AQA A-Level Chemistry Paper 3, 2021: Question 1

20 marks · Hard difficulty · Long Answer

Polyesters, Titration & Thermodynamics with Ethanedioic Acid and its Salts This question covers the formation of polyesters from ethanedioic acid and propane-1,3-diol, including structural repetition. It explores biodegradability of polymers, calculation of concentrations using redox titrations involving sodium ethanedioate and potassium manganate(VII), correct use of lab glassware, and health and safety precautions. Finally, students must explain, using entropy and enthalpy changes, why ligand substitution reactions involving ethanedioate ions are thermodynamically favourable.

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AQA A-Level Chemistry Paper 3, 2021: Question 1
Question text

01 This question is about ethanedioic acid (HOOCCOOH) and the

ethanedioate ion (−OOCCOO−).

01.1 Ethanedioic acid reacts with propane-1,3-diol (HOCH2CH2CH2OH) to form a polyester.

Draw the repeating unit of this polyester.

[2 marks]

01.2 Explain why polyesters are biodegradable but polyalkenes are not biodegradable.

[2 marks]

01.3 Sodium ethanedioate is used to find the concentration of solutions of

potassium manganate(VII) by titration. The equation for this reaction is

2 MnO – + 16 H+ + 5 C O 2– → 2 Mn2+ + 8 H O + 10 CO

42 4 2 2

A standard solution is made by dissolving 162 mg of Na2C2O4 (Mr = 134.0) in water

and making up to 250 cm3 in a volumetric flask.

25.0 cm3 of this solution and an excess of sulfuric acid are added to a conical flask.

The mixture is warmed and titrated with potassium manganate(VII) solution.

The titration is repeated until concordant results are obtained.

The mean titre is 23.85 cm3

Calculate the concentration, in mol dm–3, of the potassium manganate(VII) solution.

[4 marks]

Concentration4 mol dm−3

01.4 Figure 1 shows the 25.0 cm3 pipette used to measure the

sodium ethanedioate solution.

*03* Figure 1

On Figure 1, draw the meniscus of the solution when the pipette is ready to transfer

25.0 cm3 of the sodium ethanedioate solution.

[1 mark]

01.5 Potassium manganate(VII) is oxidising and harmful.

Sodium ethanedioate is toxic.

Suggest safety precautions, other than eye protection, that should be taken when:

• filling the burette with potassium manganate(VII) solution

• dissolving the solid sodium ethanedioate in water.

[2 marks]

Filling the burette

Dissolving the solid

01.6 State the colour change seen at the end point of each titration.

[1 mark]

01.7 Figure 2 shows the burette containing potassium manganate(VII) solution.

Figure 2

Give two practical steps needed before recording the initial burette reading.

[2 marks]

01.8 When Na C O (aq) is added to a solution containing [Fe(H O) ]3+ ions, a reaction

22 4 2 6

occurs in which all six water ligands are replaced by ethanedioate ions.

Explain why the replacement of the water ligands by ethanedioate ions is favourable.

In your answer refer to:

• the enthalpy and entropy changes for the reaction

• how the enthalpy and entropy changes influence the free-energy change for the

reaction.

[6 marks]

Mark scheme

Show the mark scheme Mark scheme for AQA A-Level Chemistry Paper 3, 2021: Question 1

Question Answers Additional comments/Guidelines Mark

1.1 ignore brackets and ‘n’

allow (CH2)3

−O− at either end but not both

M1 ester link including C−O−C 1

M2 rest of structure including trailing bonds not M2 if more than one repeating unit 1

allow for one mark −OOCCOOCH2CH2CH2− as

long as trailing bonds included

1.2 polyesters: C=O/C−O OR polar bonds / chain AND not just ‘polyesters are polar’

polyalkenes: (only) C−C OR non-polar bonds / chain not M1 if C=C mentioned 1

(polyesters) susceptible to nucleophilic attack / can be hydrolysed 1

1.3 0.162 1

M1 amount of Na2C2O4 = = 0.00121 mol

134.0

2 −4

M2 stoichiometry ( ) (4.84 x 10 ) 2 1

5 M1 x

M3 scaling (÷10) M2 ÷ 10 (conc/40) 1

2 -5

= 0.00121 x ÷ 10 = 4.84 x 10 mol

5 M3 x 1000 1

M4 concentration of MnO – =

4 23.85

4.84 x 10−5

= 0.00203 mol dm−3

23.85 Min 2 sig figs

1000

– A-LEVEL CHEMISTRY – –

Question Answers Additional comments/Guidelines Mark 11

1.4 Meniscus curved with the bottom of the curve on 1

the horizontal line

1.5 (burette) fill below/at eye level ignore make sure tap closed / funnel / gloves 1

