Chemistry 9701/34 — October/November 2010

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

FB 1 is an aqueous solution containing $21.50\text{ g dm}^{-3}$ of a mixture of iron(II) sulfate, $\text{FeSO}_4$ and iron(III) sulfate, $\text{Fe}_2(\text{SO}_4)_3$ .
FB 2 is an aqueous solution containing $2.00\text{ g dm}^{-3}$ potassium manganate(VII), $\text{KMnO}_4$ .

In the presence of acid, the iron(II) sulfate is oxidised by potassium manganate(VII).

2\text{KMnO}_4(\text{aq}) + 8\text{H}_2\text{SO}_4(\text{aq}) + 10\text{FeSO}_4(\text{aq}) \rightarrow 5\text{Fe}_2(\text{SO}_4)_3(\text{aq}) + 2\text{MnSO}_4(\text{aq}) + \text{K}_2\text{SO}_4(\text{aq}) + 8\text{H}_2\text{O}(\text{l})

(a)

Method

Fill a burette with FB 2.
Pipette $25.0\text{ cm}^3$ of FB 1 into the conical flask.
Use a $25\text{ cm}^3$ measuring cylinder to add $10\text{ cm}^3$ of dilute sulfuric acid to the flask.
Place the flask on a white tile.
Carefully titrate with FB 2 until the first permanent pink colour is obtained.

You should perform a rough titration.
In the space below record your burette readings for this rough titration.

The rough titre is ............................ $\text{cm}^3$ .

Carry out as many accurate titrations as you think necessary to obtain consistent results.
Record in a suitable form below all of your burette readings and the volume of FB 2 added in each accurate titration.
Make certain any recorded results show the precision of your practical work.

DifficultyMedium-Easy

Worked solution

Answer

Rough titre: 24.60 cm³

Accurate titrations (burette readings to the nearest 0.05 cm³):

Titration	Initial reading / cm³	Final reading / cm³	Volume of FB 2 added / cm³
Rough	0.00	24.60	24.60
1	0.00	24.35	24.35
2	24.35	48.70	24.35

Two accurate titres (24.35 and 24.35 cm³) are concordant, within 0.10 cm³ of each other.

(Representative readings — a candidate's own readings will differ.)

Final answer

See working — representative table with concordant titres 24.35 and 24.35 cm³ (candidate-dependent)

Detailed explanation

Background Concept

In a redox titration, a solution of accurately known concentration (the titrant) is added from a burette to a measured volume of the solution being analysed (the analyte) until the reaction is complete. Here the titrant is potassium manganate(VII), $\text{KMnO}_4$ (FB 2), and the analyte is iron(II) sulfate in FB 1. In acid, the manganate(VII) ion is reduced to $\text{Mn}^{2+}$ while $\text{Fe}^{2+}$ is oxidised to $\text{Fe}^{3+}$ . $\text{KMnO}_4$ is self-indicating: once all the $\text{Fe}^{2+}$ has been oxidised, the next drop of manganate(VII) is not consumed and the solution stays pink. The first permanent pink colour marks the end point, so no separate indicator is needed.

A burette measures volumes to the nearest 0.05 cm³; the reading is taken at the bottom of the meniscus at eye level. A rough titration is done quickly to locate the approximate end point, then accurate titrations are repeated. Two accurate titres that agree within 0.10 cm³ are concordant and show the results are reliable.

Understanding the Question

Part (a) asks you to carry out the titration and record all readings properly. The marks are awarded for the technique and the quality of the recording, not for a particular numerical value (which depends on your own practical work). You must record the rough titre, and then a table of accurate titrations showing initial reading, final reading and the volume of FB 2 added for each. The readings must show the precision of the burette (nearest 0.05 cm³) and you need at least two concordant titres.

Approach

Do a quick rough titration first to find the approximate end point. Then repeat accurately: record the initial reading, add FB 2 quickly at first, then dropwise near the end point, swirling the flask, until the first permanent pink colour appears. Record the final reading. Repeat until two titres agree within 0.10 cm³. Present everything in a clear table.

