Edexcel A-Level Chemistry Paper 3, June 2019: Question 7
17 marks · Hard difficulty · Calculations
Determine the values of y and z in the hydrated salt KH3(C2O4)y.zH2O using titration data and mass measurements from thermal decomposition.
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Question text
7 A group of students analysed a hydrated salt with the formula KH3(C2O4)y.zH2O
where y and z are whole numbers.
The students carried out experiments to determine the values of y and z.
(a) Experiment 1 – to determine the value of y
One student was provided with a 0.0235 mol dm–3 solution of the salt.
25.0 cm3 portions of the salt solution were acidified with excess dilute sulfuric acid
and heated to about 60 °C.
Each portion was titrated with 0.0203 mol dm–3 potassium manganate(VII).
The results of four titrations are shown in the table.
Titration number 1 2 3 4
Final burette reading / cm3 23.85 47.20 24.05 48.10
Initial burette reading / cm3 0.00 24.00 0.50 25.00
Titre / cm3 23.85 23.20 23.55 23.10
(i) Complete the diagram to show the final burette reading in Titration 1.
(2)
(ii) Explain why this student should use a mean titre of 23.15 cm3 and not
23.43 cm3 in the calculation.
(2)
(iii) The uncertainty in each burette reading is ±0.05 cm*P58308A01632*3.
Calculate the percentage uncertainty in the titre volume of
potassium manganate(VII) solution used in Titration 2.
(1)
(iv) The equation for the reaction is
2MnO– + 5C O2– + 16H+ o 2Mn2+ + 10CO + 8H O
42 4 2 2
Deduce, by calculation, the value of y, to the nearest whole number, in the
formula KH3(C2O4)y.zH2O.
Use the mean titre of 23.15 cm3 and other data from Experiment 1.
You must show your working.
(4)
(b) Experiment 2 – to determine the value of z
Another student wrote an account of the method for this experiment.
A crucible was weighed.
A sample of the hydrated salt was added to the crucible and it was reweighed.
The crucible and salt were heated to remove the water of crystallisation and
then allowed to cool.
The crucible and contents were weighed again.
Results
Mass of crucible = 19.56 g
Mass of crucible + KH3(C2O4)y.zH2O = 22.97 g
Mass of crucible + KH3(C2O4)y = 22.52 g
(i) Deduce, by calculation, the value of z, to the nearest whole number, in the
formula KH3(C2O4)y.zH2O.
You must use the data from Experiment 2 and your value of y in (a)(iv).
You must show your working.
(3)
(ii) A third student carried out Experiment 2 and calculated a value of z that was
lower than expected.
This student evaluated the experiment and gave two suggestions for*P58308A01732* z being lower.
Suggestion 1
“Some of the crystals jumped out of the crucible while it was being heated.”
Suggestion 2
“It was difficult to tell when all the water of crystallisation had been lost.”
Evaluate these two suggestions to decide whether they could account for the
lower value of z obtained from the experimental results.
Include an explanation of the effect each suggestion would have on the calculated
value of z and how the method could be improved to prevent these errors.
(5)
… *P58308A01832*
(Total for Question 7 = 17 marks)
Mark scheme
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How to answer it
Analysis of a Hydrated Salt Titration & Gravimetric Analysis
What this question tests
This multi-step synoptic question tests core practical and calculation skills: reading precision laboratory instruments (burettes), processing titration data (identifying concordant titres and calculating means), calculating percentage uncertainties, performing redox stoichiometry calculations to deduce crystal formulas, analyzing mass loss (gravimetric analysis) data, and critically evaluating experimental errors and improvements.
Burette Reading & Recording
✅ Correct Answer
The bottom of the meniscus must be drawn cleanly touching or positioned precisely between 23.8 and 23.9 cm³ (e.g. showing a meniscus curving downwards at 23.85 cm³ ).
❌ Common Errors
- Shading below or above the meniscus line incorrectly.
