Chemistry 9701/34 — 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

Hydrogen peroxide, $\text{H}_2\text{O}_2$ , is used in hair bleach and for skin therapies. In this experiment you will determine the concentration of a solution of hydrogen peroxide by titration with acidified potassium manganate(VII).

FB 1 is $0.0250\text{ mol dm}^{-3}$ potassium manganate(VII), $\text{KMnO}_4$ .
FB 2 is dilute sulfuric acid, $\text{H}_2\text{SO}_4$ .
FB 3 is aqueous hydrogen peroxide, $\text{H}_2\text{O}_2$ .

(a)

Method

Dilution of FB 3

Pipette $25.0\text{ cm}^3$ of FB 3 into the volumetric (graduated) flask.
Make the solution up to the mark using distilled water.
Shake the flask thoroughly.
This diluted solution of hydrogen peroxide is FB 4.

Titration

Fill the burette with FB 1.
Pipette $10.0\text{ cm}^3$ of FB 4 into a conical flask.
Use a measuring cylinder to add $25\text{ cm}^3$ of FB 2 into the same flask.
Add FB 1 until a permanent pale pink colour is seen.
Perform 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 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.

Keep solution FB 2 for use in Question 3 and solution FB 3 for use in Questions 2 and 3.

DifficultyMedium-Easy

Worked solution

Answer

A suitable results table using representative readings (your own readings should be substituted):

Titration	final burette reading / $\text{cm}^3$	volume of FB 1 added / $\text{cm}^3$
rough	25.10	25.10
accurate 1	25.00	25.00
accurate 2	25.00	25.00
accurate 3	25.00	25.00

All burette readings are recorded to the nearest $0.05\text{ cm}^3$ . The three accurate titres are concordant (all $25.00\text{ cm}^3$ , within $0.10\text{ cm}^3$ ).

Final answer

Representative titres shown; all accurate titres concordant at 25.00 cm3

Detailed explanation

Background Concept

Acidified potassium manganate(VII), $\text{KMnO}_4$ , is a powerful oxidising agent. In acidic solution the purple $\text{MnO}_4^-$ ion is reduced to almost colourless $\text{Mn}^{2+}$ . Hydrogen peroxide is oxidised to oxygen gas. Because the titrant itself is strongly coloured, the end-point is a permanent pale pink colour caused by a tiny excess of $\text{KMnO}_4$ ; no separate indicator is needed.

Understanding the Question

This part asks for the practical record of the titration. You must show a rough titre and at least two accurate titrations, with all burette readings tabulated in a suitable form. The marks reward correct headings and units, readings to the nearest $0.05\text{ cm}^3$ , and concordant accurate titres within $0.10\text{ cm}^3$ . The exact numbers depend on your own experiment, so the solution above uses representative values.

Approach

Use the rough titration to find the approximate end-point. Then repeat the titration carefully until you obtain at least two accurate titres that agree. Record each initial and final burette reading immediately, and calculate the volume of FB 1 added as final minus initial. Build a clear table with quantity, unit and consistent decimal places.

Step-by-Step Reasoning

Fill the burette with FB 1 and note the initial reading. With a pipette place $10.0\text{ cm}^3$ of FB 4 in a conical flask and add $25\text{ cm}^3$ of FB 2 using a measuring cylinder.
Add FB 1 until a permanent pale pink colour appears; record the rough titration.
Repeat accurately, taking readings to $0.05\text{ cm}^3$ : for example, initial $0.00$ and final $25.00\text{ cm}^3$ gives a titre of $25.00\text{ cm}^3$ .
Repeat until at least two accurate titres agree within $0.10\text{ cm}^3$ . The representative table shows three accurate titres of $25.00\text{ cm}^3$ .
Label the rough row and do not use it in the average.

Key Takeaways

A good titration record is precise and clearly presented. Concordant means that the accurate titres differ by no more than $0.10\text{ cm}^3$ . Burette readings must be recorded to the nearest $0.05\text{ cm}^3$ , but the titre itself may be quoted to one decimal place.

