Chemistry 9701/36 — October/November 2014

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

You are to determine the concentration of a solution of sodium thiosulfate, $\text{Na}_2\text{S}_2\text{O}_3$ .

To do this you will first produce a known amount of iodine by reacting iodate(V) ions, $\text{IO}_3^-$ , with an excess of iodide ions, $\text{I}^-$ . The equation for this reaction is below.

\text{IO}_3^- + 5\text{I}^- + 6\text{H}^+ \rightarrow 3\text{I}_2 + 3\text{H}_2\text{O}

The amount of iodine produced in this reaction can be found by titrating with thiosulfate ions. The equation for this reaction is below.

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

FB 1 is aqueous sodium thiosulfate, $\text{Na}_2\text{S}_2\text{O}_3$ .
FB 2 is aqueous potassium iodate(V) containing $3.60\text{ g dm}^{-3}$ $\text{KIO}_3$ .
FB 3 is sulfuric acid, $\text{H}_2\text{SO}_4$ .
FB 4 is aqueous potassium iodide, $\text{KI}$ .
starch indicator

(a)

Method

Fill a burette with FB 1.
Pipette $25.0\text{ cm}^3$ of FB 2 into the conical flask.
Use the measuring cylinder to add $25\text{ cm}^3$ of FB 3 into the conical flask.
Use the measuring cylinder to add $10\text{ cm}^3$ of FB 4 into the conical flask. Brown iodine solution is produced.
Add FB 1 from the burette until most of the iodine has been removed and the solution in the conical flask is yellow.
Add 10 drops of starch indicator to the contents of the conical flask. The solution will turn blue-black.
Continue adding FB 1, from the burette, until the blue-black colour just disappears.
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 certain any recorded results show the precision of your practical work.
Record, in a suitable form below, all of your burette readings and the volume of FB 1 added in each accurate titration.

DifficultyMedium-Easy

Worked solution

Answer

Example record (candidate-dependent):

Titration	Rough	1	2
Initial burette reading / cm $^3$	0.00	0.00	0.00
Final burette reading / cm $^3$	25.55	25.25	25.30
Volume of FB 1 added / cm $^3$	25.55	25.25	25.30

Use separate column headings: Initial burette reading / cm$^3$, Final burette reading / cm$^3$ and Volume of FB 1 added / cm$^3$.
Record every burette reading to the nearest 0.05 cm $^3$ .
The accurate titres 25.25 cm $^3$ and 25.30 cm $^3$ differ by 0.05 cm $^3$ , so two concordant titres within 0.10 cm $^3$ are obtained.

Final answer

Example record: concordant titres 25.25 cm3 and 25.30 cm3, all burette readings to nearest 0.05 cm3 (candidate-dependent).

Detailed explanation

Background Concept

This is an iodine–thiosulfate titration. Iodine is produced in the flask and then titrated with sodium thiosulfate. Starch is used as the indicator: it forms an intense blue-black complex with iodine, so it is added only when most of the iodine has been consumed. The end-point is the point at which the blue-black colour just disappears. A burette measures the volume of thiosulfate solution delivered, and the titre is the difference between the final and initial burette readings.

In Paper 3, technique marks are awarded not just for getting sensible numbers, but for recording them correctly: with proper headings, units, precision and enough repeat titrations to show concordance.

Understanding the Question

The method tells you to fill a burette with FB 1, pipette 25.0 cm $^3$ of FB 2 into the flask, and add FB 3 and FB 4. It then asks you to perform a rough titration, record it, and then perform as many accurate titrations as needed to obtain consistent results. The marks are for the quality of the practical record: two burette readings for the rough titration, initial and final readings for at least two accurate titrations, correct table headings and units, readings to 0.05 cm $^3$ , and two accurate titres within 0.1 cm $^3$ of each other.

Approach

First perform a rough titration to find the approximate end-point. Then repeat the titration accurately. For each accurate titration, record the initial burette reading, the final burette reading and the titre. Use a table with clear headings and units, and record all burette readings to the nearest 0.05 cm $^3$ . Finally check that at least two accurate titres agree within 0.1 cm $^3$ .

