WJEC A-Level Chemistry AS Unit 1, June 2025: Question 13

13 marks · Medium difficulty · Structured Questions

Describe a back titration method to determine the relative formula mass of calcium carbonate, explain acid-base properties, and carry out dilution and pH calculations.

Practise this question

Question

Question 13 is a multi-part exam question. Part (a) is a 6-mark QER question asking to describe an experimental method for a back titration using 0.500 g of calcium carbonate powder, 25.0 cm³ of 0.500 mol dm⁻³ HCl, and standard 0.200 mol dm⁻³ NaOH, including how to calculate the relative formula mass. Part (b)(i) asks for the definition of a strong acid (1 mark). Part (b)(ii) asks whether ethanoic acid turns pH paper red and neutralises 25.0 cm³ of 1.00 mol dm⁻³ NaOH identically to hydrochloric acid (2 marks). Part (c) asks to calculate the volume of water added to dilute 100 cm³ of 2.00 mol dm⁻³ HCl to 1.60 mol dm⁻³ (2 marks). Part (d) asks to calculate the minimum hydrogen ion concentration for bacteria thriving in a pH range of 5.5 to 8.1 (2 marks).

Mark scheme

Show the mark scheme The mark scheme details points for Question 13: Part (a) uses banded criteria (5-6, 3-4, 1-2 marks) covering practical steps (reaction until fizzing stops, indicator, burette use, titration to end point) and mathematical processing steps (initial moles of acid, excess acid moles, reacted acid, moles of CaCO3, Mr). Part (b)(i) gives 1 mark for 'acid that fully dissociates in solution'. Part (b)(ii) gives 1 mark for stating pH paper turns orange/yellow because it is weak, and 1 mark for stating the neutralisation property is correct because both contain equal moles of acid. Part (c) awards 2 marks for calculating n(HCl) = 0.200 mol, new volume = 125 cm³, and water added = 25 cm³. Part (d) awards 2 marks for using [H+] = 10^-8.1 to obtain 7.9 x 10^-9 mol dm⁻³.

How to answer it

Quantitative Acid-Base Chemistry & Titration Analysis

📋 WHAT THIS QUESTION TESTS

This question covers Unit 1: The Language of Chemistry, Structure of Matter and Simple Reactions. It tests practical and mathematical competence in:

  • Designing and evaluating a 6-mark QER back titration experiment to determine molar mass (Mr).
  • Distinguishing between strong and weak acids regarding dissociation vs stoichiometric neutralisation capacity.
  • Executing dilution calculations ( n = c × V ) and finding volume of solvent added.
  • Relating pH to hydrogen ion concentration [ H⁺ ] using the logarithmic relationship [H⁺] = 10−pH .
Part (a) • 6 Marks (QER)

Experimental Method & Calculation Steps for Back Titration

Determining the relative formula mass (Mr) of calcium carbonate

✅ Model Response (Indicative Content)

Method & Measurements:

  1. Add the 0.500 g of powder to the conical flask containing 25.0 cm³ of 0.500 mol dm⁻³ HCl. Wait until effervescence stops (no more bubbles).
  2. Add a few drops of a suitable indicator (e.g. phenolphthalein or methyl orange).
  3. Rinse and fill a burette with standard 0.200 mol dm⁻³ NaOH solution using a funnel; ensure the funnel is removed and the jet is full of liquid with no air bubbles.
  4. Record initial burette reading to 2 decimal places (ending in .00 or .05).
  5. Add NaOH to the conical flask while swirling continuously. Wash down flask sides with deionised water near the end-point.
  6. Add NaOH dropwise until the indicator shows a permanent colour change. Record final reading and calculate titre volume.

Processing the Results:

  1. Calculate initial moles of HCl: n(HCl)initial = (25.0/1000) × 0.500 = 0.0125 mol
  2. Calculate moles of excess HCl: n(HCl)excess = n(NaOH) = (titre/1000) × 0.200
  3. Calculate moles of reacted HCl: n(HCl)reacted = 0.0125 − n(HCl)excess
  4. Calculate moles of CaCO₃: From equation 1:2 ratio, n(CaCO₃) = n(HCl)reacted / 2
  5. Calculate Mr of CaCO₃: Mr = mass / moles = 0.500 / n(CaCO₃)

🧠 QER Banding & Exam Technique

  • Top Band (5–6 marks): Requires both a logical, clear practical method (including practical rigor: removing funnel, swirling, dropwise at end-point) and a complete, correct algebraic/numerical pathway from raw titre to Mr.
  • Middle Band (3–4 marks): Good practical method with incomplete calculation steps, OR complete calculations with significant practical gaps.
  • Bottom Band (1–2 marks): Basic points only (e.g. mentions adding indicator and Mr = m/n ).

