Chemistry 9701/35 — May/June 2015

Cambridge AS Level · Advanced Practical Skills · worked solutions for every part, with the mark scheme

questions

marks

120

minutes

Topics Analysis, Conclusions and Evaluation · Manipulation, Measurement and Observation · Presentation of Data and Observations · Qualitative Analysis

Q1Manipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and EvaluationFree sample

In this experiment you will determine the formula of iron(III) ammonium sulfate, $\text{FeNH}_4(\text{SO}_4)_2 \cdot x\text{H}_2\text{O}$ , where $x$ is the number of molecules of water of crystallisation.

A known mass of this iron(III) compound reacted with excess acidified potassium iodide to produce iodine. You will determine the amount of iodine produced by titrating the mixture with sodium thiosulfate.

FA 1 is $0.900 \text{ mol dm}^{-3}$ sodium thiosulfate, $\text{Na}_2\text{S}_2\text{O}_3$ .
FA 2 is a solution of iodine, $\text{I}_2$ , produced as outlined in the paragraph above.
starch indicator

(a)

Method

Diluting FA 1

Pipette $25.0 \text{ cm}^3$ of FA 1 into the $250 \text{ cm}^3$ volumetric (graduated) flask.
Make the solution up to the mark using distilled water.
Shake the flask to mix the solution thoroughly before using it for your titrations.
Label this diluted solution of sodium thiosulfate FA 3.
Rinse the pipette with distilled water.

Keep FA 1 for use in Question 3.

Titration

Fill the burette with FA 3.
Use the pipette to transfer $25.0 \text{ cm}^3$ of FA 2 into a conical flask.
Add FA 3 from the burette into the conical flask until the mixture becomes pale yellow.
Then add 10 drops of starch indicator to give a blue-black colour.
Continue adding FA 3 until this blue-black colour disappears. This is the end-point of the titration.
Perform a rough titration and record your burette readings in the space below.

The rough titre is ................ $\text{cm}^3$ .

Carry out as many accurate titrations as you think necessary to obtain consistent results.
Make sure any recorded results show the precision of your practical work.
Record in a suitable table below, all of your burette readings and the volume of FA 3 added in each accurate titration.

DifficultyMedium-Easy

Worked solution

Answer

Rough titration: record the initial and final burette readings and the titre.

Accurate titrations: record all readings in a table with correct headings and units, e.g.

Initial burette reading / cm³	Final burette reading / cm³	Titre (volume of FA 3 added) / cm³
0.00	26.00	26.00
26.00	52.00	26.00

All burette readings recorded to 0.05 cm³.
Two accurate titres within 0.10 cm³ of each other (concordant).

Final answer

Candidate-dependent: results table with correct headings and units, all readings to 0.05 cm³, two concordant titres within 0.10 cm³.

Detailed explanation

Background Concept

In a redox titration the end-point is detected by a colour change. Here iodine (FA 2) is titrated with sodium thiosulfate (FA 3). Iodine forms a blue-black complex with starch indicator; when all the iodine has been reduced to iodide, the blue-black colour disappears — that is the end-point. The quality of the data is judged by precision (burette readings to 0.05 cm³) and concordance (titres within 0.10 cm³ of each other).

Understanding the Question

You must perform a rough titration, then as many accurate titrations as needed, and record all readings in a table. The marks reward: (I) the rough titre and a 2 × 2 box of accurate readings, (II) correct headings with units, (III) readings to 0.05 cm³, (IV) two concordant titres within 0.10 cm³, and (V–VII) agreement with the supervisor's titre.

Approach

Do a rough titration first to locate the end-point quickly. Then repeat accurately, adding FA 3 dropwise near the end-point. Record everything in a properly headed table. Aim for at least two titres within 0.10 cm³.

Step-by-Step Reasoning

Rough titration: add FA 3 until the blue-black colour just disappears; record initial and final burette readings and the titre.
Accurate titrations: repeat the procedure. Add starch only when the solution is pale yellow, so the blue-black colour appears just before the end-point and the end-point is not masked.
Table: three columns — initial burette reading, final burette reading, titre (volume of FA 3 added) — each with the unit cm³.
Precision: every reading to 0.05 cm³. A burette is graduated in 0.1 cm³ divisions, so you estimate to half a division.
Concordance: at least two titres within 0.10 cm³ of each other.

