Chemistry 9701/35 — October/November 2020

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

questions

marks

120

minutes

Topics Manipulation, Measurement and Observation · Analysis, Conclusions and Evaluation · 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 value of $x$ in the formula of hydrated sodium thiosulfate, $\text{Na}_2\text{S}_2\text{O}_3 \cdot x\text{H}_2\text{O}$ , where $x$ is an integer. You will first prepare a solution of the salt and then use this solution in a titration with aqueous iodine. The thiosulfate ions react with iodine as shown.

2\text{S}_2\text{O}_3^{2-}(\text{aq}) + \text{I}_2(\text{aq}) \rightarrow \text{S}_4\text{O}_6^{2-}(\text{aq}) + 2\text{I}^-(\text{aq})

FA 1 is hydrated sodium thiosulfate, $\text{Na}_2\text{S}_2\text{O}_3 \cdot x\text{H}_2\text{O}$ .
FA 3 is $0.0500\text{ mol dm}^{-3}$ iodine, $\text{I}_2$ .
starch indicator

(a)

Method

Preparation of salt solution

Weigh the container containing FA 1.
Tip the contents of the container into the $250\text{ cm}^3$ beaker.
Weigh the container with any residue.
Record all your readings in the space below.
Add approximately $200\text{ cm}^3$ of distilled water to the salt in the beaker and stir until the salt has dissolved.
Pour the contents carefully into the $250\text{ cm}^3$ volumetric flask.
Rinse the beaker with a little distilled water and add these washings to the flask.
Fill the flask to the mark with distilled water and shake to ensure thorough mixing.
Label this solution FA 2.

Titration

Fill a burette with FA 2.
Pipette $25.0\text{ cm}^3$ of FA 3 into the conical flask.
Add FA 2 from the burette until the solution in the flask turns yellow.
Add 10 drops of starch indicator to the conical flask. The solution will turn blue-black.
Continue to add more FA 2 from the burette until the blue-black colour just disappears. This is the end-point of the titration.
Carry out 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 your recorded results show the precision of your practical work.
Record, in a suitable form in the space below, all of your burette readings and the volume of FA 2 added in each accurate titration.

DifficultyMedium

Worked solution

Answer

Weighing

Record the initial mass of the container + FA 1 and the final mass of the container after tipping out the salt, both to the same precision (e.g. 2 dp) with units (g).
Mass of FA 1 used = initial mass − final mass.

Rough titration

Record initial and final burette readings and the rough titre.

Accurate titrations

Carry out accurate titrations until at least two concordant results (within 0.10 cm³) are obtained.
Record all burette readings to 0.05 cm³ (e.g. 0.00, 24.50 cm³).

Results table

Headings with units: initial (burette) reading / cm³, final (burette) reading / cm³, titre (volume of FA 2 added) / cm³.

Final answer

See working — candidate-dependent practical data; technique as described.

Detailed explanation

Background Concept

This is a gravimetric–titrimetric determination. The candidate weighs a sample of hydrated sodium thiosulfate, $\text{Na}_2\text{S}_2\text{O}_3 \cdot x\text{H}_2\text{O}$ , dissolves it to make exactly $250\text{ cm}^3$ of solution, and titrates a fixed volume of iodine solution against it. The mass of salt and the titre together give the molar mass of the hydrated salt, from which the number of water molecules per formula unit ( $x$ ) is deduced.

Understanding the Question

Part (a) is worth 8 marks and tests practical skills: weighing by difference, preparing a solution in a volumetric flask, carrying out a titration with starch indicator, and recording data to the required precision. The marks reward correct technique and recording, not the actual numerical values (those are checked against a supervisor's value).

Approach

The key steps: weigh by difference; dissolve and make up to $250\text{ cm}^3$ in a volumetric flask; titrate $25.0\text{ cm}^3$ of iodine with the thiosulfate using starch near the end-point; record all burette readings to 0.05 cm³; obtain at least two concordant titres.

