Chemistry 5070/32 — May/June 2011

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

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marks

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Topics Observations and Measurements · Experimental Contexts · Use of Techniques, Apparatus and Materials · Analysis, Conclusions and Evaluation · Qualitative Analysis

Q116MObservations and MeasurementsUse of Techniques, Apparatus and MaterialsAnalysis, Conclusions and EvaluationExperimental ContextsFree sample

Seaweed can be used as a commercial source of iodine. The amount of iodine present in a sample of seaweed is often stated in parts per million, ppm. For instance, if a sample contains 200 ppm, then there are 200 g of iodine in every 1 000 000 g of seaweed.

You are provided with an aqueous solution of iodine which has been obtained from seaweed. You are required to determine its concentration by titration with sodium thiosulfate, $\text{Na}_2\text{S}_2\text{O}_3$ , using starch as an indicator and then calculate how much iodine is present in the seaweed.

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 the aqueous solution of iodine.
$\mathbf{Q}$ is $0.100\text{ mol/dm}^3$ 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 (✓)

Summary

Tick (✓) the best titration results.

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

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

12M

DifficultyMedium-Easy

Worked solution

Answer

Complete the table with your own readings. For each titration:

record the final and initial burette readings to 1 decimal place (e.g. 24.8, 0.0)
calculate the volume used as final reading − initial reading
tick the two (or more) titres that are within 0.2 cm³ of each other as the best results
calculate the average of the ticked titres, e.g. if the two best titres are 24.8 cm³ and 24.9 cm³, the average is 24.85 cm³

Volume of solution P used is the pipette volume, either 25.0 cm³ or 20.0 cm³.

Final answer

See working: table completed with own readings, best titres ticked, average volume of Q calculated, volume of P as pipette volume (25.0 or 20.0 cm³).

Detailed explanation

Walkthrough

This part is a practical titration. You need to:

Record the readings: Write the final and initial burette readings for each titration. The initial reading is often 0.0 cm³ but could be any value. Record all readings to 1 decimal place (e.g., 24.8 cm³, not 24.85 cm³ for a burette, which typically reads to 0.05 cm³ but 5070 expects 1 decimal place).
Calculate the volume used: For each titration, subtract the initial reading from the final reading. This is the titre.
Repeat until consistent: You should repeat the titration until you have at least two titres that are within 0.2 cm³ of each other (concordant). This ensures accuracy.
Tick the best results: Tick the concordant titres. The mark scheme awards marks based on how close your titres are to the supervisor's value and how consistent they are.
Calculate the average: Add the ticked titres and divide by the number of ticked titres. For example, if you ticked 24.8 and 24.9, the average is (24.8 + 24.9) / 2 = 24.85 cm³.

Key Takeaways

Burette readings should be recorded to 1 decimal place.
The volume used is final reading minus initial reading.
Concordant results are those within 0.2 cm³ of each other.
The average of the best titres is used for calculations.

Common Mistakes

Recording readings to 2 decimal places (e.g., 24.85) – the mark scheme expects 1 decimal place for burette readings.
Not repeating the titration enough times to get concordant results.
Ticking non-concordant results.
Calculating the average of all titrations instead of only the ticked ones.

Things to Be Careful About

Ensure the burette is rinsed with the solution Q before filling.
Read the burette at eye level to avoid parallax error.
The pipette volume is fixed (25.0 cm³ or 20.0 cm³) – use the correct value in later calculations.
The average titre must be calculated correctly; the mark scheme allows an error of ±0.05 cm³.

Techniques used

record burette readings to appropriate precisioncalculate volume used as final minus initial readingrepeat titration until concordant resultstick the best (concordant) titrescalculate the average of the ticked titres

(b)

$\mathbf{Q}$ is $0.100\text{ mol/dm}^3$ sodium thiosulfate.

