Chemistry 9701/32 — May/June 2022

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

questions

marks

120

minutes

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

Q1Manipulation, Measurement and ObservationPresentation of Data and ObservationsFree sample

A bottle containing the acid salt sodium hydrogen sulfate, $\text{NaHSO}_4$ , has been contaminated. You will determine the percentage purity by mass of the sodium hydrogen sulfate by titrating a solution of the acid salt against a known concentration of sodium hydroxide.

\text{NaHSO}_4(\text{aq}) + \text{NaOH}(\text{aq}) \rightarrow \text{Na}_2\text{SO}_4(\text{aq}) + \text{H}_2\text{O}(\text{l})

The impurity in the sodium hydrogen sulfate does not react with aqueous sodium hydroxide under the conditions of the titration.

FB 1 is $0.100 \text{ mol dm}^{-3}$ sodium hydroxide, $\text{NaOH}$ .
FB 2 is $12.53 \text{ g dm}^{-3}$ impure sodium hydrogen sulfate.
FB 3 is thymol blue indicator.

(a)

Method

Fill a burette with FB 1.
Pipette $25.0 \text{ cm}^3$ of FB 2 into a conical flask.
Add approximately 10 drops of FB 3.
Perform a rough titration and record your burette readings in the space below. The end-point is shown by the appearance of a permanent blue colour.

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

Carry out as many titrations as you think necessary to obtain consistent results.
Make certain any recorded results show the precision of your practical work.
Record, in a suitable form below, all your burette readings and the volume of FB 1 added in each accurate titration.

DifficultyMedium-Easy

Worked solution

Answer

Rough titre: 24.3 cm³.

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

	Titration 1	Titration 2
initial burette reading / cm³	0.00	24.20
final burette reading / cm³	24.20	48.40
titre / cm³	24.20	24.20

The two accurate titres agree within 0.10 cm³, so they are concordant and no further titrations are needed.

Final answer

Representative example: rough titre 24.3 cm³; accurate titres 24.20 and 24.20 cm³ (readings to 0.05 cm³, concordant within 0.10 cm³). Candidate-dependent.

Detailed explanation

Background Concept

A titration is a quantitative technique used to determine the unknown concentration of a solution by reacting it with a solution of known concentration. Here, the acid salt sodium hydrogen sulfate, NaHSO $_4$ , is titrated against sodium hydroxide, NaOH, of known concentration 0.100 mol dm $^{-3}$ . The hydrogen sulfate ion, HSO $_4^-$ , behaves as a monobasic acid: it contains a single acidic hydrogen atom that is donated to the hydroxide ion. The reaction is therefore 1:1:

\text{NaHSO}_4(\text{aq}) + \text{NaOH}(\text{aq}) \rightarrow \text{Na}_2\text{SO}_4(\text{aq}) + \text{H}_2\text{O}(\text{l})

Thymol blue is the indicator. It is yellow in acidic solution and turns blue when the solution becomes alkaline. The end-point is the appearance of a permanent blue colour, which signals that exactly the stoichiometric amount of NaOH has been added.

In Paper 3, the marks in part (a) are not about obtaining 'the right answer' — they reward technique and precision. The examiner checks that all data are recorded, that headings and units are correct, that burette readings are to the nearest 0.05 cm³, and that accurate titres are concordant (within 0.10 cm³). The supervisor performs the same titration, and accuracy marks are awarded according to how close the candidate's mean titre is to the supervisor's mean titre.

Understanding the Question

You must actually carry out the titration and record your results. Part (a) asks for two things: the rough titre, and a table of all your accurate burette readings and titres. The marking scheme awards marks for: (I) recording the rough data and two or more accurate titrations; (II) correct headings and units in the table; (III) readings to the nearest 0.05 cm³; (IV) accurate titres within 0.10 cm³ of each other; and (V–VII) accuracy marks based on closeness to the supervisor's mean titre.

Approach

First perform a rough titration to find the approximate end-point. Then repeat the titration carefully, reading the burette to the nearest 0.05 cm³, until you obtain two accurate titres that agree within 0.10 cm³. Record the initial and final burette readings for each accurate titration and compute the titre as final reading minus initial reading.