Step-by-Step Reasoning

Weighing by difference: Weigh the container with FA 1, tip the salt into the beaker, then weigh the empty container. Mass used = initial − final. Both readings to the same precision with units.
Solution preparation: Dissolve in about $200\text{ cm}^3$ distilled water, transfer to the $250\text{ cm}^3$ volumetric flask, rinse the beaker with a little water and add the washings (to transfer all salt), make up to the mark and shake.
Titration: Pipette $25.0\text{ cm}^3$ of FA 3 (iodine) into the conical flask. Add FA 2 from the burette until the solution turns yellow (most iodine consumed), add 10 drops of starch (blue-black), continue until the blue-black colour just disappears — the end-point.
Recording: Burette readings to 0.05 cm³. Rough titre recorded, then accurate titrations until at least two concordant (within 0.10 cm³).
Table: Correct headings with units — initial reading / cm³, final reading / cm³, titre / cm³.

Key Takeaways

Weighing by difference, volumetric solution preparation, titration technique with starch, concordant titres, correct data recording.

Common Mistakes

Not recording units with every reading.
Burette readings not to 0.05 cm³ (e.g. 24.5 instead of 24.50).
Using 50.00 as an initial reading.
Not obtaining concordant titres.
Not rinsing the beaker into the flask.

Things to Be Careful About

Weigh both readings to the same precision.
Add starch only near the end-point (when the solution is yellow).
Record the titre as final − initial reading.

Techniques used

weigh by differenceprepare a solution in a volumetric flaskperform a titration with starch indicatorrecord burette readings to 0.05 cm³obtain concordant titres

(b)

From your accurate titration results, obtain a value for the volume of FA 2 to be used in your calculations. Show clearly how you obtained this value.

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

DifficultyMedium-Easy

Worked solution

Working

Representative example: accurate titres 25.00, 25.05, 25.10 cm³. The closest pair are 25.00 and 25.10 (difference 0.10 cm³, within 0.20 cm³).

\text{average titre} = \frac{25.00 + 25.10}{2} = 25.05\text{ cm}^3

Answer

$25.05\text{ cm}^3$ (representative example — candidate-dependent)

Final answer

25.05 cm³ (representative example — candidate-dependent)

Detailed explanation

Background Concept

Concordant titres are averaged to give the titre used in calculations. The mark scheme requires averaging two or more titres all within 0.20 cm³ of each other.

Understanding the Question

From the accurate titrations in (a), select and average the concordant ones to obtain the titre for the calculations in (c).

Approach

Identify two or more titres within 0.20 cm³, average them, and show the working (or tick the selected titres).

Step-by-Step Reasoning

Representative example: accurate titres 25.00, 25.05, 25.10 cm³. The closest pair are 25.00 and 25.10 (difference 0.10 cm³, within 0.20 cm³). Average = (25.00 + 25.10)/2 = 25.05 cm³. This value is used in (c)(iii).

Key Takeaways

Concordance and averaging.

Common Mistakes

Averaging titres that are not concordant (differ by more than 0.20 cm³).

Things to Be Careful About

Show which titres are selected (ticks or working); the mark requires the selection to be visible.

Techniques used

average concordant titresselect titres within 0.20 cm³

(c)

Calculations

(i)

Give your answers to (c)(ii) and (c)(iii) to the appropriate number of significant figures.

DifficultyEasy

Worked solution

Answer

Answers to (c)(ii) and (c)(iii) must be quoted to 3–4 significant figures.

Final answer

3–4 significant figures

Detailed explanation

Background Concept

Significant figures indicate the precision of a measured or calculated value. In this experiment, the concentration (0.0500 mol dm⁻³, 3 sf) and volume (25.0 cm³, 3 sf) limit the precision of derived values.

Understanding the Question

This is a reminder to quote the calculated values in (c)(ii) and (c)(iii) to the appropriate precision — 3 to 4 significant figures.

Approach

Ensure 3–4 sf in the answers to (c)(ii) and (c)(iii).

Step-by-Step Reasoning

For example, 1.25 × 10⁻³ (3 sf), 2.50 × 10⁻³ (3 sf), 2.50 × 10⁻² (3 sf).

Key Takeaways

Significant figures.

Common Mistakes

Quoting too many or too few significant figures (e.g. 0.00125 instead of 1.25 × 10⁻³).

