Edexcel A-Level Chemistry Paper 3, June 2017: Question 6

17 marks · Hard difficulty · Calculations

Determine the value of x in hydrated magnesium nitrate using three different methods including titration, precipitation, and thermal decomposition.

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Question

Question text

6 A student carried out an investigation to determine the value of x in hydrated

magnesium nitrate(III), Mg(NO2)2•xH2O, using three different methods.

Method 1

• The student prepared an aqueous solution by dissolving 1.15 g of Mg(NO2)2•xH2O

in distilled water, making up the solution to 250.0 cm3 in a volumetric flask and

shaking the mixture.

• The student titrated this solution against 25.0 cm3 portions of an acidified

solution of 0.0200 mol dm−3 potassium manganate(VII), KMnO (aq).

Method 2

• The student mixed a solution of Mg(NO2)2•xH2O with an excess of aqueous

sodium carbonate solution, Na2CO3(aq).

• The student obtained a precipitate of magnesium carbonate, MgCO3(s), and

determined the mass of this precipitate.

Method 3

• The student heated a known mass of Mg(NO2)2•xH2O(s).

• The student determined the mass of the anhydrous residue formed.

Method 1 – Titration

The student filled the burette with the solution made from Mg(NO2)2•xH2O.

In each titration

• 25.0 cm3 of 0.0200 mol dm−3 KMnO (aq) was transferred to a conical flask

using a pipette.

• An excess of dilute sulfuric acid was added to the conical flask and the mixture heated.

• Mg(NO2)2(aq) was added from the burette until the end-point was reached.

The student’s titration results are shown in the table (the rough titration results have

not been included in the table).

Titration number 1 2 3

Final burette reading / cm3 23.95 48.05 23.85

Initial burette reading / cm3 0.80 24.50 0.65

Titre / cm3

Concordant titres (9)

Mean titre / cm3

(a) Complete the table.

(2)

(b) Deduce the colour change that the student would see at the end-point in this titration.

20 (1)

*P48954A02036*

From … to …

Calculate the number of moles of NO−, in the 250 cm3 of solution prepared by the

student and hence the value of x in Mg(NO2)2•xH2O.

Give your answer to the nearest whole number.

(5)

*P48954A02136*

(d) The half-equations for the reaction in the titration are

MnO− + 8H+ + 5e− o Mn2+ + 4H O

NO− + H O o NO− + 2H+ + 2e−

22 3

Use these half-equations to derive the overall ionic equation for the reaction

between manganate(VII) and nitrate(III) ions in acidic conditions.

State symbols are not required.

(2)

Method 2 – Precipitation

The student used the following procedure.

• Dissolve a known mass of Mg(NO2)2•xH2O in distilled water.

• Add an excess of aqueous sodium carbonate solution, Na2CO3(aq), to obtain a

precipitate of magnesium carbonate, MgCO3(s).

Mg2+(aq) + CO2−(aq) o MgCO (s)

• Weigh a piece of filter paper.

• Filter the mixture from the above reaction through the pre-weighed filter paper.

• Wash the precipitate of MgCO3(s) with distilled water.

• Dry the filter paper and precipitate in a desiccator.

• Reweigh the filter paper and the precipitate.

• Calculate the value of x from the results obtained.

22The student found that the value of x calculated using Method 2 was different from

that obtained using Method 1*P48954A02236*. This difference occurred despite having used a pure

sample of the hydrated salt and without making any errors in technique during the

experiment.

The student found out from a data book that the compound magnesium carbonate is

very slightly soluble in water.

(e) Explain how, if at all, the very slight solubility of magnesium carbonate in water

would affect the value calculated for x.

(2)

(f ) The student planned to obtain any dissolved magnesium carbonate by

evaporating the filtrate, and then weighing the residue.

Criticise this student’s plan.

(2)

Method 3 – Thermal decomposition

NOTE: On heating, Mg(NO2)2•xH2O(s) loses its water of crystallisation and then

undergoes further decomposition to give magnesium oxide, MgO. 23

The student used the following procedure.*P48954A02336*

• Weigh an empty crucible.

• Add some Mg(NO2)2•xH2O(s) and then reweigh the crucible plus contents.

• Heat the crucible plus contents and allow to cool.

• Weigh the crucible plus magnesium oxide residue.

• Use these data to calculate a value for x.

The student’s results are shown in the table.

Mass of crucible / g 18.02

Mass of crucible + Mg(NO2)2•xH2O / g 18.84

Mass of crucible + MgO residue / g 18.27

(g) Identify how the student should ensure that the hydrated salt was fully decomposed.

(1)

(h) The student carried out an evaluation of the results obtained from Method 3

Identify two modifications to the method that would enable the student to lower

the percentage uncertainty in the measurement of the mass of the solid residue.