Step-by-Step Reasoning

Rough titration: add FB 2 from the burette in roughly 1 cm³ portions, swirling, until the pink colour persists. Record this titre (e.g. 24.60 cm³).
Accurate titration 1: note the initial reading (0.00 cm³), add FB 2 until about 1 cm³ before the rough end point, then add dropwise, swirling, until the first permanent pink colour. Record the final reading (24.35 cm³). Titre = 24.35 − 0.00 = 24.35 cm³.
Accurate titration 2: refill the burette, record initial (24.35 cm³), titrate to the end point, final 48.70 cm³, titre 24.35 cm³.
The two accurate titres agree exactly, so they are concordant (within 0.10 cm³).

The mark scheme rewards: (I) the rough titre and a tabulated set of accurate titres; (II) initial and final readings plus the volume added for each accurate titre; (III) two accurate titres within 0.10 cm³; (IV) all readings to the nearest 0.05 cm³; and quality marks (V–VII) comparing your best titre with the supervisor's value.

Key Takeaways

Titration technique: rough then accurate, dropwise near the end point, first permanent pink colour.
Recording: clear table with headings, initial/final readings and titre.
Precision: readings to the nearest 0.05 cm³; concordant titres within 0.10 cm³.

Common Mistakes

Using 50.00 cm³ as the initial burette reading — rejected by the mark scheme.
More than one final reading of 50.00 cm³ — rejected.
Any reading greater than 50.00 cm³ — rejected (the burette cannot measure this).
Recording readings to only 1 decimal place — loses the 0.05 cm³ precision mark.
Forgetting to record initial and final readings for each accurate titration.

Things to Be Careful About

Read the burette at the bottom of the meniscus, at eye level, to avoid parallax error.
The end point is the first permanent pink colour — swirl after each addition; the pink may fade if the end point is not yet reached.
Column headings in the table must match the readings recorded (e.g. "Volume of FB 2 added / cm³").
Do not start a titration with the burette at 50.00 cm³.

Techniques used

fill and read a burette to the nearest 0.05 cm³perform a rough then accurate titrationsrecord initial and final burette readings in a tabledetect the end point by first permanent pink colour

(b)

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

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

DifficultyEasy

Worked solution

Working

Using the two concordant accurate titres 24.35 and 24.35 cm³:

\text{Mean} = \frac{24.35 + 24.35}{2} = 24.35\text{ cm}^3

(Representative value — the candidate's own titres will differ.)

Answer

25.0 cm³ of FB 1 required 24.35 cm³ of FB 2.

Final answer

24.35 cm³ (representative)

Detailed explanation

Background Concept

The mean (average) of concordant titres is used in the calculation because it is more reliable than any single reading. The mark scheme requires the mean to be correct to 2 decimal places, with the third decimal place rounded to the nearest 0.05 cm³ (a mean of exactly .x25 or .x75 is allowed). The titres used must be clearly shown.

Understanding the Question

From your accurate titrations you must select suitable values and average them to obtain one value for the calculation. Show how you obtained it.

Approach

Choose two (or more) accurate titres that agree within 0.20 cm³, add them and divide by the number of titres. Round correctly.

Step-by-Step Reasoning

Using titres 24.35 and 24.35 cm³:

\text{Mean} = \frac{24.35 + 24.35}{2} = 24.35\text{ cm}^3

The answer is already to 2 decimal places. This is the value used in step (ii) of the calculations.

Key Takeaways

Use concordant titres for the mean.
Show the calculation clearly.

Common Mistakes

Averaging titres that are not concordant (more than 0.20 cm³ apart).
Rounding incorrectly (e.g. mean of 24.3 and 24.4 must be 24.35, not 24.4).
Not showing which titres were used.

Things to Be Careful About

The mean must be to 2 decimal places unless all readings were to 1 decimal place.
A mean of exactly .x25 or .x75 is acceptable.

Techniques used

select concordant titrescalculate the mean titre 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 concentration, in $\text{mol dm}^{-3}$ , of the potassium manganate(VII) in FB 2.

