Chemistry 9701/34 — October/November 2024

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

Q1MediumManipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and Evaluation

Iodine reacts with propanone in the presence of an acid catalyst. The rate of this reaction can be measured by determining how the concentration of iodine in the reaction mixture changes with time.

\text{CH}_3\text{COCH}_3(\text{aq}) + \text{I}_2(\text{aq}) \rightarrow \text{CH}_3\text{COCH}_2\text{I}(\text{aq}) + \text{HI}(\text{aq})

A student studies this reaction by mixing together the following three solutions and immediately starting the stop-clock.

$100.0\text{ cm}^3$ of $0.1000\text{ mol dm}^{-3}$ iodine, $\text{I}_2$
$50.0\text{ cm}^3$ of $1.00\text{ mol dm}^{-3}$ propanone, $\text{CH}_3\text{COCH}_3$
$50.0\text{ cm}^3$ of $1.00\text{ mol dm}^{-3}$ sulfuric acid, $\text{H}_2\text{SO}_4$

The student removes $25.0\text{ cm}^3$ of the solution. After 80 seconds, $50.0\text{ cm}^3$ of a solution of sodium hydrogencarbonate, $\text{NaHCO}_3$ , is added which reacts with all the sulfuric acid in the sample. Distilled water is added until the volume of the solution is $150.0\text{ cm}^3$ . This solution is FB 1.

In this experiment you will determine the concentration of iodine in FB 1 and so determine the average rate of reaction during the first 80 seconds. You will do this by titration using sodium thiosulfate solution.

\text{I}_2(\text{aq}) + 2\text{S}_2\text{O}_3^{2-}(\text{aq}) \rightarrow 2\text{I}^-(\text{aq}) + \text{S}_4\text{O}_6^{2-}(\text{aq})

FB 1 is a sample of the solution prepared by the student.
FB 2 is $0.0100\text{ mol dm}^{-3}$ sodium thiosulfate, $\text{Na}_2\text{S}_2\text{O}_3$ .
FB 3 is starch indicator.

(a)

Method

Fill the burette with FB 2.
Pipette $25.0\text{ cm}^3$ of FB 1 into a conical flask.
Run FB 2 into the conical flask until the colour of the solution turns yellow. Then add 10 drops of FB 3. The solution will turn blue-black.
The end-point of the titration is when the solution turns colourless.
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 all your burette readings and the volume of FB 2 added in each accurate titration.

Results

7M Similar

DifficultyMedium-Easy

Worked solution

Answer

Record the rough titre and at least two accurate titrations. Use a table with headings and units:

Titration	final burette reading / cm3	titre / cm3
rough	26.40	26.40
accurate 1	26.20	26.20
accurate 2	26.30	26.30

All burette readings are recorded to the nearest 0.05 cm3. The accurate titres agree within 0.10 cm3.

Final answer

See working: example table with rough and accurate titres, precise to 0.05 cm3 and concordant within 0.10 cm3.

Detailed explanation

Background Concept

In an iodine–thiosulfate titration, iodine is reduced to iodide by thiosulfate while thiosulfate is oxidised to tetrathionate. Starch is used as an indicator because it forms an intense blue-black complex with iodine. The titration must be performed carefully so that the end-point is sharp and the volume of thiosulfate needed is precise.

Understanding the Question

This part asks you to carry out the titration and record your results in a way that earns full credit: a rough titre, at least two accurate titres, correct table headings with units, readings to the nearest 0.05 cm3, and concordant results.

Approach

First do a rough titration to find the approximate end-point. Then repeat accurately. Record initial and final burette readings for each accurate titration and calculate the titre. Keep all readings to 0.05 cm3 and continue until two titres agree within 0.10 cm3.

Step-by-Step Reasoning

Fill the burette with FB2 and remove the air jet.
Pipette 25.0 cm3 of FB1 into a conical flask.
Add thiosulfate until the colour turns yellow, then add starch. The blue-black colour appears.
Continue adding thiosulfate dropwise until the solution becomes colourless. This is the end-point.
Record the rough titre, then repeat. For each accurate titration, record initial and final readings and the titre.
Use a table with headings such as 'initial burette reading / cm3', 'final burette reading / cm3' and 'titre / cm3'.
All burette readings must be to the nearest 0.05 cm3. Accurate titres should agree within 0.10 cm3.

Key Takeaways

Good titration technique and clear recording are essential. The marks are for precision, concordance, and correct presentation of data.

Common Mistakes

Recording burette readings to only 0.1 cm3 instead of 0.05 cm3.
Omitting units from table headings.
Using only one accurate titre or titres that are not concordant.
Forgetting to record the rough titre.

Things to Be Careful About

Read the burette at eye level to avoid parallax error.
The end-point is the first permanent colourless solution.
Make sure the table headings include the unit, e.g. / cm3.

Techniques used

perform a rough titrationrecord burette readings to 0.05 cm3carry out concordant accurate titrationstabulate initial and final readings with units

(b)

From your accurate titration results, calculate a suitable mean value to be used in your calculations.
Show clearly how you obtained this value.

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

1M Similar

(c)

Calculations

(i)

Give your answers to each part of (c)(ii) and (c)(iii) to an appropriate number of significant figures.

1M Similar

(ii)

Use your answer to (b) to calculate the amount, in mol, of thiosulfate ions in your mean titre.

amount of $\text{S}_2\text{O}_3^{2-}$ = .............................. mol

Hence calculate the amount, in mol, of iodine present in the total volume of $150.0\text{ cm}^3$ that the student prepares.

amount of $\text{I}_2$ = .............................. mol

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Amount of thiosulfate:

n(\text{S}_2\text{O}_3^{2-}) = \frac{26.25}{1000} \times 0.0100 = 2.625 \times 10^{-4}\,\text{mol}

From the equation $\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}$ , 1 mol I2 reacts with 2 mol thiosulfate.