Common Mistakes

Mixing up initial and final headings, or omitting units.
Recording burette readings to 1 dp only, e.g. $25.0$ instead of $25.00$ .
Using the rough titration in the mean.
Having more than one final burette reading of $50.00\text{ cm}^3$ ; this is penalised.
Using $50.00\text{ cm}^3$ as an initial reading.

Things to Be Careful About

Write each heading with both a quantity and its unit, e.g. initial burette reading / $\text{cm}^3$ . Keep the same decimal places down each column. Make sure all accurate readings are in the table, not just the titres. The mark scheme does not award the concordance mark if any accurate titre is recorded to zero decimal places.

Techniques used

perform a rough and accurate titrationsrecord burette readings to 0.05 cm3tabulate initial/final readings and titres with unitsselect concordant titres

(b)

From your accurate titration results, obtain a suitable value for the volume of FB 1 to be used in your calculations.
Show clearly how you have obtained this value.

$10.0\text{ cm}^3$ of FB 4 required ................ $\text{cm}^3$ of FB 1.

DifficultyEasy

Worked solution

Answer

Using the three concordant accurate titres, $25.00\text{ cm}^3$ , $25.00\text{ cm}^3$ and $25.00\text{ cm}^3$ :

\text{mean titre} = \frac{25.00 + 25.00 + 25.00}{3} = 25.00\text{ cm}^3

The suitable volume of FB 1 used in calculations is $25.00\text{ cm}^3$ (quoted to 2 dp).

Final answer

25.00 cm3

Detailed explanation

Background Concept

After choosing concordant accurate titres, the mean titre is usually the volume used in later calculations. A mean is only meaningful if the individual titres agree closely; the mark scheme expects a total spread of no more than $0.20\text{ cm}^3$ , and ideally within $0.10\text{ cm}^3$ .

Understanding the Question

Use your accurate titres to obtain a single suitable volume of FB 1. Show how you obtained it and quote it to 2 dp unless a special convention allows otherwise.

Approach

Select the accurate titres that agree, average them, and round the mean to 2 dp. Ticks next to the selected readings or a short calculation show which readings were used.

Step-by-Step Reasoning

With representative titres $25.00$ , $25.00$ and $25.00\text{ cm}^3$ :
$\text{mean} = \frac{25.00 + 25.00 + 25.00}{3} = 25.00\text{ cm}^3$
The mean is already exact and is quoted to 2 dp. If the mean had been $25.025\text{ cm}^3$ , it may be written as $25.025\text{ cm}^3$ at 3 dp. If all accurate readings were recorded to 1 dp and the mean is exactly correct at 1 dp, that is also allowed.

Key Takeaways

The volume taken forward is the mean of concordant accurate titres, not the rough titrearning. Showing working is part of the mark.

Common Mistakes

Including the rough titre in the average.
Averaging titres whose spread is greater than $0.20\text{ cm}^3$ .
Rounding the mean to 1 dp although readings were at 2 dp.

Things to Be Careful About

Always state the readings you have selected VOWh, either by a calculation or by ticking them. Quote the final mean to 2 dp unless one of the special cases in the mark scheme allows 3 dp or 1 dp.

Techniques used

identify concordant accurate titrescalculate a mean titreround the mean to 2 decimal places

(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 potassium manganate(VII) present in the volume calculated in (b).

moles of $\text{KMnO}_4$ = ....................... mol

DifficultyMedium-Easy

Worked solution

Working

Volume of FB 1 used = $25.00\text{ cm}^3 = 0.02500\text{ dm}^3$ .

\text{moles KMnO}_4 = 0.0250 \times \frac{25.00}{1000} = 6.25 \times 10^{-4}\text{ mol}

Answer

$6.25 \times 10^{-4}\text{ mol}$

Final answer

6.25 x 10^-4 mol

Detailed explanation

Background Concept

The amount of a solute in moles is given by concentration multiplied by volume in cubic decimetres: moles = concentration ( $\text{mol dm}^{-3}$ ) × volume ( $\text{dm}^3$ ). Because burette readings are in $\text{cm}^3$ , divide by 1000 before multiplying.

