OCR A-Level Chemistry Unified chemistry (03), June 2023: Question 5
11 marks · Hard difficulty · Structured Questions
Determine the mass of aspirin in a tablet using back-titration data, draw a reflux apparatus, and complete a chemical equation for the reaction of aspirin with hot sodium hydroxide.
Practise this questionQuestion
Question text
5 Aspirin tablets are used for pain relief.
The structure of aspirin is shown below.
O OH
O
O
Aspirin
(a) A student uses the reaction of aspirin with cold NaOH(aq) to determine the mass of aspirin in
one tablet.
In this reaction, 1 mol of aspirin reacts with 1 mol of cold NaOH(aq).
The student’s method is outlined below.
Step 1 The student reacts three aspirin tablets with 100 cm3 of 0.500 mol dm–3 NaOH(aq).
The NaOH is in excess. A colourless solution forms.
Step 2 The colourless solution from Step 1 is made up to 250.0 cm3 with distilled water.
Step 3 A 25.00 cm3 sample of the diluted solution from Step 2 is titrated with
0.200 mol dm–3 HCl(aq) in the burette.
The HCl(aq) reacts with excess NaOH(aq) that remains in Step 1:
NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l)
The student repeats the titration to obtain concordant (consistent) titres.
Titration results
The trial titre has been omitted.
The burette readings have been read to the nearest 0.05 cm3.
12 3
Final reading / cm3 23.10 45.40 27.40
Initial reading / cm3 0.00 23.10 5.00
Analysis of results
From the results, the student can determine the following.
1. The amount, in mol, of excess NaOH(aq) that remains after the reaction of aspirin with
NaOH(aq).
2. The amount, in mol, of NaOH(aq) that reacted with the aspirin.
Use the results to determine the mass, in mg, of aspirin in one aspirin tablet.
mass of aspirin in one tablet = … mg [6]
(b) Aspirin reacts with hot NaOH(aq), under reflux.
(i) Draw a labelled diagram of suitable apparatus for reflux.
[2]
(ii) In this reaction, 1 mol of aspirin reacts with 3 mol of hot NaOH(aq).
Complete the equation for the reaction of aspirin with an excess of hot NaOH(aq).
Show structures for organic compounds.
O OH
O
+ …
O
[3]
Mark scheme
Show the mark scheme
AO
Question Answer Marks Guidance
element
5 (a) FIRST CHECK THE ANSWER ON ANSWER LINE 6 AO2.8 FULL ANNOTATIONS MUST BE USED
If Mass = 318 (mg) award 6 marks 6 ----------------------------------------------------------------
---------------------------------------------------------------------------- Common error:
Mean titre 1 mark Incorrect mean from all 3 titres = 22.6 cm3
(22.30 + 22.40) 3 CHECK BELOW TITRATION TABLE
= 2 = 22.35(0) (cm ) ✓
Analysis of results 5 marks Use ECF throughout
22.35 –3 Intermediate values for working to at least 3 SF.