Amount of I2 in 25.0 cm3 of FB1:

n(\text{I}_2) = \frac{2.625 \times 10^{-4}}{2} = 1.3125 \times 10^{-4}\,\text{mol}

Amount of I2 in total 150.0 cm3 FB1:

n(\text{I}_2) = 1.3125 \times 10^{-4} \times \frac{150}{25} = 7.875 \times 10^{-4}\,\text{mol}

Answer

amount of $\text{S}_2\text{O}_3^{2-}$ = $2.63 \times 10^{-4}$ mol

amount of $\text{I}_2$ = $7.88 \times 10^{-4}$ mol

Final answer

amount of S2O3^2- = 2.63 × 10^-4 mol; amount of I2 = 7.88 × 10^-4 mol

Detailed explanation

Background Concept

The titration reaction is $\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}$ . One mole of iodine reacts with two moles of thiosulfate. The amount of thiosulfate used in the titre gives the amount of iodine in the aliquot of FB1 titrated. Because only 25.0 cm3 of the 150.0 cm3 FB1 was titrated, the amount in the aliquot must be scaled up by 150/25 to find the total amount in FB1.

Understanding the Question

Use the mean titre from (b) to calculate moles of thiosulfate, then moles of iodine in the 25.0 cm3 aliquot, then moles of iodine in the whole 150.0 cm3 FB1.

Approach

Convert the titre volume to dm3, multiply by concentration to get moles of thiosulfate. Halve for moles of iodine in the aliquot. Multiply by 150/25 for the total volume.

Step-by-Step Reasoning

Volume of thiosulfate = 26.25 cm3 = 0.02625 dm3.
$n(\text{S}_2\text{O}_3^{2-}) = 0.02625 \times 0.0100 = 2.625 \times 10^{-4}$ mol.
From the 2:1 ratio, $n(\text{I}_2)$ in 25.0 cm3 FB1 = $2.625 \times 10^{-4} / 2 = 1.3125 \times 10^{-4}$ mol.
Total volume is 150.0 cm3, so $n(\text{I}_2)$ in FB1 = $1.3125 \times 10^{-4} \times 150/25 = 7.875 \times 10^{-4}$ mol.
Round to 3 significant figures: $2.63 \times 10^{-4}$ mol and $7.88 \times 10^{-4}$ mol.

Key Takeaways

Always use the stoichiometric ratio and scale from the aliquot to the total volume.

Common Mistakes

Forgetting the 2:1 ratio and using 1:1.
Forgetting to scale up by 150/25.
Using the rough titre in the calculation.

Things to Be Careful About

Use the mean titre from (b), not the rough titre.
Keep intermediate values unrounded until the final answer.

Techniques used

calculate moles from concentration and volumeapply the 1:2 stoichiometric ratioscale from the aliquot to the total volume

(iii)

Calculate the concentration, in $\text{mol dm}^{-3}$ , of iodine in the sample that the student removes from the reaction mixture.

concentration of $\text{I}_2$ = .............................. $\text{mol dm}^{-3}$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

The amount of I2 in the 150.0 cm3 FB1 equals the amount in the 25.0 cm3 sample removed from the reaction mixture.

[\text{I}_2] = \frac{7.875 \times 10^{-4}}{0.0250} = 0.0315\,\text{mol dm}^{-3}

Answer

concentration of $\text{I}_2$ = $0.0315\,\text{mol dm}^{-3}$

Final answer

0.0315 mol dm^-3

Detailed explanation

Background Concept

The 25.0 cm3 sample removed from the reaction mixture was diluted to 150.0 cm3 to make FB1, but dilution does not change the amount of iodine. Therefore the amount of iodine in 150.0 cm3 FB1 equals the amount of iodine in the original 25.0 cm3 sample. To find the concentration in the original sample, divide by the original sample volume, 0.0250 dm3.

Understanding the Question

Use the amount of I2 in the total 150.0 cm3 FB1 from (c)(ii) to calculate the concentration of iodine in the sample removed from the reaction mixture.

Approach

Divide the amount of I2 by the volume of the original sample in dm3.

Step-by-Step Reasoning

Amount of I2 in 150.0 cm3 FB1 = $7.875 \times 10^{-4}$ mol.
This amount was originally in 25.0 cm3 = 0.0250 dm3 of reaction mixture.
$[\text{I}_2] = 7.875 \times 10^{-4} / 0.0250 = 0.0315$ mol dm-3.

Key Takeaways

Dilution changes concentration but not amount. Use the volume of the original sample, not the diluted volume, when finding the concentration in the reaction mixture.

Common Mistakes

Dividing by 0.150 dm3 instead of 0.0250 dm3, which gives the concentration in FB1, not in the reaction mixture.
Confusing amount and concentration.

Things to Be Careful About

The volume must be in dm3.
Quote the answer to 3 significant figures.

Techniques used

divide amount by the original sample volumeconvert cm3 to dm3quote concentration to 3 significant figures

(iv)

Calculate the initial concentration of iodine in the reaction mixture.

(If you were unable to determine an answer to (c)(iii) use $0.0315\text{ mol dm}^{-3}$ as the concentration of $\text{I}_2$ in the sample the student removed.)

initial concentration of $\text{I}_2$ = .............................. $\text{mol dm}^{-3}$

Hence calculate the average rate of reaction during the first 80 seconds using the formula shown.

