OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2023: Question 4

1 mark · Medium difficulty · Multiple Choice

Calculate the concentration of the resulting solution after mixing specified volumes and concentrations of hydrochloric acid and sodium hydroxide.

Practise this question

Question

Multiple choice question 4 asking for the concentration of the resulting solution when 40.0 cm3 of 0.200 mol dm-3 HCl is added to 60.0 cm3 of 0.100 mol dm-3 NaOH. Four options A, B, C, and D provide different combinations of HCl and NaCl concentrations in mol dm-3, with an answer box at the bottom.
Question text

4 40.0 cm3 of 0.200 mol dm–3 HCl is added to 60.0 cm3 of 0.100 mol dm–3 NaOH.

What is the concentration of the resulting solution?

A 0.0200 mol dm–3 HCl and 0.0200 mol dm–3 NaCl

B 0.0200 mol dm–3 HCl and 0.0400 mol dm–3 NaCl

C 0.0200 mol dm–3 HCl and 0.0600 mol dm–3 NaCl

D 0.0600 mol dm–3 HCl and 0.0200 mol dm–3 NaCl

Your answer [1]

Mark scheme

Show the mark scheme Mark scheme indicating the correct answer for question 4 is C, worth 1 mark.

4 C 1 AO2.2

How to answer it

Calculating Concentration of Resulting Solutions

📚 What this question tests

This question assesses your understanding of moles, stoichiometry, neutralisation reactions, and calculating the final concentration of species left in solution after mixing excess acid or alkali. It tests Application of Knowledge (AO2.2) by requiring you to calculate moles before and after reaction, determine limiting reagents, and account for the total combined volume of the final solution.

Question 4 (Multiple Choice)

Determining the composition and concentration of a mixed ionic solution

✅ Correct Answer

C (0.0200 mol dm⁻³ HCl and 0.0600 mol dm⁻³ NaCl)

HCl is in excess, meaning both unreacted HCl and formed NaCl remain in the final total volume.

💡 Key Knowledge

  • Equation: HCl + NaOH -> NaCl + H₂O
  • Moles formula: Moles = (Volume in cm³ × Concentration) / 1000
  • Total volume = Volume of acid + Volume of alkali
  • Concentration = Moles remaining / Total volume in dm³

🧠 Exam Technique

Always calculate initial moles of both reactants first. Identify the limiting reagent to see which reactant is completely consumed and which one is in excess.

❌ Common Errors

  • Volume Trap: Forgetting to add the two volumes together to find the new total volume ( 40.0 + 60.0 = 100.0 cm³ ).
  • Stoichiometry Mix-up: Assuming the concentration of NaCl equals the initial concentration of NaOH without scaling for the new total volume.

📐 Step-by-Step Calculation Guide

  1. Calculate initial moles of HCl:
    (40.0 / 1000) × 0.200 = 0.00800 mol
  2. Calculate initial moles of NaOH:
    (60.0 / 1000) × 0.100 = 0.00600 mol
  3. Determine moles reacting and moles remaining:
    NaOH is the limiting reagent (0.00600 mol). It reacts completely with 0.00600 mol of HCl.
    Unreacted HCl remaining = 0.00800 - 0.00600 = 0.00200 mol
    Moles of NaCl formed = 0.00600 mol (in a 1:1 ratio with NaOH)
  4. Calculate total final volume:
    40.0 cm³ + 60.0 cm³ = 100.0 cm³ = 0.100 dm³
  5. Calculate concentrations in the final mixture:
    [HCl] = 0.00200 mol / 0.100 dm³ = 0.0200 mol dm⁻³
    [NaCl] = 0.00600 mol / 0.100 dm³ = 0.0600 mol dm⁻³
Mark Scheme Note: 1 mark awarded for selecting option C. Tested under AO2.2.

Topics

Module 2: Foundations in chemistry · 2.1 Atoms and reactions

Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.