OCR A-Level Chemistry Unified chemistry (03), November 2021: Question 3

12 marks · Hard difficulty · Calculations

Describe the preparation of a standard barium hydroxide solution and use titration results involving an unsaturated carboxylic acid to determine two possible cis stereoisomer structures.

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Question text

3 This question is about carboxylic acids.

Compound D is a cis stereoisomer of an unsaturated organic acid with the general formula

CnH2n–1COOH.

A student plans to analyse acid D by carrying out a titration.

(a) A student first prepares 250.0 cm3 of a standard solution of 0.150 mol dm–3 Ba(OH) for the

titration.

The student is provided with solid Ba(OH)2 and usual laboratory apparatus and equipment.

Describe how the student would prepare the standard solution, giving full details of quantities,

apparatus and method.

… [5]

(b) The student prepares a 100.0 cm3 solution containing 3.215 g of acid D.

The student titrates 25.0 cm3 samples of the solution of D with 0.150 mol dm–3 Ba(OH) (aq) in

the burette.

1 mol Ba(OH)2 reacts with 2 mol of D.

The mean titre of Ba(OH) (aq) is 23.50 cm3.

Analyse the titration results to determine two possible structures for the cis stereoisomer of

organic acid D.

Structures of 2 possible cis stereoisomers of acid D

[7]

Mark scheme

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Question Answer Marks Guidance

element

3 (a) 250 5

n(Ba(OH)2) = 0.150 × 1000 OR 0.0375 (mol) AO2.4

×2 ALLOW ECF from incorrect n(Ba(OH)2)

Mass Ba(OH)2 = 0.0375 × 171.3 = 6.42375 (g) ALLOW 6.42 up to 6.42375 correctly rounded

6.42 g subsumes 1st mark

Dissolve solid in (distilled) water (less than 250 cm3) in

beaker AO1.2 ALLOW conical flask for beaker

×3

Transfer (solution) to volumetric flask ALLOW graduated flask

AND

Transfer washings (from beaker) to flask DO NOT ALLOW round-bottom or conical flask

Make up to mark/up to 250 cm3 with (distilled) water

AND

Invert flask (several times to ensure mixing)

(b) 23.50 7 Use ECF throughout

n(Ba(OH)2) = 0.150 × 1000 Intermediate values for working to at least 3 SF.

= 3.525 × 10–3 (mol) AO2.8

×4 TAKE CARE as value written down may be

n(D) in 25.0 cm3 = 2 × 3.525 × 10–3 truncated value stored in calculator.

= 7.05 × 10–3 (mol) Depending on rounding, either can be credited.

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

3 –3 100

n(D) in 100 cm = 7.05 × 10 ×

25.0

= 0.0282 (mol) 3 3.215

ALLOW Mass D in 25.0 cm = 4 = 0.80375 g

3.215 –1

Molar mass (D) = = 114 (g mol ) 0.80375

0.0282 Molar mass (D) = –3 = 114

7.05 × 10

Formula: = C H COOH AO3.2

OR CnH2n–1: M(C5H9) = 114 – 45 = 69 ×1

If not stated, could be credited from structure

Question Answer 12 Marks Guidance

element

cis stereoisomers. COMMON ERRORS:

The drawn stereoisomers must have AO3.2 Up to Molar mass = 114 (1st 4 marks)

• Different groups attached to each C atom of C=C ×2 M = 456 → 3/4 marks (mol in 100 cm3 omitted)

• Each C of C=C has the same group on the same side 3.215

M = –3 = 456

7.05 × 10

Any 2 cis isomers Many possibilities, e.g.

M = 228 → 3/4 marks (No × 2 for n(D))

–3 100

3.525 × 10 × = 0.0141

25.0

3.215

M = = 228

0.0141

M = 100.8 → 3/4 marks

23.50 instead of 25.00 and scaling by × 23.50

0.150 –3

25.0 × = 3.75 × 10

1000

→ 2 × 3.75 × 10–3 = 7.5 × 10–3

–3 100

→ 7.5 × 10 × = 0.0319

23.50

3.215

→ → 100.8

0.0319

ALLOW correct structural, with ‘cis’ part displayed

OR skeletal THEN ALLOW ECF for carboxylic acid closest to

OR displayed formula calculated M(alkyl group) but must be CnH2n–1

OR mixture of above as long as non-ambiguous e.g. For M(alkyl) = 100, ALLOW C4H7 (55)

For M(alkyl) = 411, ALLOW C29H57 (405)

ALLOW side chains as molecular formula, OR C30H59 (419)

e.g. C3H7 for (CH3)2CH OR CH3CH2CH2 THEN judge cis isomers with closest match

e.g. C3H5O2 for CH2CH2COOH

ALLOW 1 mark for 2 trans isomers shown

IGNORE poor connectivity to all groups instead of 2 cis isomers

ECF for Same error made twice.

