OCR A-Level Chemistry AS Depth in chemistry (02), June 2018: Question 1

13 marks · Medium difficulty · Practical Questions

Perform calculations and analysis based on a titration of a weak acid with sodium hydroxide, including determining the molar mass and structure of the acid.

Practise this question

Question

A multi-part chemistry exam question about an acid-base titration. It includes definitions for standard solution and alkali, diagrams of burette readings for three titrations, a table to record final and initial burette readings and titres, percentage uncertainty calculations, identifying errors in solution preparation, and calculation of the molar mass and suggested structure of a dicarboxylic acid A.
Question text

1 A student carries out a titration to determine the molar mass and structure of a weak acid A.

The student follows the method below.

• Dissolve a weighed mass of A in 100 cm3 of distilled water and make the solution up to

250 cm3 in a beaker.

• Add the solution of A to a burette.

• Titrate the solution of A with a standard solution of sodium hydroxide, NaOH.

(a) What is meant by the term standard solution?

… [1]

(b) Sodium hydroxide is an alkali.

What is meant by the term alkali?

… [1]

(c) The student carries out a trial, followed by three further titrations.

The diagram shows the initial and final burette readings for the three further titrations.

The student measures all burette readings 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 readings and the titres in the table below.

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

analysing the results.

Titration 1 Titration 2 Titration 3

Final reading/cm3

Initial reading/cm3

Titre/cm3

mean titre = … cm3 [4]

(ii) The uncertainty in each burette reading is ± 0.05 cm3.

Calculate the percentage uncertainty for the titre in Titration 1.

percentage uncertainty = … % [1]

(iii) The student realised that the solution of A had not been prepared correctly.

How should the student have made up the solution?

… [1]

(d) A student repeats the titration to determine the molar mass and structure of A.

• The student prepares a 250.0 cm3 solution from 1.513 g of A.

• The solution of A is added to the burette and titrated with 25.0 cm3 volumes of

0.112 mol dm–3 NaOH(aq).

• 1 mol of A reacts with 2 mol of NaOH.

• The student obtains a mean titre of 27.30 cm3.

(i) Calculate the molar mass of A from these results.

Give your answer to the nearest whole number.

Show your working.

molar mass of A = … g mol–1 [4]

(ii) A is an organic acid, containing C, H and O only.

One molecule of A contains two COOH groups.

Suggest the structure of A.

[1]

Mark scheme

Show the mark scheme The mark scheme provides acceptable answers and guidance for all parts of question 1. It details the required definitions, correct readings and calculated mean for the burette table, percentage uncertainty calculation, correct use of a volumetric flask, stoichiometric mole calculations leading to the molar mass of 118 g mol-1, and acceptable structural or skeletal formulas for the dicarboxylic acid.

Question Answer Marks Guidance

1 (a) A solution of known concentration 1 ALLOW description of concentration

1 (b) Releases OH– (ions in aqueous solution) 1 ALLOW containing OH– ions

IGNORE mention of pH

1 (c) (i) 4 ANNOTATE ANSWER WITH TICKS AND CROSSES

Final ETC

3 27.30 27.00 27.75

reading/cm

Initial

30.45 0.60 1.25

reading/cm

Titre/cm3 26.85 26.40 26.50

ALLOW missing zeroes for burette readings

i.e. 0.6 for 0.60

Initial and final readings 27 OR 27.0 for 27.00

All burette readings ( 6) correct

ALLOW ECF from incorrect burette readings

Titres recorded to two decimal places with the last figure

either 0 or 5

Correct subtractions to obtain final titre values

IF MEAN IS CALCULATED FROM ECF, IT MUST BE

Mean titre calculated from concordant results FROM CLOSEST TITRES

Correct mean titre = 26.45 (cm3)

ALLOW ecf from incorrect mean

DO NOT ALLOW 26.5 cm3

Question asks for nearest 0.05 cm3

Mean titre recorded to accuracy of burette

Final answer recorded to two decimal places with the

last figure either 0 or 5

1 (c) (ii) 2 0.05 1 ALLOW 0.4 up to full calculation display of

26.85 100 = 0.37(2) (%) 0.372439478

ALLOW ECF FOR CORRECT CALCULATION FROM

1 (c) (i) OR USE OF ANY TITRE

H032/02 Mark scheme June 2018

1 (c) (iii) Use a (250 cm3) volumetric flask (instead of a beaker) 1 IGNORE graduated flask

1 (d) (i) FIRST CHECK ANSWER ON ANSWER LINE 4 ANNOTATE ANSWER WITH TICKS AND CROSSES

If answer = 118 (g mol–1) award 4 marks ETC

–1 6

If answer = 108 (g mol ) award 3 marks Throughout: IGNORE trailing zeroes in intermediate

----------------------------------------------------------------------- working,

n(NaOH) e.g. For n(NaOH) ALLOW 0.0028 for 0.00280

25.0

= 0.112 1000 = 0.00280 (mol)

n(A) in 25.0 cm3

0.00280 ALLOW ECF from incorrect n(NaOH)

= 2 = 0.00140 (mol)

n(A) in 250 cm3 ALLOW ECF from incorrect n(A) OR n(NaOH)

