OCR A-Level Chemistry AS Depth in chemistry (02), November 2021: Question 3

12 marks · Medium difficulty · Practical Questions

Perform calculations and analysis based on an acid-base titration of glutaric acid with sodium hydroxide.

Practise this question

Question

A structured examination question about an acid-base titration of glutaric acid (HOOC(CH2)nCOOH) with NaOH. It includes experimental details, a titration results table with trial and three further titrations, and parts (a) to (d) asking for colour changes, titre completion, mean titre calculation, percentage uncertainty, a multi-step molar mass calculation for n, and an error analysis for washing a pipette with water.
Question text

3 Glutaric acid is used in the production of polymers.

The formula of glutaric acid can be represented as HOOC(CH2)nCOOH, where n is a whole

number.

A student carries out a titration to find the value of n.

1. The student dissolves 2.891 g of glutaric acid in water and makes up the solution to 250.0 cm3

in a volumetric flask.

2. The student transfers 25.0 cm3 of this solution into a conical flask.

3. The student titrates the solution with 0.240 mol dm–3 NaOH(aq) in the burette.

Equation:

HOOC(CH2)nCOOH(aq) + 2NaOH(aq) NaOOC(CH2)nCOONa(aq) + 2H2O(l)

The student uses phenolphthalein as the indicator.

Phenolphthalein is colourless in acid and pink in alkali.

(a) State the colour change observed at the end point of the titration.

Colour from … to … [1]

(b) The student carries out a trial titration followed by three further titrations, 1, 2 and 3.

The results are shown in the table below.

Titration Trial 1 2 3

Final reading / cm3 18.70 36.55 18.30 36.60

Initial reading / cm3 0.20 18.50 0.10 18.30

Titre / cm3

(i) Complete the table to show the titre in each titration. [1]

(ii) Why does the student carry out a trial titration?

… [1]

(iii) Calculate the mean titre of NaOH(aq) that the student should use for analysing the

results.

mean titre = … cm3 [1]

(iv) In the titration, the uncertainty in each burette reading is ± 0.05 cm3.

Calculate the percentage uncertainty in the titre for Titration 1.

percentage uncertainty = … % [1]

(c) Calculate the value of n in HOOC(CH2)nCOOH.

Give your answer to the nearest whole number.

n = … [5]

(d) A 25.0 cm3 pipette was used to measure out the 25.0 cm3 of glutaric acid solution for each

titration.

Before use, one student washed the pipette out with water instead of the glutaric acid solution.

State the effect of this mistake on the titre.

Explain your answer.

Effect …

Explanation …

[2]

Mark scheme

Show the mark scheme The official mark scheme showing answers and guidance for all parts of question 3. Part (a) awards 1 mark for 'colourless to pink'. Part (b)(i) awards 1 mark for completed titres. Part (b)(ii) awards 1 mark for estimating the titre. Part (b)(iii) awards 1 mark for the mean titre value of 18.25 cm3. Part (b)(iv) awards 1 mark for percentage uncertainty calculation (0.55%). Part (c) awards 5 marks for calculating the moles and molar mass to find n = 3. Part (d) awards 2 marks for stating the titre would be less and explaining why the glutaric acid solution becomes more dilute.

How to answer it

Glutaric Acid Titration Analysis

What this question tests

This multi-step question assesses core AS practical chemistry and stoichiometry skills. You are tested on experimental technique (titre data processing, indicator colour changes, and titration errors), percentage uncertainty calculations, and multistep moles calculations to determine an unknown formula component (n).

Part (a)

Indicator Colour Change

✅ Correct Answer

From colourless to pink

💡 Key Knowledge

Phenolphthalein is colourless in acidic solutions and pink/magenta in alkaline solutions. Because the acid is in the conical flask and alkali is added from the burette, the titration starts acidic and ends slightly alkaline.

❌ Common Errors

Reversing the colours (writing "pink to colourless") by forgetting that alkali is running into the acid flask rather than the other way around.

