OCR A-Level Chemistry AS Depth in chemistry (02), June 2025: Question 3

11 marks · Medium difficulty · Structured Questions

Determine the mass of malic acid in a tablet from titration data, evaluate the effect of rinsing a burette with water, and draw the cis and trans stereoisomers formed by dehydration.

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Question

Question text

3 Malic acid tablets are sold as a health supplement.

A student carries out a titration to determine the mass of malic acid, C4H6O5, in one tablet.

The student follows the method below:

Step 1 Crush three tablets, transfer the powder into a beaker and dissolve in distilled water.

Step 2 Transfer the solution into a 250.0 cm3 volumetric flask and make up to the mark with

distilled water.

Step 3 Pipette 25.0 cm3 of the solution from Step 2 into a conical flask and add a few drops of

indicator.

Step 4 Titrate this solution with 0.0800 mol dm–3 sodium hydroxide, NaOH, in the burette.

The equation for the neutralisation reaction is:

C4H6O5 + 2NaOH C4H4O5Na2 + 2H2O

(a) The student takes burette readings to the nearest 0.05 cm3.

The student’s readings are shown in the table.

The rough titre has been omitted.

(i) Complete the table below.

Titration 1 2 3

Final reading / cm3 23.60 46.70 25.65

Initial reading / cm3 0.30 23.65 2.50

Titre / cm3

[1]

(ii) Calculate the mean titre of NaOH that the student should use to analyse the results.

Mean titre = … cm3 [1]

(iii) Calculate the mass, in mg, of malic acid in one tablet.

Assume that malic acid (molar mass 134.0 g mol–1) is the only acid in the tablets.

Give your answer to 3 significant figures.

Mass of malic acid … mg [5]

(b) Another student carries out the same experiment.

Instead of rinsing the burette used in Step 4 with 0.0800 mol dm–3 NaOH, the student rinses with

water before filling it up with 0.0800 mol dm–3 NaOH.

State and explain how this error would affect the titre.

… [2]

OH H

HOOC C C COOH

H H

When malic acid is heated with an acid catalyst, a mixture of cis and trans stereoisomers is

produced.

Draw the structures of the cis and trans isomers.

cis trans

[2]

Mark scheme

Show the mark scheme

1.51 –3 ALLOW 3SF up to calculator value 7.653319818 x 10

n(BaCO3) = 197.3 OR 7.6533…. × 10 IGNORE rounding errors past 3SF

vCO = 7.6533…. × 10–3 x 24000 = 184 cm3 ALLOW ECF from incorrect n(BaCO )

Answer must be 3 SF and in cm3

ALLOW use of ideal gas equation with an appropriate

temperature (290–298K) and pressure (100/101/101325 kPa)

e.g.

e.g. At 293K and 100 kPa,

V = (7.65… x 10-3 x 8.314 x 293) / 100 x 103

V = 186 cm3

(ii) 2 IGNORE Incomplete reaction

IGNORE human errors e.g. unsealed apparatus, gas escaping

the measuring cylinder

Any two from

Question Answer Mark Guidance

• (Gas lost /escapes) when the bung is replaced ALLOW ‘when BaCO3 is added to flask’

• Some CO2 / gas dissolves (in the water)

• Conditions / T/ P not at RTP ALLOW not standard conditions

(iii) Use a gas syringe (in place of the upturned measuring 1 IGNORE gas cylinder

cylinder)

Place the barium carbonate inside a small tube and release it ALLOW use a water bath set at room temperature

into the acid without removing the stopper OWTTE

3 (a) (i) 1 DO NOT ALLOW 23.3

23.30 23.05 23.15

(ii) 23.05 + 23.15 1 ALLOW ECF from incorrect titres from 3a(i) but values must

Mean titre = 2 = 23.1(0) be concordant (i.e. within 0.1 cm3 of each other)

(iii) FIRST CHECK THE ANSWER ON THE ANSWER LINE 5 ALLOW ECF from incorrect mean titre in b(ii) and throughout

If answer = 413 (mg) award 5 marks

ALLOW 3SF up to calculator value throughout

23.1 × 0.08 –3 TAKE CARE: values shown may be truncated calculator

n(NaOH) = = 1.848 × 10

1000 values.

IGNORE rounding errors past 3SF

1.848 x 10–3

n(acid) in 25 cm3 = = 9.24× 10–4

2 Steps can be calculated in any order which will change the

intermediate answers. Marks are for the processing

n(acid) in 250 cm3 for 3 tablets = 9.24 × 10–4 × 10 of the data.

