AQA A-Level Chemistry Paper 1, 2018: Question 10

7 marks · Hard difficulty · State/Explain/Numerical

Calculate the value of x in hydrated sodium carbonate using back-titration data.

Practise this question

Question

Question 10.1 outlines a back titration experiment where 627 mg of hydrated sodium carbonate (Na2CO3.xH2O) is reacted with 200 cm3 of 0.250 mol dm-3 hydrochloric acid. The resulting solution is diluted to 250 cm3 in a volumetric flask. 25.0 cm3 portions are then titrated against 0.150 mol dm-3 aqueous sodium hydroxide, requiring a mean titre of 26.60 cm3. The question asks to calculate the integer value of x in Na2CO3.xH2O, showing all working, worth 7 marks.
Question text

. A student added 627 mg of hydrated sodium carbonate (Na CO .xH O) to 200 cm3 of

10 1 2 3 2

0.250 mol dm–3 hydrochloric acid in a beaker and stirred the mixture.

After the reaction was complete, the resulting solution was transferred to a volumetric

flask, made up to 250 cm3 with deionised water and mixed thoroughly.

Several 25.0 cm3 portions of the resulting solution were titrated with 0.150 mol dm–3

aqueous sodium hydroxide. The mean titre was 26.60 cm3 of aqueous sodium

hydroxide.

Calculate the value of x in Na2CO3.xH2O

Show your working.

Give your answer as an integer.

[7 marks]

Value of x

Mark scheme

Show the mark scheme Mark scheme for question 10.1 outlining 7 marks: M1 for calculating moles of HCl added (0.050 mol) and NaOH used (3.99 x 10^-3 mol); M2 for scaling moles of excess HCl to the 250 cm3 flask (3.99 x 10^-2 mol); M3 for moles of HCl reacted with sodium carbonate (0.0101 mol); M4 for moles of Na2CO3.xH2O reacted (5.05 x 10^-3 mol); M5 for converting 627 mg to 0.627 g; M6 for calculating the mass or Mr of xH2O (18); M7 for finding x = 1.

Question Answers Additional Comments/Guidance Mark

M1 HCl added = 0.050 mol and 1

NaOH used in titration = 3.99 x 10-3 mol

M2 So moles that would be needed to neutralise total excess HCl = Alternative: divide moles HCl by 10 = 0.005 and 1

3.99 x 10-3 x 10 = 3.99 x 10-2 mol 0.005 - 3.99 x 10-3 = 0.00101

M3 Therefore the moles of HCl reacted with the Na2CO3.xH2O = Alternative: 0.00101 x 10 to produce 0.0101 1

0.050 - 3.99 x 10-2 = 0.0101 mol

M4 So moles Na2CO3.xH2O reacted with the HCl = 0.0101 /2 = 5.05 x 1

10.1 10-3 mol

M5 Conversion of mg to g = 0.627 (g) or 627 x 10-3 (g) 1

M6 xH O = 0.627/5.05 x 10-3 -106.0 = 18 (.16) Alternative: mass Na CO that reacted with the HCl 1

22 3

5.05 x 10-3 x106.0 = 0.5353 g and mass H O =

0.627- 0.5353 = 0.0917 g

M7 so x =1 Alternative: 0.0917 /18.0 = 5.094 x 10-3 so ratio 1

Na2CO3 to H2O = 1:1.009 ie 1:1 so x = 1

Total 7

How to answer it

Determination of Water of Crystallisation via Back Titration

📌 What this question tests

This 7-mark question assesses quantitative analysis involving an indirect (back) titration. You must demonstrate mastery of:

  • Calculating amount of substance in moles using n = c × V .
  • Scaling up moles across volumetric dilution steps (aliquot vs total flask volume).
  • Applying reacting stoichiometry (1:1 for HCl + NaOH; 2:1 for HCl + Na₂CO₃).
  • Converting mass units accurately ( mg to g ).
  • Determining water of crystallisation ( x ) and expressing the final answer as an integer.
Exam Part: Question 10.1 (7 Marks)

