OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2025: Question 5

1 mark · Medium difficulty · Multiple Choice

Calculate the number of water molecules formed when 0.200 mol of butanedioic acid reacts with 0.300 mol of sodium hydroxide.

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Question

Multiple choice question 5 asks: '0.200 mol of HOOCCH2CH2COOH is reacted with 0.300 mol NaOH. How many water molecules are formed?' Four options are given: A: 1.204 × 10^23, B: 1.806 × 10^23, C: 2.408 × 10^23, and D: 3.612 × 10^23, with an answer box provided.
Question text

5 0.200 mol of HOOCCH2CH2COOH is reacted with 0.300 mol NaOH.

How many water molecules are formed?

A 1.204 × 1023

B 1.806 × 1023

C 2.408 × 1023

D 3.612 × 1023

Your answer

[1]

Mark scheme

Show the mark scheme Mark scheme excerpt showing question number 5 with correct answer B and an allocation of 1 mark.

5 B 1

How to answer it

Neutralisation Stoichiometry & Avogadro's Constant

📋 What This Question Tests
  • Dicarboxylic acid behaviour: Recognising that butanedioic acid, HOOCCH₂CH₂COOH , is diprotic and can donate 2 moles of H⁺ ions per mole of acid.
  • Limiting reactant concept: Identifying which reactant runs out first when stoichiometric amounts are not provided in exact molar proportions.
  • Avogadro calculations: Using Avogadro's constant ( NA = 6.02 × 10²³ mol⁻¹ ) to calculate the actual number of molecules produced.

Question 5 Walkthrough

Multiple Choice Question (1 Mark)

✅ Correct Answer

B: 1.806 × 10²³

Award [1] for selecting option B.

💡 Key Knowledge

  • Diprotic acid reaction:
    HOOCCH₂CH₂COOH + 2NaOH → NaOOCCH₂CH₂COONa + 2H₂O
  • Ionic neutralisation:
    H⁺(aq) + OH⁻(aq) → H₂O(l)
  • Each mole of OH⁻ ions consumed produces exactly 1 mole of H₂O.
  • Number of particles = moles (n) × Avogadro's constant (NA)

📐 Step-by-Step Calculation

1 Identify the reacting mole ratio:
1 mole of dicarboxylic acid requires 2 moles of NaOH for complete neutralisation:
Acid : NaOH = 1 : 2

2 Determine the limiting reagent:
• Moles of acid present = 0.200 mol (contains 0.400 mol H⁺ )
• Moles of NaOH required for full reaction = 0.200 × 2 = 0.400 mol
• Moles of NaOH actually supplied = 0.300 mol
Since 0.300 mol < 0.400 mol , NaOH is the limiting reagent, and the acid is in excess.

3 Determine moles of water formed:
From the ionic equation, every mole of OH⁻ reacts with H⁺ to produce 1 mole of H₂O :
n(H₂O) = n(NaOH reacted) = 0.300 mol

4 Calculate the number of water molecules:
Number of molecules = n × NA
= 0.300 mol × (6.02 × 10²³ mol⁻¹) = 1.806 × 10²³ molecules

❌ Distractor Traps & Common Errors

  • Option C (2.408 × 10²³): Forgetting to check the limiting reagent! Students assumed all 0.200 mol of acid reacted to produce 0.400 mol H₂O ( 0.400 × 6.02 × 10²³ = 2.408 × 10²³ ).
  • Option A (1.204 × 10²³): Assuming a 1:1 reaction where 0.200 mol of acid makes 0.200 mol H₂O ( 0.200 × 6.02 × 10²³ ), completely overlooking that it is a dicarboxylic acid.
  • Option D (3.612 × 10²³): Doubling the moles of NaOH unnecessarily ( 0.600 × 6.02 × 10²³ ).

🧠 Exam Technique & Examiner Advice

  • Two quantities given? Stop and check! Whenever an exam question gives amounts for both reactants, it is almost certainly a limiting reactant problem.
  • Count reactive functional groups: Look carefully at formulas like HOOCCH₂CH₂COOH . Two -COOH groups mean twice as many protons!
  • Use the fundamental ionic equation: In any neutralisation, H⁺ + OH⁻ → H₂O . If OH⁻ is limiting, the moles of water formed always equals the moles of OH⁻ added.

Topics

Module 2: Foundations in chemistry · Module 6: Organic chemistry and analysis · 2.1 Atoms and reactions · 6.1 Aromatic compounds, carbonyls and acids

Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.