Things to Be Careful About

3–4 sf is the requirement.

Techniques used

quote values to the correct number of significant figures

(ii)

Calculate the number of moles of iodine in $25.0\text{ cm}^3$ of FA 3.

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

DifficultyMedium-Easy

Worked solution

Working

n(\text{I}_2) = cV = 0.0500 \times \frac{25.0}{1000} = 1.25 \times 10^{-3}\text{ mol}

Answer

$1.25 \times 10^{-3}$ mol

Final answer

1.25 × 10⁻³ mol

Detailed explanation

Background Concept

Moles = concentration × volume, with volume in dm³. The iodine solution is 0.0500 mol dm⁻³ and 25.0 cm³ is pipetted.

Understanding the Question

Calculate the moles of I₂ in 25.0 cm³ of 0.0500 mol dm⁻³ solution.

Approach

Use $n = cV$ , converting cm³ to dm³ by dividing by 1000.

Step-by-Step Reasoning

$n(\text{I}_2) = 0.0500 \times 25.0/1000 = 1.25 \times 10^{-3}$ mol. The answer is quoted to 3 sf, consistent with the given data.

Key Takeaways

$n = cV$ .

Common Mistakes

Forgetting to convert cm³ to dm³ (giving 1.25 mol instead of 1.25 × 10⁻³ mol).

Things to Be Careful About

Quote to 3–4 sf.

Techniques used

calculate moles from concentration and volume

(iii)

Calculate the number of moles of thiosulfate ions in the volume recorded in (b).

moles of $\text{S}_2\text{O}_3^{2-}$ = .............................. mol

Hence calculate the number of moles of hydrated sodium thiosulfate in the mass weighed in (a).

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

DifficultyMedium

Worked solution

Working

From the equation, $2\text{S}_2\text{O}_3^{2-} : 1\text{I}_2$ , so:

n(\text{S}_2\text{O}_3^{2-} \text{ in titre}) = 2 \times 1.25 \times 10^{-3} = 2.50 \times 10^{-3}\text{ mol}

Total moles in the $250\text{ cm}^3$ flask (using titre $25.0\text{ cm}^3$ ):

n(\text{S}_2\text{O}_3^{2-}) = 2.50 \times 10^{-3} \times \frac{250}{25.0} = 2.50 \times 10^{-2}\text{ mol}

1 mol of $\text{Na}_2\text{S}_2\text{O}_3 \cdot x\text{H}_2\text{O}$ gives 1 mol of $\text{S}_2\text{O}_3^{2-}$ , so:

n(\text{Na}_2\text{S}_2\text{O}_3 \cdot x\text{H}_2\text{O}) = 2.50 \times 10^{-2}\text{ mol}

Answer

$2.50 \times 10^{-2}$ mol

Final answer

2.50 × 10⁻² mol

Detailed explanation

Background Concept

The stoichiometry of the reaction is $2\text{S}_2\text{O}_3^{2-} + \text{I}_2 \rightarrow \text{S}_4\text{O}_6^{2-} + 2\text{I}^-$ , so 2 mol of thiosulfate react with 1 mol of iodine.

Understanding the Question

Find the moles of thiosulfate in the titre volume, then scale up to the 250 cm³ flask. Since each formula unit of $\text{Na}_2\text{S}_2\text{O}_3 \cdot x\text{H}_2\text{O}$ contains one $\text{S}_2\text{O}_3^{2-}$ ion, the moles of salt equal the moles of thiosulfate.

Approach

First use the 2:1 ratio to find moles of thiosulfate in the titre. Then scale by $250/(b)$ to find the total moles in the flask.

Step-by-Step Reasoning

$n(\text{S}_2\text{O}_3^{2-} \text{ in titre}) = 2 \times 1.25 \times 10^{-3} = 2.50 \times 10^{-3}$ mol.
Total in 250 cm³ = $2.50 \times 10^{-3} \times (250/25.0) = 2.50 \times 10^{-2}$ mol.
Since 1 mol salt gives 1 mol thiosulfate, moles of salt = $2.50 \times 10^{-2}$ mol.

Key Takeaways

Stoichiometric ratio and scaling factor.