n(HCl) = 0.200 1000 = 4.47 10 (mol) ✓
n(NaOH) remaining in 25.0 cm3 = n(HCl)
n(NaOH) remaining in 250 cm3
= 4.47 10–3 10 = 4.47 10–2 OR 0.0447 (mol) ✓
n(NaOH) that reacted with aspirin
= 0.0500 – 4.47 10–2 = 5.30 10–3 (mol) ✓ ALLOW scaling for 1 aspirin tablet early in calc,
e.g. for final 2 marks:
5.30 10–3
mass in 3 tablets = 5.30 10–3 180 = 0.954 g ✓ n(aspirin) in 1 tablet = = 1.77….. 10–3 (mol) ✓
Mass in 1 tablet = 1.77….. 10–3 180 = 0.318 g
Mass in 1 tablet = 318 mg ✓ = 318 mg ✓
COMMON ERRORS: No subtraction from 0.05 5 marks
No scaling 10 → 4.47 10–2 180 → 8.046 → 2682/2680 mg in 1 tablet
0.05 – 4.47 10–3 → 4.553 10–2 ✓ -------------------------------------------------
4.553 10–2 180 → 8.1954 g in 3 tablets ✓ Omitting initial titration calculation 2 marks
→ 2731.8/2732/2730 mg in 1 tablet ✓ 5 marks 0.05 180 → 9 g in 3 tablets ✓ → 3000 mg in 1 tablet ✓
------------------------------------------------- -------------------------------------------------
No scaling 10 before subtraction but scaling after Mean of 22.60 (use of all 3 titres) 5 marks
4 marks Mean = 67.8/3 = 22.60 ✘→ 4.52 10–3 ✓ 10 → 4.52 10–2 ✓
0.05 – 4.47 10–3 → 4.553 10–2 ✓ 0.05 – 4.52 10–2 → 4.80 10–3 ✓
4.553 10–2 10 180 → 81954 g in 3 tablets 4.80 10–3 180 → 0.864 g in 3 tablets ✓
→ 27318 / 27320 / 27300 mg in 1 tablet ✓ → 288 mg in 1 tablet ✓
20 AO
element
(b) (i) 2 AO3.3 For open system,
2 DO NOT ALLOW
Water out
Condenser
Water in
For open system, ALLOW label. e.g. ‘open at top’
Pear-shaped/
Round-bottom
flask
Heat
Reaction apparatus (Labels NOT required)
flask
AND upright condenser
AND open system at top ✓ (Could be labelled) ALLOW line across flask
Labels AND direction of water flow
Pear-shaped/round-bottom flask
AND condenser
AND water in at bottom and out at top
Heat NOT required
DO NOT ALLOW flask, conical flask, volumetric flask
DO NOT ALLOW thermometer
DO NOT ALLOW condensing tube as label
21 AO
element
ALLOW small gap between flask and condenser
BOD, e.g.
If in doubt, ask Team Leader
AO
element
(b) (ii) O OH 3 ALLOW any combination of skeletal OR structural
AO2.6 OR displayed formula as long as unambiguous
O IGNORE annotations of provided structure of
+ 3NaOH
aspirin at top left
O ALLOW equation with 3OH– OR 3NaOH giving
22 anions for organic products, i.e.
i.e.
O OH O O–
O ONa
O O–
+ 3OH– –
+ CH3COO + 2H2O
ONa O
+ CH3COONa + 2H2O
OR
O OH O O–
O O–
+ 3NaOH + CH COO– + 2H O
Organic products ✓ ✓ 2 marks 3 + 3Na + 2
O
3NaOH AND 2H2O ✓ 1 mark
ALLOW 1 of the 2 organic products mark for BOTH
NOTE: ALLOW O–Na+ for ONa throughout structures as COOH and OH (or mixture) e.g
O OH
SCROLL DOWN FOR PRODUCTS
OH
+ CH3COOH
O ONa
OH
+ CH3COOH
How to answer it
Aspirin Analysis & Hydrolysis Study Guide
What this question tests
This question assesses multi-step back-titration calculations involving moles, concentrations, dilutions, and stoichiometric ratios, alongside practical organic chemistry techniques (reflux apparatus setup) and organic reaction mechanisms involving the alkaline hydrolysis of esters and carboxylic acids.
Question (a): Back-Titration Analysis of Aspirin Tablets
Determining the mass of aspirin per tablet using back-titration data [6 Marks]
✅ Final Correct Answer
Mass of aspirin in 1 tablet = 318 mg (or 0.318 g )
💡 Key Knowledge
- Back-titrations: Used when a reaction is too slow or involves an insoluble solid (like aspirin). Excess reactant is added, and the unreacted excess is titrated.
- Concordant Titres: Only use titres within 0.10 cm³ of each other to calculate the mean. Here, titres 2 and 3 ( 22.30 cm³ and 22.40 cm³ ) give a mean of 22.35 cm³ . Titre 1 is the rough titre and must be omitted.
🧠 Exam Technique
Always work methodically through back-titrations using sequential sub-steps: find moles of titrant used, scale up for dilutions, find unreacted moles, subtract from initial moles to find reacted moles, use stoichiometry, and finally convert moles to mass with correct units (mg vs g).
