Chemistry 9701/33 — May/June 2021

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

Bleach is made by reacting chlorine with a cold solution of sodium hydroxide. This reaction produces sodium chlorate(I), $\text{NaClO}$ .

\text{Cl}_2(\text{aq}) + 2\text{NaOH}(\text{aq}) \rightarrow \text{NaClO}(\text{aq}) + \text{NaCl}(\text{aq}) + \text{H}_2\text{O}(\text{l})

In this experiment you will determine the concentration of sodium chlorate(I) in a sample of bleach, FA 1.

To do this, you will react an acidified dilute solution of the bleach with iodide ions, $\text{I}^-$ . This reaction produces iodine, $\text{I}_2$ .

\text{ClO}^-(\text{aq}) + 2\text{I}^-(\text{aq}) + 2\text{H}^+(\text{aq}) \rightarrow \text{I}_2(\text{aq}) + \text{Cl}^-(\text{aq}) + \text{H}_2\text{O}(\text{l})

The amount of iodine produced will then be determined by titration with thiosulfate ions, $\text{S}_2\text{O}_3^{2-}$ .

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

FA 1 is a solution of bleach.
FA 3 is dilute sulfuric acid, $\text{H}_2\text{SO}_4$ .
FA 4 is $0.500\text{ mol dm}^{-3}$ potassium iodide, $\text{KI}$ .
FA 5 is $0.100\text{ mol dm}^{-3}$ sodium thiosulfate, $\text{Na}_2\text{S}_2\text{O}_3$ .
starch indicator

(a)

Method

Dilution

Pipette $25.0\text{ cm}^3$ of FA 1 into the $250\text{ cm}^3$ volumetric flask.
Add distilled water to make $250\text{ cm}^3$ of solution and shake the flask thoroughly.
Label this flask FA 2.

Titration

Fill a burette with FA 5.
Rinse the pipette thoroughly with distilled water and then with a little FA 2.
Pipette $25.0\text{ cm}^3$ of FA 2 into a conical flask.
Use the measuring cylinder to add $20\text{ cm}^3$ of FA 3 to the conical flask.
Use the measuring cylinder to add $15\text{ cm}^3$ of FA 4 to the conical flask. The solution will turn brown as iodine is produced.
Add FA 5 from the burette until the solution has turned yellow.
Add 10 drops of starch indicator to the conical flask. The solution will turn blue-black.
Continue to add more FA 5 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 provided.

\text{The rough titre} = \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\text{ cm}^3

Carry out as many accurate titrations as you think necessary to obtain consistent results.
Make sure that your recorded results show the precision of your practical work.
Record in a suitable form in the space below all your burette readings and the volume of FA 5 added in each accurate titration.

Keep FA 4 and FA 5 for use in Question 3.

DifficultyMedium-Easy

Worked solution

Answer

Record the rough titre and all accurate burette readings in a table.

For each accurate titration record:

the final burette reading / cm³
the initial burette reading / cm³
the titre / cm³

Read the burette to the nearest 0.05 cm³. Repeat the titration until at least two accurate titres agree within 0.10 cm³.

Example table (your values will differ):

Titration	Rough	1	2
final reading / cm³	25.10	25.00	49.95
initial reading / cm³	0.00	0.00	24.95
titre / cm³	25.10	25.00	25.00

Final answer

Candidate-dependent: rough titre and accurate titration table with concordant readings to 0.05 cm³.

Detailed explanation

Background Concept

This experiment uses an iodine–thiosulfate titration. Chlorate(I) ions, ClO⁻, oxidise iodide ions to iodine, I₂. The iodine is then titrated with sodium thiosulfate, S₂O₃²⁻. Starch is used as an indicator because it forms a deep blue-black complex with iodine. The end-point is the sudden disappearance of the blue-black colour. For a reliable result, the titration must be repeated until at least two titres agree closely.

Understanding the Question

Part (a) asks you to carry out the dilution and titration and to record all your readings. The marks are not for a particular value but for the quality of your data: correct headings and units, readings to the correct precision, and concordant accurate titres.

Approach

Start with a rough titration to find the approximate end-point. Then perform accurate titrations, recording initial and final burette readings each time. Continue until at least two accurate titres agree within 0.10 cm³. Present the data in a clear table.