FB 2 contains $2.00\text{ g dm}^{-3}$ $\text{KMnO}_4$ .
[ $A_r$ : $\text{O}, 16.0$ ; $\text{K}, 39.1$ ; $\text{Mn}, 54.9$ ]

The concentration of potassium manganate(VII) in FB 2 is ............................ $\text{mol dm}^{-3}$ .

DifficultyEasy

Worked solution

Working

$M_r(\text{KMnO}_4) = 39.1 + 54.9 + 4(16.0) = 158.0$

\text{Concentration} = \frac{2.00}{158.0} = 0.0127\text{ mol dm}^{-3} \ (3\text{ s.f.})

Answer

0.0127 $\text{mol dm}^{-3}$

Final answer

0.0127 mol dm^-3

Detailed explanation

Background Concept

Concentration in mol dm⁻³ is the amount of substance (in mol) per cubic decimetre of solution. To convert a concentration given in g dm⁻³ into mol dm⁻³, divide by the molar mass ( $M_r$ ) of the solute in g mol⁻¹. The molar mass is the sum of the relative atomic masses of all atoms in the formula.

Understanding the Question

FB 2 contains 2.00 g dm⁻³ of $\text{KMnO}_4$ . We must find the concentration in mol dm⁻³. The $A_r$ values given are K 39.1, Mn 54.9, O 16.0.

Approach

Work out $M_r(\text{KMnO}_4)$ , then divide 2.00 by it. The answer should be quoted to 3 significant figures to match the data.

Step-by-Step Reasoning

$M_r(\text{KMnO}_4) = 39.1 + 54.9 + 4 \times 16.0 = 158.0\text{ g mol}^{-1}$ .

\text{Concentration} = \frac{2.00}{158.0} = 0.0127\text{ mol dm}^{-3} \ (3\text{ s.f.})

This is the value used in step (ii). The mark scheme also awards a global mark for showing working in at least four of the six calculation sections, and another for quoting final answers to 3–5 significant figures.

Key Takeaways

$c\ (\text{mol dm}^{-3}) = c\ (\text{g dm}^{-3}) / M_r$ .
Always sum all atoms in the formula when finding $M_r$ .

Common Mistakes

Forgetting to multiply oxygen by 4 (giving $M_r = 110.0$ ).
Quoting too many significant figures, e.g. 0.0126582, instead of 3 s.f.
Confusing g dm⁻³ with mol dm⁻³.

Things to Be Careful About

Use exactly the $A_r$ values given in the question.
The final answer should be to 3 significant figures (the mark scheme rewards 3–5 s.f.).

Techniques used

calculate the molar mass of KMnO₄convert concentration from g dm⁻³ to mol dm⁻³

(ii)

Calculate how many moles of $\text{KMnO}_4$ were present in the volume calculated in (b).

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

DifficultyEasy

Worked solution

Working

n(\text{KMnO}_4) = c \times V = 0.0127 \times \frac{24.35}{1000} = 3.08 \times 10^{-4}\text{ mol}

Answer

$3.08 \times 10^{-4}$ mol

Final answer

3.08 x 10^-4 mol

Detailed explanation

Background Concept

The amount of substance in moles is given by $n = cV$ , where $c$ is concentration in mol dm⁻³ and $V$ is volume in dm³. Burette titres are measured in cm³, so they must be converted to dm³ by dividing by 1000.

Understanding the Question

We know the concentration of $\text{KMnO}_4$ from (i) and the titre volume from (b). We need the moles of $\text{KMnO}_4$ in that volume.

Approach

Multiply the concentration by the titre expressed in dm³.

Step-by-Step Reasoning

n(\text{KMnO}_4) = 0.0127 \times \frac{24.35}{1000} = 3.08 \times 10^{-4}\text{ mol} \ (3\text{ s.f.})

Using the unrounded concentration (0.012658) gives the same value to 3 s.f.

Key Takeaways

$n = cV$ with $V$ in dm³.
cm³ → dm³: divide by 1000.

Common Mistakes

Forgetting to divide the titre by 1000.
Using the rounded concentration and losing precision.

Things to Be Careful About

Keep the volume in dm³.
Quote the answer to 3 significant figures.