Common Mistakes

Forgetting the 2:1 ratio; not scaling to 250 cm³; using 25.0 cm³ instead of the titre volume (b).

Things to Be Careful About

Use the titre volume (b) in the scaling factor, not 25.0 cm³.

Techniques used

apply stoichiometric ratio from the balanced equationscale moles up to the full flask volume

(iv)

Calculate the value for $x$ in the formula of hydrated sodium thiosulfate, $\text{Na}_2\text{S}_2\text{O}_3 \cdot x\text{H}_2\text{O}$ .
Show your working.

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

DifficultyMedium

Worked solution

Working

Using mass of FA 1 = 6.2 g (representative example):

M_r = \frac{\text{mass}}{\text{moles}} = \frac{6.2}{2.50 \times 10^{-2}} = 248

x = \frac{M_r - M_r(\text{Na}_2\text{S}_2\text{O}_3)}{18} = \frac{248 - 158.2}{18} = \frac{89.8}{18} = 4.99

Answer

$x = 5$

Final answer

x = 5

Detailed explanation

Background Concept

$M_r = \text{mass}/\text{moles}$ . The water of crystallisation is found from $x = (M_r - M_r(\text{Na}_2\text{S}_2\text{O}_3))/18$ , where 18 is the $M_r$ of water and $M_r(\text{Na}_2\text{S}_2\text{O}_3) = 158.2$ .

Understanding the Question

Use the mass of FA 1 weighed in (a) and the moles calculated in (c)(iii) to find the molar mass, then deduce $x$ .

Approach

$M_r = \text{mass}/\text{moles}$ ; then $x = (M_r - 158.2)/18$ , rounding to the nearest integer.

Step-by-Step Reasoning

$M_r(\text{Na}_2\text{S}_2\text{O}_3) = 2(23) + 2(32) + 3(16) = 46 + 64 + 48 = 158$ .
$M_r(\text{hydrate}) = 6.2/2.50 \times 10^{-2} = 248$ .
$x = (248 - 158)/18 = 90/18 = 5$ .
So the formula is $\text{Na}_2\text{S}_2\text{O}_3 \cdot 5\text{H}_2\text{O}$ .

Key Takeaways

$M_r$ from mass and moles; water of crystallisation.

Common Mistakes

Using the wrong $M_r$ of the anhydrous salt; not rounding $x$ to the nearest integer.

Things to Be Careful About

The mark scheme requires display of $(M_r - 158.2)/18$ ; use the precise value 158.2.

Techniques used

calculate Mr from mass and molesdeduce water of crystallisation from Mr

(d)

(i)

State the maximum error in a single reading on the balance used in (a).

maximum error = $\pm$ .............................. g

Calculate the maximum percentage error in the mass of FA 1 used in (a).
Show your working.

maximum percentage error = $\pm$ .............................. %

DifficultyMedium-Easy

Worked solution

Working

Balance uncertainty (1 dp balance): $\pm 0.05$ g per reading.
Two readings (initial and final) → total uncertainty $= 2 \times 0.05 = 0.10$ g.

\text{percentage error} = \frac{2 \times 0.05}{6.2} \times 100 = 1.6\%

Answer

$\pm 0.05$ g; $\pm 1.6\%$

Final answer

±0.05 g; ±1.6%

Detailed explanation

Background Concept

Weighing by difference involves two readings (initial and final), so the total uncertainty is 2 × the single reading uncertainty.

Understanding the Question

State the balance uncertainty and calculate the percentage error in the mass of FA 1.

Approach

Percentage error = (2 × single uncertainty / mass) × 100.

Step-by-Step Reasoning

For a 1 dp balance, single reading uncertainty = ±0.05 g. Two readings → ±0.10 g.
% = (0.10/6.2) × 100 = 1.6%.

Key Takeaways

Doubling for two readings.

Common Mistakes

Not doubling the uncertainty (using ±0.05 g instead of ±0.10 g).

Things to Be Careful About

Use the mass actually used in (a).

Techniques used

calculate percentage error from balance uncertaintydouble uncertainty for two readings

(ii)

Assume that the uncertainty in the mass of FA 1 is the only source of error in your experiment.