Part (b)

Titration Data Processing & Uncertainty

(i) Completing the Table

✅ Correct Answers

Trial: 18.50 | Titration 1: 18.05 | Titration 2: 18.20 | Titration 3: 18.30

🧠 Exam Technique

All burette readings and titres must be recorded to 2 decimal places, ending in either .0 or .5 (reflecting the true precision of a school burette).

(ii) & (iii) Trial Titres and Mean Titre

✅ Correct Answers

(ii) To estimate the titre (or get a rough idea of the end-point volume).

(iii) Mean titre = 18.25 cm³

🧠 Exam Technique for Mean Titres

Always select only the concordant titres (titres within 0.10 cm³ of each other) to calculate your mean. Here, titrations 2 ( 18.20 ) and 3 ( 18.30 ) average to 18.25 cm³ .

(iv) Percentage Uncertainty

📐 Calculation Steps

  1. Identify total error: A burette uses two readings (initial and final), doubling the uncertainty: 0.05 × 2 = ±0.10 cm³.
  2. Apply formula: Percentage uncertainty = (Absolute uncertainty / Measured value) × 100
  3. Calculation: ( 0.10 / 18.05 ) × 100 = 0.55 %

❌ Common Errors

Forgetting to multiply the burette uncertainty by 2. A burette reading is a difference between two points, so both points carry potential error.

Part (c)

Calculating the Value of n

📐 Step-by-Step Calculation

  1. Moles of NaOH:
    n(NaOH) = (18.25 × 0.240) / 1000 = 4.38 × 10⁻³ mol
  2. Moles of Glutaric Acid in 25 cm³:
    Using the 1:2 stoichiometric ratio from the equation, divide NaOH moles by 2:
    n(acid in 25 cm³) = 4.38 × 10⁻³ / 2 = 2.19 × 10⁻³ mol
  3. Moles of Glutaric Acid in whole 250 cm³ flask:
    Multiply by factor of 10 (250 / 25):
    n(acid in 250 cm³) = 2.19 × 10⁻³ × 10 = 2.19 × 10⁻² mol
  4. Molar Mass of Glutaric Acid:
    M = mass / moles = 2.891 / (2.19 × 10⁻²) = 132.0 g mol⁻¹
  5. Determine value of n:
    Formula is HOOC(CH₂)ₙCOOH.
    Total Mr = (2 × 12.0) + (4 × 1.0) + (4 × 16.0) + n(12.0 + 2.0) = 90 + 14n
    132 - 90 = 42
    n = 42 / 14 = 3

💡 Examiner Guidance

Top-level responses laid out their stoichiometry clearly with explicit subheadings or labels for each step. Examiners allowed error carried forward (ECF) from minor arithmetic slips in earlier parts.

❌ Common Calculation Traps

Forgetting to account for the 1:2 acid-to-alkali ratio in the balanced equation, or failing to scale up from the 25 cm³ pipette sample to the 250 cm³ volumetric flask.

Part (d)

Practical Error Analysis

✅ Correct Answer

Effect: The titre would be less (smaller volume).

Explanation: Washing the pipette with water instead of glutaric acid solution dilutes the acid, making it less concentrated. Therefore, fewer moles of acid are delivered into the conical flask, requiring less NaOH to neutralise it.

🧠 Exam Technique

Always split your answer into two distinct parts when an exam question asks "State the effect..." followed by "Explain your answer...". Never assume dilution effects are intuitive—always explicitly link dilution to fewer moles and therefore a smaller required neutralisation volume.

Topics

Module 1: Development of practical skills in chemistry · Module 2: Foundations in chemistry · Practical Activity Groups · 1.1 Practical skills assessed in a written examination · PAG 2: Acid-base titration · 2.1 Atoms and reactions

Question and mark scheme from the OCR A-Level Chemistry examination, AS Depth in chemistry (02), November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.