(solution) wear gloves allow wash/rinse hands after any spillage 1

not fume cupboard

ignore lab coat / stir carefully

1.6 colourless to pink/pale purple not just purple 1

not ‘clear’ for ‘colourless’

– A-LEVEL CHEMISTRY – –

1.7 remove funnel 1

ensure jet is filled / no (air) bubbles allow open tap to fill space below tap 1

1.8 This question is marked using Levels of Response. Refer to the Mark Stage 1 - H 6

Scheme Instructions for Examiners for guidance. 1a H negligible

Level 3 All stages are covered and each stage is generally correct 1b make & break same number of bonds

5-6 marks and virtually complete. 1c make & break same type of bonds /

Answer is communicated coherently and shows a logical

progression from Stage 1 to Stages 2 and 3 bonds have similar enthalpies

Covers at least 2 point for stage 1, 1 for stage 2 and 2 for

stage 3. Stage 2 - S

If given equation must show correct stoichiometry for six 2a increase in entropy

marks 2b increase in particles in solution / from 4 to

Level 2 All stages are covered but stage(s) may be incomplete or 7 particles (ecf from incorrect equation

3-4 marks may contain inaccuracies showing increase in no. of moles)

OR two stages are covered and are generally correct and

virtually complete. Stage 3 - G

Answer is communicated mainly coherently and shows a

logical progression from Stage 1 to Stages 2 and 3. 3a G = H –T S

3b G negative (for forward reaction)

3c correct discussion of why G is negative

Level 1 Two stages are covered but stage(s) may be incomplete based on H and T S

1-2 marks or may contain inaccuracies

OR only one stage is covered but is generally correct and

virtually complete.

Answer includes isolated statements but these are not

presented in a logical order.

0 mark Insufficient correct chemistry to gain a mark.

How to answer it

Study Guide: Polyesters, Redox Titrations and Ligand Substitution

Q1. Drawing the Repeating Unit of the Polyester

Key idea: Join ethanedioic acid and propane-1,3-diol by forming two ester links (–COO–)

Each link comes from an –OH on the diol and a –COOH on the dicarboxylic acid.

Avoid drawing more than one repeat unit. Include trailing bonds and only one full unit!

Q2. Why Are Polyesters Biodegradable but Polyalkenes Aren't?

Polyesters: Contain polar C=O and C–O bonds which are susceptible to nucleophilic attack and hydrolysis.

Polyalkenes: Only contain non-polar C–C and C–H bonds , which do not hydrolyse.

Don’t just say “polyesters are polar” – specify the bond types and their role in hydrolysis.

Q3. Titration: Calculating the Concentration of KMnO₄

Steps:

  1. Calculate moles of Na2C2O4: 0.162 ÷ 134.0 = 0.00121 mol
  2. Use the ratio from the equation: MnO₄⁻ : C₂O₄²⁻ = 2:5 → moles MnO₄⁻ = 0.00121 × (2/5)
  3. Scale up to 250 cm³ sample from 25 cm³: multiply by 10
  4. Use volume from mean titre (23.85 cm³) to find concentration:

Answer: 0.00203 mol dm–3

Watch your stoichiometry, volume scaling, and significant figures!

Q4. Drawing the Meniscus

The bottom of the curved meniscus must sit on the graduation line .

Almost 25% of students drew this incorrectly. Always place the lowest point of the curve on the line.

Q5. Safety Precautions

When filling the burette: Do it below eye level .

When handling ethanedioate: Wear gloves to avoid skin contact as it’s toxic.

Also good to rinse after spills, but gloves alone are sufficient for this question.

Q6. Colour Change at End Point

Colour change: Colourless to pink/pale purple

“Just purple” or “clear” aren't enough. You must give the full before-and-after change.

Q7. Preparing a Burette for Use

Steps:

  1. Remove the funnel from the top of the burette.
  2. Ensure the jet is filled – let solution run through before taking a reading.

Q8. Explaining Why Ligand Exchange Is Favourable

Step 1 – ΔH: The enthalpy change is close to zero because 6 Fe–O bonds are broken and 6 new Fe–O bonds are formed.

Step 2 – ΔS: Entropy increases – going from 4 species (reactants) to 7 species (products).

Step 3 – ΔG: ΔG = ΔH – TΔS → Since ΔS is positive and ΔH is ≈ 0, ΔG is negative , so the reaction is feasible.

Link the bond-making/breaking idea to ΔH ≈ 0. You need all three ideas (ΔH, ΔS, and ΔG) clearly explained for Level 3.

Topics

Organic Chemistry · Inorganic Chemistry · Physical Chemistry · 3.3.12 Polymers · 3.2.5 Transition Metals · 3.1.8 Thermodynamics · 3.1.2 Amount of Substance

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.