AQA A-Level Chemistry AS Paper 1, June 2025: Question 5

8 marks · Medium difficulty · Practical Techniques & Data Analysis

Determine the relative molecular mass of a diprotic acid from titration data, calculate the burette percentage uncertainty, and suggest how to reduce it.

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Question

Question 05 describes an experiment to determine the relative molecular mass of a solid acid H2A reacting with NaOH according to H2A + 2NaOH -> Na2A + 2H2O. 2.09 g of H2A is dissolved to make 250.0 cm3 of solution; 25.0 cm3 portions are titrated with 0.380 mol dm-3 NaOH. Table 3 lists burette readings: Rough (11.05 cm3), 1 (10.65 cm3), 2 (10.85 cm3), and 3 (10.55 cm3). Part 05.1 asks to calculate the relative molecular mass of H2A (5 marks). Part 05.2 gives an uncertainty of ±0.15 cm3 for the burette and asks for the percentage uncertainty in titration 1 (1 mark). Part 05.3 asks for a suggestion and justification to reduce percentage uncertainty without changing apparatus (2 marks).
Question text

05 An experiment is done to determine the relative molecular mass of a solid acid, H2A

H2A reacts with sodium hydroxide as shown.

H2A + 2NaOH → Na2A + 2H2O

Method

• 2.09 g of H A are dissolved in water and the solution is made up to 250.0 cm3 in

a volumetric flask.

• 25.0 cm3 of this solution are transferred to a conical flask using a pipette.

• A few drops of phenolphthalein are added.

• 0.380 mol dm–3 sodium hydroxide solution is added from a burette until the

indicator changes colour.

Table 3 shows the results of the experiment.

Table 3

Rough 1 2 3

Final burette reading / cm3 11.05 23.40 34.25 44.80

Initial burette reading / cm3 0.00 12.75 23.40 34.25

Titre / cm3 11.05 10.65 10.85 10.55

05.1 Calculate the relative molecular mass of H2A

[5 marks]

Relative molecular mass of H2A

05.2 The uncertainty in the use of the burette in each titration is ±0.15 cm3

Calculate the percentage uncertainty in the use of the burette in titration 1.

[1 mark]

Percentage uncertainty

05.3 Suggest one way to reduce the percentage uncertainty in each titration without

*12changing the apparatus.*

Justify your answer.

[2 marks]

Suggestion

Justification

Mark scheme

Show the mark scheme Mark scheme for question 05: 05.1 awards 5 marks for M1 (mean concordant titre = 10.60 cm3), M2 (moles of NaOH = 4.028 x 10^-3 mol), M3 (moles of H2A in 25 cm3 = 2.014 x 10^-3 mol), M4 (moles in 250 cm3 = 2.014 x 10^-2 mol), and M5 (Mr = 103.8 or 104). 05.2 gives 1 mark for (0.15 / 10.65) x 100 = 1.4(1)%. 05.3 gives 2 marks: M1 for using more dilute NaOH, a larger mass of solid H2A, or larger concentration of H2A; M2 for justification that this leads to a larger titre value or larger volume of NaOH required.

Question Marking guidance Additional Comments/Guidelines Mark

M1: Mean titre = 10.60 cm3 M1: Calculation of mean titre using concordant

titres

0.380×M1

M2: n(NaOH) = 1000 –3

M2: n(NaOH) = 4.028 × 10 (mol)

3 M2 5

M3: n(H2A in 25.0 cm ) = M3: n(H A in 25.0 cm3) = 2.014 × 10–3 (mol)

05.1 2 2 (4 x AO2,

33 –3 1 x AO3)

M4: n(H2A in 250 cm ) = M3 × 10 M4: n(H2A in 250 cm ) = 2.014 × 10 × 10

= 2.014 × 10–2 (mol)

Mass 2.09

M5: Mr = Moles = M4

M5: Mr = 103.8 / 104

0.15 1

05.2 10.65 × 100 = 1.4(1)%

(1 x AO2)

M1: Use more dilute NaOH / use a larger mass of solid H2A / use a

larger concentration of H2A

05.3

(2 x AO3)

M2: Would lead to a larger titre value / larger volume of NaOH

required

How to answer it

Titration of a Diprotic Acid & Percentage Uncertainty

What this question tests

  • Concordant Titres: Selecting values within 0.10 cm³ of each other and calculating an accurate mean titre (excluding rough titration).
  • Multi-Step Stoichiometric Calculations: Handling the 1 : 2 reacting mole ratio of a diprotic acid (H₂A) to NaOH, scaling up from an aliquot (25.0 cm³) to a standard solution (250.0 cm³), and determining relative molecular mass (Mr).
  • Apparatus Uncertainty: Calculating percentage apparatus uncertainty for volumetric equipment.
  • Practical Procedural Improvements: Modifying reagent concentrations or masses to reduce overall percentage uncertainty without changing equipment.
Question 05.1 • 5 Marks