\text{average rate of reaction} = \frac{\text{change in the concentration of } \text{I}_2}{\text{time}}

average rate = ........................ .........................
(value) (units)

2M Similar

DifficultyMedium

Worked solution

Working

Initial concentration of I2:

[\text{I}_2]_0 = 0.1000 \times \frac{100.0}{200.0} = 0.0500\,\text{mol dm}^{-3}

Average rate:

\text{rate} = \frac{0.0500 - 0.0315}{80} = 2.31 \times 10^{-4}\,\text{mol dm}^{-3}\,\text{s}^{-1}

Answer

initial concentration of $\text{I}_2$ = $0.0500\,\text{mol dm}^{-3}$

average rate = $2.31 \times 10^{-4}\,\text{mol dm}^{-3}\,\text{s}^{-1}$

Final answer

initial concentration = 0.0500 mol dm^-3; average rate = 2.31 × 10^-4 mol dm^-3 s^-1

Detailed explanation

Background Concept

The initial concentration of iodine is found by dilution: 100.0 cm3 of 0.1000 mol dm-3 iodine is mixed with other solutions to a total volume of 200.0 cm3. The average rate is the change in iodine concentration divided by the time interval.

Understanding the Question

Calculate the initial iodine concentration, then use the concentration at 80 s from (c)(iii) to find the average rate over the first 80 seconds.

Approach

Use dilution: initial concentration = $0.1000 \times 100.0/200.0 = 0.0500$ mol dm-3. Then rate = $(0.0500 - 0.0315)/80$ .

Step-by-Step Reasoning

Total volume after mixing = 100.0 + 50.0 + 50.0 = 200.0 cm3.
$[\text{I}_2]_0 = 0.1000 \times 100.0/200.0 = 0.0500$ mol dm-3.
$[\text{I}_2]$ at 80 s = 0.0315 mol dm-3.
Change in concentration = 0.0500 - 0.0315 = 0.0185 mol dm-3.
Average rate = 0.0185 / 80 = $2.3125 \times 10^{-4}$ mol dm-3 s-1 ≈ $2.31 \times 10^{-4}$ mol dm-3 s-1.

Key Takeaways

The rate is a positive quantity: initial minus final concentration divided by time. Units are mol dm-3 s-1.

Common Mistakes

Using the total volume as 150 cm3 instead of 200 cm3.
Forgetting to subtract the final concentration from the initial.
Omitting the units or using mol dm-3 instead of mol dm-3 s-1.

Things to Be Careful About

The initial concentration is 0.0500 mol dm-3, not 0.1000 mol dm-3.
Include the time in seconds and the correct units.

Techniques used

calculate initial concentration by dilutionsubtract final concentration from initial concentrationdivide by time to calculate average ratestate units of rate

(d)

Suggest why the starch solution is added when the solution in the conical flask turns yellow and not added at the start of the titration.

1M Similar

(e)

A student suggests that the experimental procedure is incorrect. The student says that the sample should be removed from the reaction mixture at 80 seconds rather than the sodium hydrogencarbonate being added at 80 seconds.
State if you agree with this student. Explain your answer.

1M Similar

Q2MediumManipulation, Measurement and ObservationPresentation of Data and ObservationsQualitative AnalysisAnalysis, Conclusions and Evaluation

In this experiment you will identify the ions in the hydrated salt $\text{MA}_2\cdot 2\text{H}_2\text{O}$ , where $\text{M}$ is a Group 2 metal. You will first determine the relative formula mass of the salt by measuring the mass loss when the sample is heated. Heating the sample produces the anhydrous salt and water of crystallisation. You will then select reagents to determine the identity of the ion $\text{A}^-$ .

FB 4 is the salt, $\text{MA}_2\cdot 2\text{H}_2\text{O}$ .

(a)

Method

Weigh the empty crucible with its lid. Record the mass.
Transfer all of FB 4 into the crucible.
Weigh the crucible, lid and FB 4. Record the mass.
Calculate and record the mass of FB 4 used.
Place the crucible and contents on a pipe-clay triangle.
Heat the crucible gently, with the lid on, for approximately 1 minute.
Heat strongly, with the lid off, for a further 4 minutes.
Replace the lid and leave the crucible to cool for at least 5 minutes.

While the crucible is cooling, you may wish to begin work on Question 3.

When the crucible has cooled, weigh the crucible with its lid and contents. Record the mass.
Heat strongly, with the lid off, for a further 2 minutes.
Replace the lid and leave the crucible to cool for at least 5 minutes.
When the crucible has cooled, reweigh the crucible with its lid and contents. Record the mass.
Calculate and record the mass of residue obtained.

Results

5M Similar

DifficultyMedium-Easy

Worked solution

Answer

Record all four masses to the same number of decimal places, with units.

Weighing	Mass / g
crucible + lid	20.00
crucible + lid + FB 4	22.40
crucible + lid + contents after first heating	22.06
crucible + lid + contents after second heating	22.05

Calculated masses:

mass of FB 4 used = $22.40 - 20.00 = 2.40$ g
mass of residue = $22.05 - 20.00 = 2.05$ g

(Values shown are illustrative; use your own readings.)

Final answer

See working (candidate-dependent readings)

Detailed explanation

Background Concept

This is a gravimetric experiment to determine the water of crystallisation in a hydrated salt. When a hydrated salt such as $\text{MA}_2 \cdot 2\text{H}_2\text{O}$ is heated, the water molecules are driven off as steam, leaving the anhydrous salt. The loss in mass of the sample equals the mass of water lost. To be sure all the water has been removed, the sample is heated, cooled and weighed, then heated again and reweighed until two consecutive masses agree closely (constant mass).