Calculating the Value of x in Na₂CO₃·xH₂O

Multi-Step Back Titration Calculation

📐 Step-by-Step Calculation Breakdown

  1. Initial moles of HCl added:
    n(HCl)initial = (200 / 1000) dm³ × 0.250 mol dm⁻³ = 0.0500 mol
  2. Moles of NaOH used in titration (reacts with excess unreacted HCl in 25.0 cm³):
    Reaction: HCl + NaOH → NaCl + H₂O (1:1 ratio)
    n(NaOH) = (26.60 / 1000) dm³ × 0.150 mol dm⁻³ = 3.99 × 10⁻³ mol
    Therefore, n(HCl) in 25.0 cm³ aliquot = 3.99 × 10⁻³ mol
    🎯 M1 awarded: Correctly finding both initial moles of HCl (0.050 mol) AND moles of NaOH (3.99 × 10⁻³ mol).
  3. Total excess moles of HCl in the 250 cm³ volumetric flask:
    Dilution factor = 250 cm³ / 25.0 cm³ = 10
    Total excess n(HCl) = 3.99 × 10⁻³ mol × 10 = 3.99 × 10⁻² mol (0.0399 mol)
    🎯 M2 awarded: Multiplying titre moles by 10 to find total unreacted HCl in 250 cm³.
  4. Moles of HCl that reacted with Na₂CO₃·xH₂O:
    n(HCl)reacted = n(HCl)initial - n(HCl)excess
    n(HCl)reacted = 0.0500 - 0.0399 = 0.0101 mol (1.01 × 10⁻² mol)
    🎯 M3 awarded: Subtracting excess moles from initial moles to find moles of HCl that reacted.
  5. Moles of Na₂CO₃·xH₂O:
    Equation: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂
    Molar ratio: 1 mol Na₂CO₃ reacts with 2 mol HCl
    n(Na₂CO₃·xH₂O) = 0.0101 / 2 = 5.05 × 10⁻³ mol
    🎯 M4 awarded: Dividing moles of reacted HCl by 2.
  6. Convert sample mass to grams:
    Mass = 627 mg = 627 × 10⁻³ g = 0.627 g
    🎯 M5 awarded: Converting 627 mg to 0.627 g.
  7. Find the value of x (Two alternative valid methods):

    Method A (via Molar Mass):
    Mr(Na₂CO₃·xH₂O) = mass / moles = 0.627 g / (5.05 × 10⁻³ mol) = 124.16 g mol⁻¹
    Mr(Na₂CO₃) = (2 × 23.0) + 12.0 + (3 × 16.0) = 106.0 g mol⁻¹
    Mass of water of crystallisation = 124.16 - 106.0 = 18.16 g mol⁻¹

    Method B (via Mass & Moles of Water):
    Mass of Na₂CO₃ = 5.05 × 10⁻³ mol × 106.0 g mol⁻¹ = 0.5353 g
    Mass of H₂O = 0.627 g - 0.5353 g = 0.0917 g
    n(H₂O) = 0.0917 / 18.0 = 5.094 × 10⁻³ mol

    🎯 M6 awarded: Correct mathematical setup to isolate the contribution of water (e.g. 18.16 or 0.0917 g).
  8. Final answer for x:
    x = 18.16 / 18.0 = 1.01 ≈ 1 (or ratio of n(H₂O) : n(Na₂CO₃) = 5.094 × 10⁻³ : 5.05 × 10⁻³ ≈ 1 : 1)
    🎯 M7 awarded: Final integer value 1.

✅ Final Answer

Value of x = 1

The formula is sodium carbonate monohydrate, Na₂CO₃·H₂O .

💡 Key Knowledge

  • Back titration: An excess of a known reagent is added to an analyte; the unreacted excess is titrated to determine how much reacted.
  • Stoichiometry: Carbonates are dibasic bases: CO₃²⁻ + 2H⁺ → H₂O + CO₂ . The ratio of HCl to Na₂CO₃ is 2 : 1.
  • Mr values: Na₂CO₃ = 106.0, H₂O = 18.0.

🧠 Exam Technique

  • Map the sequence backwards: Titre gives moles in aliquot (25 cm³) → scale up to full flask (250 cm³) → subtract from starting moles to find moles reacted.
  • Keep unrounded numbers in your calculator: Intermediate rounding can cause significant rounding errors that might distort the final integer.
  • Check question instructions: Notice the command "Give your answer as an integer". Leaving x as 1.01 will lose M7.

❌ Common Calculation Traps

  • Forgetting the 10× dilution factor: Forgetting to scale from 25.0 cm³ to 250 cm³ makes excess moles 10× too small, yielding an impossible negative value or non-sensical ratio.
  • Missing the 2:1 stoichiometric ratio: Forgetting that 1 mole of carbonate reacts with 2 moles of acid (failing to divide by 2 at M4).
  • Unit conversion error: Using 627 g instead of 0.627 g (failing to divide mg by 1000).
  • Assuming x = 10: Students often guess x = 10 because washing soda is commonly decahydrate (Na₂CO₃·10H₂O), rather than trusting their calculated data!

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, Paper 1, 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.