Calculate the relative molecular mass of H₂A

Multi-step titration calculation involving dilution and stoichiometry

📐 Step-by-Step Calculation

  1. Identify concordant titres & calculate the mean titre:
    Concordant titres are within 0.10 cm³ of each other: Titre 1 = 10.65 cm³ , Titre 3 = 10.55 cm³ (Difference = 0.10 cm³).
    Note: Titre 2 (10.85 cm³) is non-concordant; ignore the Rough titre (11.05 cm³).
    Mean titre = (10.65 + 10.55) / 2 = 10.60 cm³
  2. Calculate moles of NaOH used in the titration:
    n(NaOH) = (concentration × volume) / 1000
    n(NaOH) = (0.380 × 10.60) / 1000 = 4.028 × 10⁻³ mol
  3. Calculate moles of H₂A in the 25.0 cm³ aliquot:
    Equation: H₂A + 2NaOH → Na₂A + 2H₂O (Mole ratio is 1 H₂A : 2 NaOH)
    n(H₂A in 25.0 cm³) = 4.028 × 10⁻³ / 2 = 2.014 × 10⁻³ mol
  4. Scale up to find moles of H₂A in the total 250.0 cm³ volumetric flask:
    Scaling factor = 250.0 / 25.0 = 10
    n(H₂A in 250 cm³) = 2.014 × 10⁻³ × 10 = 2.014 × 10⁻² mol
  5. Calculate the relative molecular mass (Mr) of H₂A:
    Mr = mass / moles = 2.09 / 2.014 × 10⁻² = 103.77 g mol⁻¹
    Answer: 103.8 or 104

✅ Correct Answer

Relative molecular mass of H₂A: 103.8 (or 104 )

❌ Common Errors

  • Averaging all titres: Including Titre 2 or the Rough titre in the mean will lose M1 immediately.
  • Inverting the ratio: Multiplying NaOH moles by 2 instead of dividing by 2 (missing the 1:2 diprotic stoichiometry).
  • Forgetting the dilution factor: Omitting the ×10 step to convert 25.0 cm³ back to the full 250.0 cm³ solution.
Mark Breakdown:
M1: Mean titre = 10.60 cm³ calculated from concordant titres (1 and 3).
M2: Moles of NaOH = 4.028 × 10⁻³ mol.
M3: Moles of H₂A in 25.0 cm³ = M2 / 2 = 2.014 × 10⁻³ mol.
M4: Moles of H₂A in 250 cm³ = M3 × 10 = 2.014 × 10⁻² mol.
M5: Mr = 2.09 / M4 = 103.8 (or 104).
Question 05.2 • 1 Mark

Calculate the percentage uncertainty in Titration 1

Determining percentage apparatus uncertainty for a single run

📐 Calculation

% uncertainty = (uncertainty / titre volume) × 100

% uncertainty = (0.15 / 10.65) × 100 = 1.408... %

Answer: 1.4% or 1.41%

🧠 Exam Technique

  • Read carefully: the question specifically asks for titration 1 ( 10.65 cm³ ), not the mean titre.
  • The total burette uncertainty ( ±0.15 cm³ ) is given directly in the question prompt. Do NOT multiply it by 2.
Mark Breakdown:
M1: (0.15 / 10.65) × 100 = 1.4% or 1.41%
Question 05.3 • 2 Marks

Reducing percentage uncertainty without changing apparatus

Practical analysis: modifying experimental design to increase titre volume

✅ Correct Answer

Suggestion (Any one):

  • Use a more dilute solution of sodium hydroxide (lower [NaOH]).
  • Use a larger mass of solid H₂A (or a higher concentration of H₂A).

Justification:

  • This would lead to a larger titre value (a greater volume of NaOH solution required to neutralize the acid).

💡 Key Knowledge

  • Percentage uncertainty = (Apparatus Uncertainty / Reading) × 100 .
  • Because you cannot change the apparatus (the numerator is fixed), you must increase the reading (the denominator) to reduce the overall percentage uncertainty.
  • A larger titre volume means the fixed burette error represents a smaller fraction of the measured volume.

❌ Common Errors

  • Changing the apparatus: Suggesting a 3-decimal place balance, Grade A pipette, or digital burette scores zero because the question explicitly specifies "without changing the apparatus".
  • Confusing concentration effects: Stating "use more concentrated NaOH" will make the titre even smaller, which increases percentage uncertainty.
  • Missing the link: Providing a valid suggestion but failing to justify it by stating that it results in a larger titre / volume.
Mark Breakdown:
M1 (Suggestion): Use more dilute NaOH / use a larger mass of solid H₂A / use a larger concentration of H₂A.
M2 (Justification): Would lead to a larger titre value / larger volume of NaOH required.

Topics

Physical Chemistry · Required Practicals · 3.1.2 Amount of Substance · Required Practical 1: Making up a volumetric solution

Question and mark scheme from the AQA A-Level Chemistry examination, AS Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.