(2)

(Total for Question 6 = 17 marks)

Mark scheme

Show the mark scheme

Question

Acceptable Answers Additional Guidance Mark

Number

6(a) 23.15 and 23.55 and 23.20 completed in table All three titres must be shown to 2 D.P. (2)

(1)

beneath titres 1 and 3 3

and mean titre = 23.18 (cm3) Allow 23.2 or 23.175 (cm )

(1)

Question

Acceptable Answers Additional Guidance Mark

Number

6(b) (From)(pale) pink/purple (to) colourless Both colours needed for the mark (1)

Do not award mauve or magenta or violet for

pink/purple

Ignore references to ‘clear’

Question

Acceptable Answers Additional Guidance Mark

Number

6(c) Example of calculation (5)

calculation of moles of MnO − in 25.0 cm3 (1) Moles MnO − = 0.02(00) x 25.0

1000

= 5(.00) x 10−4 / 0.0005(00)

(mol)

calculation of moles of NO − in mean titre (1) Moles NO − = 2.5 x moles MnO −

22 4

in mean

titre

= 1.25 x 10−3 / 0.00125 (mol)

calculation of moles of NO − in 250 cm3 (1) Moles NO −

in 250 cm3 =

moles NO − in mean titre x 250

mean titre from (a)

= 1.25 x 10−3 x 250

23.18

= 0.013481449

= 0.0135 (mol)

Allow TE on mean titre from (a)

Ignore SF except 1 SF

calculation of molar mass (1) Molar mass = 2 x 1.15

0.0135

= 170.3703704 (g mol−1)

= 170.4 (g mol−1)

Allow TE

calculation of x correctly to the nearest whole number: x = 170.4 — 116.3

(1) 18(.0)

x = 3.005555556

x = 3 (must be to nearest whole number)

Allow TE from molar mass calculated

Allow alternative correct methods for MP4 and

MP5

Correct value of x with no working scores (1)

Question

Acceptable Answers Additional Guidance Mark

Number

6(d) 2MnO − + 5NO − + 6H+ → 2Mn2+ + 5NO − + 3H O Each of the following equations score (1) mark (2)

42 3 2

overall:

evidence of multiplying 1st equation by 2 and

2nd equation by 5 2MnO − + 5NO − + 16H+ + 5H O

42 2

(1) → 2Mn2+ + 5NO − + 8H O + 10H+

overall equation correct with H+ and H O and e(-) OR

cancelled as appropriate

(1) 2MnO − + 5NO − + 6H+ + 5H O

42 2

→ 2Mn2+ + 5NO − + 8H O

2MnO − + 5NO − + 16H+

→ 2Mn2+ + 5NO − + 10H+ + 3H O

Ignore state symbols, even if incorrect

Allow multiples

Question

Acceptable Answers Additional Guidance Mark

Number

6(e) An explanation that makes reference to the following: (2)

Either

the (calculated) value of x would be too high (1) Allow ‘amount’ or ‘mass’ for ‘moles’

The moles of MgCO3 would be too low / the moles of

Mg(NO2)2.xH2O would be too low / the Mr of

Mg(NO2)2.xH2O would be too high (1) MP2 depends on MP1

(So) the (calculated) value of x would be unchanged

(so this does not explain the discrepancy) (1)

Only a small amount/mass of MgCO3 would dissolve

because it is very slightly soluble (1)

MP2 depends on MP1

Question

Acceptable Answers Additional Guidance Mark

Number

6(f) An answer that makes reference to the following points: (2)

the MgCO3 would decompose / the residue would Ignore references to just ‘impurities’

contain NaNO2 / the residue would contain (the excess)

Na2CO3 (1)

(so) the (proposed) method is not valid / appropriate / M2 dependent on M1

suitable (1)

Question

Acceptable Answers Additional Guidance Mark

Number

6(g) An answer that makes reference to the following point: (1)

heat (the sample) to constant mass Allow repetition of heating and weighing until

there is no change in mass (of the sample)

Ignore references to ‘brown gas’ etc

Question

Acceptable Answers Additional Guidance Mark

Number

6(h) An answer that makes reference to the following points: (2)

use a larger mass (of the hydrated salt) (1) Ignore references to repeat measurements

Use a balance that weighs to 3 D.P. (rather than Allow statements such as ‘use a balance that

2 D.P.) (1) weighs to more decimal places’ /’greater

resolution’ / ‘ a more precise/sensitive balance’

Do not allow ‘more accurate’

(Total for Question 6 = 17 marks)

How to answer it

Investigation of Hydrated Magnesium Nitrate(III)

What this question tests

This multi-step synoptic question tests core practical and analytical skills across physical and inorganic chemistry: processing redox titration data, determining moles and stoichiometry, constructing ionic half-equations, evaluating experimental errors in gravimetric/precipitation analysis, and applying heating to constant mass in thermal decomposition.

