Edexcel A-Level Chemistry AS Paper 1, November 2021: Question 2

17 marks · Medium difficulty · Calculations

Describe the preparation of a standard solution, determine titration end-point colours and mean titres, and perform stoichiometric calculations involving hydrates, acid-base neutralisation, and percentage yield.

Practise this question

Question

Exam question about sodium carbonate hydrates containing multiple parts. Part (a) asks to describe how to make a 250 cm3 standard solution from 10.0 g of the hydrate. Part (b) presents a titration table with three titrations of sodium carbonate against hydrochloric acid, asking multiple-choice for methyl orange indicator colour change, reasons for discarding a titration, calculation of mean titre, and relative formula mass of the hydrate. Part (c) asks for Mr of anhydrous sodium carbonate and x in the formula. Part (d) provides a two-stage manufacturing process and asks to calculate the maximum mass of sodium carbonate obtainable from 500 kg of sodium chloride.
Question text

2 This question is about sodium carbonate.

(a) Sodium carbonate forms a number of hydrates with the general formula Na2CO3.xH2O.

A 250 cm3 standard solution of one of these hydrates contained 10.0 g of the compound.

Describe, including the names of any relevant apparatus, how to make this standard

solution when provided with 10.0g of the hydrate in a beaker.

(5)

(b) 25.0 cm3 portions of the standard solution described in (a) are titrated with

hydrochloric acid solution of concentration 0.300 mol dm–3, using methyl orange

as an indicator.

The table shows the results for this titration.

Titration 1 Titration 2 Titration 3

Final volume / cm3 30.25 29.75 31.25

Initial volume / cm3 0.30 0.90 2.60

Total titre / cm3 29.95 28.85 28.65

4 (i) What is the colour change at the end-point of the reaction?

*P67083A0424* (1)

From To

A red orange

B red yellow

C yellow orange

D yellow red

(ii) State why the value for the total titre in Titration 1 should not be used to

calculate the mean titre.

(1)

(iii) Calculate the mean titre.

(1)

(iv) Calculate the relative formula mass, Mr, of the hydrated sodium carbonate,

Na2CO3.xH2O.

The equation for the reaction in the titration is

Na2CO3 + 2HCl → 2NaCl + H2O + CO2

(4)

*P67083A0524*

(c) In an experiment, the Mr of a different hydrated sodium carbonate was found to

be 286 g mol–1.

(i) Calculate the relative formula mass of anhydrous sodium carbonate, Na2CO3.

(1)

(ii) Calculate the number of molecules of water of crystallisation, x, for this

hydrated sodium carbonate, Na2CO3.xH2O.

(1)

(d) Sodium carbonate is manufactured from sodium chloride in a two-stage process.

NaCl + NH3 + CO2 + H2O → NaHCO3 + NH4Cl

2NaHCO3 → Na2CO3 + H2O + CO2

Calculate the maximum mass of sodium carbonate, Na2CO3, which could be

obtained from 500kg of sodium chloride.

6 (3)

*P67083A0624*

(Total for Question 2 = 17 marks)

Mark scheme

Show the mark scheme Mark scheme providing detailed marking points for all parts of Question 2, including standard solution preparation steps, correct multiple choice answer C for indicator colour change, titration concordance criteria, calculation steps for moles, relative molecular mass, water of crystallisation, and stoichiometric mass calculations for the industrial process.

Question Answer Additional Guidance Mark

Number

2(a) An answer that makes reference to the following points: Marks 1 to 3 can be scored on one of two routes (5)

Marks 4 and 5 are scored by either route

Route 1 – Dissolve the solid in a beaker

• dissolve the solid in distilled / deionised water Do not award if the solid is dissolved in 250 cm3

(using a glass rod) (1) Award with distilled / deionised water anywhere in

the answer

• pour the solution into a volumetric flask (using a

funnel) (1)

• rinse the beaker and transfer the washings to the

conical flask (and rinse the funnel and glass rod) (1)

Route 2 – Transfer the solid to the volumetric flask

• transfer the solid to a volumetric flask (through a

solids funnel) (1)

• rinse the container (and funnel) with distilled /

deionised water and transfer washings (1) Award with distilled / deionised water anywhere in

the answer

• dissolve the solid (in less than 250 cm3)