Step-by-Step Reasoning

Rough titration: Add NaOH from the burette fairly quickly while swirling the flask, until the blue colour appears. Record the rough titre, e.g. 24.3 cm³. This tells you roughly where the end-point is.
Accurate titrations: Refill the burette to 0.00 cm³. Add NaOH rapidly at first, then dropwise as you approach the end-point, swirling continuously. Stop at the first permanent blue colour. Record the initial and final readings.
Precision: Read the burette to the nearest 0.05 cm³ — that is, half of the smallest scale division (0.1 cm³).
Concordance: Two accurate titres within 0.10 cm³ are concordant. If your first two are not within 0.10 cm³, perform a third (and further) titration until two agree.
Recording: Use a table with proper headings that include both the quantity and its unit, e.g. 'initial burette reading / cm³', 'final burette reading / cm³', 'titre / cm³'.

Key Takeaways

Burette readings are recorded to the nearest 0.05 cm³.
Titre = final reading − initial reading.
Concordant titres agree within 0.10 cm³.
Table headings must state the quantity and its unit.
The rough titre is for orientation only and is not used in the mean.

Common Mistakes

Recording burette readings to 0.1 cm³ instead of 0.05 cm³ — the marking scheme requires 0.05 cm³.
Omitting units from table headings.
Including the rough titre in the mean titre calculation.
Stopping after one accurate titration even though two concordant titres are required.

Things to Be Careful About

Read the bottom of the meniscus at eye level to avoid parallax error.
Swirl the flask continuously so the alkali is well mixed.
Add NaOH dropwise near the end-point so you do not overshoot.
The end-point is the first permanent blue colour — a colour that fades on swirling is not the end-point.

Techniques used

fill and read a burette to the nearest 0.05 cm³perform a rough titration to locate the end-pointrepeat titrations until concordant within 0.10 cm³record readings in a table with headings and units

(b)

From your accurate titration results, calculate a suitable mean value to use in your calculations. Show clearly how you obtain the mean value.

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

DifficultyEasy

Worked solution

Working

Mean titre $= \frac{24.20 + 24.20}{2} = 24.20 \text{ cm}^3$

Answer

24.20 cm³

Final answer

24.20 cm³ (representative example; candidate-dependent)

Detailed explanation

Background Concept

The mean titre is the average of the concordant accurate titres. It is the single value used in all subsequent calculations, so it must be reliable — hence only the concordant accurate titres are averaged, and the rough titre is excluded. The mean is expressed to 2 decimal places.

Understanding the Question

From the accurate titration results in part (a), calculate a suitable mean titre to use in the calculations. The mark is awarded for correctly averaging accurate titres that are within 0.20 cm³ total spread, expressed to 2 dp.

Approach

Sum the concordant accurate titres and divide by the number of titres used. Express the result to 2 decimal places.

Step-by-Step Reasoning

With the two concordant titres 24.20 cm³ and 24.20 cm³:

\text{mean titre} = \frac{24.20 + 24.20}{2} = 24.20 \text{ cm}^3

If the titres were 24.20 cm³ and 24.25 cm³, the mean would be 24.225 cm³, rounded to 24.23 cm³ (2 dp).

Key Takeaways

Mean titre = average of concordant accurate titres, to 2 dp.
Exclude the rough titre.

Common Mistakes

Including the rough titre in the mean.
Not rounding the mean to 2 dp.
Averaging non-concordant titres.

Things to Be Careful About

The marking scheme requires the mean to be calculated from accurate titres within 0.20 cm³ total spread.
Keep the mean to 2 dp for use in the calculations.

Techniques used

average concordant titresexpress the mean to 2 decimal places

(c)

Calculations

(i)

Give all your answers to (c)(ii), (c)(iii) and (c)(iv) to an appropriate number of significant figures.

DifficultyEasy

Worked solution

Answer

All final answers in (c)(ii), (c)(iii) and (c)(iv) are reported to 3 significant figures:

(c)(ii) $2.42 \times 10^{-3}$ mol
(c)(iii) 11.6 g
(c)(iv) 92.8%

Final answer

3 significant figures

Detailed explanation

Background Concept

Significant figures indicate the precision of a measured or calculated value. The titre, 24.20 cm³, has four significant figures, but the concentration of NaOH, 0.100 mol dm $^{-3}$ , has three. The marking scheme requires the final answers in (c)(ii), (c)(iii) and (c)(iv) to be given to 3–4 significant figures.

Understanding the Question

This part instructs you to report your final answers to an appropriate number of significant figures. The mark is awarded only if all three final answers are to 3–4 sf.

