OCR A-Level Chemistry Unified chemistry (03), June 2022: Question 6

13 marks · Hard difficulty · Structured Questions

Determine optical isomers, explain water solubility, deduce ester formula from carboxylic acid reaction, draw acid-base proton transfer curly arrow mechanism, calculate pH of weak acid solution, and determine vitamin C mass in orange juice via titration calculation.

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Question

Question text

6 A student carries out an investigation on vitamin C, C6H8O6.

(a) The structure of vitamin C is shown below. Vitamin C is an optical isomer.

O O H

Vitamin C

What is the total number of optical isomers with the structure of vitamin C?

total number of optical isomers = … [1]

(b) Vitamin C is extremely soluble in water. This means that vitamin C is removed rapidly from

the body. ‘Vitamin C ester’ is available in tablet form as a less soluble source of vitamin C

which stays in the body for longer.

(i) Suggest why vitamin C is extremely soluble in water.

… [1]

(ii) A ‘vitamin C ester’ tablet contains an ester with the molecular formula C22H38O7.

This ester can be prepared by reacting vitamin C with a long chain carboxylic acid,

CxHyCOOH, in the presence of an acid catalyst.

Vitamin C and the long chain carboxylic acid react in a 1:1 molar ratio.

Determine x and y in the formula of this carboxylic acid.

x = … y = … [2]

68 6 a

as ascorbic acid.

(i) In aqueous solution, vitamin C donates a proton to water:

C H O + H O C H O – + H O+

68 6 2 6 7 6 3

Add curly arrows to the diagram to suggest the mechanism for this process.

H2O:

O O H O O

H H

H H +

+ H3O

O – O

O O

H H

[2]

(ii) The student dissolves 0.150 mol of vitamin C in water and makes the solution up to

250 cm3 in a volumetric flask.

Calculate the pH of this solution of vitamin C.

Give your answer to 2 decimal places.

pH = … [3]

(d) The label on a carton of orange juice lists the mass of vitamin C, in mg, in a typical serving

of 150 cm3.

The student carries out an investigation to check the vitamin C content in the orange juice.

Vitamin C can be oxidised by iodine:

C H O (aq) + I (aq) C H O (aq) + 2I–(aq) + 2H+(aq)

68 6 2 6 6 6

The student dilutes 150 cm3 of the orange juice with water to 250.0 cm3 in a volumetric flask.

The student then titrates 25.0 cm3 volume of this solution with 9.60 × 10–4 mol dm–3 iodine

solution, I2(aq).

The mean titre of I (aq) is 22.50 cm3.

Determine the mass, in mg, of vitamin C in a 150 cm3 serving of the orange juice.

mass of vitamin C in the 150 cm3 serving of orange juice = … mg [4]

Mark scheme

Show the mark scheme

Question Answer Marks AO Guidance

element

6 (a) 1 AO2.1

Number of optical isomers = 4

(b) (i) 1 AO2.1

Hydrogen bonding

AND ALLOW 4 OH

Many OH/hydroxyl / hydroxy / alcohol

DO NOT ALLOW OH–

(ii) 2 AO3.2

x = 15 y = 31 ×2

×2 IGNORE incorrect curly arrows

IGNORE ‘double’ curly arrows such as:

3 OR 4 curly arrows correct → 2 marks

1 curly arrow correct → 1 mark

H2O Curly arrow must

• start from, OR be traced back to any point

across width of lone pair on H2O:

27 element

(ii) FIRST CHECK ANSWER ON THE ANSWER LINE 3

If answer = 2.16 award 3 marks

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

[Vitamin C] = 0.150 × 4 = 0.600 (mol dm–3) AO2.4

0.6 seen anywhere ×2

+] = √ (K × [Vitamin C]) For [H+]

[H a

= √ (7.94 × 10–5 × 0.600) ALLOW ECF from incorrect [vitamin C]

= 6.90 × 10–3 (mol dm–3)

for pH

+ ALLOW ECF ONLY if [H+] has been derived

pH = –log [H ]

= –log 6.90 × 10–3 from Ka AND [vitamin C]