Step-by-Step Reasoning

Rough titration: add FB 1 quickly until the blue-black colour just disappears. Record the initial and final burette readings and calculate the rough titre.
Accurate titrations: refill the burette, use an initial reading such as 0.00 cm $^3$ , and add FB 1 dropwise near the end-point. When the blue-black colour just disappears, record the final reading. Repeat at least twice.
Calculate each titre as final reading minus initial reading.
In the table, use full headings, not abbreviations. Initial burette reading, Final burette reading and Volume of FB 1 added are acceptable; V, vol, difference, total or change are not accepted by the mark scheme.
Give every burette reading to the nearest 0.05 cm $^3$ , e.g. 25.25, not 25.2 or 25.3.
Choose two accurate titres that agree within 0.1 cm $^3$ for later use. In the example, 25.25 and 25.30 differ by 0.05 cm $^3$ , so they are concordant.

Key Takeaways

A clear results table with correct headings and units is a major part of the marks.
Burette readings must be recorded to 0.05 cm $^3$ .
Concordant accurate titres are those within 0.1 cm $^3$ of one another.
The rough titre is a guide only; it is not used for the accurate mean.

Common Mistakes

Writing V or vol instead of initial burette reading and final burette reading.
Writing difference, total or change for the titre.
Omitting the unit / cm $^3$ .
Recording readings to 1 decimal place, e.g. 25.3 instead of 25.30.
Using 50.00 as an initial burette reading, or recording any reading greater than 50.00.
Counting the rough titre as one of the concordant accurate titres.
Recording only one accurate titre, which is insufficient for a reliable mean.

Things to Be Careful About

The mark scheme is strict on precision: all burette readings must be to the nearest 0.05 cm $^3$ , and 0.00 must be written as 0.00, not 0. It also disallows more than one final reading of 50.00 and any reading above 50.00. Use distinct, complete column headings and put the unit in the heading rather than with each number.

Techniques used

use a burette correctlyrecord initial and final burette readings to 0.05 cm3perform a rough titration followed by accurate repeat titrationsidentify the starch/iodine end-point

(b)

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

$25.0\text{ cm}^3$ of FB 2 required .......................... $\text{cm}^3$ of FB 1.

DifficultyEasy

Worked solution

Working

Select the two concordant accurate titres: 25.25 cm $^3$ and 25.30 cm $^3$ .

\text{mean titre} = \frac{25.25 + 25.30}{2} = 25.275 \text{ cm}^3

Rounded to the nearest 0.01 cm $^3$ : 25.28 cm $^3$ .

Answer

25.0 cm $^3$ of FB 2 required 25.28 cm $^3$ of FB 1 (using the example accurate titres).

Final answer

25.28 cm3 (mean of example titres 25.25 cm3 and 25.30 cm3)

Detailed explanation

Background Concept

A mean titre is used to reduce the effect of random errors. Only concordant accurate titres should be averaged; the rough titre is only an estimate and must not be included. The mean is normally quoted to two decimal places because burette readings are recorded to 0.05 cm $^3$ and averaging can produce a result needing 0.01 cm $^3$ precision.

Understanding the Question

Part (b) asks you to choose a suitable volume of FB 1 to use in the calculations. The mark is awarded for selecting two or more accurate titres with a total spread of no more than 0.2 cm $^3$ , showing how the mean was obtained, and giving the mean to the nearest 0.01 cm $^3$ .

Approach

Look at all the accurate titres. Choose the two best concordant ones, usually those closest together. Add them and divide by two. Round the result to two decimal places, unless one of the special cases allowed by the mark scheme applies. Show the calculation clearly or tick the readings used.

Step-by-Step Reasoning

Ignore the rough titre.
The accurate titres in the example are 25.25 cm $^3$ and 25.30 cm $^3$ .
Their spread is 0.05 cm $^3$ , which is within 0.1 cm $^3$ , so they are concordant.
Calculate the mean:

\frac{25.25 + 25.30}{2} = 25.275

Round 25.275 to two decimal places: the third decimal digit is 5, so 25.275 rounds to 25.28 cm $^3$ .
State clearly that 25.0 cm $^3$ of FB 2 required 25.28 cm $^3$ of FB 1.