Approach

Carry each calculation to full precision using unrounded intermediate values, then round the final answer to 3 significant figures — consistent with the 3 sf of the concentration given.

Step-by-Step Reasoning

(c)(ii): $n = 0.100 \times 24.20/1000 = 2.420 \times 10^{-3}$ mol → 2.42 × 10⁻³ mol (3 sf).
(c)(iii): mass = 2.420 × 10⁻³ × 120.1 × 40 = 11.6257 g → 11.6 g (3 sf).
(c)(iv): purity = (11.6257/12.53) × 100 = 92.78% → 92.8% (3 sf).

Key Takeaways

Report final answers to 3–4 sf.
Do not round intermediate values; round only the final answer.

Common Mistakes

Rounding intermediate values, which introduces error into the final answer.
Giving answers to 2 sf or to too many figures.

Things to Be Careful About

The mark scheme explicitly requires all three final answers to be to 3–4 sf.
Use the unrounded values in each step.

Techniques used

report final answers to 3 significant figures

(ii)

Use your answer to (b) to calculate the amount, in mol, of sodium hydroxide, FB 1, titrated.

\text{amount of NaOH} = \text{.............................. mol}

Hence, deduce the amount, in mol, of sodium hydrogen sulfate present in $25.0 \text{ cm}^3$ of FB 2.

\text{amount of NaHSO}_4 = \text{.............................. mol}

DifficultyMedium-Easy

Worked solution

Working

$n(\text{NaOH}) = c \times V = 0.100 \times \frac{24.20}{1000} = 2.42 \times 10^{-3} \text{ mol}$

From the equation, $\text{NaHSO}_4 : \text{NaOH} = 1 : 1$ , so:

$n(\text{NaHSO}_4) = 2.42 \times 10^{-3} \text{ mol}$

Answer

amount of NaOH = $2.42 \times 10^{-3}$ mol; amount of NaHSO $_4$ = $2.42 \times 10^{-3}$ mol

Final answer

2.42 × 10⁻³ 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 dm $^3$ . The titre is measured in cm³, so it must be divided by 1000 to convert to dm³. The balanced equation shows that one mole of NaHSO $_4$ reacts with one mole of NaOH, so the amount of NaHSO $_4$ in the 25.0 cm³ sample equals the amount of NaOH titrated.

Understanding the Question

Use the mean titre from part (b) to calculate the amount of NaOH titrated, then deduce the amount of NaHSO $_4$ present in the 25.0 cm³ sample of FB 2.

Approach

Convert the titre volume to dm³, multiply by the concentration to find moles of NaOH, then use the 1:1 stoichiometric ratio to find moles of NaHSO $_4$ .

Step-by-Step Reasoning

With mean titre 24.20 cm³:

n(\text{NaOH}) = 0.100 \times \frac{24.20}{1000} = 2.42 \times 10^{-3} \text{ mol}

From the equation, $\text{NaHSO}_4 : \text{NaOH} = 1 : 1$ , so:

n(\text{NaHSO}_4) = 2.42 \times 10^{-3} \text{ mol}

This is the amount of NaHSO $_4$ in the 25.0 cm³ pipetted sample.

Key Takeaways

$n = cV$ with volume in dm³.
Use the balanced equation to establish the stoichiometric ratio.

Common Mistakes

Forgetting to divide the titre volume by 1000.
Using an incorrect stoichiometric ratio.

Things to Be Careful About

Volume must be in dm³: divide cm³ by 1000.
The 1:1 ratio comes directly from the balanced equation.

Techniques used

convert titre volume from cm³ to dm³calculate moles from concentration and volumeapply the 1:1 stoichiometric ratio

(iii)

Use your final answer to (c)(ii) to calculate the mass of sodium hydrogen sulfate present in $1.00 \text{ dm}^3$ of FB 2.