= 9.24× 10–3

Alternative Approach

13 -4 –3

mass of acid in 3 tablets = 9.24 × 10–3 × 134 n(acid) in one tablet = 9.24 x 10 / 3 = 3.08 × 10

= 1.23(816) g

mass of acid in one tablet = 3.08 × 10–3 x 134 × 1000

1.23816 x 1000 = 413 (mg)

mass of acid in one tablet = = 413 (mg)

Final answer must be to 3 SF AND mg

COMMON ERRORS

0.413, 4.13, 41.3 4 marks Incorrect conversion to mg

(check for x 10 in method)

1240 / 1238 4 marks Not divided by 3 for 1 tablet

Question Answer 14 Mark Guidance

3 (b) The titre would be greater 2 ALLOW AW

(NaOH) would be more dilute / less concentrated (so a bigger

volume would be needed to neutralise the acid).

3 (c) 2 ALLOW any combination of skeletal OR structural OR

displayed formula as long as unambiguous

ALLOW one mark if both stereoisomers of are shown but in

the incorrect columns

DO NOT ALLOW incorrect connectivity e.g. or - HOOC

on first occasion but allow ECF in second structure.

How to answer it

Titration Analysis & Stereoisomerism of Malic Acid

📋 What This Question Tests

This question evaluates your core practical and theoretical competence in quantitative analysis and organic stereochemistry:

Burette Data Processing: Recording titre values to 2 decimal places and selecting concordant titres (within 0.10 cm³) to calculate a mean titre.
Multi-Step Quantitative Chemistry: Linking stoichiometry (1:2 reacting ratio), dilution factors (25.0 cm³ into 250.0 cm³), tablet scaling (3 tablets to 1 tablet), and unit conversions (g to mg; final answer to 3 significant figures).
Practical Procedural Errors: Explaining the directional impact of incorrect rinsing techniques on titre volumes.
Alkene Stereoisomerism: Deducing acid-catalysed elimination (dehydration) products and drawing correctly bonded cis and trans (E/Z) isomers.

Part (a)(i)

Completing the Titration Table

Recording Burette Titres [1 Mark]

✅ Correct Answer

Titration	1	2	3
Final reading / cm³	23.60	46.70	25.65
Initial reading / cm³	0.30	23.65	2.50
Titre / cm³	23.30	23.05	23.15

All three values calculated correctly: 23.30, 23.05, 23.15.

❌ Common Errors

Writing 23.3 instead of 23.30 . Burette readings and titres must always be recorded to 2 decimal places, ending in either .00 or .05.
Simple subtraction arithmetic slips under exam pressure.

Mark Scheme: 1 mark for all three titres correct to 2 decimal places. DO NOT ALLOW 23.3.

Part (a)(ii)

Selecting Concordant Titres

Calculating the Mean Titre [1 Mark]

📐 Step-by-Step Calculation

Identify concordant results (within 0.10 cm³ of each other):
• Titration 2 = 23.05 cm³
• Titration 3 = 23.15 cm³
Difference = 23.15 - 23.05 = 0.10 cm³ (concordant).
Reject Titration 1 (23.30 cm³) as it is 0.15 cm³ away from Titration 3 and 0.25 cm³ from Titration 2.
Calculate the mean:
Mean = (23.05 + 23.15) / 2 = 23.10 cm³ (or 23.1 cm³)

🧠 Exam Technique & Insight

Concordancy Definition: Only average titres that agree within ±0.10 cm³. Averaging all three results loses the mark immediately.
Error Carried Forward (ECF): Allowed if your calculated titres in (a)(i) were incorrect, provided the ones you averaged were within 0.10 cm³ of each other.

Mark Scheme: 1 mark for 23.10 cm³ (or 23.1 cm³). Calculation must clearly use Titrations 2 and 3 only.

Part (a)(iii)

Multi-Step Quantitative Titration Calculation

Mass of Malic Acid in One Tablet [5 Marks]

📐 Full Step-by-Step Working

Step 1: Calculate moles of NaOH used in the titration

n(NaOH) = (volume × concentration) / 1000
n(NaOH) = (23.10 × 0.0800) / 1000 = 1.848 × 10⁻³ mol

Step 2: Use stoichiometry to find moles of malic acid in the 25.0 cm³ aliquot
From the balanced equation: C₄H₆O₅ + 2NaOH → C₄H₄O₅Na₂ + 2H₂O
Reacting ratio is 1 mol acid : 2 mol NaOH.
n(acid in 25.0 cm³) = (1.848 × 10⁻³) / 2 = 9.24 × 10⁻⁴ mol

Step 3: Scale up to the full 250.0 cm³ volumetric flask
The pipette took 25.0 cm³ out of a total 250.0 cm³ solution (a 10-fold dilution factor):
n(acid in 250.0 cm³ = 3 tablets) = 9.24 × 10⁻⁴ × (250 / 25) = 9.24 × 10⁻³ mol

