OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2018: Question 10

1 mark · Medium difficulty · Multiple Choice

Calculate the pH of a buffer solution prepared by mixing propanoic acid and sodium propanoate solutions of given volumes and concentrations, given the acid dissociation constant Ka.

Practise this question

Question

Multiple choice question 10 asking for the pH of a buffer solution made by mixing 200 cm3 of 2.00 mol dm-3 propanoic acid and 600 cm3 of 1.00 mol dm-3 sodium propanoate, with Ka = 1.32 x 10-5 mol dm-3. Four options are given: A 4.58, B 4.70, C 5.06, D 5.18, with an answer box below.
Question text

10 A buffer solution is prepared by mixing 200 cm3 of 2.00 mol dm–3 propanoic acid, CH CH COOH,

with 600 cm3 of 1.00 mol dm–3 sodium propanoate, CH CH COONa.

K for CH CH COOH = 1.32 × 10–5 mol dm–3

a 3 2

What is the pH of the buffer solution?

A 4.58

B 4.70

C 5.06

D 5.18

Your answer [1]

Mark scheme

Show the mark scheme Mark scheme indicating the correct answer is C for question 10, worth 1 mark.

10 C 1 AO2.6

How to answer it

Calculating the pH of an Acidic Buffer Solution

What this question tests

This question assesses your ability to calculate the pH of an acidic buffer solution formed by mixing a weak acid and its salt. You must demonstrate competence in calculating moles after dilution, rearranging the acid dissociation constant expression (Ka), and converting hydrogen ion concentrations into pH values using logarithms.

Question 10

Exam Breakdown & Step-by-Step Solution

✅ Correct Answer

C (5.06)

Awarded 1 mark for selecting the correct multiple-choice option.

💡 Key Knowledge

  • Buffer expression: [H⁺] = Ka × ([acid] / [salt])
  • Alternatively, use moles directly since total volume cancels out in the ratio: [H⁺] = Ka × (moles of acid / moles of salt)
  • Definition of pH: pH = -log₁₀[H⁺]

📐 Step-by-Step Calculation

  1. Calculate moles of propanoic acid:
    (200 / 1000) dm³ × 2.00 mol dm⁻³ = 0.400 mol
  2. Calculate moles of sodium propanoate:
    (600 / 1000) dm³ × 1.00 mol dm⁻³ = 0.600 mol
  3. Rearrange the Ka expression to find [H⁺]:
    [H⁺] = 1.32 × 10⁻⁵ × (0.400 / 0.600) = 8.80 × 10⁻⁶ mol dm⁻³
  4. Calculate the pH:
    pH = -log₁₀(8.80 × 10⁻⁶) = 5.0555... = 5.06 (to 2 decimal places)

❌ Common Errors & Traps

  • Volume Omission: Forgetting that concentrations change upon mixing volumes ( 200 cm³ + 600 cm³ = 800 cm³ total volume). Though concentrations divide by 0.800 for both, skipping this step can lead to confusion if ratios aren't used.
  • Inverted Ratios: Putting the salt over the acid instead of acid over the salt in the [H⁺] expression.
  • Logarithm Errors: Forgetting the negative sign when converting [H⁺] to pH.

🧠 Exam Technique & Tutor Tips

In multiple-choice buffer questions, look at whether volumes need to be converted into concentrations. Because both species exist in the same total volume, you can use the ratio of their moles directly, saving you time and avoiding intermediate rounding errors with decimal concentrations!

Topics

Module 5: Physical chemistry and transition elements · 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 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.