\text{mass of NaHSO}_4 = \text{.............................. g}

DifficultyMedium-Easy

Worked solution

Working

$M_r(\text{NaHSO}_4) = 23.0 + 1.0 + 32.1 + 4(16.0) = 120.1 \text{ g mol}^{-1}$

Mass of NaHSO $_4$ in 25.0 cm³ = $n \times M_r = 2.42 \times 10^{-3} \times 120.1 = 0.291 \text{ g}$

Mass in 1.00 dm³ = $0.291 \times \frac{1000}{25.0} = 0.291 \times 40 = 11.6 \text{ g}$

Answer

mass of NaHSO $_4$ = 11.6 g per dm³

Final answer

11.6 g dm⁻³

Detailed explanation

Background Concept

The molar mass of NaHSO $_4$ is:

M_r = 23.0 + 1.0 + 32.1 + 4(16.0) = 120.1 \text{ g mol}^{-1}

Mass is related to amount by $m = nM_r$ . The amount found in (c)(ii) is in 25.0 cm³ of FB 2; to find the mass in 1.00 dm³, multiply by the scaling factor 1000/25.0 = 40.

Understanding the Question

Use the amount of NaHSO $_4$ from (c)(ii) to calculate the mass of NaHSO $_4$ present in 1.00 dm³ of FB 2.

Approach

First find the mass in the 25.0 cm³ sample using $m = nM_r$ , then scale up to 1 dm³ by multiplying by 40.

Step-by-Step Reasoning

Mass in 25.0 cm³ = $2.42 \times 10^{-3} \times 120.1 = 0.291$ g.

Mass in 1.00 dm³ = $0.291 \times \frac{1000}{25.0} = 0.291 \times 40 = 11.6$ g.

(Using the unrounded amount gives 11.63 g, reported as 11.6 g to 3 sf.)

Key Takeaways

$m = nM_r$ .
Scaling from 25.0 cm³ to 1 dm³ uses the factor 40.

Common Mistakes

Using the molar mass of Na $_2$ SO $_4$ (142.1) instead of NaHSO $_4$ (120.1).
Forgetting the scaling factor from 25 cm³ to 1 dm³.

Things to Be Careful About

$M_r(\text{NaHSO}_4) = 120.1$ g mol⁻¹.
The mark scheme route is: concentration = (c)(ii) × 40, then mass = concentration × 120.1.

Techniques used

calculate mass from moles and molar massscale the mass from 25.0 cm³ to 1.00 dm³

(iv)

Use your answer to (c)(iii) and the information on page 2 to calculate the percentage purity by mass of the sodium hydrogen sulfate.

\text{percentage purity} = \text{.............................. \%}

DifficultyEasy

Worked solution

Working

$\text{percentage purity} = \frac{\text{mass of NaHSO}_4 \text{ in 1 dm}^3}{\text{mass of impure sample in 1 dm}^3} \times 100$

$= \frac{11.6257}{12.53} \times 100 = 92.78\% = 92.8\%$

Answer

percentage purity = 92.8%

Final answer

92.8%

Detailed explanation

Background Concept

Percentage purity by mass is defined as:

\% \text{ purity} = \frac{\text{mass of pure substance}}{\text{mass of impure sample}} \times 100

FB 2 contains 12.53 g of impure NaHSO $_4$ per dm³. The impurity does not react with NaOH, so the titre (and hence the mass found in (c)(iii)) reflects only the pure NaHSO $_4$ . Comparing the pure mass to the total impure mass gives the percentage purity.

Understanding the Question

Use the mass of NaHSO $_4$ in 1 dm³ from (c)(iii) and the stated 12.53 g dm⁻³ of impure solid to calculate the percentage purity by mass.

Approach

Divide the mass of pure NaHSO $_4$ by the total mass of impure solid per dm³ and multiply by 100.

Step-by-Step Reasoning

Using the unrounded mass 11.6257 g:

\% \text{ purity} = \frac{11.6257}{12.53} \times 100 = 92.78\% \approx 92.8\%

This means about 92.8% of the solid is NaHSO $_4$ ; the remaining ~7.2% is non-reactive impurity.

Key Takeaways

% purity = (pure mass / impure mass) × 100.
Because the impurity does not react, the titration measures only the pure NaHSO $_4$ .

Common Mistakes

Dividing by the wrong mass (e.g. by 100 or by the mass in 25 cm³).
Forgetting to multiply by 100.

Things to Be Careful About

Use the unrounded mass from (c)(iii) in the division to avoid rounding error; round only the final answer.
The mark scheme route is % = (final answer (c)(iii) / 12.53) × 100.

Techniques used

calculate percentage purity by mass

The rest of this paper

2 more questions

Q2Manipulation, Measurement and Observation · Presentation of Data and Observations · Analysis, Conclusions and Evaluation16M
Q3Qualitative Analysis · Manipulation, Measurement and Observation · Analysis, Conclusions and Evaluation12M

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