Key Takeaways

Burette readings must be recorded to 0.05 cm³.
Concordant titres (within 0.10 cm³) confirm a reliable end-point.
A results table needs correct headings, each with a unit.

Common Mistakes

Using 50.00 as an initial burette reading (rejected by the mark scheme).
Recording readings to 0 decimal places instead of 0.05 cm³.
Titres more than 0.10 cm³ apart.
Labelling the titre column "difference" or "total" (rejected).

Things to Be Careful About

Headings must match the readings recorded beneath them.
Every entry in the table must carry the unit.
Add the starch indicator near the end-point, not at the start, or the colour change is masked.

Techniques used

perform a rough titration to locate the end-pointrecord burette readings to 0.05 cm³obtain concordant titres within 0.10 cm³construct a results table with correct headings and units

(b)

From your accurate titration results, obtain a suitable value to be used in your calculations. Show clearly how you obtained this value.

$25.0 \text{ cm}^3$ of FA 2 required ................ $\text{cm}^3$ of FA 3.

DifficultyEasy

Worked solution

Working

Select two (or more) accurate titres that are within 0.20 cm³ of each other, e.g. 26.00 and 26.00 cm³.

$\text{mean titre} = \frac{26.00 + 26.00}{2} = 26.00 \text{ cm}^3$

Answer

25.0 cm³ of FA 2 required 26.00 cm³ of FA 3.

Final answer

26.00 cm³ (representative mean of concordant titres)

Detailed explanation

Background Concept

The mean titre is the single value used in all subsequent calculations. It is obtained by averaging two or more accurate titres that agree closely (within 0.20 cm³). The mean is normally quoted to 2 decimal places.

Understanding the Question

From your accurate titration results, select the concordant readings, show how you chose them (ticks or working), and calculate the mean. This mean is the volume of FA 3 that reacts with 25.0 cm³ of FA 2.

Approach

Identify the two (or more) titres that are within 0.20 cm³ of each other, average them, and round to 2 decimal places.

Step-by-Step Reasoning

Mark the concordant titres with ticks.
Add them and divide by the number of readings.
Example: titres 26.00 and 26.00 cm³ give a mean of 26.00 cm³. If the titres were 26.60 and 26.70 cm³, the mean would be 26.65 cm³.
Round to 2 dp: e.g. 26.667 rounds to 26.67.

Key Takeaways

The mean titre is the average of concordant readings.
The mean is normally quoted to 2 decimal places.

Common Mistakes

Averaging titres that are not concordant (more than 0.20 cm³ apart).
Not showing which readings were selected.
Quoting the mean to too many decimal places.

Things to Be Careful About

The mark scheme allows a mean to 3 dp only for endings of 0.025 or 0.075 (e.g. 26.325).
A mean to 1 dp is allowed only if all accurate readings were given to 1 dp and the mean is exactly correct.

Techniques used

select concordant titresaverage the selected titresround the mean titre to 2 decimal places

(c)

Calculations

Show your working and appropriate significant figures in the final answer to each step of your calculations.

(i)

Using information on page 2, calculate the concentration, in $\text{mol dm}^{-3}$ , of sodium thiosulfate in FA 3.

concentration of $\text{Na}_2\text{S}_2\text{O}_3$ in FA 3 = ............................. $\text{mol dm}^{-3}$

DifficultyEasy

Worked solution

Working

25.0 cm³ of FA 1 is diluted to 250 cm³, a dilution factor of 25.0/250 = 0.1.

$\text{concentration of FA 3} = 0.900 \times \frac{25.0}{250} = 0.0900 \text{ mol dm}^{-3}$

Answer

0.0900 mol dm⁻³

Final answer

0.0900 mol dm⁻³

Detailed explanation

Background Concept

When a solution is diluted, the number of moles of solute is unchanged, so $C_1V_1 = C_2V_2$ . Here 25.0 cm³ of 0.900 mol dm⁻³ FA 1 is made up to 250 cm³, a ten-fold dilution.

Understanding the Question

FA 3 is the diluted sodium thiosulfate. You must find its concentration in mol dm⁻³ using the volumes given in the method.

Approach

Multiply the original concentration by the dilution factor (volume taken ÷ final volume).