- Misreading the scale by reading upwards instead of downwards past 23 cm³.
Choosing Concordant Titres
✅ Correct Answer
Use 23.15 cm³ because it is the mean of the concordant titres (titres 2 and 4: 23.20 and 23.10 cm³ ). Do not use 23.43 cm³ because it includes rough or non-concordant/inaccurate values (titres 1 and 3: 23.85 and 23.55 cm³ ).
💡 Key Knowledge
Concordant titres are those that lie within 0.10 cm³ of each other. Rough titres are always omitted from mean calculations.
Percentage Uncertainty Calculation
📐 Step-by-Step Calculation
- Identify total burette error: A titration uses two burette readings (initial and final), each with an uncertainty of ±0.05 cm³. Total error = 2 × 0.05 = ±0.10 cm³ .
- Apply formula: (Total Error / Titre Value) × 100
- Calculation: ( 0.10 / 23.20 ) × 100 = 0.431% (or 0.43% / 0.4%).
❌ Common Calculation Traps
Forgetting to multiply the single burette uncertainty ( ±0.05 ) by 2. You always account for both the initial and final volume readings in a titration!
Stoichiometry and Deducing Value y
📐 Step-by-Step Calculation
- Moles of MnO₄⁻:
(23.15 × 0.0203) / 1000 = 4.6995 × 10⁻⁴ mol - Moles of C₂O₄²⁻ in 25.0 cm³ portion:
Using equation ratio ( 2 MnO₄⁻ : 5 C₂O₄²⁻ ):
4.6995 × 10⁻⁴ × (5/2) = 1.1749 × 10⁻³ mol - Moles of C₂O₄²⁻ in 1.00 dm³ solution:
1.1749 × 10⁻³ × (1000 / 25.0) = 0.046995 mol - Deduce y:
Concentration of salt provided = 0.0235 mol dm⁻³ .
Ratio of salt to C₂O₄²⁻ = 0.0235 : 0.046995 = 1 : 1.9998 , so y = 2 .
🧠 Exam Technique
Always state your rounded whole number clearly at the end. Carry unrounded intermediate answers through your calculator to prevent rounding errors.
Gravimetric Analysis and Deducing Value z
📐 Step-by-Step Calculation
- Mass of anhydrous salt:
22.52 - 19.56 = 2.96 g - Mass of H₂O lost:
22.97 - 22.52 = 0.45 g - Molar mass of anhydrous salt KH₃(C₂O₄)₂ (y=2):
Mr = 39.1 + 3(1.0) + 4(12.0) + 8(16.0) = 218.1 g mol⁻¹ - Moles of anhydrous salt:
2.96 / 218.1 = 0.01357 mol - Moles of H₂O:
0.45 / 18.0 = 0.025 mol - Ratio (salt : H₂O):
0.01357 : 0.025 = 1 : 1.84 ... wait, accounting for experimental loss, to nearest whole number z = 2 .
❌ Common Errors
Subtracting crucible masses incorrectly or using the wrong Mr by omitting the value of y correctly established in part (a).
Evaluating Experimental Suggestions
💡 Evaluation & Explanation
- Suggestion 1 (Crystals jumped out): Causes a higher calculated value of z because more mass/water appears to be lost than expected. Improvement: Place a lid loosely on the crucible or heat more gently.
- Suggestion 2 (Incomplete dehydration): Causes a lower calculated value of z because less mass of water is recorded as lost. Improvement: Heat to constant mass.
🧠 Exam Technique
To secure all 5 marks, structure your answer clearly for both suggestions by addressing: (1) whether it accounts for the error, (2) the direction of the effect on z , and (3) a valid practical method to prevent it.
Topics
Physical Chemistry · Core Practicals · Core Practical 3: Find the concentration of a solution of hydrochloric acid · Topic 5: Formulae, Equations and Amounts of Substance
Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 3, June 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.