Understanding the Question

The method tells you to weigh an empty crucible with its lid, add the hydrated salt FB 4, weigh again, heat, cool, weigh, then reheat and reweigh. You need to record all four masses in a clear table with headings and units, and then calculate the mass of FB 4 used and the mass of residue. The marks are awarded for correct table headings, consistent precision, correct subtractions and sensible accuracy.

Approach

Set up a table with four rows: crucible + lid; crucible + lid + FB 4; crucible + lid + contents after first heating; crucible + lid + contents after second heating. Record every mass to the same number of decimal places (two or three). After the second heating the mass should be very close to the first heating mass, showing that all water has been removed. Calculate the mass of FB 4 and the mass of residue by subtracting the mass of the empty crucible + lid.

Step-by-Step Reasoning

Weigh the empty crucible with its lid, for example 20.00 g.
Add all of FB 4 and weigh again, for example 22.40 g. The mass of FB 4 is $22.40 - 20.00 = 2.40$ g.
Heat gently with the lid on, then strongly with the lid off, cool and weigh, for example 22.06 g.
Reheat strongly for a further 2 minutes, cool and weigh, for example 22.05 g. The two masses are close, so constant mass has been reached.
The mass of residue is $22.05 - 20.00 = 2.05$ g.
These example values are only illustrative; in the exam you would record your own readings, but the layout and calculations would be the same.

Key Takeaways

Always give each heading a quantity and a unit, e.g. Mass / g.
Record all readings to the same number of decimal places.
Heat to constant mass so that all water of crystallisation is removed.
Calculate masses by subtraction from the empty crucible + lid mass.

Common Mistakes

Omitting units from table headings.
Recording readings to different numbers of decimal places.
Using the mass after the first heating instead of the final constant mass.
Subtracting the wrong way round or using the wrong pair of readings.

Things to Be Careful About

Keep the lid on while the crucible cools so that the anhydrous salt does not absorb moisture from the air.
Use tongs to handle the hot crucible.
The fourth weighing should be within +0.02 and -0.05 g of the third; if it is not, heat again until constant mass is achieved.

Techniques used

weigh by differencerecord results to a consistent number of decimal placesheat to constant masscalculate masses by subtraction

(b)

Calculate the amount, in mol, of water lost.

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

Hence calculate the relative formula mass, $M_r$ , of $\text{MA}_2$ .

$M_r$ of $\text{MA}_2$ = ..............................

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Using the illustrative results from (a):

Mass of water lost = $2.40 - 2.05 = 0.35$ g

Amount of $\text{H}_2\text{O}$ = $\frac{0.35}{18.0} = 0.0194$ mol (3 s.f.)

$M_r$ of $\text{MA}_2$ = $\frac{2 \times 2.05}{0.0194} = 211$ (3 s.f.)

Answer

amount of $\text{H}_2\text{O}$ = 0.0194 mol

$M_r$ of $\text{MA}_2$ = 211

Final answer

amount of H2O = 0.0194 mol; Mr(MA2) = 211 (using illustrative data)

Detailed explanation

Background Concept

The hydrated salt has the formula $\text{MA}_2 \cdot 2\text{H}_2\text{O}$ , so each mole of the salt contains two moles of water of crystallisation. When heated, the water is driven off. The mass lost by the sample is therefore the mass of water. The amount of water in moles is found by dividing the mass of water by its molar mass (18.0 g mol $^{-1}$ ). Because the salt contains two water molecules per formula unit, the amount of anhydrous salt $\text{MA}_2$ is half the amount of water. The relative formula mass of $\text{MA}_2$ is then the mass of residue divided by the amount of $\text{MA}_2$ .

Understanding the Question

You need to use your results from part (a): the mass of FB 4 used and the mass of residue. First calculate the mass of water lost, then the amount of water, then the relative formula mass of $\text{MA}_2$ .

Approach

Mass of water lost = mass of FB 4 - mass of residue.
Amount of water = mass of water / 18.0.
Amount of $\text{MA}_2$ = half the amount of water.
$M_r(\text{MA}_2)$ = mass of residue / amount of $\text{MA}_2$ = $2 \times$ mass of residue / amount of water.

Step-by-Step Reasoning

Using the illustrative results from (a):

Mass of water lost = $2.40 - 2.05 = 0.35$ g.
Amount of water = $0.35 / 18.0 = 0.0194$ mol.
Amount of $\text{MA}_2$ = $0.0194 / 2 = 0.00972$ mol.
$M_r(\text{MA}_2)$ = $2.05 / 0.00972 = 211$ (3 s.f.).
The mark scheme accepts the equivalent formula $M_r = 2 \times$ mass of residue / amount of water. Give your answer to 2-4 significant figures.

Key Takeaways

Mass loss on heating equals mass of water of crystallisation.
Use the stoichiometry of the formula to relate moles of water to moles of anhydrous salt.
$M_r$ = mass / amount, with units g mol $^{-1}$ .

Common Mistakes

Forgetting the factor 2 in the formula, because each formula unit contains two water molecules.
Using the mass of FB 4 instead of the mass of residue in the $M_r$ calculation.
Giving too few or too many significant figures.

Things to Be Careful About

Use 18.0 (or 18) for the molar mass of water.
Keep the same units throughout (grams).
The answer should be to 2-4 s.f., not rounded to one significant figure.

Techniques used

calculate mass of water lostconvert mass to molesuse the stoichiometric ratio to find Mr

(c)

FB 5 is a solution of the hydrated salt $\text{MA}_2$ . The $\text{A}^-$ ion is a halide.
Carry out tests to identify the halide present in $\text{MA}_2$ .
Record the reagents used, the results of your tests and the identity of $\text{A}^-$ .

$\text{A}^-$ is ..............................