Step-by-Step Reasoning

The rough titre tells you roughly how much FA5 is needed, so you can add most of it quickly in the accurate titrations. For each accurate titration, record the initial and final burette readings and calculate the titre by subtraction. Burettes can be read to the nearest 0.05 cm³, so all readings should be recorded to that precision. Two titres are concordant if they differ by no more than 0.10 cm³. The examiner will check that your final accurate titre agrees with another accurate titre. The table must have headings that include the quantity and its unit, for example 'final burette reading / cm³'.

Key Takeaways

Good titration technique and clear recording are essential. The examiner checks precision, concordance, and correct table headings.

Common Mistakes

Recording readings to only 0.1 cm³ instead of 0.05 cm³.
Omitting units from table headings.
Using non-concordant titres in the mean.
Forgetting to record the rough titre.

Things to Be Careful About

Read the burette at eye level to avoid parallax error. Make sure the jet of the burette is filled before taking the initial reading. Use the same volume of FA2 in every titration. Add the starch indicator only when the solution is yellow, not at the start.

Techniques used

record burette readings to 0.05 cm³calculate titres from initial and final readingsrepeat titrations to obtain concordant resultsconstruct a results table with headings and units

(b)

From your accurate titration results, obtain a value for the volume of FA 5 to be used in your calculations. Show clearly how you obtained this value.

25.0\text{ cm}^3\text{ of FA 2 required }\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\text{ cm}^3\text{ of FA 5.}

DifficultyEasy

Worked solution

Working

Select two accurate titres that agree within 0.20 cm³. For example, if the accurate titres are 25.00, 25.00 and 25.05 cm³, use the two 25.00 cm³ values.

Mean titre = (25.00 + 25.00) / 2 = 25.00 cm³

Answer

25.00 cm³ (example)

Final answer

25.00 cm³ (example mean titre)

Detailed explanation

Background Concept

A mean titre is only meaningful if the individual accurate titres are concordant. The mark scheme requires the selected titres to be within a total spread of 0.20 cm³. The mean is quoted to 2 decimal places.

Understanding the Question

Part (b) asks you to choose the volume of FA5 to use in the later calculations. You must show how you obtained it, either by writing the calculation or by ticking the readings you selected.

Approach

Look at your accurate titres. Select two or more that agree within 0.20 cm³, calculate their mean, and round to 2 decimal places.

Step-by-Step Reasoning

For example, if your accurate titres are 25.00, 25.00 and 25.05 cm³, select the two 25.00 cm³ values. Mean = (25.00 + 25.00)/2 = 25.00 cm³. If your titres were 24.95, 25.00 and 25.05, you might select 24.95 and 25.00 (spread 0.05) and average to 24.975, which rounds to 24.98 cm³. The important point is that the selected values are close and the mean is quoted to 2 dp.

Key Takeaways

Always show which readings you averaged and quote the mean to 2 decimal places.

Common Mistakes

Averaging titres that differ by more than 0.20 cm³.
Quoting the mean to 3 decimal places.
Not showing which readings were selected.

Things to Be Careful About

The mean titre is used in all subsequent calculations, so make sure it is correct and clearly written.

Techniques used

select concordant titrescalculate a mean titreround a mean to two decimal places

(c)

Calculations

(i)

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

DifficultyEasy

Worked solution

Answer

Give the answers to (c)(ii) and (c)(iii) to 3 or 4 significant figures.

Final answer

3 or 4 significant figures

Detailed explanation

Background Concept

Significant figures indicate the precision of a measurement. In this experiment, the concentration 0.100 mol dm⁻³ has 3 significant figures and the volume 25.00 cm³ has 4 significant figures. The mark scheme accepts answers to 3 or 4 significant figures.

Understanding the Question

Part (c)(i) asks you to give the answers to (c)(ii) and (c)(iii) to the appropriate number of significant figures.

Approach

Use 3 or 4 significant figures for both calculated values.

Step-by-Step Reasoning

The titre is measured to 0.05 cm³, so a titre such as 25.00 cm³ has 4 significant figures. The concentration of FA5 is given as 0.100 mol dm⁻³, which has 3 significant figures. The final answers should therefore be quoted to 3 or 4 significant figures to match the precision of the data.

Key Takeaways

The number of significant figures in a calculated answer should reflect the least precise measurement used.

Common Mistakes

Giving too many significant figures, such as 0.0012500 mol.
Giving too few, such as 0.001 mol.

Things to Be Careful About

Do not confuse significant figures with decimal places. 0.00125 has 3 significant figures but 5 decimal places.

Techniques used

determine appropriate significant figures

(ii)

Use your answer to (b) and the relevant equation on page 2 to calculate the number of moles of iodine that formed when $25.0\text{ cm}^3$ of FA 2 reacted with FA 4.

