Edexcel A-Level Chemistry AS Paper 2, June 2024: Question 1

9 marks · Medium difficulty · Practical Techniques and Data Analysis

Use titration data for an unknown dicarboxylic acid to determine the mean titre, the relative molecular mass of the acid, and the value of n in its molecular formula.

Practise this question

Question

Question 1 outlines a titration experiment where a solid dicarboxylic acid, HOOC(CH2)nCOOH (represented as H2A), is dissolved to make 250 cm³ of solution and titrated against 0.100 mol dm⁻³ sodium hydroxide. Part (a) is a multiple-choice question on selecting the correct apparatus to make exactly 250 cm³ of solution. Part (b) includes a diagram of a burette showing a meniscus between 2.0 and 3.0 cm³, asking for the initial reading, the rough titre calculation, and an explanation of how the rough titration is used during accurate titrations. Part (c) provides the balanced equation H2A + 2NaOH -> Na2A + 2H2O and a table of three accurate titration results (titres of 23.30, 23.35, and 23.25 cm³), followed by sub-questions to calculate the mean titre, moles of NaOH, and given 1.54 g of acid was dissolved, the relative molecular mass of H2A and the integer value of n.
Question text

1 The relative molecular mass of a solid dicarboxylic acid, HOOC(CH2)nCOOH, can be

found using a titration. The acid, which can be represented as H2A, was dissolved in

deionised water and the solution made up to 250 cm3.

(a) Which piece of apparatus should be used for making a solution with a volume of

exactly 250 cm3 ?

(1)

A burette

B measuring cylinder

C pipette

D volumetric flask

(b) A solution of 0.100 mol dm–3 sodium hydroxide solution was added to a burette.

A rough titration was carried out on a 25.0 cm3 portion of the acid solution.

(i) The diagram shows the burette before the rough titration.

What is the initial burette reading for this titration?

(1)

2.0

3.0

A 2.40 cm3

B 2.45 cm3

C 3.55 cm3

D 3.60 cm3

(ii) The final burette reading for the rough titration was 26.50 cm3.

Calculate the volume of sodium hydroxide solution added in the

rough titration, using your answer to (b)(i).

(1)

*P76894A0232*

(iii) Describe how you would use the rough titration value when carrying out the

accurate titrations.

(1)

(c) 25.0 cm3 portions of the acid solution were titrated with

0.100 mol dm–3 sodium hydroxide solution.

The equation for the reaction is shown.

H2A(aq) + 2NaOH(aq) → Na2A(aq) + 2H2O(l)

The acid solution was pipetted into a conical flask and titrated.

The accurate titrations were carried out three times.

The following results were recorded for the accurate titrations.

Titration number 1 2 3

Burette reading (final) / cm3 47.80 24.35 47.60

Burette reading (initial) / cm3 24.50 1.00 24.35

Volume of NaOH used / cm3 23.30 23.35 23.25

(i) Calculate the mean titre for these accurate titrations.

(1)

*P76894A0332*

(ii) Calculate the number of moles of sodium hydroxide in the mean titre.

(1)

(iii) The mass of H A used to make up 250 cm3 of solution in the experiment

was 1.54 g.

Calculate the relative molecular mass of H2A and therefore the value of n in

HOOC(CH2)nCOOH.

You must show your working.

(3)

*P76894A0432*

(Total for Question 1 = 9 marks)

Mark scheme

Show the mark scheme Mark scheme for Question 1: (a) D (volumetric flask). (b)(i) B (2.45 cm³). (b)(ii) Calculation: 26.50 - 2.45 = 24.05 cm³, with TE allowed. (b)(iii) Explains adding NaOH rapidly until near the rough value, then dropwise. (c)(i) Mean titre: (23.30 + 23.35 + 23.25) / 3 = 23.30 cm³. (c)(ii) Moles of NaOH = 23.30 x 0.100 / 1000 = 0.00233 mol. (c)(iii) 3 marks: moles of H2A in 25.0 cm³ = moles of NaOH / 2 = 0.001165 mol; Mr = (1.54 / 10) / moles of H2A = 132; n = (132 - 90) / 14 = 3.