2M Similar

DifficultyEasy

Worked solution

Answer

Add dilute nitric acid, then aqueous silver nitrate to FB 5.

A white precipitate forms.

Add excess aqueous ammonia: the white precipitate dissolves.

$\text{A}^-$ is $\text{Cl}^-$ (chloride).

Final answer

Cl- (chloride)

Detailed explanation

Background Concept

Halide ions can be identified by their reaction with aqueous silver nitrate. Silver chloride is a white precipitate, silver bromide is cream and silver iodide is yellow. The precipitates differ in their solubility in ammonia: AgCl dissolves in dilute or excess ammonia, AgBr dissolves only in concentrated ammonia, and AgI is insoluble. The test solution is usually acidified with dilute nitric acid first to remove carbonate or sulfite ions that would also give a precipitate with silver nitrate.

Understanding the Question

FB 5 is a solution of the hydrated salt $\text{MA}_2$ , and you are told that $\text{A}^-$ is a halide. You must carry out a test to identify which halide is present, record the reagents and results, and state the identity of $\text{A}^-$ .

Approach

Add dilute nitric acid to FB 5, then add aqueous silver nitrate. Observe the colour of any precipitate. Then add excess aqueous ammonia to see whether the precipitate dissolves. Use the colour and solubility to identify the halide.

Step-by-Step Reasoning

Acidify a sample of FB 5 with dilute nitric acid.
Add aqueous silver nitrate. A white precipitate shows the presence of chloride, cream suggests bromide, yellow suggests iodide.
Add excess aqueous ammonia. A white precipitate that dissolves in excess ammonia confirms chloride, because AgCl forms a soluble complex with ammonia. AgBr is only sparingly soluble in concentrated ammonia, and AgI is insoluble.
Therefore $\text{A}^-$ is $\text{Cl}^-$ .

Key Takeaways

Silver nitrate is the key reagent for halide tests.
Precipitate colour and solubility in ammonia distinguish Cl, Br and I.
Acidifying with dilute nitric acid removes interfering ions.

Common Mistakes

Not acidifying with dilute nitric acid, so carbonate could also give a white precipitate.
Confusing the colours: chloride white, bromide cream, iodide yellow.
Using dilute ammonia to test AgBr; it needs concentrated ammonia.

Things to Be Careful About

Record both the reagent and the observation.
The mark scheme requires 'white precipitate' and 'soluble in excess ammonia' to score both marks.
If you see a cream precipitate, the halide would be bromide, not chloride.

Techniques used

carry out a halide test with silver nitratetest the solubility of the precipitate in ammoniaidentify the halide from precipitate behaviour

(d)

Using your answers to (b) and (c), identify the ion $\text{M}^{2+}$ .

$\text{M}^{2+}$ is ..............................

1M Similar

DifficultyMedium-Easy

Worked solution

Working

$M_r(\text{MA}_2) = 211$

$A_r(\text{M}) = 211 - 2(35.5) = 211 - 71 = 140$

The Group 2 metal with $A_r$ closest to 140 is barium ( $A_r = 137.3$ ).

Answer

$\text{M}^{2+}$ is $\text{Ba}^{2+}$ .

Final answer

Ba2+ (barium)

Detailed explanation

Background Concept

The salt is $\text{MA}_2$ , where M is a Group 2 metal and A is the halide identified in part (c). Since there are two halide ions per formula unit, the relative formula mass of $\text{MA}_2$ is the relative atomic mass of M plus twice the relative atomic mass of the halide. For chloride, $A_r(\text{Cl}) = 35.5$ , so $M_r(\text{MA}_2) = A_r(\text{M}) + 2(35.5) = A_r(\text{M}) + 71$ . Rearranging, $A_r(\text{M}) = M_r(\text{MA}_2) - 71$ . Compare this value with the relative atomic masses of the Group 2 metals.

Understanding the Question

You are asked to use your calculated $M_r$ from part (b) and the identity of $\text{A}^-$ from part (c) to identify the metal ion $\text{M}^{2+}$ .

Approach

Subtract the total mass of the two halide ions from $M_r(\text{MA}_2)$ to find $A_r(\text{M})$ . Then choose the Group 2 metal whose relative atomic mass is closest to this value.

Step-by-Step Reasoning

Using the illustrative $M_r = 211$ and $\text{A}^- = \text{Cl}^-$ :

$A_r(\text{M}) = 211 - 2(35.5) = 211 - 71 = 140$ .
Group 2 metals have $A_r$ values: Be 9, Mg 24, Ca 40, Sr 88, Ba 137, Ra 226.
The closest to 140 is barium, $A_r = 137.3$ .
Therefore $\text{M}^{2+}$ is $\text{Ba}^{2+}$ .

Key Takeaways

Use the stoichiometry of the formula to relate $M_r$ to atomic masses.
Compare calculated $A_r$ with known Group 2 values.
The ion has a 2+ charge because it is a Group 2 metal.

Common Mistakes

Forgetting to multiply the halide atomic mass by 2.
Using the wrong halide atomic mass (e.g. 35.5 for Cl, 80 for Br, 127 for I).
Not comparing with Group 2 values, or choosing the wrong metal.

Things to Be Careful About

The answer must be a Group 2 metal ion with a 2+ charge.
Show the subtraction clearly to earn the method mark.
The calculated $A_r$ may not match exactly; choose the closest.

Techniques used

subtract the halide mass from the Mrcompare the calculated Ar with Group 2 metalsdeduce the identity of M2+

(e)

A student correctly identifies $\text{A}^-$ but did not heat the sample of FB 4 for long enough to remove all the water of crystallisation.
Despite this error, the student still correctly identifies $\text{M}^{2+}$ .