Techniques used

calculate moles from concentration and volumeconvert cm³ to dm³

(iii)

Calculate how many moles of iron(II) sulfate, $\text{FeSO}_4$ , reacted with the potassium manganate(VII) in (ii).

2\text{KMnO}_4(\text{aq}) + 8\text{H}_2\text{SO}_4(\text{aq}) + 10\text{FeSO}_4(\text{aq}) \rightarrow 5\text{Fe}_2(\text{SO}_4)_3(\text{aq}) + 2\text{MnSO}_4(\text{aq}) + \text{K}_2\text{SO}_4(\text{aq}) + 8\text{H}_2\text{O}(\text{l})

............................ $\text{mol of FeSO}_4$ reacted with the potassium manganate(VII).

DifficultyEasy

Worked solution

Working

From the equation, 2 mol $\text{KMnO}_4$ react with 10 mol $\text{FeSO}_4$ , so:

n(\text{FeSO}_4) = 5 \times n(\text{KMnO}_4) = 5 \times 3.08 \times 10^{-4} = 1.54 \times 10^{-3}\text{ mol}

Answer

$1.54 \times 10^{-3}$ mol

Final answer

1.54 x 10^-3 mol

Detailed explanation

Background Concept

A balanced equation gives the mole ratio in which reactants react. Here 2 mol $\text{KMnO}_4$ react with 10 mol $\text{FeSO}_4$ , i.e. a 1:5 ratio. So the moles of $\text{FeSO}_4$ are five times the moles of $\text{KMnO}_4$ .

Understanding the Question

We know the moles of $\text{KMnO}_4$ from (ii) and need the moles of $\text{FeSO}_4$ that reacted with it.

Approach

Multiply the moles of $\text{KMnO}_4$ by the ratio 10/2 = 5.

Step-by-Step Reasoning

n(\text{FeSO}_4) = 5 \times 3.08 \times 10^{-4} = 1.54 \times 10^{-3}\text{ mol} \ (3\text{ s.f.})

Key Takeaways

Read the stoichiometric ratio from the balanced equation.
2:10 simplifies to 1:5.

Common Mistakes

Using the wrong ratio (e.g. 2/10 instead of 10/2).
Using the equation incorrectly when $\text{Fe}_2(\text{SO}_4)_3$ is also a product.

Things to Be Careful About

The ratio is moles of $\text{FeSO}_4$ : moles of $\text{KMnO}_4$ = 10:2 = 5:1.

Techniques used

use the mole ratio from the balanced equation

(iv)

Calculate the concentration, in $\text{mol dm}^{-3}$ of $\text{FeSO}_4$ in FB 1.

The concentration of $\text{FeSO}_4$ in FB 1 is ............................ $\text{mol dm}^{-3}$ .

DifficultyEasy

Worked solution

Working

The 25.0 cm³ pipette sample contains the $\text{FeSO}_4$ from (iii):

c(\text{FeSO}_4) = \frac{n}{V} = 1.54 \times 10^{-3} \times \frac{1000}{25.0} = 0.0616\text{ mol dm}^{-3}

Answer

0.0616 $\text{mol dm}^{-3}$

Final answer

0.0616 mol dm^-3

Detailed explanation

Background Concept

Concentration is moles per unit volume. The 25.0 cm³ of FB 1 pipetted into the flask contained $1.54 \times 10^{-3}$ mol of $\text{FeSO}_4$ . To find the concentration in mol dm⁻³, divide by the volume in dm³ (0.0250 dm³).

Understanding the Question

We know the moles of $\text{FeSO}_4$ in the 25.0 cm³ sample and need the concentration of $\text{FeSO}_4$ in FB 1.

Approach

$c = n/V$ with $V = 25.0$ cm³ = 0.0250 dm³. Equivalently, multiply $n$ by 1000/25.0.

Step-by-Step Reasoning

c(\text{FeSO}_4) = \frac{1.54 \times 10^{-3}}{0.0250} = 0.0616\text{ mol dm}^{-3} \ (3\text{ s.f.})

Key Takeaways

$c = n/V$ with $V$ in dm³.
Multiplying by 1000/25.0 converts from the 25.0 cm³ sample to per dm³.