Step-by-Step Reasoning

Volume taken = 25.0 cm³; final volume = 250 cm³.
Dilution factor = 25.0/250 = 0.1.
Concentration of FA 3 = 0.900 × 0.1 = 0.0900 mol dm⁻³.
The answer is quoted to 3 significant figures (0.0900), consistent with the data.

Key Takeaways

Dilution factor = volume taken ÷ final volume.
Moles of solute are conserved on dilution.

Common Mistakes

Using 250/25 instead of 25/250, giving 9.00 mol dm⁻³.
Quoting 0.09 instead of 0.0900 (losing significant figures).

Things to Be Careful About

Keep the answer to 3 significant figures to match the given concentration.
The mark scheme accepts 0.09(00), so the trailing zeros are expected.

Techniques used

calculate a dilution factorcalculate concentration after dilution

(ii)

Calculate the number of moles of sodium thiosulfate present in the volume of FA 3 calculated in (b).

moles of $\text{Na}_2\text{S}_2\text{O}_3$ = ............................. $\text{mol}$

DifficultyEasy

Worked solution

Working

$\text{moles of Na}_2\text{S}_2\text{O}_3 = 0.0900 \times \frac{26.00}{1000} = 2.34 \times 10^{-3} \text{ mol}$

Answer

2.34 × 10⁻³ mol

Final answer

2.34 × 10⁻³ mol

Detailed explanation

Background Concept

Moles = concentration × volume, where volume must be in dm³. A volume in cm³ is converted by dividing by 1000.

Understanding the Question

Using the concentration of FA 3 from (i) and the mean titre from (b), find the number of moles of sodium thiosulfate that reacted.

Approach

Substitute the concentration and the titre (converted to dm³) into moles = c × V.

Step-by-Step Reasoning

Concentration of FA 3 = 0.0900 mol dm⁻³.
Volume = 26.00 cm³ = 26.00/1000 = 0.02600 dm³.
Moles = 0.0900 × 0.02600 = 2.34 × 10⁻³ mol.
The answer is quoted to 3 significant figures.

Key Takeaways

moles = c × V with V in dm³.
cm³ → dm³ by dividing by 1000.

Common Mistakes

Forgetting to convert cm³ to dm³, giving 2.34 mol.
Quoting too many significant figures.

Things to Be Careful About

This part is marked jointly with (i): both the concentration and the moles must be correct for the mark.

Techniques used

calculate moles from concentration and volumeconvert cm³ to dm³

(iii)

Use the equation below to calculate the number of moles of iodine that reacted with the sodium thiosulfate in (ii).

\text{I}_2 + 2\text{Na}_2\text{S}_2\text{O}_3 \rightarrow 2\text{NaI} + \text{Na}_2\text{S}_4\text{O}_6

moles of $\text{I}_2$ = ............................. $\text{mol}$

DifficultyMedium-Easy

Worked solution

Working

From $\text{I}_2 + 2\text{Na}_2\text{S}_2\text{O}_3 \rightarrow 2\text{NaI} + \text{Na}_2\text{S}_4\text{O}_6$ , the mole ratio I₂ : S₂O₃²⁻ is 1 : 2.

$\text{moles of I}_2 = \frac{2.34 \times 10^{-3}}{2} = 1.17 \times 10^{-3} \text{ mol}$

Answer

1.17 × 10⁻³ mol

Final answer

1.17 × 10⁻³ mol

Detailed explanation

Background Concept

The titration reaction is $\text{I}_2 + 2\text{Na}_2\text{S}_2\text{O}_3 \rightarrow 2\text{NaI} + \text{Na}_2\text{S}_4\text{O}_6$ . One mole of iodine reacts with two moles of thiosulfate, so moles of I₂ = ½ × moles of S₂O₃²⁻.

Understanding the Question

Use the moles of sodium thiosulfate from (ii) and the stoichiometry of the titration equation to find the moles of iodine in the 25.0 cm³ of FA 2 that was titrated.

Approach

Read the mole ratio directly from the balanced equation and halve the moles of thiosulfate.

Step-by-Step Reasoning

The equation shows 1 mol I₂ : 2 mol Na₂S₂O₃.
Moles of I₂ = 2.34 × 10⁻³ ÷ 2 = 1.17 × 10⁻³ mol.
This is the amount of iodine in the 25.0 cm³ portion of FA 2 titrated.

Key Takeaways

The stoichiometric ratio comes from the coefficients of the balanced equation.
I₂ : S₂O₃²⁻ = 1 : 2 in this titration.