❌ Common Mistakes in Back Titrations

  • Forgetting that HCl is in excess: students often try to equate moles of NaOH directly to CaCO₃.
  • Missing the 1 : 2 stoichiometry between CaCO₃ and HCl ( CaCO₃ + 2HCl ).
  • Leaving the funnel in the top of the burette (leads to dripping and incorrect volume readings).
  • Not specifying that the reaction between CaCO₃ and HCl must finish before titrating.
Mark Scheme: AO1 = 2 marks, AO2 = 2 marks, AO3 = 2 marks | Total: 6 marks [Practical Skills Tested]
Part (b)(i) & (b)(ii) • 3 Marks

Strong vs Weak Acids: Definition & Properties

Comparing 1.00 mol dm⁻³ HCl and 1.00 mol dm⁻³ CH₃COOH

💡 (b)(i) Key Definition: Strong Acid

Answer: An acid that fully dissociates (or ionises) into ions in aqueous solution.

HCl(aq) → H⁺(aq) + Cl⁻(aq) [100% dissociation]

Do not say "contains many H⁺ ions" or "has a low pH" – it must be defined by extent of dissociation.

✅ (b)(ii) Property Evaluation

  • Second Property (pH paper turns red):
    Does not apply (Incorrect). Ethanoic acid is a weak acid, meaning it only partially dissociates into ions, giving a lower [H⁺] and a higher pH (typically pH 2–4). It turns pH paper orange/yellow, not red. [1 mark]
  • Third Property (Neutralises 25.0 cm³ of 1.00 mol dm⁻³ NaOH):
    Applies (Correct). Both solutions have the same concentration and volume, so they contain the same number of moles of acid (0.025 mol). Because CH₃COOH is monoprotic, complete neutralisation requires the same stoichiometric volume of alkali. [1 mark]

❌ Critical Conceptual Distinction

pH depends on [H⁺] at equilibrium (rate and extent of dissociation), whereas volume of base required for neutralisation depends strictly on stoichiometry (moles of available H⁺). As NaOH neutralises H⁺, the weak acid equilibrium shifts completely to the right ( CH₃COOH ⇌ CH₃COO⁻ + H⁺ ) until all molecules are reacted.

Mark Scheme: (b)(i) 1 mark [AO1] | (b)(ii) 2 marks [AO1: 1, AO3: 1]
Part (c) • 2 Marks

Dilution Calculation

Calculate volume of water that should be added to dilute 100 cm³ of 2.00 mol dm⁻³ HCl to 1.60 mol dm⁻³

📐 Step-by-Step Calculation

  1. Find the moles of HCl present:
    n = c × V = 2.00 × (100 / 1000) = 0.200 mol
    [1 mark awarded for finding 0.200 mol]
  2. Calculate the total volume required for 1.60 mol dm⁻³:
    Vtotal = n / c = 0.200 / 1.60 = 0.125 dm³ = 125 cm³
  3. Calculate the volume of water ADDED:
    Vwater added = Vtotal − Vinitial
    Vwater added = 125 cm³ − 100 cm³ = 25 cm³
    Final Answer: 25 cm³ [1 mark]

❌ Common Calculation Trap

Stopping at 125 cm³:

The question asks for the volume of water that should be added, not the final total volume.

  • Total volume = 125 cm³
  • Water added = 125 − 100 = 25 cm³

Always re-read the final line of the question to confirm what quantity is requested.

Mark Scheme: AO2 = 2 marks (1 method mark for moles/total volume, 1 mark for 25 cm³)
Part (d) • 2 Marks

pH and Hydrogen Ion Concentration

Calculate the minimum hydrogen ion concentration for legionella (pH range 5.5–8.1)

📐 Step-by-Step Calculation

  1. Select the correct pH: Because pH = −log₁₀[H⁺] , pH and [H⁺] are inversely related:
    • Lowest pH (5.5) = maximum [H⁺]
    • Highest pH (8.1) = minimum [H⁺]
    Therefore, use pH = 8.1.
  2. Calculate [H⁺]:
    [H⁺] = 10−pH = 10−8.1
    [1 mark awarded for writing 10−8.1]
  3. Evaluate on calculator:
    [H⁺] = 7.943 × 10−9 ≈ 7.9 × 10−9 mol dm⁻³
    Final Answer: 7.9 × 10−9 mol dm⁻³ [1 mark]

❌ Inverse Relationship Pitfall

Many candidates automatically substitute the smallest number (5.5) because the question asks for the "minimum" concentration:

10−5.5 = 3.16 × 10−6 mol dm⁻³ (INCORRECT - this is the MAXIMUM [H⁺])

Remember: higher pH always means lower hydrogen ion concentration!

Mark Scheme: AO3 = 1 mark, AO2 = 1 mark | Total: 2 marks

Topics

Physical Chemistry · Practical · 1.3 Chemical calculations · 1.7 Simple equilibria and acid-base reactions · AS Unit 1 practical work

Question and mark scheme from the WJEC A-Level Chemistry examination, AS Unit 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.