(1) Do not award if the solid is dissolved in 250 cm3

Both routes

• make up to the mark / line / 250 cm3

(1)

• shake / mix / swirl the flask

(1) For an answer making no reference to deionised /

distilled water max (4)

Question Answer

Mark

Number

2(b)(i) The only correct answer is C (yellow orange) (1)

A is not correct because this colour change would be for the acid in the flask and the carbonate in the burette

B is not correct because this colour change would be for the acid in the flask and the carbonate in the burette and going

beyond the end point

D is not correct because this colour change would go beyond the end-point

Number

2(b)(ii) (1)

• it is not concordant with the other two Allow it is not within ± 0.2 cm3 of the results of Titrations

results / it is not within 0.2 cm3 of the other 2 and 3

results Allow values less than 0.2 cm3 apart

Allow ‘the volume is a lot larger than the other results’ but

not ‘the volume is a bit larger than the other results’

Do not award just ‘it is not close to the other results’

Number

2(b)(iii) Example of calculation (1)

• calculation of mean titre ( 28.85 + 28.65) ÷ 2 = 28.75 (cm3)

Number

2(b)(iv) Example of calculation (4)

• calculation of moles of hydrochloric acid (1) = 28.75 × 0.300 = 0.008625 / 8.625 × 10-3 (mol)

1000

3 (1) = 0.008625 = 0.0043125 / 4.3125 × 10-3 (mol)

• calculation of moles of Na2CO3 in 25 cm

3 (1) = 0.0043125 × 10 = 0.043125 / 4.3125 × 10-2 (mol)

• calculation of moles of Na2CO3 in 250 cm

(1) = 10.0 = 232 (g mol-1)

• calculation of Mr of Na2CO3.xH2O

0.043125

Correct answer with no working scores (4)

Allow TE throughout

Ignore SF except 1SF

Number

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

• calculation of Mr of Na2CO3 = (2 x 23) + 12 + (3 x 16) = 106

Correct answer with no working scores (1)

Number

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

• calculation of mass of water in 1 mole of = 286 – 106 = 10

Na2CO3.xH2O and calculation of x. 18

Correct answer with no working scores (1)

Ignore SF

Allow TE on incorrect Mr from 2(c)(i)

Number

2(d) Example of calculation (3)

• calculation of the number of moles of sodium chloride (1) = 500 000 = 8547 / 8.547 x 103 (mol)

58.5

= 8547 = 4273.5 / 4.2735 x 103 (mol)

• calculation of number of possible moles of sodium

carbonate (1) 2

• calculation of the mass of sodium carbonate (1) = 4273.5 x 106 = 452991 (g) / 453000 (g) /

452.99 (kg) / 453 (kg)

Correct answer with no working scores 3

Ignore SF except 1 SF throughout

Allow TE throughout

Allow TE on incorrect Mr of sodium carbonate

from 2(c)(i)

(Total for Question 2 = 17 marks)

How to answer it

Edexcel AS Level Chemistry Study Guide: Sodium Carbonate Hydrates

What this question tests

This multi-step question assesses core practical skills (making standard solutions, identifying titration end-point indicators, handling concordant/non-concordant titration data) alongside quantitative chemistry calculations including stoichiometry, reacting ratios, moles, molar mass (Mr), water of crystallisation, and industrial percentage yield/scaling calculations.

Question 2 (a) - Making a Standard Solution

Preparation of a 250 cm³ Standard Solution

✅ Full-Mark Method (Route 1)

  • Dissolve the 10.0 g solid in distilled/deionised water using a beaker and glass rod.
  • Pour the solution carefully through a funnel into a 250 cm³ volumetric flask.
  • Rinse the beaker, glass rod, and funnel with distilled water and transfer all washings into the volumetric flask.
  • Make up to the mark (calibration line) using distilled water (add dropwise near the bottom of the meniscus).
  • Stopper and invert/shake/swirl the flask to ensure thorough mixing.

💡 Key Apparatus & Terminology

  • Volumetric flask (250 cm³): Precisely calibrated for a single fixed volume.
  • Distilled / Deionised water: Prevents introduction of interfering ions found in tap water.
  • Meniscus: The bottom of the liquid curve must sit exactly on the graduation line at eye level.

