Edexcel A-Level Chemistry AS Paper 2, June 2016: Question 8

14 marks · Medium difficulty · Calculations

Prepare ethanedioic acid from ethane-1,2-diol, describe the preparation of a standard solution, state indicator colour changes, calculate the value of n in ethanedioic acid crystals from titration data, and explain the effect of damp crystals.

Practise this question

Question

Question 8 features parts a and b about ethanedioic acid. Part (a) shows a reaction scheme converting ethane-1,2-diol to ethanedioic acid with chemical structures and asks for reagents and conditions. Part (b) outlines a titration procedure using ethanedioic acid crystals and sodium hydroxide with phenolphthalein indicator, asking students to describe making a standard solution, state indicator colour changes, calculate water of crystallisation n, and explain the effect of damp crystals.
Question text

8 Ethanedioic acid has two carboxylic acid groups.

(a) Ethanedioic acid, H2C2O4, can be prepared from ethane-1,2-diol.

H H

O O

H O C C O H C C

H H HO OH

Give the reagents and condition required for this reaction.

(2)

Reagents …

Condition …

(b) The formula for ethanedioic acid crystals is H2C2O4.nH2O.

To determine the number of moles of water of crystallisation, n, in 1 mol of

ethanedioic acid crystals, a student carried out the following procedure.

• Prepare 250.0cm3 of a solution containing a known mass of about 1g of

ethanedioic acid crystals.

• Titrate 25.0cm3 portions of the ethanedioic acid solution with 0.103 mol dm−3

sodium hydroxide solution, using phenolphthalein as indicator.

The student obtained these results:

mass of ethanedioic acid crystals = 1.09g

mean titre = 16.20cm3

The equation for the reaction is

H2C2O4 + 2NaOH o Na2C2O4 + 2H2O

(i) Describe how the student should prepare the 250.0 cm3 of ethanedioic acid

solution.

(4)

… *P49838A02428*

(ii) Give the colour change at the end-point in this titration.

(1)

From … to …

(iii) Calculate a value of n in the formula H2C2O4.nH2O from these data.

(5)

(iv) The student thought that the ethanedioic acid crystals used may have been

slightly damp.

Explain the effect of using damp crystals on the titre and on the value of n.

(2)

(Total for Question 8 = 14 marks)

Mark scheme

Show the mark scheme Mark scheme for question 8 detailing acceptable answers and marks for reagents (acidified potassium dichromate and reflux), preparation of standard solution steps, colour change from colourless to pink, multi-step titration calculation to find n, and explanation of damp crystals increasing titre and decreasing calculated n.

How to answer it

Preparation and Titration of Ethanedioic Acid

What this question tests

This question assesses core organic oxidation reactions (converting a diol to a carboxylic acid), standard laboratory techniques for preparing a standard solution, indicator colour changes, multi-step stoichiometry calculations involving water of crystallisation, and the evaluation of experimental errors.

Part (a) • 2 Marks

Oxidation of Ethane-1,2-diol

✅ Correct Answer

  • Reagents: Potassium dichromate(VI) ( K₂Cr₂O₇ ) AND sulfuric acid ( H₂SO₄ ). (Both name and formula must match if both are given, or use correct ions: Cr₂O₇²⁻ and H⁺ ).
  • Condition: Heat under reflux .

❌ Common Errors

  • Forgetting to include the acid alongside potassium dichromate.
  • Suggesting hydrochloric or nitric acid (only sulfuric acid is suitable as it does not participate in unwanted redox reactions here).
  • Omitting "reflux" or stating simple distillation, which would fail to fully oxidise the intermediate groups to carboxylic acids.
Mark breakdown: 1 mark for correct oxidising agent system; 1 mark for heating under reflux.
Part (b)(i) • 4 Marks

Preparation of a Standard Solution

💡 Key Knowledge: 4-Step Preparation Method

  1. Weighing: Weigh the ethanedioic acid crystals in a suitable container (e.g. a weighing boat) on a balance, and record the exact mass. Accurately transfer and re-weigh (or wash container).
  2. Dissolving: Dissolve the solid in a beaker using a smaller volume of distilled/deionised water (never tap water). Stir with a glass rod.
  3. Transfer & Washings: Transfer the solution completely into a 250 cm³ volumetric flask . Rinse the beaker and glass rod with distilled water and add washings into the flask.
  4. Make up to mark & Mix: Carefully add distilled water dropwise until the bottom of the meniscus sits exactly on the graduation mark. Stopper the flask and invert/swirl to mix thoroughly.

