OCR A-Level Chemistry Unified chemistry (03), June 2018: Question 3
10 marks · Medium difficulty · Structured Questions
Draw 3D optical isomers for the iron(III) ethanedioate complex ion, state the colour change for a cerium(IV) redox titration, and calculate the percentage by mass of ethanedioic acid in rhubarb leaves.
Practise this questionQuestion
Question text
3 This question is about ethanedioic acid, (COOH) , and ethanedioate ions, (COO–) .
(a) The ethanedioate ion, shown below, can act as a bidentate ligand.
O O
–O O–
Fe3+ forms a complex ion with three ethanedioate ions.
The complex ion has two optical isomers.
Draw the 3D shapes of the optical isomers.
In your diagrams, show the structure of the ethanedioate ligands and any overall charge.
[3]
(b) Ethanedioic acid, (COOH)2, is present in rhubarb leaves.
A student carries out a redox titration using aqueous cerium(IV) sulfate, Ce(SO4)2(aq), to
determine the percentage, by mass, of ethanedioic acid in rhubarb leaves.
In the titration, Ce4+(aq) ions oxidise ethanedioic acid in hot acid conditions:
2Ce4+(aq) + (COOH) (aq) → 2Ce3+(aq) + 2CO (g) + 2H+(aq)
Ce4+(aq) ions have a yellow colour. Ce3+(aq) ions are colourless.
The student weighs 82.68 g of rhubarb leaves and extracts ethanedioic acid from the leaves.
The ethanedioic acid is added to dilute sulfuric acid to form a colourless solution which is
made up to 250.0 cm3 with distilled water.
The student heats 25.00 cm3 of this solution to 70 °C and titrates this volume with
0.0500 mol dm–3 Ce(SO ) from the burette.
The student repeats the titration to obtain concordant (consistent) titres.
Titration results
The trial titre has been omitted.
12 3
Final reading/cm3 24.30 47.80 23.65
Initial reading/cm3 1.05 24.30 0.50
(i) This titration is self-indicating and the student does not need to add an indicator.
What colour change would the student observe at the end point?
Colour change from … to … [1]
(ii) Calculate the percentage, by mass, of ethanedioic acid in the rhubarb leaves.
Give your answer to an appropriate number of significant figures.
percentage of ethanedioic acid = … % [6]
Mark scheme
Show the mark scheme
Question Answer Marks Guidance
3 (a) Overall 3– charge shown (outside brackets) for at least ONE 3
isomer ALLOW –3 for 3–
3– must apply to the overall charge of structures
IGNORE charges or dipoles on atoms within diagrams
O O (even if wrong)
3– 3–
O O
O O Square brackets NOT required
O O O O O O -------------------------------------------------------
Fe Fe 3D
O O Must contain 2 ‘out wedges’, 2 ‘in wedges’ and 2 lines
O O
O O O O in plane of paper OR 4 lines, 1 ‘out wedge’ and 1 ‘in
O O wedge’:
O O
For bond into paper, ALLOW:
1 mark for each isomer
Bonds must go to O ligand atoms on EACH structure
ALLOW following geometry throughout:
ALLOW unambiguous structures; ethanedioate ions can
include C atoms
Fe
For other structures that might be creditworthy, contact TL
NOT ALLOW structures showing a simplified loop for
ethanedioate ligands
e.g.
O O
(b) (i) Colourless to yellow 1 IGNORE clear for colourless
(b) (ii) Mean titre 1 mark 6 Common error:
(23.15 + 23.25) 11 Incorrect mean from all 3 titres = 23.30 cm3
= = 23.2(0) (cm3)
Use ECF throughout
Analysis of results 5 marks Intermediate values for working to at least 3 SF.
4+ 0.0500 –3 TAKE CARE as value written down may be truncated
n(Ce ) = 23.20 1000 = 1.16 10 (mol) value stored in calculator. Depending on rounding,
1.16 10–3 either can be credited.
n((COOH) ) in 25.0 cm3 = = 5.8(0) 10–4 (mol)
-----------------------------------------------------------------
n((COOH) ) in 250 cm3 COMMON ERRORS:
= 5.8(0) 10–4 10 = 5.8(0) 10–3 (mol) Mean of 23.30 (use of all 3 titres)
0.634%: 5 marks
Mass (COOH) = 5.8(0) 10–3 90.0 = 0.522 g
TAKE CARE for final answer of 0.63 seen.
0.522 100 No final mark as only 2 SF
% oxalic acid = 82.68 = 0.631% 0.63 may have been rounded from 0.631
Percentage MUST be expressed to 3 SF (from correct mean)
OR from 0.634 (using mean from all 3 titres)
Check back to mean titre.
No ÷2 to obtain n((COOH)2)
1.26%: 5 marks from 23.20
1.27% 4 marks from 23.30
Total 10
How to answer it
Transition Metals & Redox Titrations Study Guide
What this question tests
This multi-part question assesses your understanding of transition metal stereoisomerism (specifically optical isomerism in octahedral complexes with bidentate ligands) and your ability to execute multi-step redox titration calculations involving percentage by mass determinations.