Key Takeaways

Use only accurate, concordant titres to find the mean.
The mean titre should normally be quoted to 2 decimal places.
Show which readings were selected, either by ticks or by writing the averaging calculation.

Common Mistakes

Using the rough titre in the mean calculation.
Averaging only one accurate titre.
Selecting titres that differ by more than 0.2 cm $^3$ .
Incorrectly subtracting initial and final readings to obtain a titre.
Rounding incorrectly, e.g. writing 26.667 as 26.7 instead of 26.67.
Giving all burette readings as integers; this prevents the mean mark from being awarded.

Things to Be Careful About

The mark scheme allows a mean to 1 decimal place only if all the accurate readings were recorded to 1 decimal place and the mean is exactly correct. It also allows 3 decimal places in the special cases of 0.025 and 0.075. For the example here, two decimal places is expected.

Techniques used

select two concordant accurate titrescalculate the mean titreround the mean to the nearest 0.01 cm3

(c)

Calculations

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

(i)

Calculate the number of moles of $\text{KIO}_3$ present in $25.0\text{ cm}^3$ of FB 2.
[ $A_r$ : O, 16.0; K, 39.1; I, 126.9]

moles of $\text{KIO}_3$ = .......................... mol

DifficultyMedium-Easy

Worked solution

Working

M_r(\text{KIO}_3)=39.1+126.9+3(16.0)=214

Amount in 25.0 cm $^3$ :

n = \frac{3.60}{214} \times \frac{25.0}{1000} = \frac{3.60}{214 \times 40} = 4.21 \times 10^{-4} \text{ mol}

Answer

moles of $\text{KIO}_3 = 4.21 \times 10^{-4}$ mol

Final answer

4.21 × 10^-4 mol

Detailed explanation

Background Concept

Concentration in g dm $^{-3}$ is converted to concentration in mol dm $^{-3}$ by dividing by molar mass:

[\text{KIO}_3] = \frac{3.60 \text{ g dm}^{-3}}{214 \text{ g mol}^{-1}}

The number of moles in a sample is then:

n = cV

where $V$ is in dm $^3$ . Since 25.0 cm $^3$ = 0.0250 dm $^3$ = $\frac{1}{40}$ dm $^3$ , the calculation can be shortened to $3.60/(214 \times 40)$ .

Understanding the Question

Part (c)(i) asks for the number of moles of KIO $_3$ in 25.0 cm $^3$ of FB 2. The concentration of FB 2 is given as 3.60 g dm $^{-3}$ , and the atomic masses allow the molar mass to be calculated. The final answer should be given to an appropriate number of significant figures, usually 3.

Approach

Calculate $M_r(\text{KIO}_3)$ .
Convert the concentration from g dm $^{-3}$ to mol dm $^{-3}$ .
Multiply by the volume in dm $^3$ to find moles.
Round the final answer to 3 significant figures.

Step-by-Step Reasoning

$M_r(\text{KIO}_3)$ :

39.1 + 126.9 + 3(16.0) = 214

Concentration in mol dm $^{-3}$ :

\frac{3.60}{214} = 0.01682 \text{ mol dm}^{-3}

Volume: $25.0 \text{ cm}^3 = 0.0250 \text{ dm}^3$ .
Moles:

0.01682 \times 0.0250 = 4.2056 \times 10^{-4} \text{ mol}

Rounded to 3 significant figures this is $4.21 \times 10^{-4}$ mol.

Key Takeaways

The units g dm $^{-3}$ must be divided by g mol $^{-1}$ to give mol dm $^{-3}$ .
Volume must be converted from cm $^3$ to dm $^3$ before multiplying by concentration.
Using the equivalence $25.0 \text{ cm}^3 = 1/40 \text{ dm}^3$ gives a convenient short cut.

