Chemistry 5070/31 — May/June 2017

Cambridge O-Level · Practical Test · worked solutions for every part, with the mark scheme

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Topics Observations and Measurements · Experimental Contexts · Analysis, Conclusions and Evaluation · Qualitative Analysis

Q118MObservations and MeasurementsExperimental ContextsAnalysis, Conclusions and EvaluationFree sample

Chlorine water is an aqueous solution of chlorine made by bubbling the gas through water. The amount of chlorine present in the solution can be estimated by reacting the chlorine with aqueous potassium iodide.

\text{Cl}_2 + 2\text{KI} \rightarrow 2\text{KCl} + \text{I}_2

The amount of iodine produced by the above reaction can then be determined by titration with aqueous sodium thiosulfate, $\text{Na}_2\text{S}_2\text{O}_3$ , using starch as an indicator.

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

$\mathbf{P}$ is an aqueous solution of iodine produced by mixing $50\text{ cm}^3$ of chlorine water with $200\text{ cm}^3$ of aqueous potassium iodide, an excess.

$\mathbf{Q}$ is $0.0230\text{ mol / dm}^3$ sodium thiosulfate.

(a)

Put $\mathbf{Q}$ into the burette.

Pipette a $25.0\text{ cm}^3$ (or $20.0\text{ cm}^3$ ) portion of $\mathbf{P}$ into a flask.

Add $\mathbf{Q}$ from the burette until the red-brown colour fades to pale yellow, then add a few drops of the starch indicator. This will give a dark blue solution. Continue adding $\mathbf{Q}$ slowly from the burette until one drop of $\mathbf{Q}$ causes the blue colour to disappear, leaving a colourless solution.

Record your results in the table, repeating the titration as many times as you consider necessary to achieve consistent results.

Results

Burette readings

titration number	1	2
final reading / $\text{cm}^3$
initial reading / $\text{cm}^3$
volume of $\mathbf{Q}$ used / $\text{cm}^3$
best titration results $(\checkmark)$

Summary

Tick $(\checkmark)$ the best titration results.

Using these results, the average volume of $\mathbf{Q}$ required was ______ $\text{cm}^3$ .

Volume of $\mathbf{P}$ used was ______ $\text{cm}^3$ .

12M

DifficultyMedium

Worked solution

Answer

Complete the results table with:

initial and final burette readings for each titration, both recorded to one decimal place
volume of Q used = final − initial, calculated correctly for each titration
tick the two most consistent (concordant) titres
average volume of Q = mean of the ticked titres
volume of P used = 25.0 cm3 (the pipette volume)

Final answer

See working — candidate's own readings required

Detailed explanation

Walkthrough

This part is the practical titration itself. Chlorine water was reacted with excess potassium iodide to produce iodine, giving solution P. The iodine is then titrated with sodium thiosulfate Q. Starch is added near the end (when the colour is pale yellow) because starch forms an intense blue-black complex with iodine, making the end point much sharper. The end point is when one drop of Q turns the blue solution colourless.

The candidate must:

Fill the burette with Q and record the initial reading (to 1 decimal place).
Pipette 25.0 cm3 of P into the flask.
Titrate until the red-brown colour fades to pale yellow, add a few drops of starch, then continue until the blue colour disappears.
Record the final reading and calculate the volume used (final − initial).
Repeat until consistent (concordant) results are obtained.
Tick the best titres and calculate their average.

The 12 marks are awarded for: correct readings to 1 decimal place (1), correct titre calculations (1), accuracy against the supervisor's value — within 0.2 cm3 (3), 0.3 cm3 (2) or 0.4 cm3 (1) (6), concordance of the ticked values — within 0.2 cm3 (3), 0.3 cm3 (2) or 0.4 cm3 (1) (3), and a correct average of the ticked titres (1).

Key Takeaways

Burette readings are recorded to one decimal place, including 0.0.
Volume used = final reading − initial reading.
Concordant titres are within 0.2 cm3 of each other.
Starch indicator is added near the end point, not at the start.
The average must be calculated only from the ticked (best) titres.