= 2.16 AO1.2 ----------------------------------------------------

2 DP required ×1 COMMON ERRORS

pH = 4.32 2/3 calculation marks

No square root of (7.94 × 10–5 × 0.600)

pH = 2.46 2/3 calculation marks

No × 4 (7.94 × 10–5 × 0.150)

pH = 2.76 2/3 calculation marks

÷ 4 (7.94 × 10–5 × 0.0375)

pH = 4.92 1/3 calculation mark

No square root AND 0.150

pH = 5.53 1/3 calculation mark

No square root AND 0.0375

28 element

(d) FIRST CHECK ANSWER ON THE ANSWER LINE 4 AO2.8 Use ECF throughout

If answer = 38 (mg) award 4 marks ×4 Intermediate values for working to at least 3

-------------------------------------------------------------------------- SF.

TAKE CARE as value written down may be

9.60 × 10–4 truncated value stored in calculator.

n(I ) = 22.50 × = 2.16 × 10–5 (mol)

2 1000 Depending on rounding, either can be

credited.

n(vitamin C) in 250 cm3 volumetric flask

= 10 × 2.16 × 10–5 = 2.16 × 10–4 (mol) -----------------------------------------------------------

COMMON ERRORS:

M(Vitamin C: C6H8O6) = 176 OR (12 × 6) + (1 × 8) + (16 × 6)

Seen anywhere 22.81 mg scaling by 150/250 → 3 marks

FINAL MARK LOST BY SCALING

Mass vitamin C in 150 cm3 of orange

= 2.16 × 10–4 × 176.0 = 0.038016 g

= 38 (mg)

2 SF or more

42.24 mg using 25.0 cm3 instead of 22.50

→ 3 marks

25.34 mg using 25.0 cm3 AND scaling by

150/250 instead of 22.50

→ 2 marks

63.36 mg scaling by 250/150 → 3 marks

How to answer it

Investigation into Vitamin C (Ascorbic Acid)

OCR A-Level Chemistry • Comprehensive Study Guide

What this question tests

This synoptic organic and physical chemistry question assesses your ability to identify chiral centers, explain solubility using intermolecular forces, deduce ester formulas from molecular structures, draw curly arrow reaction mechanisms for acid dissociation, calculate pH for weak acids using Ka expressions, and perform complex multi-step titration stoichiometry calculations involving dilution and volumetric scaling.

Part (a) — Optical Isomerism

Total number of optical isomers

✅ Correct Answer

💡 Key Knowledge

An optical isomer (chiral center) requires a carbon atom bonded to 4 different groups.
Count the asymmetric carbon atoms in the vitamin C structure carefully (there are 2 chiral centers).
Use the formula 2ⁿ where n = number of chiral centers (2² = 4).

Marks: 1 mark for the correct numerical value.

Part (b) — Solubility & Esterification

(i) Solubility Explanation

✅ Correct Answer

Hydrogen bonding forms between the many OH (hydroxyl/alcohol) groups on vitamin C and water molecules.

❌ Common Errors

Writing just "hydrogen bonding" without referencing the specific functional groups ( OH groups) loses the mark. Do not write formulas like OH⁻ since vitamin C contains neutral alcohol/hydroxyl groups, not hydroxide ions.

(ii) Determining Carboxylic Acid Formula

✅ Correct Answer

x = 15 , y = 31 (Formula: C₁₅H₃₁COOH )

📐 Calculation Steps

Reacting Ratio: Vitamin C ( C₆H₈O₆ ) reacts with the long-chain carboxylic acid ( CₓHᵧCOOH ) in a 1:1 molar ratio to form the ester ( C₂₂H₃₈O₇ ) plus water ( H₂O ).
Equation Setup: C₆H₈O₆ + CₓHᵧCOOH → C₂₂H₃₈O₇ + H₂O
Balance Carbon: Total carbons on product side = 22. In vitamin C = 6. Therefore, carbons in carboxylic acid ( x + 1 ) = 22 - 6 = 16. Thus, x = 15 .
Balance Hydrogen: Total hydrogens on product side = 38 (in ester) + 2 (in H₂O from condensation) = 40. Hydrogens in vitamin C = 8. Therefore, y + 1 = 40 - 8 = 32 . Thus, y = 31 .