Step 4: Calculate total mass of acid in the 3 tablets
Given molar mass of malic acid, M = 134.0 g mol⁻¹ :
mass in 3 tablets = 9.24 × 10⁻³ mol × 134.0 g mol⁻¹ = 1.23816 g

Step 5: Determine mass of malic acid in ONE tablet in mg
• Mass in 1 tablet (in grams): 1.23816 / 3 = 0.41272 g
• Convert to mg (multiply by 1000): 0.41272 × 1000 = 412.72 mg
• Round to 3 significant figures: 413 mg

✅ Final Answer

Mass = 413 mg

(Award full 5 marks if 413 mg is written on the answer line with clear evidence of method)

❌ Common Calculation Pitfalls

Ignoring the 1:2 ratio: Forgetting to divide n(NaOH) by 2 gives 825 mg.
Forgetting to divide by 3: Failing to divide by 3 tablets gives 1240 mg (or 1238 mg) [loses 1 mark].
Unit conversion error (g vs mg): Forgetting to multiply by 1000 leaves the answer as 0.413 g [loses 1 mark].
Significant figures: Quoting 412.7 mg or 410 mg instead of exactly 3 significant figures (413 mg).

Mark Scheme Breakdown (5 Marks):
• M1: n(NaOH) = 1.848 × 10⁻³ mol
• M2: n(acid in 25 cm³) = 9.24 × 10⁻⁴ mol (dividing by 2)
• M3: n(acid in 250 cm³) = 9.24 × 10⁻³ mol (multiplying by 10)
• M4: Mass of acid in 3 tablets = 1.238 g (multiplying by 134.0)
• M5: Final answer 413 (mg) (divided by 3, converted to mg, given to 3 SF).

Part (b)

Procedural Errors in Titration

Rinsing the Burette with Water [2 Marks]

✅ Model Answer

Effect: The titre would be greater / larger / increase. [1 mark]

Explanation: Water left in the burette dilutes the NaOH solution (lowers its concentration), meaning a larger volume of NaOH is required to neutralise the same amount of acid. [1 mark]

💡 Key Laboratory Principles

Burette & Pipette: Must always be rinsed with the solution they are going to contain. Rinsing with water dilutes the reagent.
Conical Flask: Can be rinsed with distilled water because adding extra water does not change the number of moles of acid transferred by the pipette.

Mark Scheme:
• Mark 1: Titre would be greater / larger.
• Mark 2: (NaOH) would be diluted / less concentrated (so a bigger volume is needed).

Part (c)

Stereoisomerism from Elimination Reaction

Drawing cis and trans Isomers [2 Marks]

💡 Reaction Mechanism Insight

Heating malic acid, HOOC-CH(OH)-CH₂-COOH , with an acid catalyst causes an elimination (dehydration) reaction. An -OH group is lost from one carbon and an -H from the adjacent carbon, forming a C=C double bond (producing but-2-enedioic acid):

HOOC-CH(OH)-CH₂-COOH → HOOC-CH=CH-COOH + H₂O

cis isomer

High priority groups / identical groups on the same side of the C=C double bond

HOOC COOH \ / C == C / \ H H

Also known as (Z)-butenedioic acid / maleic acid

trans isomer

High priority groups / identical groups on opposite sides across the C=C double bond

HOOC H \ / C == C / \ H COOH

Also known as (E)-butenedioic acid / fumaric acid

🧠 Connectivity & Drawing Rules

Ensure the bond connects to the carbon atom of the carboxylic acid: write HOOC-C on the left rather than COOH-C or C-HOOC .
Show clear 120° trigonal planar bond angles around each alkene carbon atom.

❌ Common Mistakes to Avoid

Swapping the columns: drawing the trans isomer under the cis label and vice versa (in this case, examiners only award a maximum of 1 mark).
Incorrect connectivity: bonding hydrogen directly to the alkene carbon as C-OOC or drawing bad bonds to text.

Mark Scheme:
• 1 mark for correct cis structure.
• 1 mark for correct trans structure.
Note: Allow skeletal, displayed, or structural formulas as long as unambiguous. If both structures are correct but placed in inverted boxes, award 1 mark total. DO NOT ALLOW incorrect connectivity on first occasion.

Topics

Module 1: Development of practical skills in chemistry · Module 2: Foundations in chemistry · Module 4: Core organic chemistry · Practical Activity Groups · 1.1 Practical skills assessed in a written examination · PAG 2: Acid-base titration · 2.1 Atoms and reactions · 4.1 Basic concepts and hydrocarbons · 4.2 Alcohols, haloalkanes and analysis

Question and mark scheme from the OCR A-Level Chemistry examination, AS Depth in chemistry (02), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.