OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2025: Question 8

1 mark · Medium difficulty · Multiple Choice

Calculate the percentage uncertainty of a titre using initial and final burette readings.

Practise this question

Question

Question 8 shows a table of burette readings: Final reading is 29.35 cm³ and Initial reading is 2.60 cm³. Below the table, it states that the burette has an uncertainty of ±0.05 cm³. Students are asked to select the percentage uncertainty of the titre from four options: A (0.17%), B (0.19%), C (0.34%), and D (0.37%).

Mark scheme

Show the mark scheme Mark scheme table indicating for question number 8 that the correct answer is D, with 1 mark awarded.

How to answer it

Burette Titre & Percentage Uncertainty

📌 What this question tests

This question assesses your ability to calculate experimental uncertainty in volumetric titrations. Specifically, it tests whether you recognize that a titre is determined from two separate readings (initial and final), requiring the equipment uncertainty to be doubled when calculating the percentage uncertainty of the delivered volume.

Question 8 • Multiple Choice [1 Mark]

Percentage Uncertainty in a Titration

Quantitative chemistry: Practical skills & uncertainties

✅ Correct Answer

D — 0.37%

The total uncertainty for the titre is 2 × (±0.05 cm³) = ±0.10 cm³. Divided by the titre volume (26.75 cm³) and multiplied by 100 gives 0.37%.

💡 Key Knowledge

  • Two-reading instruments: A burette requires both an initial reading and a final reading to calculate the volume delivered ( titre = final − initial ).
  • Adding absolute uncertainties: Each individual reading has an uncertainty of ±0.05 cm³. For two readings, total absolute uncertainty = 2 × 0.05 = ±0.10 cm³.
  • Percentage uncertainty formula:
    % uncertainty = (Total uncertainty / Measured value) × 100

📐 Step-by-Step Calculation

1 Calculate the titre volume:

Titre = Final reading − Initial reading
Titre = 29.35 cm³ − 2.60 cm³ = 26.75 cm³

2 Determine the total absolute uncertainty:

Because two readings were taken using the burette (initial and final):
Total uncertainty = 2 × (±0.05 cm³) = ±0.10 cm³

3 Calculate percentage uncertainty:

% uncertainty = (0.10 cm³ / 26.75 cm³) × 100 = 0.3738...%
Rounding to 2 significant figures gives: 0.37% (Option D)

❌ Distractor Breakdown & Traps

Each incorrect option represents a classic student error:

  • Option A (0.17%): Used only 1 reading uncertainty and divided by the final volume: (0.05 / 29.35) × 100 .
  • Option B (0.19%): The most common error! Calculated the titre correctly but forgot to multiply uncertainty by 2: (0.05 / 26.75) × 100 .
  • Option C (0.34%): Correctly doubled the uncertainty but divided by the final reading instead of the titre: (0.10 / 29.35) × 100 .

🧠 Exam Technique & Examiner Insights

  • Always identify "difference" measurements: Whenever you subtract two values from the same instrument (e.g. burettes, mass by difference on a balance, temperature change ΔT on a thermometer), you must double the apparatus uncertainty.
  • Single-reading apparatus: Contrast this with a volumetric pipette or volumetric flask, which are filled to a single calibration mark and therefore only carry a single uncertainty value.
  • Check your denominator: Always ensure the denominator is the actual quantity measured in the experiment (the titre = 26.75 cm³), never just the raw endpoint reading on the scale.
Mark Scheme Reference: Question 8 • Correct response: D • Total: 1 mark

Topics

Module 1: Development of practical skills in chemistry · Practical Activity Groups · 1.1 Practical skills assessed in a written examination · PAG 2: Acid-base titration

Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.