Common Mistakes

Recording readings to two decimal places or as whole numbers.
Subtraction errors when calculating the titre.
Adding starch at the start of the titration — it binds all the iodine and obscures the end point.
Not repeating the titration enough to obtain concordant results.
Averaging all titres instead of only the ticked ones.

Things to Be Careful About

No reading may exceed 50.0 cm3 and no initial reading may be given as 50.0 cm3.
All readings must be to one decimal place.
The volume of P used is the pipette volume, 25.0 cm3 (or 20.0 cm3 if that pipette was used).

Techniques used

record burette readings to one decimal placecalculate titre volumes from initial and final readingsselect concordant results for averagingcalculate the average titre

(b)

$\mathbf{Q}$ is $0.0230\text{ mol / dm}^3$ sodium thiosulfate.

Calculate the number of moles of sodium thiosulfate in the average volume of $\mathbf{Q}$ used in the titration.

moles of sodium thiosulfate = ______

DifficultyMedium-Easy

Worked solution

Working

Using the average titre from part (a) — shown here with the example value of 20.3 cm3:

\text{moles of } \text{Na}_2\text{S}_2\text{O}_3 = \frac{20.3}{1000} \times 0.0230

= 0.000467 \text{ mol}

Answer

0.000467 mol

Final answer

0.000467 mol

Detailed explanation

Walkthrough

The concentration of Q is 0.0230 mol / dm3. To find the number of moles in the average titre volume, convert the volume from cm3 to dm3 (divide by 1000) and multiply by the concentration:

\text{moles} = \text{concentration} \times \text{volume in dm}^3

Using the example average titre of 20.3 cm3:

\text{moles of } \text{Na}_2\text{S}_2\text{O}_3 = \frac{20.3}{1000} \times 0.0230 = 0.000467 \text{ mol}

Key Takeaways

moles = concentration (mol / dm3) × volume (dm3).
Convert cm3 to dm3 by dividing by 1000.

Common Mistakes

Forgetting to convert cm3 to dm3 — this would give 0.467 mol instead of 0.000467 mol.
Using the wrong volume (e.g. the pipette volume instead of the titre).

Things to Be Careful About

The volume used must be the average titre from part (a), divided by 1000.
Give the answer to 3 significant figures.

Techniques used

convert volume from cm3 to dm3apply moles = concentration × volume

(c)

Using your answer from (b), deduce the number of moles of iodine in the volume of $\mathbf{P}$ used in the titration.

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

moles of iodine = ______

DifficultyMedium-Easy

Worked solution

Working

From the equation:

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

2 mol thiosulfate reacts with 1 mol iodine, so:

\text{moles of iodine} = \frac{0.000467}{2} = 0.000234 \text{ mol}

Answer

0.000234 mol

Final answer

0.000234 mol

Detailed explanation

Walkthrough

The balanced equation shows the stoichiometric relationship:

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

2 mol of sodium thiosulfate react with 1 mol of iodine, so the moles of iodine are half the moles of thiosulfate:

\text{moles of iodine} = \frac{0.000467}{2} = 0.000234 \text{ mol}

Key Takeaways

Use the mole ratio from the balanced equation to convert between reactants.

Common Mistakes

Using a 1:1 ratio instead of the correct 2:1 ratio.

Things to Be Careful About

The ratio is thiosulfate : iodine = 2 : 1, so the moles of iodine are half the moles of thiosulfate.

Techniques used

use the stoichiometric ratio from the balanced equation

(d)

Using your answer from (c), calculate the number of moles of iodine in $250\text{ cm}^3$ of $\mathbf{P}$ .

moles of iodine in $250\text{ cm}^3$ of $\mathbf{P}$ = ______

DifficultyMedium-Easy

Worked solution

Working

The 25.0 cm3 portion of P contains 0.000234 mol iodine. The total volume of P is 250 cm3, so:

\text{moles of iodine in } 250\text{ cm}^3 = 0.000234 \times \frac{250}{25}

= 0.00234 \text{ mol}

Answer

0.00234 mol

Final answer

0.00234 mol

Detailed explanation

Walkthrough

The 25.0 cm3 portion of P that was pipetted contained 0.000234 mol of iodine. The total volume of P is 250 cm3 (50 cm3 chlorine water + 200 cm3 potassium iodide solution). Because the solution is uniform, the iodine is spread evenly throughout, so the moles in the whole 250 cm3 are:

\text{moles of iodine in } 250\text{ cm}^3 = 0.000234 \times \frac{250}{25} = 0.00234 \text{ mol}

Key Takeaways

Scaling a portion to the whole volume uses the ratio of the volumes.

