OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2019: Question 4

1 mark · Medium difficulty · Multiple Choice

Calculate the minimum mass of anhydrous sodium carbonate required to neutralise a given volume and concentration of hydrochloric acid.

Practise this question

Question

Multiple-choice question 4 asks for the minimum mass of anhydrous sodium carbonate required to neutralise 50.0 cm3 of 2.00 mol dm-3 HCl(aq). Four options are provided: A 4.15 g, B 5.30 g, C 8.30 g, and D 10.6 g. An answer box is shown at the bottom with a mark allocation of [1].
Question text

4 In the laboratory, acid spills can be cleaned up and made safe by spreading anhydrous sodium

carbonate over the spill to neutralise the acid.

A student accidentally spills 50.0 cm3 of 2.00 mol dm−3 HCl (aq) on the bench.

What is the minimum mass of anhydrous sodium carbonate required to neutralise the acid?

A 4.15 g

B 5.30 g

C 8.30 g

D 10.6 g

Your answer [1]

Mark scheme

Show the mark scheme The mark scheme indicates that the correct answer for question 4 is option B, worth 1 mark.

4 B 1 AO2.8

How to answer it

Calculating Mass for Neutralisation Spills

📌 What this question tests

This question assesses your ability to apply stoichiometry and concentration calculations to a practical scenario. You must write a balanced neutralisation equation, calculate moles from volume and concentration, use reacting molar ratios, and convert moles into mass using molar mass.

Question 4: Multiple Choice Solution

Full Mark Breakdown & Step-by-Step Guide

✅ Correct Answer: B (5.30 g)

The correct option is B. Following the stoichiometric steps yields exactly 0.0500 moles of sodium carbonate, giving a mass of 5.30 g.

💡 Key Knowledge

  • Moles formula: Moles = (Concentration × Volume) / 1000
  • Reacting Ratios: Acid-base stoichiometry must be balanced (2HCl : 1Na₂CO₃).
  • Molar Mass: Na₂CO₃ = (23.0 × 2) + 12.0 + (16.0 × 3) = 106.0 g mol⁻¹.

🧠 Exam Technique

Always write out the balanced symbol equation first before performing any calculations. Missing the 2:1 stoichiometric ratio is the most common pitfall leading to distractor options like D (10.6 g).

❌ Common Errors

  • Ratio oversight: Assuming a 1:1 reaction between HCl and Na₂CO₃ (leading to option D).
  • Volume conversion errors: Forgetting to divide cm³ by 1000 to convert into dm³.
  • Molar mass calculation: Forgetting to multiply the relative atomic mass of sodium (23.0) by 2.

📐 Step-by-Step Calculation Breakdown

Step 1: Write the balanced equation for the neutralisation reaction

2HCl(aq) + Na₂CO₃(s) ➔ 2NaCl(aq) + H₂O(l) + CO₂(g)

Note the 2:1 stoichiometric ratio between hydrochloric acid and sodium carbonate.

Step 2: Calculate the number of moles of HCl spilled

Moles of HCl = Concentration × Volume (in dm³)

Moles of HCl = 2.00 mol dm⁻³ × (50.0 / 1000) dm³ = 0.100 mol

Step 3: Use the molar ratio to find required moles of Na₂CO₃

Ratio HCl : Na₂CO₃ is 2 : 1

Moles of Na₂CO₃ = 0.100 / 2 = 0.0500 mol

Step 4: Calculate the molar mass (M) of anhydrous sodium carbonate (Na₂CO₃)

M(Na₂CO₃) = (23.0 × 2) + 12.0 + (16.0 × 3) = 46.0 + 12.0 + 48.0 = 106.0 g mol⁻¹

Step 5: Calculate the minimum mass required

Mass = Moles × Molar Mass

Mass = 0.0500 mol × 106.0 g mol⁻¹ = 5.30 g

Examiner note: Option D (10.6 g) represents the student error of skipping the 1:1 vs 2:1 ratio adjustment, calculating 0.100 mol × 106.0 g mol⁻¹ instead.

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 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.