OCR A-Level Chemistry AS Depth in chemistry (02), June 2022: Question 1

15 marks · Medium difficulty · Structured Questions

Calculate the mass of citric acid in a lime using titration results, and write an oxidation equation for malic acid.

Practise this question

Question

Question text

1 Lime is a citrus fruit containing citric acid, C6H8O7.

(a) Citric acid is a weak organic acid.

(i) What is meant by an acid?

… [1]

(ii) What is meant by an acid that is weak?

… [1]

(b) A student carries out a titration to determine the mass of citric acid in a lime.

The student follows the method below:

• Squeeze the juice out of two limes.

• Transfer the juice into a 250.0 cm3 volumetric flask and make up to the mark with

distilled water.

• Pipette 25.0 cm3 of the diluted lime juice into a conical flask and add a few drops of

phenolphthalein indicator.

• Titrate this solution with 0.800 mol dm–3 NaOH(aq).

The student carries out a trial titration, followed by three further titrations.

The diagram shows the burette readings for the three further titrations.

Each reading is measured to the nearest 0.05 cm3.

Titration 1 Titration 2 Titration 3

Initial reading Final reading Initial reading Final reading Initial reading Final reading

0 27 0 27 0 27

1 28 1 28 1 28

2 29 2 29 2 29

(i) Record the student’s burette readings in the table below.

Calculate the mean titre, to the nearest 0.05 cm3, that the student should use to analyse

the results.

Titration 1 Titration 2 Titration 3

Final reading / cm3

Initial reading / cm3

Titre / cm3

mean titre … cm3 [4]

(ii) Citric acid, C6H8O7, is neutralised by NaOH as shown in the equation below.

C6H8O7 + 3NaOH Na3C6H5O7 + 3H2O

Calculate the mass, in g, of citric acid in one lime.

Assume that citric acid (Mr = 192.0) is the only acid in lime juice.

mass of citric acid in one lime = … g [5]

hydroxide concentration of 0.800 mol dm–3 for the titration.

Suggest how the student could modify the method using a sodium hydroxide concentration

of 0.200 mol dm–3 instead of 0.800 mol dm–3.

The student should aim to have the same titre as in the original method.

Justify your answer.

… [2]

(d) Other fruits contain different organic acids.

Apple juice contains malic acid which has the following structure.

H OH

HOOC C C COOH

H H

Malic acid can be oxidised by heating with acidified potassium dichromate(VI).

Write a balanced equation for the reaction, showing the structure of the organic product.

Use [O] to represent the oxidising agent.

[2]

Mark scheme

Show the mark scheme

Question Answer Marks Guidance

element

1 (a) (i) (Acid) releases H+ ions/ H+ donor 1 AO1.1 ALLOW H+ OR proton

(ii) (weak acid) partially dissociates/ionises 1 AO1.1 IGNORE vague responses that do not imply

a number, e.g.

• poor proton donor

IGNORE ‘doesn’t easily dissociate’

IGNORE ‘strong acid completely dissociates’

(b) (i) 4 ANNOTATE ANSWER WITH TICKS AND

Titration Titration Titration CROSSES ETC

12 3

Final reading/

3 27.35 27.65 27.85 ALLOW missing zeroes throughout except

Initial for last marking point

30.05 0.10 0.45

reading/cm

3 e.g. 0.1 for 0.10

Titre/cm 27.30 27.55 27.40

Initial and final readings ALLOW ECF from incorrect burette readings

All titration readings (×6) correct AO1.2

×4

Titres

Correct subtractions to obtain final titre values

Mean titre calculated from concordant results IF MEAN IS CALCULATED FROM ECF, IT

Correct mean titre = 27.35 (cm3) MUST BE FROM CLOSEST TITRES

ALLOW any number of decimal places for

mean titre for this mark

Reading recorded to accuracy of burette

All values including mean titre recorded to two decimal Note: Question asks for mean titre to

nearest 0.05 cm3

places with the last figure either 0 or 5

element

(ii) FIRST CHECK THE ANSWER ON ANSWER LINE 5 ANNOTATE ANSWER WITH TICKS AND

If answer = 7.(00) award 5 marks CROSSES ETC

------------------------------------------------------------------------ ----------------------------------------------------

ALLOW ECF from incorrect titre calculated

n(NaOH) in 1(b)(i)

27.35 × 0.800 Throughout:

= = 0.02188 AO2.8 ALLOW 3 SF or more, correctly rounded

1000

×4 e.g. n(NaOH) = 0.0219 for 0.02188

n(A) in 25.0 cm3

0.02188

= = 0.00729(33) ALLOW ECF from incorrect n(NaOH)

n(A) in 250 cm3 ALLOW ECF for all subsequent steps

= 10 × 0.00729(33) = 0.0729(33)

mass citric acid in 250 cm3

From n(NaOH) = 0.0219,

= 0.0729 × 192 = 14(.0032) (g)

n(A) = 0.073(0)

mass citric acid in one lime mass citric acid = 14(.016)

14.0 mass in 1 lime = 7(.008)

= = 7.(00) (g) AO2.4

Use half a lime ×2 a dilution factor of 4

Make up lime juice (solution) in 1 dm3 volumetric flask

Dilution ratio to justify

4 times less citric acid/lime juice ALLOW quartered

NaOH is 4 times more dilute (giving same titre)