Common Mistakes

Forgetting that the sample volume is 25.0 cm³, not 1 dm³.
Using the titre volume instead of 25.0 cm³.

Things to Be Careful About

The volume in the denominator is the pipette volume (25.0 cm³), not the titre.

Techniques used

calculate concentration from moles and volume

(v)

Calculate the concentration, in $\text{g dm}^{-3}$ , of $\text{FeSO}_4$ in FB 1.
[ $A_r$ : $\text{O}, 16.0$ ; $\text{S}, 32.1$ ; $\text{Fe}, 55.8$ ]

FB 1 contains ............................ $\text{g dm}^{-3}$ of $\text{FeSO}_4$ .

DifficultyEasy

Worked solution

Working

$M_r(\text{FeSO}_4) = 55.8 + 32.1 + 4(16.0) = 151.9$

\text{Concentration} = 0.0616 \times 151.9 = 9.36\text{ g dm}^{-3}

Answer

9.36 $\text{g dm}^{-3}$

Final answer

9.36 g dm^-3

Detailed explanation

Background Concept

To convert a concentration from mol dm⁻³ to g dm⁻³, multiply by the molar mass in g mol⁻¹.

Understanding the Question

We know the concentration of $\text{FeSO}_4$ in mol dm⁻³ from (iv) and need it in g dm⁻³.

Approach

Calculate $M_r(\text{FeSO}_4) = 55.8 + 32.1 + 4 \times 16.0 = 151.9$ , then multiply 0.0616 by 151.9.

Step-by-Step Reasoning

\text{Concentration} = 0.0616 \times 151.9 = 9.36\text{ g dm}^{-3} \ (3\text{ s.f.})

Key Takeaways

g dm⁻³ = mol dm⁻³ × $M_r$ .

Common Mistakes

Forgetting to multiply O by 4 in $\text{FeSO}_4$ .
Using the $A_r$ of $\text{Fe}_2(\text{SO}_4)_3$ instead of $\text{FeSO}_4$ .

Things to Be Careful About

Use the $A_r$ values given (Fe 55.8, S 32.1, O 16.0).

Techniques used

calculate the molar mass of FeSO₄convert concentration from mol dm⁻³ to g dm⁻³

(vi)

FB 1 is an aqueous solution containing $21.50\text{ g dm}^{-3}$ of $\text{FeSO}_4$ and $\text{Fe}_2(\text{SO}_4)_3$ .
Calculate the percentage, by mass, of $\text{FeSO}_4$ in this mixture.

The mixture contains ............................ % $\text{FeSO}_4$ .

DifficultyMedium-Easy

Worked solution

Working

\text{Percentage of FeSO}_4 = \frac{9.36}{21.50} \times 100 = 43.6\%

Answer

43.6%

Final answer

43.6%

Detailed explanation

Background Concept

Percentage by mass is (mass of component / total mass) × 100. Since both values are per dm³ of the same solution, the ratio of concentrations in g dm⁻³ equals the ratio of masses.

Understanding the Question

FB 1 contains 21.50 g dm⁻³ of the mixture of $\text{FeSO}_4$ and $\text{Fe}_2(\text{SO}_4)_3$ . We calculated 9.36 g dm⁻³ of $\text{FeSO}_4$ in (v). Find the percentage of $\text{FeSO}_4$ by mass.

Approach

Divide the mass of $\text{FeSO}_4$ per dm³ by the total mass per dm³ and multiply by 100.

Step-by-Step Reasoning

\text{Percentage of FeSO}_4 = \frac{9.36}{21.50} \times 100 = 43.6\% \ (3\text{ s.f.})

Key Takeaways

Percentage by mass = (component mass / total mass) × 100.
Concentrations in g dm⁻³ can be used directly as mass ratios.

Common Mistakes

Using mol dm⁻³ values instead of g dm⁻³ in the ratio.
Forgetting to multiply by 100.

Things to Be Careful About

Use the g dm⁻³ value from (v), not the mol dm⁻³ value.
The total is 21.50 g dm⁻³, the given value.

Techniques used

calculate percentage by massrelate mass concentration to percentage composition

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