\text{moles of I}_2 = \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\text{ mol}

DifficultyMedium-Easy

Worked solution

Working

From the equation:

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

moles of $\text{S}_2\text{O}_3^{2-}$ used = $\frac{25.00}{1000} \times 0.100 = 2.50 \times 10^{-3}$ mol

moles of $\text{I}_2$ = $\frac{1}{2} \times 2.50 \times 10^{-3} = 1.25 \times 10^{-3}$ mol

Answer

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

Final answer

0.00125 mol

Detailed explanation

Background Concept

The titration reaction is:

I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻

One mole of iodine reacts with two moles of thiosulfate. Therefore, the moles of iodine are half the moles of thiosulfate used.

Understanding the Question

Part (c)(ii) asks you to calculate the number of moles of iodine formed when 25.0 cm³ of FA2 reacts with FA4, using your mean titre from part (b).

Approach

Calculate the moles of thiosulfate used in the titration, then halve this value to find the moles of iodine.

Step-by-Step Reasoning

Using the example mean titre of 25.00 cm³:

moles of S₂O₃²⁻ = (25.00/1000) × 0.100 = 2.50 × 10⁻³ mol

From the equation, 1 mol I₂ reacts with 2 mol S₂O₃²⁻, so:

moles of I₂ = 2.50 × 10⁻³ / 2 = 1.25 × 10⁻³ mol

This value is used in part (c)(iii) and part (d)(i).

Key Takeaways

Always write the balanced equation and use the stoichiometric ratio before calculating.

Common Mistakes

Forgetting to divide by 2, giving 2.50 × 10⁻³ mol instead of 1.25 × 10⁻³ mol.
Using the volume in cm³ without converting to dm³.

Things to Be Careful About

Use the mean titre selected in part (b), not the rough titre. Keep the answer to 3 or 4 significant figures.

Techniques used

calculate moles from concentration and volumeapply stoichiometric ratio from balanced equation

(iii)

Calculate the concentration of sodium chlorate(I) in FA 1.
Show your working.

\text{concentration of NaClO} = \dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\text{ mol dm}^{-3}

DifficultyMedium

Worked solution

Working

From the equation:

\text{ClO}^- + 2\text{I}^- + 2\text{H}^+ \rightarrow \text{I}_2 + \text{Cl}^- + \text{H}_2\text{O}

1 mol $\text{ClO}^-$ gives 1 mol $\text{I}_2$ , so moles of $\text{ClO}^-$ in 25.0 cm³ FA2 = $1.25 \times 10^{-3}$ mol.

Concentration of FA2 = $\frac{1.25 \times 10^{-3}}{0.0250} = 0.0500$ mol dm⁻³.

FA1 was diluted by a factor of 10 (25.0 cm³ to 250 cm³), so:

Concentration of FA1 = $10 \times 0.0500 = 0.500$ mol dm⁻³

Answer

$0.500$ mol dm⁻³

Final answer

0.500 mol dm^-3

Detailed explanation

Background Concept

The reaction between chlorate(I) and iodide is:

ClO⁻ + 2I⁻ + 2H⁺ → I₂ + Cl⁻ + H₂O

One mole of ClO⁻ produces one mole of I₂. The concentration of NaClO in FA1 must be found after accounting for the dilution of FA1 to FA2.

Understanding the Question

Part (c)(iii) asks for the concentration of sodium chlorate(I) in the original bleach FA1. You must use your moles of iodine from part (c)(ii) and remember that FA1 was diluted by a factor of 10.

Approach

First find the moles of ClO⁻ in the 25.0 cm³ sample of FA2. Then calculate its concentration. Finally multiply by the dilution factor to find the concentration in FA1.

Step-by-Step Reasoning

Moles of ClO⁻ in 25.0 cm³ FA2 = moles of I₂ = 1.25 × 10⁻³ mol.

Concentration of FA2 = 1.25 × 10⁻³ / 0.0250 = 0.0500 mol dm⁻³.

FA1 was diluted from 25.0 cm³ to 250 cm³, a factor of 10, so:

Concentration of FA1 = 10 × 0.0500 = 0.500 mol dm⁻³.

Key Takeaways

Always account for any dilution when calculating the concentration of the original solution.

Common Mistakes

Forgetting the dilution factor, giving 0.0500 mol dm⁻³ instead of 0.500 mol dm⁻³.
Using 200 instead of 400 as the multiplier because of confusion between I₂ and I⁻.