Question

Answer Mark

Number

1(a) The only correct answer is D (volumetric flask) (1)

A is not correct because this is used to measure volumes of liquids required in a reaction

B is not correct because this is not a very accurate way to measure volumes

C is not correct because this is used to measure fixed volumes of liquids, not to make up solutions

Question

Answer Mark

Number

1(b)(i) The only correct answer is B (2.45 cm3) (1)

A is not correct because the volume must be measured from the bottom of the meniscus

C is not correct because this is the volume reading from the bottom of the burette, not the top

D is not correct because this is reading from the bottom and not using the bottom of the meniscus

Question

Answer Additional Guidance Mark

Number

1(b)(ii) Example of calculation (1)

• calculation of volume by subtracting the answer to (b)(i) 26.50 ‒ 2.45 = 24.05 (cm3)

from 26.50 cm3

Allow TE on incorrect answers to (b)(i)

Question

Answer Additional Guidance Mark

Number

1(b)(iii) An answer that makes reference to the following point: (1)

• sodium hydroxide solution can be added rapidly / not Allow add acid dropwise / slow down as it gets close to /

drop by drop to a volume a little below that of the before the rough titration value / value in (b)(ii)

rough titration (making the titration quicker) Allow added rapidly to the volume given in (b)(ii) minus

about 2 cm3 or so

Do not award slow down when you get to the rough

titration value

Do not award the rough titration is a value that you try to

get as close to as possible

Question

Answer Additional Guidance Mark

Number

1(c)(i) Example of calculation (1)

• calculation of mean titre (23.30 + 23.35 + 23.25) ÷ 3 = 23.3(0) (cm3)

Correct answer with no working scores 1

Question

Answer Additional Guidance Mark

Number

1(c)(ii) Example of calculation (1)

• calculation of number of moles (23.30 × 0.100) ÷ 1000 = 0.00233 / 2.33 × 10‒3 (mol)

Allow TE on (c)(i)

Question

Answer Additional Guidance Mark

Number

1(c)(iii) Example of calculation (3)

• calculation of number of moles of H A in 25.0 cm3 (1) Answer to (c)(ii) ÷ 2

= 0.001165 / 1.165 × 10‒3 (mol)

(1) M = (1.54 ÷ 10) ÷ moles of H A in 25.0 cm3

• calculation of relative molecular mass of H2A r 2

= 132 / 132.188/ 132.19

• calculation of value of n giving the answer as an (1) n = (132 ‒ 90) ÷ 14 = 3

integer

Correct answer with some working scores (3)

Allow TE throughout (c) and from (c)(ii)

Failure to divide by 10 in M2 gives n = 88 and scores (2)

Ignore SF

(Total for Question 1 = 9 marks)

How to answer it

Titration Analysis of a Dicarboxylic Acid

What this question tests

This 9-mark practical-based question assesses your core laboratory and quantitative chemistry skills:

  • Apparatus selection: Choosing the correct glassware for preparing standard solutions accurately.
  • Reading a burette: Reading inverted scales from the bottom of the meniscus to two decimal places (ending in .00 or .05 cm³).
  • Titration procedure: Understanding the function of a rough (trial) titration to save time while maintaining accuracy.
  • Processing volumetric data: Selecting concordant titres to calculate an accurate mean titre.
  • Multi-step stoichiometric calculation: Using a 1:2 reacting ratio, scaling from a pipette sample (25.0 cm³) to a volumetric flask (250 cm³), and deducing the molecular formula of an unknown dicarboxylic acid chain HOOC(CH₂)ₙCOOH.
Part (a) · 1 Mark

Apparatus for Preparing a Standard Solution

Which piece of apparatus should be used for making a solution with a volume of exactly 250 cm³?

✅ Correct Answer

D — volumetric flask

💡 Key Knowledge

A volumetric flask is designed specifically to contain one precise, calibrated total volume (here 250.0 cm³) when filled until the bottom of the meniscus touches the calibration line.

❌ Common Distractors Explained

  • A (burette): Measures variable delivered volumes up to 50 cm³, not 250 cm³.
  • B (measuring cylinder): Used for approximate liquid measurements; lacks analytical accuracy.
  • C (pipette): Measures and delivers a single fixed volume (typically 10.0 or 25.0 cm³), not used for making up 250 cm³ standard solutions.

🧠 Exam Technique

Notice the keyword "making a solution". Whenever a solid is dissolved and made up to an exact total volume, the answer is always a volumetric flask.

Mark breakdown: [1 mark] for selecting D.
Part (b)(i), (b)(ii), (b)(iii) · 3 Marks

Burette Readings & Titration Technique

(b)(i) Reading the Initial Burette Level (1 Mark)

✅ Correct Answer

B — 2.45 cm³

💡 Key Knowledge

Burette scales run downwards (0.00 at the top, 50.00 at the bottom). Each small graduation is 0.10 cm³. Readings must be taken from the bottom of the meniscus.

❌ Why Other Options Are Wrong

  • A (2.40 cm³): Read from the top rim of the meniscus curve rather than the bottom.
  • C (3.55 cm³) & D (3.60 cm³): Read upside down (reading up from 3.0 instead of down from 2.0).

