Chemistry 9701/34 — May/June 2019

Cambridge AS Level · Advanced Practical Skills · worked solutions for every part, with the mark scheme

questions

marks

120

minutes

Topics Analysis, Conclusions and Evaluation · Manipulation, Measurement and Observation · Presentation of Data and Observations · Qualitative Analysis

Q1Manipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and EvaluationFree sample

Iron wire contains impurities. You will investigate the percentage by mass of iron in a sample of iron wire.

A sample of iron wire is reacted with an excess of sulfuric acid to produce a solution of iron(II) sulfate.

\text{Fe(s)} + \text{H}_2\text{SO}_4\text{(aq)} \rightarrow \text{FeSO}_4\text{(aq)} + \text{H}_2\text{(g)}

You will titrate the solution of iron(II) sulfate with potassium manganate(VII) of known concentration to determine the amount of iron(II) ions present and hence the percentage by mass of iron in the wire. You may assume the impurities do not form any products that react with potassium manganate(VII).

5\text{Fe}^{2+}(\text{aq}) + \text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) \rightarrow 5\text{Fe}^{3+}(\text{aq}) + \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O(l)}

FB 1 is $0.0200\text{ mol dm}^{-3}$ potassium manganate(VII), $\text{KMnO}_4$ .
FB 2 is a solution of $\text{FeSO}_4$ prepared by reacting $6.02\text{ g}$ of iron wire with sulfuric acid to make $1\text{ dm}^3$ of solution.
FB 3 is dilute sulfuric acid, $\text{H}_2\text{SO}_4$ .

(a)

Method

Fill a burette with FB 1.
Pipette $25.0\text{ cm}^3$ of FB 2 into a conical flask.
Use the measuring cylinder to transfer $25\text{ cm}^3$ of FB 3 into the conical flask.
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 certain that any recorded results show the precision of your practical work.
Record all of your burette readings and the volume of FB 1 added in each accurate titration.

Keep FB 3 for use in Question 2.

Results

DifficultyMedium

Worked solution

Answer

The table below is a representative record; any fair set of readings with the required precision and agreement is acceptable.

Rough titre: 25.20 cm³

initial burette reading / cm³	final burette reading / cm³	titre / cm³
0.00	25.05	25.05
25.05	50.10	25.05
0.00	25.00	25.00

The accurate titres are within 0.10 cm³ of each other and all readings are to the nearest 0.05 cm³.

Final answer

See working — representative rough and accurate titres (with headings and units) shown.

Detailed explanation

Background Concept

In a redox titration the volume of a solution of known concentration (the titrant) needed to react completely with a known volume of analyte is measured. Here potassium manganate(VII) is the oxidising agent and iron(II) is the reducing agent. The end point is the first permanent pink colour, which appears when one drop of manganate(VII) has been added in excess. Because the burette is marked in divisions of 0.1 cm³, a careful eye can read it to the nearest 0.05 cm³.

Understanding the Question

The task is to perform the titration and record the data so that the expected precision is visible: readings to 0.05 cm³, a table whose headings carry units, and at least two accurate titres that agree closely. The rough titration gives a quick estimate of the end point; the accurate titrations must be concordant, i.e. within 0.10 cm³ of each other.

Approach

Fill the burette with FB 1, pipette 25.0 cm³ of FB 2, add 25 cm³ of FB 3 with a measuring cylinder, then run the manganate(VII) in until the pink colour just persists. Do one rough run, then repeat accurately until at least two concordant results are obtained. Record every initial and final reading in a table with correct headings and units.

Step-by-Step Reasoning

First run the rough titration to find the approximate end point; record the rough titre. For each accurate run, read the burette before and after delivery, to the nearest 0.05 cm³. The titre is final reading minus initial reading. Keep repeating until two accurate titres differ by no more than 0.10 cm³. Present the data in a table with column headings such as 'initial burette reading / cm³', 'final burette reading / cm³' and 'titre / cm³'.

Key Takeaways

Accurate practical records need correct headings with units, readings to half a small division, and repeats whose agreement demonstrates reliability.

Common Mistakes

Reading the burette to only 0.1 cm³ instead of 0.05 cm³.
Omitting units in the table headings.
Averaging a non-concordant rough titre with accurate ones.
Confusing initial and final readings, so the titre comes out negative.

Things to Be Careful About

Read the burette at eye level to avoid parallax. Record the rough titre as well as the accurate ones. Make sure the final reading is not below zero. The marks require two accurate titres within 0.10 cm³.

Techniques used

fill a burette and pipette a known volumecarry out a rough titrationcarry out accurate titrations to concordancerecord burette readings to the nearest 0.05 cm3

(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 obtained this value.

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

DifficultyEasy

Worked solution

Answer

Accurate titres: 25.00 cm³ and 25.10 cm³ (agree within 0.10 cm³).

Mean titre = (25.00 + 25.10)/2 = 25.05 cm³.

Use 25.05 cm³ of FB 1.