Explain how the student's answer in (b) would differ from the true answer.

...................................................................................................................................................

Explain why the student still correctly identifies $\text{M}^{2+}$ .

...................................................................................................................................................

2M Similar

DifficultyMedium

Worked solution

Answer

If not all water was removed, the mass loss is too small, so the calculated amount of $\text{H}_2\text{O}$ is too small and the calculated $M_r$ of $\text{MA}_2$ is greater than the true value.

The student still identifies $\text{M}^{2+}$ because the $A_r$ values of the Group 2 metals are very different, so the calculated $A_r$ of M is still closest to the same metal (or most of the water was driven off, leaving only a small error).

Final answer

Calculated Mr would be greater; still identifies M2+ because Group 2 Ar values are widely spaced

Detailed explanation

Background Concept

In the experiment, the mass of water lost is found by heating the hydrated salt and measuring the loss in mass. If heating is not continued for long enough, some water of crystallisation remains in the residue. The measured mass loss is then too small, and the measured mass of residue is too large. Both effects make the calculated $M_r$ of $\text{MA}_2$ larger than the true value. However, the Group 2 metals have widely spaced relative atomic masses, so even a fairly large error in $M_r$ may still leave the calculated $A_r$ closest to the correct metal.

Understanding the Question

A student correctly identifies $\text{A}^-$ but does not remove all the water of crystallisation. You must explain how the student's answer in part (b) would differ from the true answer, and why the student still correctly identifies $\text{M}^{2+}$ .

Approach

Use the formula for $M_r$ to see the effect of incomplete heating: mass of water is too small, so amount of water is too small; mass of residue is too large; therefore $M_r$ is too large. Then explain that the $A_r$ values of Group 2 metals are far apart, so the calculated $A_r$ is still closest to the same metal.

Step-by-Step Reasoning

If water remains, the mass loss is smaller than it should be.
Amount of water = mass loss / 18, so the calculated amount of water is too small.
$M_r(\text{MA}_2) = 2 \times$ mass of residue / amount of water. The numerator is too large and the denominator is too small, so $M_r$ is greater than the true value.
Even so, the student still identifies $\text{M}^{2+}$ because the $A_r$ values of the Group 2 metals differ greatly (e.g. Ca 40, Sr 88, Ba 137). The calculated $A_r$ of M, although slightly too high, is still closest to the same metal. Alternatively, most of the water was driven off, so the error is small.

Key Takeaways

Incomplete heating makes the calculated $M_r$ too large.
Widely spaced $A_r$ values make identification robust.
Always heat to constant mass to minimise this error.

Common Mistakes

Saying the calculated $M_r$ would be smaller; the opposite is true.
Not explaining why identification is still correct.
Giving a vague answer such as 'the error is small' without linking it to the spacing of $A_r$ values.

Things to Be Careful About

The mark scheme accepts either 'big differences between the $A_r$ values' or 'most of the water was driven off' or 'the calculated $A_r$ is still closest to the same answer'.
Make sure you state that the amount of water is smaller and/or the mass of residue is greater.

Techniques used

analyse the effect of incomplete heating on the calculated Mrexplain why the identification remains robust

Q3MediumQualitative AnalysisManipulation, Measurement and ObservationAnalysis, Conclusions and Evaluation

Qualitative analysis

For each test you should record all your observations in the spaces provided.

Examples of observations include:

colour changes seen
the formation of any precipitate and its solubility (where appropriate) in an excess of the reagent added
the formation of any gas and its identification (where appropriate) by a suitable test.

You should record clearly at what stage in a test an observation is made.

Where no change is observed, you should write 'no change'.

Where reagents are selected for use in a test, the name or correct formula of the element or compound must be given.

If any solution is warmed, a boiling tube must be used. If a solid is heated, a hard-glass test-tube must be used.

Rinse and reuse test-tubes and boiling tubes where possible.

No additional tests should be attempted.

(a)

(i)

FB 6 is potassium manganate(VII). Place all the FB 6 in a hard-glass test-tube and heat gently at first and then more strongly.
Record your observations.

Leave the tube to cool and keep for use in (a)(ii).
While the tube is cooling, you may wish to begin work on (b).

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Initially: dark purple / purple-black crystals.
On heating: the crystals crackle / pop and jump around the tube.
A black residue / powder remains.
A gas is produced that relights a glowing splint — oxygen.

Final answer

Purple-black crystals; crackling/jumping on heating; black residue; oxygen relights a glowing splint

Detailed explanation

Background Concept

Potassium manganate(VII), $\text{KMnO}_4$ , is a purple-black crystalline solid. When heated strongly it undergoes thermal decomposition:

2\text{KMnO}_4(\text{s}) \rightarrow \text{K}_2\text{MnO}_4(\text{s}) + \text{MnO}_2(\text{s}) + \text{O}_2(\text{g})

The manganese is reduced from $+7$ in $\text{KMnO}_4$ to $+6$ in $\text{K}_2\text{MnO}_4$ (manganate(VI), green) and $+4$ in $\text{MnO}_2$ (black). Because oxidation states change, this is also a redox reaction, but the name "thermal decomposition" describes the physical process of one substance breaking into simpler substances on heating.

Understanding the Question

This is a qualitative observation task. You heat a sample of solid $\text{KMnO}_4$ in a hard-glass test-tube (gently at first, then strongly) and record everything you see — the initial appearance, what happens during heating, the residue left, and any gas produced. The gas must be identified by a suitable test.

Approach

Watch the solid before, during and after heating. Test any gas with a glowing splint (oxygen relights it). Note the colour of the residue left behind.