(b)(ii) Calculating the Rough Titre (1 Mark)

📐 Calculation

Titre = Final reading − Initial reading

26.50 − 2.45 = 24.05 cm³

🧠 Transfer of Error (TE)

If you incorrectly chose an answer in (b)(i), the mark scheme allows full credit here if you correctly subtracted your (b)(i) answer from 26.50 cm³.

(b)(iii) Using the Rough Titre in Subsequent Runs (1 Mark)

✅ Correct Answer

Add the sodium hydroxide solution rapidly until reaching a volume slightly below the rough titre (e.g. up to ~22 cm³), then add it dropwise / slowly near the end-point.

❌ Common Student Errors

  • Stating: "Slow down when you get to the rough titre value." — This causes overshooting, as the rough titre is already beyond the true end-point!
  • Stating: "The rough titre is a target to reach." — Rough runs are intentionally inaccurate.
Mark breakdown: [1 mark] for 2.45 cm³; [1 mark] for 24.05 cm³ (allow TE); [1 mark] for adding rapidly then drop by drop before the rough value.
Part (c)(i), (c)(ii), (c)(iii) · 5 Marks

Titration Calculations & Molecular Formula Determination

Reaction equation: H₂A(aq) + 2NaOH(aq) → Na₂A(aq) + 2H₂O(l)

(c)(i) Calculate Mean Titre (1 Mark)

📐 Calculation

Titres recorded: 23.30, 23.35, 23.25 cm³ (all concordant within ±0.10 cm³).

Mean = (23.30 + 23.35 + 23.25) ÷ 3 = 23.30 cm³

🧠 Exam Technique

Always check concordancy! Here all three titres are within 0.10 cm³ of each other (range = 23.35 − 23.25 = 0.10 cm³), so all three must be included. Always write to 2 decimal places: write 23.30, not 23.3.

(c)(ii) Calculate Moles of NaOH (1 Mark)

📐 Calculation

n(NaOH) = (volume in cm³ ÷ 1000) × concentration

n(NaOH) = (23.30 ÷ 1000) × 0.100 = 2.33 × 10⁻³ mol (or 0.00233 mol)

(c)(iii) Determine Mr of H₂A and Find Value of n (3 Marks)

Mass of solid H₂A dissolved in 250 cm³ = 1.54 g.

📐 Step-by-Step Calculation

1 Find moles of H₂A in the 25.0 cm³ sample (Mark 1):

From equation, ratio of H₂A : NaOH is 1 : 2.
n(H₂A in 25.0 cm³) = n(NaOH) ÷ 2 = (2.33 × 10⁻³) ÷ 2 = 1.165 × 10⁻³ mol

2 Find Mr of H₂A (Mark 2):

Method A (scale moles to 250 cm³):
n(H₂A in 250 cm³) = 1.165 × 10⁻³ × 10 = 1.165 × 10⁻² mol
Mr = mass ÷ total moles = 1.54 ÷ 0.01165 = 132.19 ≈ 132

Method B (scale mass to 25.0 cm³):
mass in 25.0 cm³ = 1.54 ÷ 10 = 0.154 g
Mr = 0.154 ÷ (1.165 × 10⁻³) = 132.19 ≈ 132

3 Deduce integer value of n in HOOC(CH₂)ₙCOOH (Mark 3):

Mass of two −COOH groups = 2 × [12.0 + (16.0 × 2) + 1.0] = 2 × 45.0 = 90.0
Mass of (CH₂)ₙ chain = 132 − 90 = 42
Mass of one CH₂ group = 12.0 + (2 × 1.0) = 14.0
n = 42 ÷ 14 = 3

❌ Common Calculation Traps

  • Forgetting the 1:2 ratio: Failing to divide moles of NaOH by 2.
  • Forgetting the 250 cm³ / 25 cm³ factor of 10: Students who forgot to divide/multiply by 10 calculated Mr ≈ 1320 and ended up with n = 88 (capped at 2 marks max).
  • Leaving n as a decimal: The question asks for the integer value of n, so state clearly n = 3 .

💡 Chemical Identity Check

When n = 3 , the acid is glutaric acid (pentanedioic acid): HOOC−(CH₂)₃−COOH , with molecular formula C₅H₈O₄ and exact Mr = 132.1 g mol⁻¹.

Mark breakdown: [1 mark] for n(H₂A in 25 cm³) = 1.165 × 10⁻³ mol; [1 mark] for Mr = 132 (accept 132 to 132.2); [1 mark] for integer n = 3 with clear working shown.

Topics

Physical Chemistry · Core Practicals · Topic 5: Formulae, Equations and Amounts of Substance · Core Practical 2: Preparation of a standard solution from a solid acid · Core Practical 3: Find the concentration of a solution of hydrochloric acid

Question and mark scheme from the Edexcel A-Level Chemistry examination, AS Paper 2, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.