Using your results from (a), calculate the concentration, in $\text{mol/dm}^3$ , of iodine in $\mathbf{P}$ .

concentration of iodine in $\mathbf{P}$ = ______ $\text{mol/dm}^3$

DifficultyMedium

Worked solution

Working

Let the average titre of Q be $V\text{ cm}^3$ (from part (a)).

Moles of sodium thiosulfate used:

\text{moles of } \text{Na}_2\text{S}_2\text{O}_3 = \frac{V}{1000} \times 0.100

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}

The mole ratio is 2:1, so:

\text{moles of } \text{I}_2 = \frac{\text{moles of } \text{Na}_2\text{S}_2\text{O}_3}{2}

Volume of P used is the pipette volume, $25.0\text{ cm}^3$ (or $20.0\text{ cm}^3$ ). Convert to dm³:

\text{volume of } \mathbf{P} = \frac{25.0}{1000} \text{ dm}^3 = 0.0250\text{ dm}^3

Concentration of iodine:

\text{concentration of } \text{I}_2 = \frac{\text{moles of } \text{I}_2}{\text{volume of } \mathbf{P} \text{ in dm}^3}

For example, if the average titre $V = 24.8\text{ cm}^3$ and the pipette volume is $25.0\text{ cm}^3$ :

\text{moles of } \text{Na}_2\text{S}_2\text{O}_3 = \frac{24.8}{1000} \times 0.100 = 0.00248

\text{moles of } \text{I}_2 = \frac{0.00248}{2} = 0.00124

\text{concentration of } \text{I}_2 = \frac{0.00124}{0.0250} = 0.0496\text{ mol/dm}^3

Answer

concentration of iodine in P = $0.0496\text{ mol/dm}^3$ (example, using titre 24.8 cm³ and 25.0 cm³ pipette)

Final answer

0.0496 mol/dm3 (example; depends on your titre and pipette volume)

Detailed explanation

Walkthrough

Write down the average titre from part (a). This is the volume of sodium thiosulfate used to react with the iodine in the pipetted portion of P.
Calculate moles of sodium thiosulfate: Use the formula: $\text{moles} = \text{concentration} \times \text{volume in dm}^3$ Convert the titre from cm³ to dm³ by dividing by 1000.
Use the balanced equation to find the mole ratio. The equation shows that 2 moles of sodium thiosulfate react with 1 mole of iodine. So moles of iodine = moles of sodium thiosulfate ÷ 2.
Calculate the concentration of iodine: Divide the moles of iodine by the volume of P used (in dm³). The pipette volume is given as 25.0 cm³ or 20.0 cm³, so convert it to dm³.
The result is the concentration in mol/dm³. The mark scheme expects the answer correct to ±1 in the third significant figure.

Key Takeaways

Always convert volumes to dm³ before using them in concentration calculations.
The balanced equation gives the mole ratio; here it is 2:1 (thiosulfate:iodine).
Concentration (mol/dm³) = moles ÷ volume (dm³).

Common Mistakes

Forgetting to divide by 2 to get moles of iodine.
Using the titre in cm³ without converting to dm³.
Using the wrong pipette volume (e.g., using 25.0 cm³ when the pipette was 20.0 cm³).
Rounding too early; keep intermediate values and round only at the end.

Things to Be Careful About

The mole ratio is critical: 2:1, not 1:1.
Ensure the volume of P is in dm³.
The answer should be given to 3 significant figures (e.g., 0.0496).

Techniques used

use the balanced equation to find the mole ratiocalculate moles of sodium thiosulfate from concentration and volumecalculate moles of iodine from the mole ratiocalculate concentration of iodine in mol/dm3

(c)

Using your answer from (b), calculate the mass, in g, of iodine in $1\text{ dm}^3$ of $\mathbf{P}$ .
[The relative atomic mass of iodine is 127.]

mass of iodine in $1\text{ dm}^3$ of $\mathbf{P}$ = ______ $\text{g}$

DifficultyMedium-Easy

Worked solution

Working

From part (b), the concentration of iodine in P is $C\text{ mol/dm}^3$ (e.g., $0.0496\text{ mol/dm}^3$ ).