Final answer

25.05 cm³ (mean of 25.00 and 25.10 cm³)

Detailed explanation

Background Concept

The most reliable value of a titre is the mean of at least two consistent accurate titrations. The mark scheme here requires two or more accurate titres within 0.20 cm³ of each other before they can be averaged.

Understanding the Question

Part (b) asks you to choose the titre you will use in the calculations and to show how you obtained it, normally by ticking the selected results and taking their mean.

Approach

Select two or more accurate titres that agree to within 0.20 cm³, average them, and write the mean in the blank.

Step-by-Step Reasoning

Suppose the accurate titres are 25.00 cm³ and 25.10 cm³; they agree to within 0.10 cm³. The mean is (25.00 + 25.10)/2 = 25.05 cm³, so 25.05 cm³ is used for all subsequent calculations.

Key Takeaways

Averaging concordant results improves reliability, but a spread of more than 0.20 cm³ means more titrations are needed.

Common Mistakes

Averaging results that are not concordant, or failing to show which results were averaged.

Things to Be Careful About

The mark is only given if the chosen values actually agree within 0.20 cm³ and the working is shown.

Techniques used

average concordant titres

(c)

(i)

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

DifficultyEasy

Worked solution

Answer

All final answers to parts (ii)–(v) are given to 3 significant figures.

Final answer

All four answers given to 3 significant figures.

Detailed explanation

Background Concept

Significant figures tell the reader how precisely a result is known. A value quoted as 5.01 × 10⁻⁴ has three significant figures; 0.140 has three; 92.9% has three.

Understanding the Question

The marks for (ii)–(v) are only given if at least three of the four final answers are quoted to 3 or 4 significant figures.

Approach

Carry the intermediate values with two or three extra digits and round only the final answer to 3 significant figures.

Step-by-Step Reasoning

After calculating the moles, multiply by 5 for the stoichiometric ratio, then by 55.8 for mass, then by the scaling factor 40 for the percentage. Keep at least two extra digits in the intermediate values and round the final percentage to 3 significant figures.

Key Takeaways

The number of significant figures in a final calculation should match the precision of the least precise measurement.

Common Mistakes

Rounding at each intermediate step, which can shift the final value by one or two in the last digit.

Things to Be Careful About

At least three of the four answers (ii)–(v) must be to 3 or 4 significant figures for the mark.

Techniques used

round final answers to 3 or 4 significant figures

(ii)

Use your answer to (b) to calculate the number of moles of potassium manganate(VII), FB 1, which reacted with $25.0\text{ cm}^3$ of FB 2.

\text{moles of } \text{MnO}_4^- = \text{.............................. mol}

DifficultyEasy

Worked solution

Working

$n(\text{MnO}_4^-) = \frac{0.0200 \times 25.05}{1000} = 5.01 \times 10^{-4}\, \text{mol}$

Answer

$5.01 \times 10^{-4}\, \text{mol}$

Final answer

5.01 x 10^-4 mol

Detailed explanation

Background Concept

Amount of substance in moles is concentration multiplied by volume in dm³: n = cV. Since volumes are given in cm³, divide by 1000.

Understanding the Question

V(FB 1) is the titre from part (b), 25.05 cm³, and c = 0.0200 mol dm⁻³. You must convert the volume to dm³ and multiply.

Approach

n = cV/1000.

Step-by-Step Reasoning

n(MnO₄⁻) = 0.0200 × 25.05 / 1000 = 5.01 × 10⁻⁴ mol.

Key Takeaways

Always convert cm³ to dm³ when using n = cV.

Common Mistakes

Forgetting to divide by 1000, which overestimates by a factor of 1000.

Things to Be Careful About

The answer should carry the same number of significant figures as the data, i.e. 3 significant figures.

Techniques used

convert cm3 to dm3use n = cV to calculate moles

(iii)

Use the information on page 2 to calculate the number of moles of iron(II) ions present in $25.0\text{ cm}^3$ of FB 2.

\text{moles of } \text{Fe}^{2+} = \text{.............................. mol}

DifficultyEasy

Worked solution

Working

$n(\text{Fe}^{2+}) = 5 \times 5.01 \times 10^{-4} = 2.51 \times 10^{-3}\, \text{mol}$

Answer

$2.51 \times 10^{-3}\, \text{mol}$

Final answer

2.51 x 10^-3 mol

Detailed explanation

Background Concept

The balanced equation shows 5 mol Fe²⁺ react with 1 mol MnO₄⁻.

Understanding the Question

Use the mole ratio to convert moles of manganate(VII) into moles of iron(II).

Approach

Multiply the moles of MnO₄⁻ by 5.

Step-by-Step Reasoning

n(Fe²⁺) = 5 × 5.01 × 10⁻⁴ = 2.51 × 10⁻³ mol. This is the number of Fe²⁺ ions in 25.0 cm³ of FB 2.

Key Takeaways

The stoichiometric ratio from the balanced equation links the two measured quantities.

Common Mistakes

Using the ratio 1:1 or 1:5 in the wrong direction.

Things to Be Careful About

Give the answer to 3 significant figures and keep the units mol.