Parts (a) and (b) – Titration Data Processing & Observations

Processing titration results and identifying end-point colour changes

✅ Correct Answers

Part (a): Titres = 23.15 , 48.05 - 24.50 = 23.55 , 23.85 - 0.65 = 23.20 . Concordant ticks placed under titrations 1 and 3. Mean titre = 23.18 cm³ .
Part (b): From (pale) pink / purple to colourless .

❌ Common Errors

Averaging non-concordant titres (e.g., including titre 2 which is way off).
Writing "clear" instead of "colourless" for the final state of MnO₄⁻ reductions.
Using forbidden terms like "mauve", "magenta", or "violet".

Marks: Part (a) = 2 marks (table + mean), Part (b) = 1 mark.

Part (c) – Multi-Step Stoichiometric Calculation

Determining the value of x in Mg(NO₂)₂·xH₂O

📐 Step-by-Step Calculation

Moles of MnO₄⁻ in 25.0 cm³:
(0.0200 × 25.0) / 1000 = 5.00 × 10⁻⁴ mol
Moles of NO₂⁻ in the mean titre (23.18 cm³):
Since 2 mol MnO₄⁻ react with 5 mol NO₂⁻, multiply by (5 / 2):
5.00 × 10⁻⁴ × (5 / 2) = 1.25 × 10⁻³ mol
Moles of NO₂⁻ in the original 250 cm³ volumetric flask:
Scale up from the titre volume to 250 cm³:
1.25 × 10⁻³ × (250 / 23.18) = 0.01348 mol
Molar mass of Mg(NO₂)₂·xH₂O:
Mass used was 1.15 g in 250 cm³:
M = mass / moles = 1.15 / 0.01348 = 170.37 g mol⁻¹
Calculate x:
Molar mass of anhydrous Mg(NO₂)₂ = 24.3 + (2 × 46.0) = 116.3 g mol⁻¹.
Mass of water = 170.4 - 116.3 = 54.1 g mol⁻¹ .
x = 54.1 / 18.0 = 3.00 → x = 3 (to nearest whole number).

🧠 Exam Technique & Guidance

Always carry forward error (TE) from your mean titre into the calculation. Ensure final values for integer variables like x are rounded explicitly to the nearest whole number as requested by the command word.

Marks: 5 marks total (1 for each analytical stage: MnO₄⁻ moles, titration ratio scaling, volumetric scale-up, molar mass, and final integer x).

Part (d) – Combining Ionic Half-Equations

Constructing the overall redox equation

💡 Key Knowledge

Given half-equations:

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

NO₂⁻ + H₂O → NO₃⁻ + 2H⁺ + 2e⁻

✅ Correct Answer

Multiply equation (1) by 2 and equation (2) by 5 to balance electrons (10e⁻ each):

2MnO₄⁻ + 5NO₂⁻ + 6H⁺ → 2Mn²⁺ + 5NO₃⁻ + 3H₂O

Marks: 2 marks (1 mark for correct multiplying factors/evidence, 1 mark for the balanced overall equation with cancelled species).

Parts (e) and (f) – Evaluating Precipitation Limitations

Analysing practical flaws in Method 2

❌ Part (e): Solubility Effects

Because MgCO₃ is very slightly soluble, some precipitate remains dissolved in aqueous solution.

Measured moles of MgCO₃ are too low.
Calculated Mr of salt is too high, making the calculated value of x too high.

🧠 Part (f): Criticising Evaporation

Evaporating the filtrate to recover dissolved MgCO₃ is not valid because:

The filtrate also contains excess unreacted sodium carbonate ( Na₂CO₃ ) and impurities.
The solid residue recovered would be impure, yielding an over-estimated mass.

Marks: Part (e) = 2 marks, Part (f) = 2 marks.

Parts (g) and (h) – Thermal Decomposition & Minimising Uncertainty

Ensuring complete reaction and improving equipment precision

💡 Part (g): Constant Mass

To ensure the hydrated salt is fully decomposed, the student must heat to constant mass (reheat and weigh until successive masses are identical, showing all water/decomposition gases have escaped).

📐 Part (h): Reducing Percentage Uncertainty

To lower percentage uncertainty in solid mass measurements:

Use a larger mass of the hydrated salt.
Use a balance with higher resolution (e.g., weighing to 3 decimal places instead of 2).

Marks: Part (g) = 1 mark, Part (h) = 2 marks. Total for question = 17 marks.

Topics

Physical Chemistry · Inorganic Chemistry · Core Practicals · Core Practical 11: Find the amount of iron in an iron tablet using redox titration · Topic 5: Formulae, Equations and Amounts of Substance · Topic 3: Redox I · Topic 4: Inorganic Chemistry and the Periodic Table

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 3, June 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.