❌ Common Errors

  • Mentioning a standard conical flask or beaker instead of a volumetric flask for making up the standard volume.
  • Failing to specify distilled or deionised water.
  • Forgetting to rinse the weighing container and glass rod, leaving residual solute behind.
Mark breakdown: 1 mark per key operational checkpoint: (1) Volumetric flask, (2) Accurate weighing/recording mass, (3) Dissolving and rinsing into flask, (4) Making up to mark and mixing.
Part (b)(ii) • 1 Mark

Indicator Colour Change

✅ Correct Answer

  • From colourless to pink (or red).

❌ Common Errors

  • Stating the reverse ("pink to colourless"). Remember: the acid (ethanedioic acid) is in the conical flask with phenolphthalein (making it colourless initially), and NaOH is added from the burette until the first permanent pink coloration appears.
  • Using forbidden terms like "clear" instead of "colourless".
Mark breakdown: 1 mark for correct sequence (colourless to pink/red).
Part (b)(iii) • 5 Marks

Calculation of Water of Crystallisation (n)

📐 Step-by-Step Calculation Guide

  1. Calculate moles of NaOH used in the titration:
    Moles = (Volume in cm³ ÷ 1000) × Concentration
    Moles = (16.20 ÷ 1000) × 0.103 = 1.6686 × 10⁻³ mol
  2. Determine moles of H₂C₂O₄ in the 25.0 cm³ pipette sample:
    From equation ( H₂C₂O₄ + 2NaOH → Na₂C₂O₄ + 2H₂O ), ratio is 1 mol acid : 2 mol NaOH .
    Moles in 25 cm³ = 1.6686 × 10⁻³ ÷ 2 = 8.343 × 10⁻⁴ mol
  3. Scale up to the full 250 cm³ volumetric flask:
    Moles in 250 cm³ = 8.343 × 10⁻⁴ × (250 ÷ 25) = 8.343 × 10⁻³ mol
  4. Find the molar mass ( Mᵣ ) of the hydrated crystals:
    Mᵣ = Mass ÷ Moles = 1.09 g ÷ 8.343 × 10⁻³ mol = 130.6 g mol⁻¹
  5. Calculate the value of n:
    Formula mass = Mᵣ(H₂C₂O₄) + n × Mᵣ(H₂O)
    130.6 = [ (2 × 1.01) + (2 × 12.01) + (4 × 16.00) ] + n(18.02)
    130.6 = 90.04 + 18.0n
    18.0n = 130.6 - 90.04 = 40.56  →  n = 40.56 ÷ 18 = 2.25 (Accept student variations rounding between 2.2 and 2.3).

🧠 Exam Technique & Consequential Marking

If you make a minor arithmetic slip in earlier steps, subsequent steps use Error Carried Forward (TE). Always write down your intermediate answers clearly so examiners can award method marks.

❌ Common Calculation Traps

  • Forgetting to divide by 2 for the stoichiometry ratio between acid and alkali.
  • Forgetting to multiply by 10 when scaling from the 25 cm³ titration aliquot up to the 250 cm³ flask.
Mark breakdown: 5 marks total across the logical calculation milestones (moles NaOH → moles acid in aliquot → moles in total volume → Mᵣ of crystals → final integer/decimal value of n).
Part (b)(iv) • 2 Marks

Evaluating Experimental Errors

✅ Correct Answer

  • Effect on Titre: The titre will be lower.
  • Effect on n: The calculated value of n will be higher.

💡 Examiner Logic

  • If crystals are damp, they contain extra unbonded water ( H₂O ). Therefore, for a weighed 1.09 g sample, there is actually less pure H₂C₂O₄ present than assumed.
  • Less acid in the flask means fewer moles of NaOH are required to neutralise it, resulting in a smaller (lower) titre.
  • Because calculated moles of acid come out lower, the apparent Mᵣ calculated is artificially inflated, which shifts the subtraction in the formula to yield a higher value for n.
Mark breakdown: 1 mark for explaining lower titre due to lower moles of acid; 1 mark for linking this to a higher calculated value of n.

Topics

Organic Chemistry · Physical Chemistry · Core Practicals · Topic 6: Organic Chemistry I · Topic 5: Formulae, Equations and Amounts of Substance · Core Practical 2: Preparation of a standard solution from a solid acid · Core Practical 3: Find the concentration of a solution of hydrochloric acid

Question and mark scheme from the Edexcel A-Level Chemistry examination, AS Paper 2, June 2016. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.