Common Mistakes

Using an incorrect molar mass, e.g. writing iodine as 127 rather than 126.9.
Forgetting to divide by 40 or to convert 25.0 cm $^3$ into dm $^3$ .
Reporting the intermediate concentration with too many significant figures and then a final answer with too few.
Giving the answer without the unit mol.

Things to Be Careful About

Keep the unrounded value, 4.2056 × 10 $^{-4}$ mol, for use in the next part. The final answer in part (c)(i) may be written as 4.21 × 10 $^{-4}$ mol, but the unrounded value should be carried forward into the calculation in part (c)(ii).

Techniques used

calculate the molar mass of KIO3convert concentration from g dm-3 to mol dm-3convert volume from cm3 to dm3calculate moles of solute

(ii)

The equations for the production of iodine and its titration with thiosulfate are shown below.

\text{IO}_3^- + 5\text{I}^- + 6\text{H}^+ \rightarrow 3\text{I}_2 + 3\text{H}_2\text{O}

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

Use these equations to calculate the number of moles of thiosulfate present in the volume of FB 1 you calculated in (b).

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

DifficultyMedium-Easy

Worked solution

Working

From the first equation: 1 mol $\text{IO}_3^-$ produces 3 mol $\text{I}_2$ .

From the second equation: 1 mol $\text{I}_2$ requires 2 mol $\text{S}_2\text{O}_3^{2-}$ .

Therefore 1 mol $\text{IO}_3^-$ requires $3 \times 2 = 6$ mol $\text{S}_2\text{O}_3^{2-}$ .

Using the unrounded value from (c)(i):

n(\text{S}_2\text{O}_3^{2-}) = 4.206 \times 10^{-4} \times 6 = 2.52 \times 10^{-3} \text{ mol}

Answer

moles of $\text{S}_2\text{O}_3^{2-} = 2.52 \times 10^{-3}$ mol

Final answer

2.52 × 10^-3 mol

Detailed explanation

Background Concept

The iodate(V) ion reacts with excess iodide to produce iodine:

\text{IO}_3^- + 5\text{I}^- + 6\text{H}^+ \rightarrow 3\text{I}_2 + 3\text{H}_2\text{O}

Each mole of iodate produces 3 moles of iodine. The iodine is then titrated with thiosulfate:

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

Each mole of iodine reacts with 2 moles of thiosulfate. Combining these ratios gives the overall stoichiometric factor: 1 mole of iodate corresponds to $3 \times 2 = 6$ moles of thiosulfate.

Understanding the Question

Part (c)(ii) asks you to use the two given equations to convert the moles of KIO $_3$ from part (c)(i) into the moles of thiosulfate that reacted with the iodine produced. The relationship is not 1:1; you must work through both equations.

Approach

Write down the conversion chain:

\text{IO}_3^- \rightarrow 3\text{I}_2 \rightarrow 3 \times 2\ \text{S}_2\text{O}_3^{2-}

Then multiply the moles of iodate by 6.

Step-by-Step Reasoning

From equation 1, $1$ mol $\text{IO}_3^-$ gives $3$ mol $\text{I}_2$ .
From equation 2, $1$ mol $\text{I}_2$ reacts with $2$ mol $\text{S}_2\text{O}_3^{2-}$ .
Therefore the overall ratio is:

1 : 6

Using the unrounded value from part (c)(i):

4.2056 \times 10^{-4} \times 6 = 2.5234 \times 10^{-3} \text{ mol}

Rounded to 3 significant figures, this is $2.52 \times 10^{-3}$ mol.

Key Takeaways

The balanced equations must be used in sequence to find the overall stoichiometric ratio.
1 mol IO $_3^-$ is equivalent to 6 mol S $_2$ O $_3^{2-}$ .
Use the unrounded value from the previous calculation to avoid rounding errors.

Common Mistakes

Using the ratio 1:3 instead of 1:6, forgetting that each I $_2$ needs two thiosulfate ions.
Using 1:2 directly without first considering the production of three I $_2$ molecules.
Using the rounded value 4.21 × 10 $^{-4}$ from part (c)(i) and then reporting an inconsistent final value.
Omitting the units or writing the answer to only 2 significant figures when 3 are required.

