OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2025: Question 19

12 marks · Medium difficulty · Structured Questions

Calculate the pH of strong and weak acids, determine oxidation numbers in a disproportionation reaction, and calculate the mass of sodium propanoate required to prepare a buffer solution and explain its action.

Practise this question

Question

Question 19 about monobasic acids and buffer solutions. Part (a) asks to calculate the pH of 0.150 mol dm⁻³ HNO3 to 2 decimal places (1 mark). Part (b)(i) asks to predict and explain whether the pH of 0.150 mol dm⁻³ HNO2 is higher, lower, or the same as HNO3 (2 marks). Part (b)(ii) presents the disproportionation equation 3HNO2(aq) → HNO3(aq) + 2NO(g) + H2O(l), asking for the definition of disproportionation and the use of oxidation numbers to prove it has occurred (3 marks). Part (c)* is an extended response question: given Ka of propanoic acid is 1.32 × 10⁻⁵ mol dm⁻³, calculate the mass of CH3CH2COONa required to make 125 cm³ of a pH 4.85 buffer using 0.180 mol dm⁻³ propanoic acid, and explain how the buffer resists changes in pH when alkali is added (6 marks).

Mark scheme

Show the mark scheme Mark scheme for Question 19. (a) pH = 0.82 (1 mark). (b)(i) Higher pH because HNO3 fully ionises and HNO2 partially dissociates, giving a lower concentration of H+ ions (2 marks). (b)(ii) Disproportionation defined as simultaneous oxidation and reduction of the same element; N changes from +3 in HNO2 to +5 in HNO3 (oxidation) and to +2 in NO (reduction) (3 marks). (c)* Level of response scheme (up to 6 marks): [H+] = 10^-4.85 = 1.41 × 10^-5 mol dm⁻³; [CH3CH2COO⁻] = Ka × [HA] / [H+] = 0.168 mol dm⁻³; moles = 0.0210 mol; mass = 0.0210 × 96.0 = 2.02 g (allow 2.00 to 2.04 g). Explanation of buffer action involves OH⁻ reacting with H⁺ (or HA) shifting equilibrium to the right to regenerate H⁺.

How to answer it

Acids, Redox Disproportionation & Buffer Solutions

What this question tests

This question assesses fundamental concepts across Physical and Inorganic Chemistry:

  • Strong vs. weak acid behaviour: Calculating pH of strong monoprotic acids and explaining dissociation differences using [H⁺].
  • Redox & Disproportionation: Defining disproportionation and tracking changes in oxidation numbers of nitrogen.
  • Buffer calculations (Extended 6-mark response): Calculating required mass of conjugate base using Ka, [H⁺], and volume; explaining buffer action on addition of alkali using Le Chatelier's principle and equilibrium equations.
Part (a) • 1 Mark

Strong Acid pH Calculation

Calculate the pH of 0.150 mol dm⁻³ HNO₃ to 2 decimal places

📐 Step-by-Step Calculation

1. Identify acid type: HNO₃ is a strong monobasic acid → fully dissociates.
[H⁺] = [HNO₃] = 0.150 mol dm⁻³
2. Apply pH formula:
pH = -log₁₀[H⁺]
pH = -log₁₀(0.150) = 0.8239...
3. Round to requested precision (2 d.p.):
pH = 0.82

❌ Common Errors & Pitfalls

  • Incorrect rounding: Writing 0.8 or 0.824 loses the mark. pH values must strictly be given to 2 decimal places.
  • Treating as weak acid: Trying to use a Ka expression when none is provided.
Mark Scheme: 0.82 (1) — Guidance: 2 decimal places strictly required.
Part (b)(i) • 2 Marks

Comparing Strong and Weak Acids

Predict and explain the pH of 0.150 mol dm⁻³ HNO₂ compared to HNO₃

✅ Correct Answer

Prediction: Higher (pH)

Explanation:

  • Nitrous acid (HNO₂) only partially dissociates / ionises (whereas HNO₃ fully dissociates).
  • Therefore, HNO₂ has a lower concentration of H⁺ ions / lower [H⁺].

🧠 Exam Technique & Examiner Commentary

  • Link pH inversely to hydrogen ion concentration: lower [H⁺] means higher pH.
  • Always state concentration of H⁺, not just "fewer H⁺ ions".
  • Do not simply say "nitric acid is a stronger acid" — that repeats information in the stem without explaining why the pH differs.
Mark Scheme: Higher (pH) AND (nitric acid fully ionises and) nitrous acid partially dissociates/ionises (1); Lower concentration of H⁺ ions (in nitrous acid) (1).
Part (b)(ii) • 3 Marks

Disproportionation Reaction

3HNO₂(aq) → HNO₃(aq) + 2NO(g) + H₂O(l)

💡 Definition of Disproportionation

The simultaneous oxidation and reduction of the same element in a single redox reaction.