Understanding the Question

You are asked to calculate the moles of $\text{KMnO}_4$ used in the titration from the concentration of FB 1 and the volume found in part (b).

Approach

Use the suitable titre from part (b) as the volume of FB 1. Convert it to $\text{dm}^3$ and multiply by the concentration.

Step-by-Step Reasoning

With a suitable titre of $25.00\text{ cm}^3$ :
$25.00\text{ cm}^3 = 0.02500\text{ dm}^3$
$\text{moles KMnO}_4 = 0.0250 \times 0.02500 = 6.25 \times 10^{-4}\text{ mol}$
The value is quoted to 3 significant figures because the concentration is given to 3 sf.

Key Takeaways

The volume from part (b) must be in $\text{dm}^3$ before calculating moles. The result is then used in the stoichiometric ratio in part (c)(iii).

Common Mistakes

Forgetting to divide the titre by 1000.
Multiplying by the volume in $\text{cm}^3$ without conversion.
Quoting too many or too few significant figures.

Things to Be Careful About

Use the exact mean titre or a rounded value that is consistent with your practical work. Keep at least 3 significant figures in intermediate steps.

Techniques used

convert cm3 to dm3calculate moles from concentration and volumeround to appropriate significant figures

(ii)

Complete the equation below for the reaction of potassium manganate(VII) with hydrogen peroxide. State symbols are not required.

\text{.........KMnO}_4 + 5\text{H}_2\text{O}_2 + 3\text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + 2\text{MnSO}_4 + \text{.........H}_2\text{O} + 5\text{O}_2

DifficultyMedium

Worked solution

Answer

2\text{KMnO}_4 + 5\text{H}_2\text{O}_2 + 3\text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + 2\text{MnSO}_4 + 8\text{H}_2\text{O} + 5\text{O}_2

Final answer

2KMnO4 + 5H2O2 + 3H2SO4 -> K2SO4 + 2MnSO4 + 8H2O + 5O2

Detailed explanation

Background Concept

In acid, permanganate is reduced: $\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$ . Hydrogen peroxide is oxidised: $\text{H}_2\text{O}_2 \rightarrow \text{O}_2 + 2\text{H}^+ + 2\text{e}^-$ . Combining equal electron transfer gives $2\text{MnO}_4^-$ with $5\text{H}_2\text{O}_2$ .

Understanding the Question

The skeleton equation is already full, except for two missing coefficients: the coefficient of $\text{KMnO}_4$ and the coefficient of $\text{H}_2\text{O}$ . You must complete the balancing.

Approach

Balance the equation by atoms. Use the given coefficient 5 on $\text{H}_2\text{O}_2$ : since one $\text{H}_2\text{O}_2$ gives one $\text{O}_2$ , the 5 on the right fixes the oxygen-gas coefficient. Balance K, Mn and then hydrogen and oxygen.

Step-by-Step Reasoning

The right has one $\text{K}_2\text{SO}_4$ , so two K atoms are needed on the left: coefficient 2 before $\text{KMnO}_4$ .
The right has two $\text{MnSO}_4$ , matching the two Mn from two $\text{KMnO}_4$ .
Count hydrogens on the left: $5\text{H}_2\text{O}_2$ gives 10 H; $3\text{H}_2\text{SO}_4$ gives 6 H, total 16 H. On the right, water must provide 16 H, so coefficient 8 for $\text{H}_2\text{O}$ .
Check oxygen: left has $2 \times 4 + 5 \times 2 + 3 \times 4 = 30$ O; right has $4 + 2 \times 4 + 8 + 5 \times 2 = 30$ O. Balanced.

Key Takeaways

In a redox titration equation, both atoms and charge must balance. Here the key ratio is $2\text{KMnO}_4 : 5\text{H}_2\text{O}_2$ , used later in the calculation.