How to answer it

Analysis of a Carboxylic Acid via Titration and Stereoisomerism

What this question tests

This multi-step synoptic question assesses practical chemistry techniques (standard solution preparation), stoichiometric calculations involving neutralisation ratios, molar mass determination, and organic structural analysis applying E/Z (cis/trans) stereoisomerism principles.

Part (a) — [5 Marks]

Preparation of a Standard Solution

📐 Step-by-Step Calculation

Calculate required moles:
n = c × V = 0.150 × (250 / 1000) = 0.0375 mol
Calculate required mass:
Molar mass of Ba(OH)₂ = 137.3 + (16.0 + 1.0) × 2 = 171.3 g mol⁻¹
Mass = 0.0375 × 171.3 = 6.42 g (or 6.42375 g)

💡 Practical Method & Apparatus

Weigh mass of solid Ba(OH)₂ accurately on a balance using a watch glass.
Dissolve solid in a beaker using distilled water (use less than 250 cm³).
Transfer solution into a 250 cm³ volumetric flask .
Rinse beaker and glass rod with distilled water and transfer washings into the flask.
Make up to the mark using distilled water (bottom of meniscus on the line).
Invert the stoppered flask several times to ensure thorough mixing.

❌ Common Errors & Examiner Guidance

Apparatus Penalties: Do not use a conical flask or round-bottom flask to make up a standard solution—it must be a volumetric flask.
Washing step omitted: Forgetting to rinse the beaker loses marking points for complete transfer.

Part (b) — [7 Marks]

Titration Analysis and Structural Determination

📐 Calculation of Molar Mass

Moles of Ba(OH)₂ in titre:
0.150 × (23.50 / 1000) = 3.525 × 10⁻³ mol
Moles of Acid D in 25.0 cm³ sample:
Ratio is 1 mol Ba(OH)₂ : 2 mol D .
2 × 3.525 × 10⁻³ = 7.05 × 10⁻³ mol
Moles of Acid D in total 100 cm³ solution:
7.05 × 10⁻³ × (100 / 25.0) = 0.0282 mol
Molar Mass of Acid D:
M = mass / moles = 3.215 / 0.0282 = 114 g mol⁻¹

💡 Dedurenching the Formula

General formula given: CₙH₂ₙ₋₁COOH , which contains one carboxylic acid group ( -COOH , mass = 45 g mol⁻¹).

Alkyl chain mass = 114 - 45 = 69 g mol⁻¹ .

For CₙH₂ₙ₋₁ : 12n + (2n - 1) = 69 → 14n = 70 → n = 5 .

Molecular formula: C₅H₉COOH (or C₆H₁₀O₂ ).

✅ Correct Answers: Cis Stereoisomers

Compound D is an unsaturated cis stereoisomer with formula C₅H₉COOH . To achieve full marks for structures, drawings must explicitly display:

A restricted rotation C=C double bond with cis (same side) arrangement of identical or priority groups.
Different groups attached to each carbon atom of the C=C double bond.
Valid examples include: CH₃CH₂CH₂CH=CHCOOH (cis-hex-2-enoic acid), CH₃CH₂CH=CHCH₂COOH , or branched isomers such as (CH₃)₂CHCH=CHCOOH arranged in the *cis* conformation around the double bond.

❌ Common Calculation Traps

Stoichiometry errors: Forgetting to multiply by 2 for the 1:2 acid-base reaction ratio yields M = 228 g mol⁻¹ (losing marks).
Dilution scaling errors: Failing to scale up from the 25.0 cm³ aliquot to the full 100.0 cm³ volumetric solution results in M = 456 g mol⁻¹ .
Stereochemistry errors: Drawing *trans* (E) isomers instead of *cis* (Z) isomers limits structural marks.

Exam Technique Tip: Always retain full calculator precision during intermediate steps and only round to appropriate significant figures (matching data given, usually 3 SF) on your final numerical answer line.

Topics

Module 2: Foundations in chemistry · Module 4: Core organic chemistry · Practical Activity Groups · PAG 2: Acid-base titration · 4.1 Basic concepts and hydrocarbons · 2.1 Atoms and reactions

Question and mark scheme from the OCR A-Level Chemistry examination, Unified chemistry (03), November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.