Examiner Note: Always specify "same element" — writing "same species" is not accepted.

📐 Oxidation Number Assignment

In HNO₂: H = +1, O = -2 → N = +3
In HNO₃: H = +1, O = -2 → N = +5 (Oxidation: +3 to +5)
In NO: O = -2 → N = +2 (Reduction: +3 to +2)

✅ Completing the Argument

Nitrogen is oxidised from +3 (in HNO₂) to +5 (in HNO₃) AND nitrogen is reduced from +3 (in HNO₂) to +2 (in NO).

❌ Common Errors

  • Writing ion charges instead of oxidation states (e.g., N³⁺ or N⁵⁺ instead of +3, +5).
  • Assigning oxidation states without explicitly stating which direction is oxidation and which is reduction.
Mark Scheme:
• (Simultaneous) oxidation and reduction of the (same) element / N (1)
• N from +3 (in HNO₂) to +5 (in HNO₃) AND N from +3 (in HNO₂) to +2 (in NO) (1)
• N is oxidised to HNO₃ AND N is reduced to NO (1)
Part (c)* • 6-Mark Level of Response

Buffer Solution: Mass Calculation & Buffer Action

Propanoic acid / sodium propanoate buffer (pH = 4.85, Ka = 1.32 × 10⁻⁵ mol dm⁻³, V = 125 cm³)

📐 Calculation: Mass of CH₃CH₂COONa Required

Step 1: Calculate [H⁺] from pH
[H⁺] = 10-pH = 10-4.85
[H⁺] = 1.4125... × 10⁻⁵ mol dm⁻³
Step 2: Calculate [CH₃CH₂COO⁻] using Ka
Ka = ([H⁺][CH₃CH₂COO⁻]) / [CH₃CH₂COOH]
[CH₃CH₂COO⁻] = (Ka × [CH₃CH₂COOH]) / [H⁺]
[CH₃CH₂COO⁻] = (1.32 × 10⁻⁵ × 0.180) / (1.4125... × 10⁻⁵)
[CH₃CH₂COO⁻] = 0.1682... mol dm⁻³
Step 3: Calculate moles in 125 cm³
n(CH₃CH₂COO⁻) = c × V
n = 0.1682 × (125 / 1000)
n = 0.02102... mol
Step 4: Calculate mass of CH₃CH₂COONa
Mr(CH₃CH₂COONa) = (3×12.0) + (5×1.0) + (2×16.0) + 23.0 = 96.0 g mol⁻¹
mass = 0.02102 × 96.0
Mass = 2.02 g (allow 2.00 g – 2.04 g)

💡 Explanation: Buffer Action on Adding Alkali (OH⁻)

State the equilibrium present:

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

Mechanism of resistance:

  • Added OH⁻ ions react with H⁺: H⁺ + OH⁻ → H₂O
    (OR react directly with the weak acid: CH₃CH₂COOH + OH⁻ → CH₃CH₂COO⁻ + H₂O ).
  • The equilibrium shifts to the right as propanoic acid dissociates to restore [H⁺].
  • Since the ratio [CH₃CH₂COOH] / [CH₃CH₂COO⁻] changes only slightly, [H⁺] and pH remain almost constant.

🧠 Securing Level 3 (5–6 Marks)

  • Both components needed: A completely correct calculation yielding 2.02 g PLUS a clear explanation of buffer action with an equilibrium equation.
  • Use proper formulae: Write out actual species ( CH₃CH₂COOH and CH₃CH₂COO⁻ ) rather than shorthand HA and A⁻ to guarantee full communication credit.
  • Molar mass check: Ensure sodium (23.0) is included in Mr calculation ( Mr = 96.0 ).
Level Descriptors:
Level 3 (5–6 marks): Correct calculation of [H⁺], correct [CH₃CH₂COO⁻], correct mass of CH₃CH₂COONa (2.00–2.04 g) AND explains buffer action on addition of OH⁻ referring to equilibrium.
Level 2 (3–4 marks): Correct [H⁺] and [CH₃CH₂COO⁻] with either mass attempt OR buffer action explained.
Level 1 (1–2 marks): Partial calculation ([H⁺] or conjugate base conc) OR qualitative explanation of buffer action.

Topics

Module 2: Foundations in chemistry · Module 5: Physical chemistry and transition elements · 2.1 Atoms and reactions · 5.1 Rates, equilibrium and pH

Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.