Common Mistakes

Putting the wrong coefficient on $\text{H}_2\text{O}$ (often 4 or 10) by miscounting hydrogens.
Forgetting to balance potassium and manganese with coefficient 2.
Adding state symbols when they are not required; they are not needed here.

Things to Be Careful About

Use the given coefficient 5 on $\text{H}_2\text{O}_2$ and $5\text{O}_2$ as anchors. Always verify the oxygen count last; it is the easiest place to make an error.

Techniques used

balance a redox equationbalance atoms of each elementidentify oxidation and reduction products

(iii)

Use your answers to (i) and (ii) to calculate the number of moles of hydrogen peroxide used in each titration.

moles of $\text{H}_2\text{O}_2$ = ...................... mol

DifficultyMedium-Easy

Worked solution

Working

From the balanced equation, the mole ratio $\text{H}_2\text{O}_2 : \text{KMnO}_4 = 5:2$ .

\text{moles H}_2\text{O}_2 = \frac{5}{2} \times 6.25 \times 10^{-4} = 1.5625 \times 10^{-3}\text{ mol}

= $1.56 \times 10^{-3}\text{ mol}$ (3 sf).

Answer

$1.56 \times 10^{-3}\text{ mol}$

Final answer

1.56 x 10^-3 mol

Detailed explanation

Background Concept

The balanced equation shows that 2 moles of $\text{KMnO}_4$ react with 5 moles of $\text{H}_2\text{O}_2$ . Therefore moles of $\text{H}_2\text{O}_2$ = 2.5 × moles of $\text{KMnO}_4$ .

Understanding the Question

Using your answer to (i), find the number of moles of hydrogen peroxide in the $10.0\text{ cm}^3$ sample of FB 4 that was titrated.

Approach

Multiply the moles of $\text{KMnO}_4$ by the ratio $5/2$ .

Step-by-Step Reasoning

$\text{moles H}_2\text{O}_2 = \frac{5}{2} \times 6.25 \times 10^{-4} = 1.5625 \times 10^{-3}\text{ mol}$
Quoted to 3 significant figures: $1.56 \times 10^{-3}\text{ mol}$ .

Key Takeaways

The mole ratio must come from the balanced equation. Keep an extra digit in intermediate working to avoid rounding errors in the next parts.

Common Mistakes

Using the ratio upside down, giving $0.4 \times$ the KMnO4 moles.
Quoting the unrounded $1.5625 \times 10^{-3}$ as final when the instruction requires 3 or 4 significant figures.
Forgetting that this value refers to only $10.0\text{ cm}^3$ of FB 4.

Things to Be Careful About

If the equation in (ii) was incorrect, the mark scheme allows error carried forward: use your own balanced ratio consistently.

Techniques used

use the stoichiometric ratio from the balanced equationcalculate moles of H2O2 from moles of KMnO4round to appropriate significant figures

(iv)

Calculate the concentration of $\text{H}_2\text{O}_2$ in FB 4, in $\text{mol dm}^{-3}$ .

concentration of $\text{H}_2\text{O}_2$ in FB 4 = ...................... $\text{mol dm}^{-3}$

DifficultyMedium-Easy

Worked solution

Working

Each titration used a $10.0\text{ cm}^3$ sample of FB 4, so volume $= 0.0100\text{ dm}^3$ .

[\text{H}_2\text{O}_2]_{\text{FB4}} = \frac{1.5625 \times 10^{-3}}{0.0100} = 0.15625\text{ mol dm}^{-3}

= $0.156\text{ mol dm}^{-3}$ (3 sf).

Answer

$0.156\text{ mol dm}^{-3}$

Final answer

0.156 mol dm^-3

Detailed explanation

Background Concept

Concentration is moles per unit volume: concentration = moles / volume in $\text{dm}^3$ . The volume used in the titration is the aliquot pipetted, $10.0\text{ cm}^3$ .