Common Mistakes

Using a 1:1 ratio, giving 2.34 × 10⁻³ mol.
Dividing by the wrong coefficient.

Things to Be Careful About

The mark scheme requires both the halving step and the subsequent concentration calculation in (iv) for the mark.

Techniques used

apply the stoichiometric ratio from the balanced equationcalculate moles of iodine from moles of thiosulfate

(iv)

Calculate the concentration of $\text{I}_2$ , in $\text{mol dm}^{-3}$ , in FA 2.

concentration of $\text{I}_2$ = ............................. $\text{mol dm}^{-3}$

DifficultyEasy

Worked solution

Working

The 1.17 × 10⁻³ mol of I₂ was present in 25.0 cm³ of FA 2.

$\text{concentration of I}_2 = 1.17 \times 10^{-3} \times \frac{1000}{25.0} = 0.0468 \text{ mol dm}^{-3}$

Answer

0.0468 mol dm⁻³

Final answer

0.0468 mol dm⁻³

Detailed explanation

Background Concept

Concentration = moles ÷ volume, with volume in dm³. Since the moles were found in 25.0 cm³, multiply by 1000/25 to express the concentration per dm³.

Understanding the Question

Find the concentration of iodine in FA 2 in mol dm⁻³, using the moles from (iii) and the 25.0 cm³ volume titrated.

Approach

Convert the 25.0 cm³ volume to dm³ and divide the moles by it.

Step-by-Step Reasoning

Volume = 25.0 cm³ = 0.0250 dm³.
Concentration = 1.17 × 10⁻³ ÷ 0.0250 = 0.0468 mol dm⁻³.
Equivalently, 1.17 × 10⁻³ × 1000/25.0 = 0.0468 mol dm⁻³.

Key Takeaways

c = n/V with V in dm³.
Multiplying by 1000/volume-in-cm³ converts moles in a portion to concentration per dm³.

Common Mistakes

Forgetting to convert the volume, giving 4.68 × 10⁻⁵ mol dm⁻³.
Mixing up the 1000/25 factor.

Things to Be Careful About

This part is marked jointly with (iii): both the moles of I₂ and the concentration must be correct.

Techniques used

calculate concentration from moles and volumeconvert cm³ to dm³

(v)

The iodine in FA 2 was produced by the reaction of iron(III) ions with excess potassium iodide. Balance the equation for this reaction.

\dots\dots \text{Fe}^{3+}(\text{aq}) + \dots\dots \text{I}^-(\text{aq}) \rightarrow \dots\dots \text{Fe}^{2+}(\text{aq}) + \dots\dots \text{I}_2(\text{aq})

Use your answer to (iv) and this equation to calculate the number of moles of iron(III) ions that reacted to produce the iodine in $1.00 \text{ dm}^3$ of FA 2.

moles of $\text{Fe}^{3+}$ = ............................. $\text{mol}$

DifficultyMedium-Easy

Worked solution

Working

$2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq})$

Mole ratio Fe³⁺ : I₂ = 2 : 1.

$\text{moles of Fe}^{3+} = 2 \times 0.0468 = 0.0936 \text{ mol}$

Answer

0.0936 mol

Final answer

0.0936 mol

Detailed explanation

Background Concept

Iron(III) oxidises iodide to iodine: Fe³⁺ is reduced to Fe²⁺ (gaining one electron) while I⁻ is oxidised to I₂ (two iodides lose two electrons). Balancing the electron transfer gives 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂, so two moles of Fe³⁺ produce one mole of I₂.

Understanding the Question

First balance the redox equation. Then, using the concentration of I₂ from (iv), calculate the moles of Fe³⁺ that produced the iodine in 1.00 dm³ of FA 2.

Approach

Balance the half-equations (Fe³⁺ + e⁻ → Fe²⁺; 2I⁻ → I₂ + 2e⁻), combine them, then double the moles of I₂ per dm³.

Step-by-Step Reasoning

Reduction: Fe³⁺ + e⁻ → Fe²⁺.
Oxidation: 2I⁻ → I₂ + 2e⁻.
To balance electrons, multiply the reduction by 2: 2Fe³⁺ + 2e⁻ → 2Fe²⁺.
Combine: 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂.
The ratio Fe³⁺ : I₂ = 2 : 1.
Moles of Fe³⁺ per dm³ = 2 × 0.0468 = 0.0936 mol.