Calculate the minimum value for the relative formula mass of FA 1.
Show your working.

minimum value for the relative formula mass of FA 1 = ..............................

DifficultyMedium

Worked solution

Working

Minimum $M_r$ uses the minimum mass (mass − total uncertainty):

\text{minimum } M_r = \frac{6.2 - 0.10}{2.50 \times 10^{-2}} = \frac{6.10}{0.0250} = 244

Answer

244

Final answer

244

Detailed explanation

Background Concept

The minimum relative formula mass corresponds to the minimum possible mass (mass − total uncertainty) divided by the moles.

Understanding the Question

Calculate the minimum $M_r$ using the mass reduced by the error, with the moles from (c)(iii).

Approach

Minimum $M_r$ = (mass − 2 × single uncertainty)/moles.

Step-by-Step Reasoning

Minimum $M_r$ = (6.2 − 0.10)/2.50 × 10⁻² = 6.10/0.0250 = 244.

Key Takeaways

Error propagation.

Common Mistakes

Using the wrong sign (adding instead of subtracting the uncertainty).

Things to Be Careful About

Consistency with the doubling in (d)(i).

Techniques used

propagate error into Mr calculation

(e)

A student prepares FA 2 using anhydrous sodium thiosulfate salt and the same mass of salt that you used in (a).

State how the student’s titre would compare with the average titre value you obtained in (b).
Explain your answer.

DifficultyMedium-Easy

Worked solution

Answer

The titre would be less. The anhydrous salt contains no water of crystallisation, so the same mass contains more moles of $\text{Na}_2\text{S}_2\text{O}_3$ and hence a higher concentration of thiosulfate ions. Less FA 2 is therefore needed to react with the same amount of iodine.

Final answer

Titre is less

Detailed explanation

Background Concept

Anhydrous sodium thiosulfate has no water of crystallisation, so the same mass contains more moles of $\text{Na}_2\text{S}_2\text{O}_3$ and hence a higher concentration of thiosulfate ions.

Understanding the Question

Compare the titre when anhydrous salt is used at the same mass.

Approach

Higher concentration → less volume needed to react with the same iodine.

Step-by-Step Reasoning

Same mass of anhydrous salt contains more moles of thiosulfate (the water contributes mass but no thiosulfate). So FA 2 has a higher concentration, and a smaller volume is needed to react with the same 25.0 cm³ of iodine. The titre is less.

Key Takeaways

Concentration effect on titre.

Common Mistakes

Saying the titre is greater.

Things to Be Careful About

Explain the reason (higher concentration).

Techniques used

compare concentrations of hydrated and anhydrous saltspredict effect on titre

(f)

In many titrations it is usual to fill the burette with the solution of known concentration.

Suggest why this was not done in (a).

DifficultyMedium-Easy

Worked solution

Answer

The dark colour of the aqueous iodine makes the burette meniscus hard to read, so the iodine is placed in the conical flask and the thiosulfate solution in the burette. Also, the blue-black → colourless end-point is easier to see against the iodine in the flask.

Final answer

Iodine is dark, making the burette meniscus hard to read

Detailed explanation

Background Concept

The iodine solution is dark brown, making the meniscus in a burette hard to read. Also, the end-point (blue-black → colourless) is easier to see with iodine in the flask.

Understanding the Question

Explain why the unknown thiosulfate is in the burette and the known iodine in the flask.

Approach

Practical visibility reason.

Step-by-Step Reasoning

If iodine were in the burette, its dark colour would obscure the meniscus, making readings inaccurate. With iodine in the flask, the colour change at the end-point (blue-black → colourless) is clearly visible.

Key Takeaways

Practical titration design.

Common Mistakes

Vague answers like "it's easier" without specifying the reason.

Things to Be Careful About

Give the specific reason (dark colour obscures meniscus, or end-point visibility).

Techniques used

explain practical titration design choice

The rest of this paper

2 more questions

Q2Manipulation, Measurement and Observation · Presentation of Data and Observations · Analysis, Conclusions and Evaluation7M
Q3Qualitative Analysis · Manipulation, Measurement and Observation · Analysis, Conclusions and Evaluation14M

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