Understanding the Question

You know the moles of hydrogen peroxide in the $10.0\text{ cm}^3$ sample of FB 4. You must convert this into a concentration for FB 4.

Approach

Convert $10.0\text{ cm}^3$ to $0.0100\text{ dm}^3$ and divide the moles from part (iii) by this volume.

Step-by-Step Reasoning

$[\text{H}_2\text{O}_2]_{\text{FB4}} = \frac{1.5625 \times 10^{-3}}{0.0100} = 0.15625\text{ mol dm}^{-3}$
With 3 significant figures, the concentration is $0.156\text{ mol dm}^{-3}$ .

Key Takeaways

The volume used must be the sample volume actually titrated, not the original FB 3 volume. This concentration refers to the diluted solution FB 4.

Common Mistakes

Dividing by $10.0$ instead of $0.0100\text{ dm}^3$ .
Confusing FB 4 with FB 3 at this stage.
Quoting $0.15625$ when the instruction asks for appropriate significant figures.

Things to Be Careful About

Keep the full intermediate value $0.15625\text{ mol dm}^{-3}$ because it is used in part (v).

Techniques used

convert the 10.0 cm3 aliquot volume to dm3calculate concentration from moles and volumeround to appropriate significant figures

(v)

Calculate the concentration of $\text{H}_2\text{O}_2$ in FB 3, in $\text{mol dm}^{-3}$ .

concentration of $\text{H}_2\text{O}_2$ in FB 3 = ...................... $\text{mol dm}^{-3}$

DifficultyMedium-Easy

Worked solution

Working

FB 3 was diluted by pipetting $25.0\text{ cm}^3$ into a $250\text{ cm}^3$ volumetric flask, so the dilution factor is 10.

[\text{H}_2\text{O}_2]_{\text{FB3}} = 0.15625 \times 10 = 1.5625\text{ mol dm}^{-3}

= $1.56\text{ mol dm}^{-3}$ (3 sf).

Answer

$1.56\text{ mol dm}^{-3}$

Final answer

1.56 mol dm^-3

Detailed explanation

Background Concept

When a solution is diluted, the number of moles of solute stays the same but the volume increases. Dilution factor = final volume / volume taken. Since $25.0\text{ cm}^3$ was diluted to $250\text{ cm}^3$ , the factor is 10; FB 3 is ten times more concentrated than FB 4.

Understanding the Question

You have found the concentration of the diluted FB 4. Now you must work backwards to find the concentration of the original FB 3.

Approach

Multiply the concentration of FB 4 by the dilution factor of 10.

Step-by-Step Reasoning

Each $1\text{ cm}^3$ of FB 3 became $10\text{ cm}^3$ of FB 4, so the concentration of FB 3 is 10 times that of FB 4:
$[\text{H}_2\text{O}_2]_{\text{FB3}} = 0.15625 \times 10 = 1.5625\text{ mol dm}^{-3}$
Quoted to 3 significant figures: $1.56\text{ mol dm}^{-3}$ .

Key Takeaways

A dilution factor relates the concentration of the diluted solution to the original solution. Work backwards by multiplying when the original is more concentrated.

Common Mistakes

Dividing by 10 instead of multiplying.
Using the volume of FB 4 in the titration instead of the dilution factor.
Forgetting that the dilution factor is 10 in this particular method.

Things to Be Careful About

Confirm the volumetric flask volume from the method; the mark scheme expects the factor of 10 for this experiment. Quote the final concentration to 3 or 4 significant figures consistently with the earlier parts.

Techniques used

use the dilution factor from 25.0 cm3 to 250 cm3calculate the original concentration of FB 3round to appropriate significant figures

The rest of this paper

2 more questions

Q2Manipulation, Measurement and Observation · Presentation of Data and Observations · Analysis, Conclusions and Evaluation14M
Q3Qualitative Analysis · Analysis, Conclusions and Evaluation · Manipulation, Measurement and Observation13M

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