Key Takeaways

Redox equations are balanced by equalising electron transfer.
The 2:1 ratio between Fe³⁺ and I₂ is essential for the next calculation.

Common Mistakes

Writing an unbalanced equation such as Fe³⁺ + I⁻ → Fe²⁺ + I₂.
Using a 1:1 ratio instead of 2:1.

Things to Be Careful About

The mark scheme requires the correctly balanced equation AND the 2 × (iv) step for the mark.

Techniques used

balance a redox equationapply the 2:1 mole ratio between Fe³⁺ and I₂

(vi)

The formula of the iron(III) compound is $\text{FeNH}_4(\text{SO}_4)_2 \cdot x\text{H}_2\text{O}$ .

$38.56 \text{ g}$ of this compound was weighed out and added to excess aqueous acidified potassium iodide.
FA 2 was made by making the resulting solution of iodine up to $1.00 \text{ dm}^3$ with distilled water.

Use this information and your answer to (v) to calculate the number of moles of water of crystallisation, $x$ , in one mole of the iron(III) compound.

[ $A_r$ : H, 1.0; N, 14.0; O, 16.0; S, 32.1; Fe, 55.8]

$x$ = .............................

DifficultyMedium

Worked solution

Working

Moles of Fe³⁺ per dm³ = 0.0936 mol, so moles of the compound in 1.00 dm³ = 0.0936 mol.

$M_r = \frac{38.56}{0.0936} = 411.97$

$M_r(\text{FeNH}_4(\text{SO}_4)_2) = 55.8 + 18.0 + 2(32.1 + 64.0) = 266.0$

$\text{mass of water} = 411.97 - 266.0 = 145.97$

$x = \frac{145.97}{18.0} = 8.11$

Answer

x = 8

Final answer

x = 8

Detailed explanation

Background Concept

Each formula unit of FeNH₄(SO₄)₂·xH₂O contains one Fe³⁺ ion, so the moles of Fe³⁺ per dm³ of FA 2 equal the moles of the compound dissolved in 1.00 dm³. The molar mass is then mass ÷ moles. The mass of water of crystallisation is the difference between the hydrated molar mass and the anhydrous molar mass (266.0). Dividing the mass of water by 18.0 gives x.

Understanding the Question

38.56 g of the compound was dissolved and the iodine produced was made up to 1.00 dm³. Using the moles of Fe³⁺ from (v), find the molar mass, then the mass of water, then x.

Approach

M_r = mass ÷ moles of compound.
Mass of water = M_r − M_r(anhydrous).
x = mass of water ÷ 18.0, rounded to the nearest integer.

Step-by-Step Reasoning

Moles of compound in 1.00 dm³ = moles of Fe³⁺ = 0.0936 mol.
M_r = 38.56 ÷ 0.0936 = 411.97.
Anhydrous M_r: Fe = 55.8; NH₄ = 14.0 + 4(1.0) = 18.0; SO₄ = 32.1 + 4(16.0) = 96.1, twice = 192.2. Total = 55.8 + 18.0 + 192.2 = 266.0.
Mass of water = 411.97 − 266.0 = 145.97.
x = 145.97 ÷ 18.0 = 8.11, which rounds to 8.
The formula is FeNH₄(SO₄)₂·8H₂O.

Key Takeaways

Moles of Fe³⁺ = moles of compound (one Fe per formula unit).
M_r = mass ÷ moles.
Water of crystallisation: x = (M_r − M_r anhydrous) ÷ 18.

Common Mistakes

Forgetting to subtract the anhydrous molar mass, giving x = 22.9.
Not rounding x to the nearest integer.
Using the wrong anhydrous M_r (e.g. omitting NH₄ or one SO₄).

Things to Be Careful About

The mark scheme awards one mark for M_r, one for the mass of water (M_r − 266), and one for x to the nearest integer.
A further mark is available for showing final answers to (i)–(v) to 2–4 significant figures.

Techniques used

calculate molar mass from mass and molessubtract the anhydrous molar mass to find mass of waterdeduce water of crystallisation to the nearest integer

The rest of this paper

2 more questions

Q2Manipulation, Measurement and Observation · Presentation of Data and Observations · Analysis, Conclusions and Evaluation13M
Q3Qualitative Analysis · Analysis, Conclusions and Evaluation

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