The molar mass of iodine molecules, $\text{I}_2$ , is:

M_r(\text{I}_2) = 2 \times 127 = 254\text{ g/mol}

Mass of iodine in $1\text{ dm}^3$ of P:

\text{mass} = C \times 254

For example, with $C = 0.0496\text{ mol/dm}^3$ :

\text{mass} = 0.0496 \times 254 = 12.6\text{ g}

Answer

mass of iodine in $1\text{ dm}^3$ of P = $12.6\text{ g}$ (example)

Final answer

12.6 g (example)

Detailed explanation

Walkthrough

Recall the concentration from part (b). This is in mol/dm³.
Find the molar mass of iodine molecules: Iodine exists as diatomic molecules, $\text{I}_2$ . Since the relative atomic mass of iodine is 127, the molar mass of $\text{I}_2$ is $2 \times 127 = 254\text{ g/mol}$ .
Calculate the mass: Multiply the concentration (mol/dm³) by the molar mass (g/mol) to get the mass in grams per dm³.

Key Takeaways

Iodine is diatomic ( $\text{I}_2$ ), so the molar mass is double the atomic mass.
Mass (g) = concentration (mol/dm³) × molar mass (g/mol) × volume (dm³). Here volume is 1 dm³, so mass = concentration × molar mass.

Common Mistakes

Using the atomic mass (127) instead of the molecular mass (254).
Forgetting to multiply by the volume (but since it's 1 dm³, it doesn't change the number).

Things to Be Careful About

The question asks for mass in 1 dm³, so the volume factor is 1.
Ensure the concentration from part (b) is used correctly.

Techniques used

calculate mass from moles and molar massuse the relative atomic mass of iodine to find molar mass of I2

(d)

If all the iodine present in $1\text{ dm}^3$ of $\mathbf{P}$ was obtained from $15\,000\text{ g}$ of seaweed, calculate the amount, in ppm, of iodine present in the seaweed.

amount of iodine present in the seaweed = ______ $\text{ppm}$

DifficultyMedium-Easy

Worked solution

Working

From part (c), the mass of iodine in $1\text{ dm}^3$ of P is $m\text{ g}$ (e.g., $12.6\text{ g}$ ).

This iodine came from $15\,000\text{ g}$ of seaweed.

ppm is defined as grams of iodine per 1,000,000 g of seaweed:

\text{ppm} = \frac{m}{15000} \times 1000000

For example, with $m = 12.6\text{ g}$ :

\text{ppm} = \frac{12.6}{15000} \times 1000000 = 840\text{ ppm}

Answer

amount of iodine present in the seaweed = $840\text{ ppm}$ (example)

Final answer

840 ppm (example)

Detailed explanation

Walkthrough

Recall the mass of iodine in 1 dm³ of P from part (c). This is the mass of iodine extracted from the given mass of seaweed.
Apply the ppm definition: ppm = (mass of iodine / mass of seaweed) × 1,000,000. The mass of seaweed is 15,000 g.
Calculate: Divide the mass of iodine by 15,000 and multiply by 1,000,000 to get the ppm.

Key Takeaways

ppm is a ratio: grams of solute per 1,000,000 grams of sample.
The calculation is straightforward once you have the mass of iodine.

Common Mistakes

Forgetting to multiply by 1,000,000.
Using the wrong mass of seaweed (e.g., 15,000 g is correct).
Mixing up units; ensure both masses are in grams.

Things to Be Careful About

The result should be a whole number or to appropriate significant figures (e.g., 840 ppm).
The mass of iodine from part (c) is for 1 dm³, which corresponds to the iodine from 15,000 g of seaweed.

Techniques used

apply the definition of ppmcalculate ppm from mass of iodine and mass of seaweed

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