OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), November 2020: Question 11

1 mark · Medium difficulty · Multiple Choice

Calculate the pH of the resulting mixture when hydrochloric acid is added to sodium hydroxide solution

Practise this question

Question

Question 11 states: 20 cm3 of 0.10 mol dm-3 hydrochloric acid is added to 10 cm3 of 0.10 mol dm-3 sodium hydroxide. What is the pH of the resulting mixture? Four multiple-choice options are provided: A, 1.00; B, 1.18; C, 1.30; D, 1.48. A box for the student's answer and the mark allocation of [1] are shown at the bottom.
Question text

11 20 cm3 of 0.10 mol dm−3 hydrochloric acid is added to 10 cm3 of 0.10 mol dm−3 sodium hydroxide.

What is the pH of the resulting mixture?

A 1.00

B 1.18

C 1.30

D 1.48

Your answer [1]

Mark scheme

Show the mark scheme The mark scheme table shows question number 11 with the correct answer D.

11 D 1 2.2

How to answer it

Calculating the pH of a Neutralisation Mixture

What this question tests

This question tests your ability to calculate the pH of a strong acid-strong base mixture where the acid is in excess. You must calculate initial moles, determine limiting reagents, find excess reactant concentrations accounting for total volume changes, and apply the logarithmic pH formula.

Question 11 (Multiple Choice)

Solution & Mark Breakdown

✅ Correct Answer: D (1.48)

Option D is the correct choice because neutralisation leaves excess HCl , and the new concentration results in a pH of 1.48 after accounting for the combined volume.

💡 Key Knowledge

  • Strong acids ( HCl ) and strong bases ( NaOH ) react in a 1:1 molar ratio.
  • Total volume equals the sum of the individual volumes added together.
  • pH = -log[H⁺]

Step-by-Step Calculation Guide

📐 Step 1: Calculate Initial Moles

Moles = Concentration × Volume (in dm³)

  • Moles of HCl = 0.020 dm³ × 0.10 mol dm⁻³ = 0.0020 mol
  • Moles of NaOH = 0.010 dm³ × 0.10 mol dm⁻³ = 0.0010 mol

📐 Step 2: Determine Excess Moles

Since NaOH is the limiting reagent:

  • Reacted moles = 0.0010 mol
  • Excess HCl = 0.0020 - 0.0010 = 0.0010 mol

📐 Step 3: Find New Concentration

Total volume = 20 cm³ + 10 cm³ = 30 cm³ = 0.030 dm³

  • [H⁺] = 0.0010 mol / 0.030 dm³ = 0.0333 mol dm⁻³

📐 Step 4: Calculate pH

  • pH = -log(0.0333...) = 1.477...
  • Round to 2 decimal places (matching standard pH convention): 1.48

❌ Common Errors & Distractor Traps

  • Distractor A (1.00): Calculated by forgetting to add volumes together (assuming volume stayed at 20 cm³).
  • Distractor B/C: Arithmetic errors with mole subtraction or failing to use correct stoichiometry.
  • Unit Trap: Forgetting to convert cm³ to dm³ when calculating concentration.

🧠 Exam Technique

Always write out the neutralization reaction first. Clearly label intermediate values (moles in excess, total volume) on your working paper before typing numbers into your calculator to avoid transcription errors.

Exam Marker Note: This is a 1-mark multiple-choice question. However, examiners report that students frequently lose subsequent structured calculation marks in paper 2/3 by failing to account for the total volume expansion after mixing solutions.

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), November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.