AQA A-Level Chemistry AS Paper 1, November 2021: Question 8
8 marks · Medium difficulty · Practical Techniques & Data Analysis
Calculate the percentage by mass of sodium ethanoate in a contaminated sample of ethanoic acid using titration data, and explain the effect of rinsing the burette with deionised water.
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Question text
08 A student is provided with a 5.60 g sample of ethanoic acid (CH3COOH) contaminated
with sodium ethanoate (CH3COONa).
The student dissolves the sample in deionised water and makes the volume up
to 200 cm3
The student removes 25.0 cm3 samples of the solution and titrates them with
0.350 mol dm–3 sodium hydroxide solution.
Table 3 shows the results of these titrations.
13 Table 3
Rough 1 2 3
Final volume / cm3 20.85 41.10 20.50 40.80
Initial volume / cm3 0.00 20.85 0.00 20.50
Titre / cm3 20.85 20.25 20.50 20.30
08.1 Use the results in Table 3 to calculate the mean titre value.
Use the mean titre to calculate the percentage by mass of sodium ethanoate in the
original sample.
[6 marks]
Mean titre value cm3
Percentage by mass
08.2 The student rinses the burette with deionised water before filling with
sodium hydroxide solution.
State and explain the effect, if any, that this rinsing will have on the value of the titre.
[2 marks]
Mark scheme
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Question Marking guidance Additional Comments/Guidelines Mark
Allow M1 = 20.28 cm3 1
20.25+20.30 3
M1: Mean titre = = 20.275 cm
M2 = M1 × 10–3 × 0.35
M2 Amount of NaOH = 0.35 × (20.275 ÷ 1000) 1
= 0.00709625 mol
Amount of ethanoic acid in 25 cm3 = 0.00709625 mol
M3 = M2 × 8
M3 Amount of ethanoic acid in 200 cm3 = 0.05677 mol
08.1 M4 Mass of ethanoic acid in sample = 60.0 × 0.05677 M4 = M3 × 60.0 1
= 3.4062 g
M5 Mass of sodium ethanoate = 5.6 – 3.4062 M5 = 5.6 – M4
= 2.1938 g 1
M6 = (M5 ÷ 5.6 ) × 100
M6 percentage CH3COONa = (2.1938 ÷ 5.6) × 100 1
= 39.1 % (39.1 – 39.2)
Accept alternative methods
M5 = (M4 ÷ 5.6) × 100) followed by M6 = 100 – M5
M1 Titre value would increase / larger value 1
08.2
M2 Because the sodium hydroxide solution would be more dilute 1
How to answer it
Contaminated Acid Titration & Percentage Purity Calculation
This multi-step quantitative chemistry problem assesses your ability to select concordant titration results, perform multi-stage mole calculations involving dilutions, calculate mass and percentage composition of impure mixtures, and understand practical titration errors regarding glassware contamination.
Question 08.1: Titration & Percentage Mass Calculation
Total: 6 Marks
✅ Correct Answers
- Mean titre: 20.28 cm³ (or 20.275 cm³ )
- Percentage by mass of sodium ethanoate: 39.1% to 39.2%
💡 Key Knowledge
- Only concordant titres (those within 0.10 cm³ of each other) are averaged. Rough titres are always discarded. Titres 1 and 3 are 20.25 cm³ and 20.30 cm³. Titre 2 (20.50 cm³) is non-concordant.
- Sodium ethanoate (CH₃COONa) is a salt and does not react with sodium hydroxide (NaOH). Only the weak acid, ethanoic acid (CH₃COOH), reacts.
🧠 Exam Technique
- Work methodically through scaling factors: titration volume (25.0 cm³) to volumetric flask volume (200 cm³). The scaling factor is 200 / 25.0 = 8.
- Keep unrounded numbers in your calculator storage until the final answer to prevent rounding errors.
❌ Common Errors
- Including Titre 2 (20.50 cm³) in the mean calculation, skewing the subsequent mole steps.
- Assuming sodium ethanoate reacts with NaOH, leading to confusion over stoichiometry.
- Forgetting to scale up the moles from the 25.0 cm³ aliquot to the full 200 cm³ solution.
📐 Step-by-Step Calculation Guide
- Step 1 (M1): Calculate the mean titre.
Select concordant titres 1 and 2 (wait, Titres 1 and 3): (20.25 + 20.30) / 2 = 20.275 cm³ (Allow 20.28 cm³). - Step 2 (M2): Calculate moles of NaOH used in titration.
Moles = concentration × volume (in dm³) = 0.350 × (20.275 / 1000) = 0.00709625 mol . - Step 3 (M3): Scale up to the total 200 cm³ solution.
Since 25.0 cm³ was titrated out of 200 cm³, multiply by (200 / 25) [factor of 8]:
0.00709625 × 8 = 0.05677 mol of CH₃COOH in the total sample. - Step 4 (M4): Calculate mass of ethanoic acid.
Molar mass of CH₃COOH = 60.0 g mol⁻¹.
Mass = moles × Mᵣ = 0.05677 × 60.0 = 3.4062 g . - Step 5 (M5): Calculate mass of sodium ethanoate contaminant.
Total sample mass = 5.60 g.
Mass of CH₃COONa = 5.60 - 3.4062 = 2.1938 g . - Step 6 (M6): Calculate percentage by mass.
(2.1938 / 5.60) × 100 = 39.17% -> 39.1% or 39.2% (to 3 sig fig).
Question 08.2: Practical Errors & Rinsing Effects
Total: 2 Marks
✅ Correct Answers
- State: The titre value would increase (or be a larger value).
- Explain: Because the sodium hydroxide solution would be diluted by residual water left in the burette.
💡 Key Knowledge
- Glassware preparation rules: Flasks and funnels can be rinsed with deionised water because it doesn't change the number of moles present. Pipettes and burettes should be rinsed with the solution they are going to contain to avoid dilution.
🧠 Exam Technique
- Structure your answer clearly: state the effect first (increase/decrease/no change), followed by the chemical/physical reasoning (dilution/concentration change). Never skip the "explain" portion.
❌ Common Errors
- Confusing burette rinsing rules with volumetric flask rinsing rules.
- Stating the titre decreases because students mistakenly think water "neutralises" the alkali.
Topics
Physical Chemistry · Required Practicals · Required Practical 1: Making up a volumetric solution · 3.1.2 Amount of Substance · 3.1.12 Acids and Bases
Question and mark scheme from the AQA A-Level Chemistry examination, AS Paper 1, November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.