1:4 ratio for NaOH concentration

10 AO

element

(d) 2 ALLOW any combination of skeletal OR

structural OR displayed formula as long as

unambiguous

AO2.5

Correct structure of product

AO2.6

Correctly balanced equation

Total 15

How to answer it

Acids, Titrations, and Organic Oxidation Study Guide

OCR AS Level Chemistry

What this question tests

This multi-step question assesses core physical and organic chemistry topics: definitions of acids and weak acids, reading burettes accurately to 0.05 cm³, performing multi-stage stoichiometric titration calculations (including scaling factors for dilutions and sample volumes), evaluating laboratory safety through concentration adjustments, and writing balanced equations for the oxidation of secondary alcohols using [O].

Question 1(a): Definitions of Acids and Weak Acids

Parts (a)(i) and (a)(ii)

✅ Correct Answers

(i) Acid: Releases H⁺ ions / Proton donor.
(ii) Weak acid: Partially dissociates / ionises.

💡 Key Knowledge

A proton donor is the standard IUPAC definition expected at AS level.
Weak acids establish an equilibrium in water where only a small fraction of molecules release H⁺ ions.

❌ Common Errors

Writing vague phrases like "poor proton donor" or "doesn't easily dissociate".
Confusing weak acids with dilute acids (concentration vs strength).
Stating "strong acids completely dissociate" instead of defining weak acids.

Mark: 1 mark for (a)(i), 1 mark for (a)(ii).

Question 1(b): Titration Analysis and Stoichiometry

Parts (b)(i) and (b)(ii)

✅ Correct Answers

Burette Readings Table:

Titration 1: Final = 27.35 , Initial = 0.05 , Titre = 27.30
Titration 2: Final = 27.65 , Initial = 0.10 , Titre = 27.55
Titration 3: Final = 27.85 , Initial = 0.45 , Titre = 27.40

Mean Titre: 27.35 cm³ (calculated using concordant results / closest titers e.g. Titration 1 and 3 depending on exact student selection, with standard OCR rules allowing 27.35).

Final Answer (b)(ii): 7.00 g of citric acid in one lime.

🧠 Exam Technique

All burette readings must be recorded to two decimal places, ending in 0 or 5 because burettes are read to the nearest 0.05 cm³.
Always check concordant titres before averaging.

📐 Step-by-Step Calculation (Part b(ii))

Moles of NaOH:
n(NaOH) = (27.35 × 0.800) / 1000 = 0.02188 mol
Moles of citric acid in 25.0 cm³ pipette sample:
Using equation ratio (1 mol acid : 3 mol NaOH):
n(acid) = 0.02188 / 3 = 0.007293 mol
Moles of citric acid in 250 cm³ volumetric flask:
Scale up by factor of 10 (250 / 25):
n(in flask) = 0.007293 × 10 = 0.07293 mol
Mass of citric acid in 250 cm³ (two limes):
Mass = moles × Mᵣ = 0.07293 × 192.0 = 14.00 g
Mass of citric acid in ONE lime:
14.00 / 2 = 7.00 g

❌ Common Calculation Traps

Forgetting to divide by 3 for the stoichiometry ratio between citric acid and NaOH.
Failing to scale up from the 25 cm³ aliquot to the 250 cm³ volumetric flask.
Dividing by 2 at the wrong stage of the calculation.

Mark: 4 marks for (b)(i), 5 marks for (b)(ii).

Question 1(c): Method Modification & Safety

Part (c)

✅ Correct Answers

Action: Use half a lime (or dilute the lime juice / make up solution in a 1 dm³ volumetric flask).
Justification: The NaOH concentration is 4 times less (0.200 mol dm⁻³ instead of 0.800 mol dm⁻³), so 4 times less acid is required to keep the titre volume the same (or a 1:4 ratio for NaOH concentration).

💡 Key Knowledge

When concentration is decreased by a factor of 4, the volume of alkali required would increase by a factor of 4 unless the amount of acid titrated is also reduced by a factor of 4 (e.g., using half a lime).

Mark: 2 marks.

Question 1(d): Organic Oxidation of Malic Acid

Part (d)

✅ Correct Answers

Organic Product Structure: The secondary alcohol group (-CH(OH)-) in malic acid oxidises into a ketone carbonyl group (-C(=O)-).
Balanced Equation:
HOOC-CH(OH)-CH₂-COOH + [O] → HOOC-C(=O)-CH₂-COOH + H₂O

🧠 Exam Technique

Secondary alcohols oxidise to form ketones and water.
Make sure the remaining carbon skeleton (the two -COOH groups and surrounding CH₂ units) remains unchanged.
Use [O] as specified in the question to represent the oxidising agent.

Mark: 2 marks. Total paper marks: 15.

Topics

Module 1: Development of practical skills in chemistry · Module 2: Foundations in chemistry · Module 4: Core organic chemistry · Practical Activity Groups · PAG 2: Acid-base titration · 2.1 Atoms and reactions · 1.2 Practical skills assessed in the practical endorsement · 4.2 Alcohols, haloalkanes and analysis

Question and mark scheme from the OCR A-Level Chemistry examination, AS Depth in chemistry (02), June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.