_w ).
Calculating the pH of pure water at non-standard temperatures (where pH ≠ 7.00 ).
Carrying out multi-step stoichiometric calculations for the reaction between a strong monoprotic acid and a strong monoacidic base.
Calculating excess hydroxide concentration using the total combined solution volume.
Using K_w at a stated temperature to calculate [H⁺] and converting to pH rounded to 2 decimal places.

Part 07.1 (3 Marks)

Expression for K_w & pH of Pure Water at 40 °C

Definition of K_w and neutral water calculation

💡 Key Knowledge

Auto-ionisation of water:
H₂O(l) ⇌ H⁺(aq) + OH⁻(aq) (endothermic).
Because water is a liquid in huge excess, [H₂O] is effectively constant and incorporated into K_w :
K_w = [H⁺][OH⁻]
In pure water, every molecule that ionises produces equal concentrations of H⁺ and OH⁻:
[H⁺] = [OH⁻] , therefore K_w = [H⁺]² .
Water at 40 °C is still strictly neutral despite having a pH < 7 because [H⁺] equals [OH⁻].

📐 Step-by-Step Calculation

1 State expression:
K_w = [H⁺][OH⁻] [Mark 1]
2 Find [H⁺]:
Since [H⁺] = [OH⁻]:
[H⁺] = √(K_w) = √(2.92 × 10⁻¹⁴)
[H⁺] = 1.7088 × 10⁻⁷ mol dm⁻³ [Mark 2]
3 Calculate pH:
pH = -log₁₀[H⁺]
pH = -log₁₀(1.7088 × 10⁻⁷) = 6.767...
pH = 6.77 (2 d.p.) [Mark 3]

✅ Final Credited Answers

K_w expression: K_w = [H⁺][OH⁻] (square brackets are mandatory to denote concentration)
pH: 6.77

❌ Common Errors & Traps

Including [H₂O]: Writing K_w = [H⁺][OH⁻]/[H₂O] is an equilibrium constant (K_c), not K_w. Scores zero.
Assuming pH = 7: Remembering standard room temperature pH and not calculating based on the given K_w value.
Rounding too early: Rounding [H⁺] to 1.7 × 10⁻⁷ gives pH = 6.769 = 6.77 (safe here, but early rounding frequently costs marks).
Wrong decimal places: Giving 6.8 or 6.767 instead of the required 2 decimal places.

Part 07.2 (5 Marks)

pH of a Strong Acid – Strong Base Mixture

Mixing 35.0 cm³ of 0.150 mol dm⁻³ NaOH with 20.0 cm³ of 0.100 mol dm⁻³ HCl at 40 °C

📐 Complete 5-Step Calculation Plan

1 Calculate initial moles of H⁺ and OH⁻:
Both HCl and NaOH are strong and monoprotic (100% dissociated).
• Moles of OH⁻ = (35.0 / 1000) × 0.150 = 5.25 × 10⁻³ mol
• Moles of H⁺ = (20.0 / 1000) × 0.100 = 2.00 × 10⁻³ mol
Award M1 for calculating both mole values correctly.
2 Determine which reagent is in excess:
Reaction: H⁺(aq) + OH⁻(aq) → H₂O(l) (1:1 stoichiometric ratio).
OH⁻ is in excess since 5.25 × 10⁻³ > 2.00 × 10⁻³.
• Excess moles of OH⁻ = 5.25 × 10⁻³ - 2.00 × 10⁻³ = 3.25 × 10⁻³ mol
Award M2 for subtracting moles of acid from moles of alkali.
3 Calculate concentration of excess OH⁻ using TOTAL volume:
• Total volume = 35.0 + 20.0 = 55.0 cm³ = 55.0 × 10⁻³ dm³
• [OH⁻] = (3.25 × 10⁻³) / (55.0 × 10⁻³) = 0.05909... mol dm⁻³ (or 0.0591 mol dm⁻³)
Award M3 for dividing excess moles by total volume (in dm³).
4 Calculate [H⁺] using K_w at 40 °C:
Rearrange K_w = [H⁺][OH⁻] to give [H⁺] = K_w / [OH⁻]
• [H⁺] = (2.92 × 10⁻¹⁴) / 0.05909... = 4.9416 × 10⁻¹³ mol dm⁻³
Award M4 for dividing 2.92 × 10⁻¹⁴ by their [OH⁻].
5 Calculate final pH:
• pH = -log₁₀[H⁺] = -log₁₀(4.9416 × 10⁻¹³) = 12.306...
• pH = 12.31 (to 2 d.p.)
Award M5 for final pH = 12.31 (strictly to 2 decimal places).

🧠 Exam Technique & Alternative Method

Alternative via pOH:

Calculate pK_w = -log₁₀(2.92 × 10⁻¹⁴) = 13.535
Calculate pOH = -log₁₀(0.05909) = 1.228
pH = pK_w - pOH = 13.535 - 1.228 = 12.31
Warning: Do NOT subtract from 14.00! pH + pOH = 14 only applies at 25 °C. At 40 °C, pH + pOH = 13.54 .

❌ Major Examiner Traps

Forgetting total volume (The #1 Error): Dividing excess moles by 35.0 cm³ instead of (35.0 + 20.0 = 55.0 cm³). This loses M3, M4, and M5.
Using standard K_w (1.00 × 10⁻¹⁴): Failing to use the value provided at 40 °C ( 2.92 × 10⁻¹⁴ ). Doing this loses M4 and M5.
Decimal Places: pH values should always be quoted to 2 decimal places in AQA exams unless stated otherwise. Writing 12.3 or 12 loses the final mark.

Topics

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