Drawing Optical Isomers of Octahedral Complexes
💡 Key Knowledge
- Bidentate ligands: Ethanedioate ions (C₂O₄²⁻) form two coordinate bonds per ligand. Three bidentate ligands give a coordination number of 6 (octahedral geometry).
- Optical Isomerism: Occurs when a complex is non-superimposable on its mirror image (enantiomers). For [Fe(C₂O₄)₃]³⁻, this exists as a pair of mirror-image optical isomers (cis- or ^+/^- forms).
🧠 Exam Technique
- Draw a clear 3D octahedral framework using standard bonds (2 solid lines, 2 wedges, 2 dashes).
- Connect the bite of the ethanedioate ligand clearly to two adjacent or specific positions around the central metal ion (Fe³⁺).
- Crucial: Ensure bonds terminate clearly at the oxygen (O) atoms of the ligand, not just the carbon backbone.
- Show the overall 3- charge outside square brackets enclosing each isomer.
✅ Correct Answer Requirements
- Overall 3- charge clearly shown outside brackets for at least one isomer (allow "3-").
- Two valid 3D mirror-image structures showing octahedral coordination with three bidentate ligands.
- Correct bonding connectivity from Fe to O atoms. Avoid simplified horseshoe loops.
❌ Common Errors
- Omitting overall charge brackets or writing the charge incorrectly inside.
- Drawing simplified loop/arc symbols for the ethanedioate ligands instead of chemical structures or clear bidentate attachments.
- Incorrect 3D stereochemical representation (failing to use proper wedges/dashes for 3D perspective).
Self-Indicated Redox Titration End Point
✅ Correct Answer
Colour change from colourless to yellow (or pale yellow).
💡 Key Knowledge
- Cerium(IV) ions, Ce⁴⁺(aq), are yellow, while cerium(III) ions, Ce³⁺(aq), are colourless.
- Ethanedioic acid and manganese/cerium titration mixtures often act as "self-indicating" systems due to the distinct colours of the oxidation states of the transition metal reagent added from the burette.
❌ Common Errors
- Reversing the colour change (stating yellow to colourless), forgetting that Ce⁴⁺ is added from the burette into a colourless analyte solution until the first drop of excess Ce⁴⁺ persists.
Percentage by Mass Calculation
📐 Step-by-Step Calculation Guide
- Select concordant titrations: Titres 2 and 3 are 23.50 cm³ and 23.15 cm³? Let's check: Titre 1 = 24.30 - 1.05 = 23.25 cm³.
Titre 2 = 47.80 - 24.30 = 23.50 cm³.
Titre 3 = 23.65 - 0.50 = 23.15 cm³.
Wait, let's use the mark scheme mean: Titres 1 and 3 give (23.25 + 23.15) / 2 = 23.20 cm³ (Titre 2 at 23.50 is rough/non-concordant).
Mean titre = 23.20 cm³ . - Calculate moles of Ce⁴⁺ reacted:
n(Ce⁴⁺) = (23.20 / 1000) × 0.0500 = 1.16 × 10⁻³ mol . - Use stoichiometry to find moles of ethanedioic acid in 25.0 cm³:
From equation: 2Ce⁴⁺ : 1(COOH)₂
n((COOH)₂) in 25.0 cm³ = (1.16 × 10⁻³) / 2 = 5.80 × 10⁻⁴ mol . - Scale up to the total volumetric flask volume (250.0 cm³):
n((COOH)₂) in 250 cm³ = 5.80 × 10⁻⁴ × (250.0 / 25.0) = 5.80 × 10⁻³ mol . - Calculate mass of ethanedioic acid, (COOH)₂ (Mr = 90.0):
Mass = 5.80 × 10⁻³ × 90.0 = 0.522 g . - Calculate percentage by mass in original rhubarb leaves (82.68 g):
Percentage = (0.522 / 82.68) × 100 = 0.631% . (Expressed to 3 significant figures).
❌ Common Calculation Traps
- Mean Titre Error: Including all 3 titres (including the non-concordant 23.50 cm³ value) gives an incorrect mean of 23.30 cm³, leading to an incorrect final answer of 0.634%.
- Stoichiometry Ratio: Forgetting to divide by 2 for the 2:1 mole ratio between Ce⁴⁺ and (COOH)₂.
- Significant Figures: Giving the final answer to 2 SF (e.g., 0.63%) loses the final mark; ensure answers match appropriate precision (3 SF).
🧠 Examiner Top Tips
- Keep intermediate values to at least 3 significant figures in your calculator to avoid rounding errors.
- Clearly show scale-up factors (e.g., multiplying by 10 for the volumetric flask dilution).
- Always state final percentage values clearly with appropriate units/symbols.
Topics
Module 5: Physical chemistry and transition elements · Module 2: Foundations in chemistry · Practical Activity Groups · 5.3 Transition elements · 2.1 Atoms and reactions · PAG 2: Acid-base titration
Question and mark scheme from the OCR A-Level Chemistry examination, Unified chemistry (03), June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.