Techniques used

apply the 5:1 stoichiometric ratio

(iv)

Calculate the mass of iron present in $25.0\text{ cm}^3$ of FB 2.

\text{mass of Fe} = \text{.............................. g}

DifficultyEasy

Worked solution

Working

$m(\text{Fe}) = 2.51 \times 10^{-3} \times 55.8 = 0.140\, \text{g}$

Answer

$0.140\, \text{g}$

Final answer

0.140 g

Detailed explanation

Background Concept

mass = moles × molar mass. For iron, M = 55.8 g mol⁻¹.

Understanding the Question

Use the moles of Fe²⁺ from part (iii) to find the mass of iron in the 25.0 cm³ portion.

Approach

m = n × M.

Step-by-Step Reasoning

m(Fe) = 2.51 × 10⁻³ × 55.8 = 0.140 g.

Key Takeaways

The molar mass converts an amount in moles into a mass in grams.

Common Mistakes

Using the molar mass of FeSO₄ instead of Fe.

Things to Be Careful About

The mass is only for the 25.0 cm³ portion, not the whole 1 dm³, so it cannot be used directly as the percentage.

Techniques used

convert moles to mass using molar mass

(v)

Calculate the percentage by mass of iron in the sample of iron wire.

\text{percentage by mass of iron in iron wire} = \text{.............................. \%}

DifficultyMedium-Easy

Worked solution

Working

$\text{percentage} = \frac{0.140 \times 40 \times 100}{6.02} = 93.0\%$

(Using unrounded values gives $92.9\%$ .)

Answer

$93.0\%$

Final answer

93.0%

Detailed explanation

Background Concept

The 6.02 g of iron wire was dissolved and made up to 1 dm³; each 25.0 cm³ portion contains 25.0/1000 = 1/40 of the total. So the mass of iron in the whole solution is the mass in the portion × 40, and the percentage is (mass of iron / 6.02) × 100.

Understanding the Question

Scale the mass of iron found in part (iv) up to the full 1 dm³, then express it as a percentage of the original 6.02 g sample.

Approach

percentage = (c(iv) × 40 × 100) / 6.02.

Step-by-Step Reasoning

percentage = (0.140 × 40 × 100) / 6.02 = 93.0%. (Using the unrounded values gives 92.9%.)

Key Takeaways

When a sample is split into aliquots, multiply by the reciprocal of the aliquot fraction to recover the whole.

Common Mistakes

Forgetting the factor of 40, which underestimates the percentage by a factor of 40.

Things to Be Careful About

The answer must be a percentage with no units and should be given to 3 significant figures.

Techniques used

scale aliquot mass to the whole samplecalculate percentage by mass

(d)

A student suggested that when a piece of iron wire was dissolved in a known volume and concentration of sulfuric acid, the number of moles of iron that reacted with the acid could be determined by working out how much acid was left after the reaction. The amount of excess acid could be determined by titrating the mixture with a known concentration of sodium hydroxide.

Explain whether the student was correct.

DifficultyMedium

Worked solution

Answer

The student was incorrect. Sodium hydroxide would also react with $\text{Fe}^{2+}$ / iron(II) / iron(II) sulfate, so the volume of NaOH consumed would not correspond to the excess acid alone. (An impurity in the wire might also react with the acid.)

Final answer

Student is incorrect — NaOH also reacts with Fe2+/iron(II) sulfate.

Detailed explanation

Background Concept

A back-titration to find excess acid is only valid if the titrant reacts with nothing except the acid. Sodium hydroxide is a strong alkali that also reacts with Fe²⁺/FeSO₄, so it cannot distinguish between acid and iron(II).

Understanding the Question

The student proposes reacting the wire with a known amount of acid, then titrating the remaining acid with NaOH. The claim is that the amount of NaOH consumed equals the excess acid. This is incorrect because NaOH also reacts with Fe²⁺.

Approach

Identify what species are in the reaction mixture: excess H₂SO₄ and FeSO₄ (and possibly dissolved impurities). Since NaOH reacts with both, it cannot measure only the acid.

Step-by-Step Reasoning

The mixture contains Fe²⁺ ions as well as H⁺ from the excess sulfuric acid. Sodium hydroxide is a base that will neutralise the acid, but it also reacts with iron(II) sulfate, forming iron(II) hydroxide. Consequently, the volume of NaOH required does not correspond solely to the excess acid. In addition, any impurity in the wire might also consume some acid, further complicating the result.

Key Takeaways

A back-titration scheme is only correct if every component that reacts with the titrant is accounted for; otherwise the calculated excess will be wrong.

Common Mistakes

Assuming that 'acid left' equals 'NaOH used' without checking what else NaOH reacts with.

Things to Be Careful About

The mark scheme credits either the Fe²⁺/NaOH reaction or the impurity point; one clear reason is enough.

Techniques used

evaluate a back-titration schemeidentify a competing reaction of the titrant

The rest of this paper

2 more questions

Q2Manipulation, Measurement and Observation · Presentation of Data and Observations · Analysis, Conclusions and Evaluation12M
Q3Qualitative Analysis · Analysis, Conclusions and Evaluation14M

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