OCR A-Level Chemistry AS Breadth in chemistry (01), June 2024: Question 22
8 marks · Medium difficulty · Calculations
Differentiate between strong and weak acids, complete titration calculations including table completion and mean titre determination, and calculate the percentage composition by mass of ethanoic acid in a descaler.
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Question text
22 This question is about the reactions of acids.
(a) What is the difference between a strong acid and a weak acid?
… [1]
(b) Ethanoic acid, CH3COOH, is found in some descalers to soften hard water.
A student carries out a titration with a standard solution of sodium carbonate, Na2CO3, to
determine the percentage composition by mass of CH3COOH in a descaler.
The equation is shown below.
2CH3COOH + Na2CO3 2CH3COONa + CO2 + H2O
(i) The method is outlined below:
• Dissolve 6.50 g of the descaler in distilled water.
• Transfer the solution into a 250.0 cm3 volumetric flask.
• Make up to the mark with distilled water and invert several times.
• Pipette 25.0 cm3 of this solution into a conical flask and add a few drops of indicator.
• Titrate this solution with 0.200 mol dm–3 Na CO (aq), in the burette.
The student carries out a trial titration, followed by further titrations.
The results are shown in the table below.
The trial titration has been omitted.
Titration 1 2 3
Final reading / cm3 48.95 24.15 48.35
Initial reading / cm3 24.55 0.00 24.10
Titre / cm3
Complete the table by adding the titres.
[1]
(ii) Calculate the mean titre, to the nearest 0.05 cm3, that the student should use for analysing these
results.
mean titre = … cm3 [1]
(iii) Calculate the percentage composition by mass of CH3COOH in the descaler.
Assume that CH3COOH is the only acid in the descaler.
Give your answer to 3 significant figures.
percentage composition by mass = … % [5]
Mark scheme
Show the mark scheme
Question Answer Marks Guidance
22 (a) strong acid: fully dissociates/ionises 1 ALLOW strong acid fully dissociates
AND weak acid dissociates/ionises less
weak acid: partially dissociates/ionises
ALLOW strong acid releases all H+ ions
weak acid partially releases H+ ions
IGNORE strrong acid dissociates more
strrong acid dissociates quicker
DO NOT ALLOW
strong acid fully dissociates
weak acid does not fully dissociate
Response does not state that weak acid dissociates
IGNORE breaks down for dissociate/ionise
DO NOT ALLOW comparison of concentrations
(b) (i) 1
Titre/cm3 24.40 24.15 24.25 ✓
DO NOT ALLOW 24.4
Correct subtractions to obtain titres to 2 DP
(b) (ii) 24.15 + 24.25 3 1 ALLOW 24.2 DP already assessed in b(i)
mean titre = 2 = 24.20 (cm ) ✓
i.e. using concordant (consistent) titres DO NOT ALLOW mean of all three titres,
24.40 + 24.15 + 24.25
i.e. 3 = 24.26/24.27
ALLOW ECF from incorrect concordant titres from 22b(i)
(b) (iii) FIRST CHECK ANSWER ON ANSWER LINE 5 ALLOW 3SF or more throughout
IF answer = 89.4 (%) award 5 marks IGNORE trailing zeroes,
e.g. ALLOW 24.2 for 24.20
CHECK mean titre from 22b(ii) first.
THEN apply ECF throughout using THIS mean titre
ALLOW ECF from incorrect mean titre in b(ii)
First 3 mark must come from the titration
ALLOW ECF from 2 incorrect n(Na2CO3)
n(Na2CO3)
24.20 –3
= 0.200 1000 = 4.84 10 (mol) ✓
ALLOW ECF from incorrect n(CH3COOH),
n(CH COOH) in 25.0 cm3 OR
= 2 4.84 10–3 = 9.68 10–3 (mol) ✓ from n(Na CO ) if n(CH COOH) stage omitted
23 3
n(CH COOH) in 250 cm3
= 10 9.68 10–3 = 9.68 10–2 (mol) ✓
ALLOW 5.81 (3 SF)
mass of CH COOH) in 250 cm3
= 60 9.68 10–2 = 5.808 (g) ✓ IF mass is rounded to 5.81, Answer is still 89.4%
Calculator = 89.38461538
% composition to 3 SF
5.808 8.94% is 4 marks (omission of 10 stage)
= 6.50 100 = 89.4 (%) ✓ 3 SF
Calculator: 89.35384615 IF incorrect mean titre of 24.26/24.27 cm3 used:
(mean of all 3 titres in b(ii)),
% composition = 89.6% to 3 SF for ALL 5 marks by ECF
----------------------------------------------------------------------------