250.0 ALLOW 3 sig fig up to full calculator display correctly

= 0.00140 27.30 = 0.0128 (mol) rounded (0.012820512)

Molar mass, M(A) to nearest whole number. ALLOW ECF from incorrect n(NaOH)

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

1.513 –1 Possible ECFs for 3 marks

= = 118 (g mol ) 1.513 ÷ (0.00140 250/25) = 108

0.0128

1.513 ÷ 0.00140 = 1081

No ÷2 for n(A)

Molar mass A = 59 (g mol–1)

Using mean titre of 26.45 cm3 from 1c(i)

Molar mass A = 114 (g mol–1)

Using 27.3 0.112 in M1 and then 25.0 in M3

Molar mass A = 99 (g mol–1)

H032/02 Mark scheme June 2018

1 (d) (ii) Structure of dicarboxylic acid 1 ALLOW correct structural OR skeletal OR displayed

HOOCCH2CH2COOH OR HOOCCH(CH3)COOH formulae OR a combination

STRUCTURE MUST MATCH Mr from answer to 1 d) i) ALLOW incorrect connectivity e.g̶ HO

(within 10 AMU)

7 ALLOW ECF from incorrect molar mass in (d)(i) but

only if 2 COOH possible and Mr is a close match to

(d) (i) within 10 AMU

Total 13

How to answer it

Overall Question Difficulty:
Medium

Titration Calculations & Practical Techniques Study Guide

What this question tests

This multi-step question assesses core AS practical and calculation skills: defining chemical terminology (standard solutions and alkalis), accurately reading burettes to 0.05 cm³, calculating percentage uncertainties, evaluating practical error correction, performing multi-stage titration stoichiometry calculations, and determining organic structures from molar mass data.

Part (a) — Standard Solution

Definition of a Standard Solution

✅ Correct Answer

A solution of known concentration.

🧠 Exam Technique

Keep definitions precise and concise. Do not confuse a standard solution with a 'standardised' solution or a buffer. Simply stating "known concentration" secures the mark instantly.

Part (b) — Alkali Definition

Definition of an Alkali

✅ Correct Answer

Releases OH⁻ ions (in aqueous solution).

❌ Common Errors

Mentioning pH values (e.g., "has a pH above 7") will not gain credit. Examiners specifically look for the chemical definition involving hydroxide ions.

Part (c) — Burette Readings & Mean Titre

Reading Burettes and Calculating Concordant Means

✅ Correct Table & Mean Values

  • Titration 1: Final = 27.30 | Initial = 0.45 | Titre = 26.85
  • Titration 2: Final = 27.00 | Initial = 0.60 | Titre = 26.40
  • Titration 3: Final = 27.75 | Initial = 1.25 | Titre = 26.50
  • Mean Titre: 26.45 cm³ (averaged from Titrations 2 and 3 as they are concordant within 0.10 cm³).

🧠 Exam Technique

All burette readings must be recorded to 2 decimal places, ending in either .00 or .05 . When calculating a mean titre, only average the concordant titrations (those within 0.10 cm³ of each other).

📐 (ii) Percentage Uncertainty Calculation

Formula: (Total burette error / Titre volume) × 100

Since a burette involves two readings (initial and final), the maximum error is 2 × 0.05 = 0.10 cm³.

Calculation: (2 × 0.05 / 26.85) × 100 = 0.372%

Mark breakdown: Correct substitution leading to ~0.37% (Allows up to full calculator display).

💡 (iii) Practical Correction

Correct Action: Use a 250 cm³ volumetric flask (instead of a beaker) to prepare the solution accurately to a known volume.

Part (d) — Molar Mass & Structure Determination

Stoichiometry and Organic Identification

📐 (i) Step-by-Step Molar Mass Calculation

  1. Moles of NaOH: 0.112 × (25.0 / 1000) = 0.00280 mol
  2. Moles of Acid A in 25.0 cm³ portion: Given a 1:2 reacting ratio, divide by 2. 0.00280 / 2 = 0.00140 mol
  3. Moles of Acid A in total 250 cm³ solution: Scale up by multiplying by (250.0 / 27.30). 0.00140 × (250.0 / 27.30) = 0.01282 mol
  4. Molar Mass of A (M): Mass / Moles = 1.513 / 0.01282 = 118 g mol⁻¹ (to nearest whole number).
Mark breakdown: 4 marks total. First check answer line: 118 awards all 4 marks immediately.

✅ (ii) Structure of Acid A

Accepted Formulae: HOOCCH₂CH₂COOH (butanedioic acid / succinic acid) or HOOCCH(CH₃)COOH (methylmalonic acid).

❌ Common Calculation Traps

  • Forgetting to factor in the 1:2 mole ratio between the acid and NaOH.
  • Failing to scale up from the 25.0 cm³ pipette volume to the full 250 cm³ volumetric solution.
  • Rounding intermediate values too early, leading to off-target final molar masses.

Topics

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

Question and mark scheme from the OCR A-Level Chemistry examination, AS Depth in chemistry (02), June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.