Step-by-Step Reasoning

Initially the solid is purple-black — the characteristic colour of $\text{KMnO}_4$ crystals.
On heating, the crystals crackle/pop and jump around the tube. This happens because oxygen gas is being released and escaping through the solid.
The residue left is black — this is $\text{MnO}_2$ (together with some $\text{K}_2\text{MnO}_4$ ).
The gas relights a glowing splint — this is the standard positive test for oxygen.

Key Takeaways

$\text{KMnO}_4$ decomposes on heating to $\text{K}_2\text{MnO}_4$ , $\text{MnO}_2$ and $\text{O}_2$ .
Oxygen is identified by relighting a glowing splint.
Always record the initial colour and any residue left after heating.

Common Mistakes

Not testing the gas with a glowing splint — the mark requires the gas identification.
Writing "bubbles" or "gas given off" without saying the gas relights a glowing splint.
Not noting the black residue that remains.

Things to Be Careful About

Use a hard-glass test-tube for heating solids, as the question requires.
Heat gently at first, then more strongly — sudden strong heating may crack the tube.
The residue is a mixture of $\text{K}_2\text{MnO}_4$ , $\text{MnO}_2$ and some unreacted $\text{KMnO}_4$ — this matters for part (a)(ii).

Techniques used

heat a solid in a hard-glass test-tubetest a gas with a glowing splintrecord colour changes and the appearance of the residue

(ii)

Complete Table 3.1 by carrying out the tests described. Record your observations.

Table 3.1

test	observations
Test 1 To a $7\text{ cm}$ depth of distilled water in a boiling tube, add approximately half of the residue from (a)(i).
Test 2 To a $4\text{ cm}$ depth of aqueous sodium hydroxide in a test-tube, add the remaining residue from (a)(i).

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Test 1 (residue + distilled water): a purple solution forms — the water turns purple.

Test 2 (residue + NaOH(aq)): a green solution forms — the NaOH turns green.

Final answer

Test 1: purple solution; Test 2: green solution

Detailed explanation

Background Concept

The residue from (a)(i) contains three species: unreacted $\text{KMnO}_4$ (purple, soluble in water), $\text{K}_2\text{MnO}_4$ (manganate(VI), green, stable in alkaline solution) and $\text{MnO}_2$ (black, insoluble).

In distilled water, the soluble $\text{KMnO}_4$ dissolves to give a purple solution.
In $\text{NaOH(aq)}$ (alkaline conditions), the manganate(VI) ion $\text{MnO}_4^{2-}$ is stable and green — the green solution is due to $\text{K}_2\text{MnO}_4$ .

Understanding the Question

Two tests are carried out on the residue from (a)(i): Test 1 adds half the residue to distilled water; Test 2 adds the remaining residue to aqueous sodium hydroxide. For each, record the colour of the solution formed.

Approach

Add the residue to each reagent and observe the colour of the solution that forms. The key is to recognise which species dissolves in each medium.

Step-by-Step Reasoning

Test 1: The residue contains unreacted $\text{KMnO}_4$ , which dissolves in water to give a purple solution. The mark scheme accepts "purple solution" or "water turns purple".

Test 2: In alkaline $\text{NaOH}$ , the manganate(VI) ion $\text{MnO}_4^{2-}$ is stable and gives a green solution. The mark scheme accepts "green solution" or "NaOH turns green".

Key Takeaways

$\text{KMnO}_4$ dissolves in water to a purple solution.
Manganate(VI), $\text{MnO}_4^{2-}$ , is green and is stable in alkaline solution.

Common Mistakes

Writing "no change" — there is a clear colour change in both tests.
Confusing the colours: purple is $\text{KMnO}_4$ , green is $\text{MnO}_4^{2-}$ .

Things to Be Careful About

Record the colour of the solution, not just the appearance of the solid.
The green colour only appears in alkaline conditions (NaOH); in water the purple $\text{KMnO}_4$ dominates.

Techniques used

dissolve a solid residue in distilled wateradd aqueous sodium hydroxide to a solid residuerecord the colour of the resulting solution

(iii)

Suggest what type of reaction takes place in (a)(i).

1M Similar

(b)

FB 7 is a solution of a salt which contains a cation and an anion from those listed in the Qualitative analysis notes.
FB 8 is $0.100\text{ mol dm}^{-3}$ sodium thiosulfate, $\text{Na}_2\text{S}_2\text{O}_3$ .

(i)

Complete Table 3.2 by carrying out the tests described. Record your observations.

Table 3.2

test	observations
Test 1 To a $1\text{ cm}$ depth of FB 7 in a test-tube, add aqueous ammonia until there is no further change, then add a few drops of hydrogen peroxide.
Test 2 To a $1\text{ cm}$ depth of FB 7 in a test-tube, add a $1\text{ cm}$ depth of aqueous potassium iodide, then add FB 8.

5M Similar

DifficultyMedium

Worked solution

Answer

Test 1

- NH $_3$ : a pale blue precipitate forms; this dissolves in excess NH $_3$ giving a deep blue solution.
- H $_2$ O $_2$ : the solution turns black / dark green; effervescence; the gas relights a glowing splint (oxygen).

Test 2

- KI: a brown / yellow-brown solution forms; an off-white / pale brown precipitate forms.
- Na $_2$ S $_2$ O $_3$ : the solution and precipitate become paler; the precipitate partially dissolves (dissolves in excess thiosulfate).

Final answer

Test 1: pale blue ppt soluble in excess NH3 to deep blue; H2O2 gives black/dark green, effervescence, O2 relights splint. Test 2: KI gives brown solution + off-white ppt; thiosulfate makes paler, ppt partially dissolves.