NOTE: Some candidates are calculating n(CH3COOH) based
on the 6.50 g sample being pure
DO NOT ALLOW 0.108(3……
6.50
n(CH3COOH) = 60 = 0.108(3……
COMMON ERRORS COMMON ERRORS
Omitting ÷ 1000 for n(Na CO ) Using 25.0 cm3 (pipette volume) instead of 24.20 cm3
Up to 3 marks are possible Up to 4 marks are possible
n(Na2CO3) n(Na2CO3)
= 0.200 24.20 = 4.84 (mol) 25.00 –3
= 0.200 1000 = 5.00 10 (mol)
n(CH COOH) in 25.0 cm3 n(CH COOH) in 25.0 cm3
= 2 4.84 = 9.68 (mol) ✓ = 2 5.00 10–3 = 1 10–2 (mol) ✓
n(CH COOH) in 250 cm3 n(CH COOH) in 250 cm3
= 10 9.68 = 96.8 (mol) ✓ = 10 1 10–2 = 1 10–1 (mol) ✓
mass of CH COOH) in 250 cm3 mass of CH COOH) in 250 cm3
= 60 96.8 = 5808 (g) ✓ = 60 1 10–2 = 6.00 (g) ✓
% composition to 3 SF % composition to 3 SF
5808 6.00
= 6.50 100 = 89400 (%) = 100 = 92.3 (%) ✓
6.50
Impossible value Calculator: 92.30769231
How to answer it
Reactions of Acids and Titration Calculations
This question assesses your understanding of acid strength definitions, practical titration processing (completing tables and selecting concordant titres), and multi-step stoichiometric calculations. Key skills include handling volumetric scaling factors, applying reacting mole ratios from balanced equations, working with standard solutions, and formatting final answers to 3 significant figures.
Part (a): Strong vs Weak Acids
Define the difference between a strong and a weak acid
✅ Correct Answer
Strong acid: Fully dissociates / ionises.
Weak acid: Partially dissociates / ionises.
💡 Key Knowledge
- Dissociation refers to the splitting of molecules into ions in water.
- Use the exact term dissociates or ionises . Avoid vague phrasing like "breaks down".
❌ Common Errors
- Writing that a weak acid "does not dissociate" (it partially dissociates).
- Confusing acid strength with concentration (e.g. talking about number of moles or molarity).
Part (b)(i): Completing the Titration Table
Calculate titres for Titration 1, 2, and 3
✅ Correct Answers
Titration 1: 48.95 - 24.55 = 24.40 cm³
Titration 2: 24.15 - 0.00 = 24.15 cm³
Titration 3: 48.35 - 24.10 = 24.25 cm³
Always subtract the initial burette reading from the final burette reading. Ensure all recorded titres are given to 2 decimal places to match the burette precision (ending in 0 or 5).
Part (b)(ii): Calculating the Mean Titre
Select concordant results and find the average
✅ Correct Answer
Concordant titres are Titration 2 ( 24.15 cm³ ) and Titration 3 ( 24.25 cm³ ).
Mean titre = (24.15 + 24.25) / 2 = 24.20 cm³
❌ Common Errors
Averaging all three titres (including the rough titration 1). Always discard the rough titre and only average concordant (consistent) values within 0.10 cm³ of each other.
Part (b)(iii): Percentage Composition Calculation
Determine the percentage by mass of CH₃COOH in the 6.50 g descaler sample
📐 Step-by-Step Calculation
- Moles of Na₂CO₃ in titre:
n(Na₂CO₃) = 0.200 × (24.20 / 1000) = 4.84 × 10⁻³ mol - Moles of CH₃COOH in 25.0 cm³ pipetted solution:
From equation (2:1 ratio), n(CH₃COOH) = 2 × 4.84 × 10⁻³ = 9.68 × 10⁻³ mol - Moles of CH₃COOH in the full 250 cm³ volumetric flask:
Scale up by factor of 10 (250 / 25):
9.68 × 10⁻³ × 10 = 9.68 × 10⁻² mol - Mass of CH₃COOH in 250 cm³:
Mr of CH₃COOH = (2 × 12.0) + (4 × 1.0) + (2 × 16.0) = 60.0 g mol⁻¹
Mass = 9.68 × 10⁻² × 60.0 = 5.808 g - Percentage composition by mass:
(5.808 / 6.50) × 100 = 89.353...%
To 3 SF: 89.4%
❌ Common Calculation Traps
- Volume conversion: Forgetting to divide by 1000 when converting cm³ to dm³.
- Ratio omission: Forgetting to multiply by 2 to account for the 2:1 stoichiometric ratio between CH₃COOH and Na₂CO₃.
- Dilution factor error: Multiplying by 25 instead of 10 when scaling from the 25.0 cm³ pipette sample to the 250 cm³ flask.
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, AS Breadth in chemistry (01), June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.