Things to Be Careful About

The final answer should be to 3 significant figures. Carrying the full calculator value, 4.2056 × 10 $^{-4}$ , gives 2.523 × 10 $^{-3}$ mol, which rounds to 2.52 × 10 $^{-3}$ mol.

Techniques used

combine two balanced equations to find the overall stoichiometric ratiocalculate moles of thiosulfate from moles of iodate

(iii)

Calculate the concentration, in $\text{mol dm}^{-3}$ , of sodium thiosulfate in FB 1.

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

DifficultyMedium-Easy

Worked solution

Working

Mean titre from part (b) = 25.28 cm $^3$ .

V(\text{S}_2\text{O}_3^{2-}) = 25.28 \text{ cm}^3 = 0.02528 \text{ dm}^3

Use the unrounded value of moles from part (c)(ii):

[\text{Na}_2\text{S}_2\text{O}_3] = \frac{n}{V} = \frac{2.523 \times 10^{-3}}{0.02528} = 0.0998 \text{ mol dm}^{-3}

Answer

concentration = 0.0998 mol dm $^{-3}$ (to 3 significant figures)

Final answer

0.0998 mol dm^-3 (using example titre 25.28 cm3)

Detailed explanation

Background Concept

The concentration of a solution is the number of moles of solute divided by the volume of solution in dm $^3$ :

c = \frac{n}{V}

In this titration, each mole of Na $_2$ S $_2$ O $_3$ contains one mole of S $_2$ O $_3^{2-}$ , so the moles of thiosulfate calculated in part (c)(ii) are also the moles of sodium thiosulfate.

Understanding the Question

Part (c)(iii) asks for the concentration of the sodium thiosulfate solution, FB 1, in mol dm $^{-3}$ . You already have the number of moles of thiosulfate from part (c)(ii) and the volume of FB 1 from part (b). The only remaining step is to divide moles by volume, with the volume converted into dm $^3$ .

Approach

Convert the mean titre from cm $^3$ to dm $^3$ .
Divide the moles of thiosulfate by this volume.
Give the final concentration to 3 or 4 significant figures with the correct unit.

Step-by-Step Reasoning

Mean titre from part (b): 25.28 cm $^3$ .
Convert to dm $^3$ :

25.28 \text{ cm}^3 = \frac{25.28}{1000} = 0.02528 \text{ dm}^3

Moles of thiosulfate from part (c)(ii), using the unrounded value:

2.523 \times 10^{-3} \text{ mol}

Concentration:

\frac{2.523 \times 10^{-3}}{0.02528} = 0.0998 \text{ mol dm}^{-3}

The answer is already to 3 significant figures, which is appropriate. If using the rounded value of 2.52 × 10 $^{-3}$ mol, the result becomes 0.0997 mol dm $^{-3}$ , which is also acceptable as 3 significant figures.

Key Takeaways

The formula $c = n/V$ requires volume in dm $^3$ .
1 mole of Na $_2$ S $_2$ O $_3$ supplies 1 mole of S $_2$ O $_3^{2-}$ , so the calculated thiosulfate concentration is the sodium thiosulfate concentration.
The final answer should be to 3 or 4 significant figures with the unit mol dm $^{-3}$ .

Common Mistakes

Forgetting to convert 25.28 cm $^3$ into dm $^3$ , leading to an answer 1000 times too large.
Using the moles of KIO $_3$ instead of the moles of thiosulfate.
Using 25.0 cm $^3$ instead of the mean titre of FB 1.
Quoting the concentration to only 1 or 2 significant figures.
Omitting the unit mol dm $^{-3}$ .

Things to Be Careful About

Carry the unrounded moles value into this calculation. The final answer depends on the candidate's actual mean titre, so a slightly different value is expected if a different concordant set of readings was obtained. The mark scheme accepts any correct working and a final answer to 3 or 4 significant figures.

Techniques used

convert the volume of thiosulfate from cm3 to dm3calculate concentration from moles and volumeround the final concentration to 3 or 4 significant figures

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 · Manipulation, Measurement and Observation10M

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