🧠 Exam Technique & Mark Allocation

This is a 5-mark procedural question. Marks are split between dissolution steps, quantitative transfer steps, rinsing steps, making up to the exact mark, and mixing. Mentioning distilled water is essential—if you omit it entirely, you cap your maximum score at 4 marks!

❌ Common Errors

  • Attempting to dissolve the solid directly inside the 250 cm³ volumetric flask from the start (impossible to mix/stir safely).
  • Forgetting to rinse the beaker and funnel, leaving residual solute behind.
  • Failing to mention distilled/deionised water.
Question 2 (b) - Titration Analysis

Titration Calculations and Indicators

(i) Indicator Colour Change

✅ Correct Answer: C (yellow to orange)

Using methyl orange with hydrochloric acid (acid in burette) added to sodium carbonate (alkali in flask): At the end-point, the solution changes from yellow to orange.

(ii) Titration 1 Exclusion

✅ Correct Answer

Titration 1 is 29.95 cm³ , whereas Titrations 2 and 3 are 28.85 cm³ and 28.65 cm³ . Titration 1 is not concordant (it is not within ±0.20 cm³ of the other titres).

(iii) Mean Titre Calculation

📐 Step-by-Step Calculation

Only average the concordant titres (Titrations 2 and 3):

(28.85 + 28.65) ÷ 2 = 28.75 cm³

(iv) Molar Mass of Hydrated Sodium Carbonate

📐 Step-by-Step Calculation (4 Marks)

  1. Moles of HCl:
    (28.75 × 0.300) ÷ 1000 = 0.008625 mol (1 mark)
  2. Moles of Na₂CO₃ in 25.0 cm³ portion:
    From equation ( Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂ ), ratio is 1:2.
    0.008625 ÷ 2 = 0.0043125 mol (1 mark)
  3. Moles of Na₂CO₃ in full 250 cm³ solution:
    0.0043125 × 10 = 0.043125 mol (1 mark)
  4. Molar mass (Mr):
    Mr = mass ÷ moles = 10.0 ÷ 0.043125 = 232 g mol⁻¹ (1 mark)

❌ Common Calculation Traps

Students often forget to scale up from the 25 cm³ pipette portion to the 250 cm³ volumetric flask volume (multiplying by 10). Another common mistake is misapplying the 1:2 stoichiometric mole ratio between the acid and carbonate.

Question 2 (c) - Water of Crystallisation

Determining the Value of x in Na₂CO₃·xH₂O

(i) M_r of Anhydrous Sodium Carbonate

📐 Calculation

Mr(Na₂CO₃) = (2 × 23.0) + 12.0 + (3 × 16.0) = 106 g mol⁻¹

(ii) Calculating x

📐 Step-by-Step Calculation

Given total molar mass of hydrate = 286 g mol⁻¹

  1. Mass of water in 1 mole = 286 - 106 = 180 g mol⁻¹
  2. Number of water molecules (x) = 180 ÷ 18.0 = 10 (So the formula is Na₂CO₃·10H₂O)
Question 2 (d) - Industrial Synthesis & Yield

Calculating Maximum Mass of Product

📐 Step-by-Step Calculation (3 Marks)

Two-stage overall stoichiometry check:
1) NaCl + NH₃ + CO₂ + H₂O → NaHCO₃ + NH₄Cl
2) 2NaHCO₃ → Na₂CO₃ + H₂O + CO₂
Combining these gives a 2:1 molar ratio of NaCl : Na₂CO₃ .

  1. Moles of NaCl:
    Mass = 500 kg = 500,000 g. Mr(NaCl) = 58.5.
    500,000 ÷ 58.5 = 8547.0 moles (1 mark)
  2. Moles of Na₂CO₃ produced:
    Apply 2:1 ratio: 8547.0 ÷ 2 = 4273.5 moles (1 mark)
  3. Mass of Na₂CO₃:
    Mr(Na₂CO₃) = 106.
    4273.5 × 106 = 452,991 g = 453 kg (to 3 sig fig) (1 mark)

🧠 Examiner Guidance on Significant Figures

Final answers should match the precision of input data (usually 3 significant figures). Intermediate rounding should be avoided to prevent rounding errors in the final mark.

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 1, November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.