Marks: 2 marks total (1 mark for x, 1 mark for y).

Part (c) — Acid Dissociation and pH

(i) Curly Arrow Mechanism

✅ Correct Answer

Curly arrow originating from the lone pair on the oxygen of H₂O attacking the acidic H of the -OH group. A second curly arrow originates from the O-H bond and goes to the oxygen atom, forming a negative charge on the conjugate base.

🧠 Exam Technique

Ensure the arrow from water starts precisely from a lone pair.
Ensure the bond-breaking arrow starts exactly from the covalent bond and points towards the oxygen atom.

(ii) pH Calculation of Vitamin C Solution

✅ Correct Answer

pH = 2.16 (3 marks)

📐 Step-by-Step Calculation

Find Concentration of Vitamin C:
0.150 mol in 250 cm³
Concentration = (0.150 / 250) × 1000 = 0.600 mol dm⁻³ (or 0.6 seen anywhere).
Calculate Hydrogen Ion Concentration ([H⁺]):
For a weak acid: Ka = [H⁺]² / [HA] ⇒ [H⁺] = √(Ka × [HA])
[H⁺] = √(7.94 × 10⁻⁵ × 0.600) = √(4.764 × 10⁻⁵) = 6.90 × 10⁻³ mol dm⁻³ .
Calculate pH:
pH = -log[H⁺] = -log(6.90 × 10⁻³) = 2.1616...
Round to 2 decimal places as requested: 2.16 .

❌ Common Calculation Traps

Forgetting to scale the concentration to 1 dm³ (using 0.150 directly without dividing by volume gives pH = 4.32 ).
Forgetting to take the square root of ( Ka × [HA] ).

Marks: 3 marks. Examiner checks answer line first. Correct answer yields full marks instantly. ECF applies where appropriate.

Part (d) — Titration & Scaling Calculation

Determining Mass of Vitamin C in Orange Juice

✅ Correct Answer

Mass = 38 mg (4 marks)

📐 Step-by-Step Calculation

Calculate moles of I₂ reacted in titration:
n(I₂) = (22.50 cm³ / 1000) × (9.60 × 10⁻⁴ mol dm⁻³) = 2.16 × 10⁻⁵ mol .
Determine moles of Vitamin C in the 25.0 cm³ titration sample:
From equation, stoichiometry is 1:1. So n(Vitamin C in 25 cm³) = 2.16 × 10⁻⁵ mol .
Scale up to the 250 cm³ volumetric flask:
n(Vitamin C in 250 cm³) = 2.16 × 10⁻⁵ × (250 / 25.0) = 2.16 × 10⁻⁴ mol .
Convert moles to mass using Molar Mass:
M(Vitamin C, C₆H₈O₆ ) = (6 × 12.0) + (8 × 1.0) + (6 × 16.0) = 176 g mol⁻³ .
Mass in 150 cm³ serving = 2.16 × 10⁻⁴ mol × 176 g mol⁻¹ = 0.038016 g .
Convert grams to milligrams (mg):
0.038016 g × 1000 = 38.016 mg = 38 mg (to 2 significant figures or more).

❌ Common Errors & Examiner Notes

Incorrect scaling ratios: Inverting or missing the 150/250 dilution scale factor or titration aliquot factor ( 25.0 / 22.50 confusion) is the most frequent place students lose marks.
Always check intermediate rounding: keep at least 3 significant figures in working out.

Marks: 4 marks. First check answer line: if 38 (mg), award 4 marks. Uses Error Carried Forward (ECF) throughout.

Topics

Module 2: Foundations in chemistry · Module 5: Physical chemistry and transition elements · Module 6: Organic chemistry and analysis · 6.1 Aromatic compounds, carbonyls and acids · 6.2 Nitrogen compounds, polymers and synthesis · 5.1 Rates, equilibrium and pH · 2.1 Atoms and reactions

Question and mark scheme from the OCR A-Level Chemistry examination, Unified chemistry (03), June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.