OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2018: Question 4

1 mark · Medium difficulty · Multiple Choice

Calculate the concentration of nitric acid in mol dm⁻³ required to neutralise a given volume and concentration of barium hydroxide solution.

Practise this question

Question

Multiple choice question asking for the concentration in mol dm⁻³ of nitric acid when 23.35 cm³ of it neutralises 25.00 cm³ of 0.0450 mol dm⁻³ barium hydroxide. Four options are given: A 0.0241, B 0.0482, C 0.0900, D 0.0964, alongside an answer box and 1 mark allocation.
Question text

4 A student titrates a standard solution of barium hydroxide, Ba(OH)2, with nitric acid, HNO3.

25.00 cm3 of 0.0450 mol dm–3 Ba(OH) are needed to neutralise 23.35 cm3 of HNO (aq).

What is the concentration, in mol dm–3, of the nitric acid?

A 0.0241

B 0.0482

C 0.0900

D 0.0964

Your answer [1]

Mark scheme

Show the mark scheme Mark scheme indicating the correct answer is D.

4 D 1 AO2.4

How to answer it

Neutralisation Titration Calculation

📚 What this question tests

This question assesses core quantitative chemistry skills: constructing a balanced neutralisation equation, calculating moles from concentration and volume, applying reacting molar ratios, and determining unknown concentrations with correct significant figures.

Question 4 Analysis & Solution

Determining the Concentration of Nitric Acid

✅ Correct Answer

D (0.0964 mol dm⁻³)

Awarded 1 mark for correct selection under AO2.4 (Application of quantitative chemistry).

💡 Key Knowledge

  • Barium hydroxide, Ba(OH)₂ , is a strong dibasic (diprotic) base releasing two moles of OH⁻ per mole of compound.
  • Nitric acid, HNO₃ , is a monobasic acid.
  • Mole formula: Moles = (Concentration × Volume in cm³) / 1000

🧠 Exam Technique

  • Never skip writing the balanced equation! Missing the stoichiometry (1:2 ratio) is the #1 cause of lost marks in titration questions.
  • Keep unrounded numbers in your calculator until the very final step to prevent rounding errors.

❌ Common Errors

  • The 1:1 Trap (Option C - 0.0900): Forgetting that Ba(OH)₂ reacts with two moles of HNO₃ . This leads to treating the ratio as 1:1 ( 0.001125 × 2 / 0.02335 ).
  • Inverted Ratios (Option A): Multiplying or dividing by the wrong stoichiometric coefficients.

📐 Step-by-Step Calculation Guide

  1. Write the balanced equation:
    Ba(OH)₂ (aq) + 2HNO₃ (aq) → Ba(NO₃)₂ (aq) + 2H₂O (l)
    Notice the 1:2 molar ratio between Ba(OH)₂ and HNO₃ .
  2. Calculate the moles of barium hydroxide:
    Moles = 25.00 × 0.0450 / 1000 = 1.125 × 10⁻³ mol
  3. Use the stoichiometric ratio to find moles of nitric acid:
    Since 1 mole of Ba(OH)₂ reacts with 2 moles of HNO₃ :
    Moles of HNO₃ = 1.125 × 10⁻³ × 2 = 2.25 × 10⁻³ mol
  4. Calculate the concentration of nitric acid:
    Concentration = (Moles × 1000) / Volume in cm³
    Concentration = (2.25 × 10⁻³ × 1000) / 23.35 = 0.096359... mol dm⁻³
  5. Round to appropriate significant figures:
    The input data ( 25.00 , 0.0450 , 23.35 ) are given to 3 and 4 significant figures, so round your final answer to 3 significant figures: 0.0964 mol dm⁻³.

Topics

Module 2: Foundations in chemistry · Practical Activity Groups · PAG 2: Acid-base titration · 2.1 Atoms and reactions

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.