Things to Be Careful About

The mark scheme accepts 400 × (c)(ii). Check that your units are mol dm⁻³.

Techniques used

relate moles of iodine to moles of chlorate(I)calculate concentration from moles and volumeapply dilution factor

(d)

(i)

In this method an excess of potassium iodide must be added to $25.0\text{ cm}^3$ of FA 2.

Use your answer to (c)(ii) to show by calculation that the potassium iodide is in excess.

DifficultyMedium-Easy

Worked solution

Working

moles of KI added = $\frac{15.0}{1000} \times 0.500 = 7.5 \times 10^{-3}$ mol

moles of $\text{I}^-$ needed = $2 \times$ (c)(ii) $= 2 \times 1.25 \times 10^{-3} = 2.5 \times 10^{-3}$ mol

Since $2.5 \times 10^{-3} < 7.5 \times 10^{-3}$ , the potassium iodide is in excess.

Answer

KI is in excess.

Final answer

KI is in excess

Detailed explanation

Background Concept

For a reagent to be in excess, the amount added must be greater than the amount required by the stoichiometry of the reaction. Here, each mole of ClO⁻ requires 2 moles of I⁻.

Understanding the Question

Part (d)(i) asks you to show by calculation that the potassium iodide is in excess, using your answer to (c)(ii).

Approach

Calculate the moles of KI added, calculate the moles of I⁻ required, and compare the two values.

Step-by-Step Reasoning

Volume of FA4 added = 15 cm³, concentration = 0.500 mol dm⁻³.

Moles of KI added = (15/1000) × 0.500 = 7.5 × 10⁻³ mol.

Moles of I⁻ required = 2 × moles of I₂ = 2 × 1.25 × 10⁻³ = 2.5 × 10⁻³ mol.

Since 2.5 × 10⁻³ < 7.5 × 10⁻³, the KI is in excess.

Key Takeaways

Excess is shown by comparing the amount available with the amount required by the balanced equation.

Common Mistakes

Comparing moles of KI directly with moles of I₂ without multiplying by 2.
Forgetting to convert cm³ to dm³.

Things to Be Careful About

Use the same value of (c)(ii) as in the rest of the calculation so that the comparison is consistent.

Techniques used

calculate moles of iodide addedcalculate moles of iodide requiredcompare amounts to confirm excess

(ii)

A student carries out the same method but the concentration of the potassium iodide solution is not stated.

What change to the practical procedure could the student make to check that the potassium iodide is in excess?
Explain your answer.

DifficultyMedium-Easy

Worked solution

Answer

Repeat the titration using a greater volume of the potassium iodide solution, for example 20 cm³ instead of 15 cm³.

If the titre of sodium thiosulfate is the same as before, the potassium iodide was in excess: the amount of iodine produced is limited by the chlorate(I), not by the iodide.

Final answer

Repeat with a greater volume of KI; if the titre is unchanged, KI was in excess.

Detailed explanation

Background Concept

If iodide is in excess, the amount of iodine produced is limited by the chlorate(I), not by the iodide. Therefore, increasing the amount of iodide should not change the titre of thiosulfate.

Understanding the Question

Part (d)(ii) asks for a change to the practical procedure that would allow the student to check that the potassium iodide is in excess, and an explanation of how the result would show this.

Approach

Repeat the titration using a greater volume of KI. If the titre is unchanged, the KI was already in excess. Alternatively, add more KI at the end-point and see if more iodine is produced.

Step-by-Step Reasoning

If the original KI was in excess, all the ClO⁻ has already reacted. Adding more KI cannot produce more iodine, so the thiosulfate titre stays the same. If the KI was not in excess, adding more KI would produce more iodine and the titre would increase.

Another valid check is to add more KI at the end-point: if the mixture turns blue-black again, iodine is still being produced, showing the KI was not in excess.

Key Takeaways

A controlled change to one reagent can reveal which reagent is limiting the reaction.

Common Mistakes

Suggesting a change without explaining what result would show.
Confusing the limiting reagent with the excess reagent.

Things to Be Careful About

The explanation must link the result (same titre or colour change) to the conclusion about excess.

Techniques used

design a procedural checkpredict the effect on the titredraw a conclusion from the result

The rest of this paper

2 more questions

Q2Manipulation, Measurement and Observation · Presentation of Data and Observations · Analysis, Conclusions and Evaluation10M
Q3Manipulation, Measurement and Observation · Qualitative Analysis · Analysis, Conclusions and Evaluation15M

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