❌ Common Errors
- Ignoring dilution factors: Forgetting to scale up the titration sample ( 25.0 cm³ out of 250 cm³ is a factor of 10 ).
- Titration errors: Including the rough titre in the mean calculation, or incorrectly reading burettes (omitting initial readings).
- Unit mix-ups: Failing to convert grams to milligrams ( × 1000 ) at the end.
📐 Step-by-Step Calculation Breakdown
- Calculate Mean Titre:
(22.30 + 22.40) / 2 = 22.35 cm³ (1 mark) - Moles of HCl in titre:
0.200 mol dm⁻³ × (22.35 / 1000) = 4.47 × 10⁻³ mol - Moles of excess NaOH in 250 cm³ volumetric flask:
Since 25.0 cm³ contained 4.47 × 10⁻³ mol, scale up by ×10:
4.47 × 10⁻³ × 10 = 4.47 × 10⁻² mol (or 0.0447 mol) - Initial moles of NaOH added in Step 1:
100 cm³ of 0.500 mol dm⁻³ = 0.500 × (100 / 1000) = 0.0500 mol - Moles of NaOH that reacted with aspirin:
Initial NaOH − Excess NaOH = 0.0500 − 0.0447 = 5.30 × 10⁻³ mol - Moles and mass of aspirin in 3 tablets:
1:1 molar ratio means 5.30 × 10⁻³ mol of aspirin in 3 tablets.
Mr of aspirin = 180.0 g mol⁻¹.
Mass in 3 tablets = 5.30 × 10⁻³ × 180 = 0.954 g - Mass in 1 tablet:
0.954 g / 3 = 0.318 g = 318 mg (5 marks for analysis steps)
Question (b)(i): Reflux Apparatus Diagram
Drawing a labelled diagram of suitable apparatus for reflux [2 Marks]
✅ Correct Drawing Requirements
- Apparatus Setup: Pear-shaped or round-bottom flask connected to an upright/vertical Liebig condenser. An open system at the top (must not be sealed).
- Labels & Water Flow: Condenser labelled (or clearly identifiable), water in at the bottom, water out at the top.
❌ Common Errors & Penalties
- Sealed system: Drawing a delivery tube, bung, or closed top on the condenser (instant loss of marks for safety hazard).
- Incorrect glassware: Using a conical flask, volumetric flask, or beaker instead of a round-bottom/pear-shaped flask.
- Water flow reversal: Running water in from the top and out the bottom (leaves air pockets in the condenser jacket, reducing cooling efficiency).
- Labelling errors: Labelling the condenser as a "condensing tube" or including a thermometer.
Question (b)(ii): Alkaline Hydrolysis Equation
Completing the equation for the reaction of aspirin with hot NaOH(aq) [3 Marks]
✅ Correct Equation Products
- Organic Products:
1. Disodium salt of the benzene ring: 2-hydroxybenzenedicarboxylate ion (both −OH and −COOH groups converted to −O⁻ Na⁺ and −COO⁻ Na⁺ respectively).
2. Sodium ethanoate: CH₃COONa (from the ester cleavage). - Inorganic Byproducts: 2H₂O (Note: 3NaOH is shown reacting on the left-hand side).
💡 Key Knowledge
- Hot aqueous sodium hydroxide causes alkaline hydrolysis of both the ester bond and neutralisation of the phenolic −OH and carboxylic acid −COOH groups.
- Stoichiometry: 1 mol of aspirin reacts with 3 moles of NaOH in total.
❌ Common Errors
- Failing to hydrolyse the phenolic −OH group or carboxylic acid group into their respective sodium salt forms (−O⁻Na⁺ and −COO⁻Na⁺).
- Incorrect balance coefficients for NaOH or water byproducts.
Topics
Module 2: Foundations in chemistry · Module 6: Organic chemistry and analysis · Practical Activity Groups · PAG 2: Acid-base titration · PAG 5: Synthesis of an organic liquid · 6.1 Aromatic compounds, carbonyls and acids · 2.1 Atoms and reactions
Question and mark scheme from the OCR A-Level Chemistry examination, Unified chemistry (03), June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.