Common Mistakes

Forgetting to scale up and giving the portion value.
Using the wrong total volume.

Things to Be Careful About

The total volume of P is 250 cm3.
If a 20.0 cm3 pipette was used, the scaling factor is 250/20.

Techniques used

scale moles from the pipetted portion to the total volume

(e)

Using your answer from (d), deduce the number of moles of chlorine in $50\text{ cm}^3$ of the chlorine water.

\text{Cl}_2 + 2\text{KI} \rightarrow 2\text{KCl} + \text{I}_2

moles of chlorine in $50\text{ cm}^3$ of the chlorine water = ______

DifficultyMedium-Easy

Worked solution

Working

From the equation:

\text{Cl}_2 + 2\text{KI} \rightarrow 2\text{KCl} + \text{I}_2

1 mol Cl2 produces 1 mol I2, so moles of chlorine = moles of iodine:

\text{moles of chlorine} = 0.00234 \text{ mol}

Answer

0.00234 mol

Final answer

0.00234 mol

Detailed explanation

Walkthrough

The equation for the production of iodine from chlorine water is:

\text{Cl}_2 + 2\text{KI} \rightarrow 2\text{KCl} + \text{I}_2

1 mol of chlorine produces 1 mol of iodine, so the moles of chlorine in the original 50 cm3 of chlorine water equal the moles of iodine in the whole 250 cm3 of P:

\text{moles of chlorine} = 0.00234 \text{ mol}

Key Takeaways

The mole ratio between chlorine and iodine is 1 : 1.

Common Mistakes

Using a 1:2 ratio (from the 2KI).

Things to Be Careful About

Chlorine is the limiting reactant; KI is in excess, so all the chlorine is converted to iodine.

Techniques used

use the stoichiometric ratio from the balanced equation

(f)

Using your answer from (e), calculate the mass, in $\text{g}$ , of chlorine in $1\text{ dm}^3$ of the chlorine water.
[ $A_r$ : $\text{Cl}$ , 35.5]

mass of chlorine in $1\text{ dm}^3$ of the chlorine water = ______ $\text{g}$

DifficultyMedium

Worked solution

Working

Mass of chlorine in 50 cm3 = moles × $M_r$ :

0.00234 \times 71 = 0.166 \text{ g}

Mass in 1 dm3 (1000 cm3):

0.166 \times \frac{1000}{50} = 3.32 \text{ g}

Answer

3.32 g

Final answer

3.32 g

Detailed explanation

Walkthrough

First convert moles of chlorine to mass. Chlorine exists as Cl2 molecules, so $M_r = 2 \times 35.5 = 71$ .

Mass of chlorine in 50 cm3 of chlorine water:

\text{mass} = 0.00234 \times 71 = 0.166 \text{ g}

Now scale up to 1 dm3 (1000 cm3):

\text{mass in } 1\text{ dm}^3 = 0.166 \times \frac{1000}{50} = 3.32 \text{ g}

Key Takeaways

mass = moles × $M_r$ .
Scaling from 50 cm3 to 1 dm3 multiplies by 1000/50 = 20.

Common Mistakes

Using $A_r$ of Cl (35.5) instead of $M_r$ of Cl2 (71).
Forgetting to scale up to 1 dm3.

Things to Be Careful About

Chlorine is a diatomic molecule, so $M_r$ = 71.
The final answer is in g per dm3, with 3 significant figures.

Techniques used

convert moles to mass using Mrscale from 50 cm3 to 1 dm3

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