Detailed explanation

Background Concept

FB 7 is copper(II) sulfate, $\text{CuSO}_4(\text{aq})$ , containing pale blue $\text{Cu}^{2+}(\text{aq})$ ions.

Test 1 — ammonia then hydrogen peroxide:
Adding aqueous ammonia to $\text{Cu}^{2+}$ first gives a pale blue precipitate of copper(II) hydroxide:

\text{Cu}^{2+}(\text{aq}) + 2\text{NH}_3(\text{aq}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow \text{Cu}(\text{OH})_2(\text{s}) + 2\text{NH}_4^+(\text{aq})

With excess ammonia the precipitate dissolves to form the deep blue tetraamminecopper(II) complex:

\text{Cu}(\text{OH})_2(\text{s}) + 4\text{NH}_3(\text{aq}) \rightarrow [\text{Cu}(\text{NH}_3)_4]^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq})

Hydrogen peroxide then decomposes, catalysed by the copper(II) ions:

2\text{H}_2\text{O}_2(\text{aq}) \rightarrow 2\text{H}_2\text{O}(\text{l}) + \text{O}_2(\text{g})

This gives effervescence, and the oxygen relights a glowing splint. The solution may turn black or dark green.

Test 2 — potassium iodide then sodium thiosulfate:
Copper(II) oxidises iodide ions to iodine, forming copper(I) iodide:

2\text{Cu}^{2+}(\text{aq}) + 4\text{I}^-(\text{aq}) \rightarrow 2\text{CuI}(\text{s}) + \text{I}_2(\text{aq})

The iodine gives a brown / yellow-brown solution; $\text{CuI}$ is an off-white / pale brown precipitate.

Sodium thiosulfate reduces iodine back to iodide ions:

\text{I}_2(\text{aq}) + 2\text{S}_2\text{O}_3^{2-}(\text{aq}) \rightarrow 2\text{I}^-(\text{aq}) + \text{S}_4\text{O}_6^{2-}(\text{aq})

so the brown colour fades. Thiosulfate also dissolves the $\text{CuI}$ precipitate (forming a soluble complex), so the precipitate partially dissolves.

Understanding the Question

Two tests are carried out on FB 7, each with two reagents added in a set order. You must record all observations at each stage — precipitate formation, colour changes, solubility in excess reagent, and any gas produced.

Approach

Add each reagent in the stated order and record the observation after each addition. Test any gas with a glowing splint. Note whether precipitates dissolve in excess reagent.

Step-by-Step Reasoning

Test 1: Add ammonia dropwise until no further change. A pale blue precipitate of $\text{Cu(OH)}_2$ forms. With excess ammonia it dissolves to a deep blue solution — this is diagnostic for $\text{Cu}^{2+}$ . Then add a few drops of hydrogen peroxide: the solution turns black/dark green, effervesces, and the gas relights a glowing splint (oxygen from catalytic decomposition of $\text{H}_2\text{O}_2$ ).

Test 2: Add potassium iodide. A brown/yellow-brown solution forms (iodine) together with an off-white/pale brown precipitate ( $\text{CuI}$ ). Then add sodium thiosulfate: the solution and precipitate become paler, and the precipitate partially dissolves (dissolves in excess thiosulfate).

Key Takeaways

$\text{Cu}^{2+}$ gives a pale blue $\text{Cu(OH)}_2$ precipitate with $\text{NH}_3$ , soluble in excess to a deep blue $[\text{Cu(NH}_3)_4]^{2+}$ complex.
$\text{Cu}^{2+}$ oxidises $\text{I}^-$ to $\text{I}_2$ (brown) while forming $\text{CuI}$ (off-white precipitate).
Thiosulfate removes iodine (colour fades) and dissolves $\text{CuI}$ .

Common Mistakes

Not noting the solubility of the precipitate in excess ammonia — this is a key diagnostic observation.
Missing the gas test for oxygen from the hydrogen peroxide decomposition.
Writing "white precipitate" for $\text{CuI}$ — it is off-white / pale brown.

Things to Be Careful About

Record the stage at which each observation is made (before and after each addition).
The deep blue solution in excess ammonia is the key diagnostic for $\text{Cu}^{2+}$ .

Techniques used

add aqueous ammonia to a cation solution until no further changeadd hydrogen peroxide and test any gas with a glowing splintadd potassium iodide to a cation solutionadd sodium thiosulfate and observe colour and precipitate changes

(ii)

The anion in FB 7 is either the sulfite ion or the sulfate ion.
Carry out tests to identify which ion is present.
Record the reagents used and the results of your tests.

2M Similar

DifficultyMedium

Worked solution

Answer

Add barium chloride (or barium nitrate) followed by dilute nitric acid (or hydrochloric acid).

A white precipitate that is insoluble in the acid shows the sulfate ion is present (a sulfite would give a white precipitate that dissolves in the acid).

Final answer

BaCl2/Ba(NO3)2 + dilute acid: white ppt insoluble in acid → sulfate present

Detailed explanation

Background Concept

Sulfate ( $\text{SO}_4^{2-}$ ) and sulfite ( $\text{SO}_3^{2-}$ ) both give white precipitates with barium ions:

\text{Ba}^{2+}(\text{aq}) + \text{SO}_4^{2-}(\text{aq}) \rightarrow \text{BaSO}_4(\text{s})

\text{Ba}^{2+}(\text{aq}) + \text{SO}_3^{2-}(\text{aq}) \rightarrow \text{BaSO}_3(\text{s})

The key difference is that $\text{BaSO}_4$ is insoluble in dilute acid, whereas $\text{BaSO}_3$ dissolves in dilute acid. So adding barium chloride (or barium nitrate) followed by dilute acid discriminates between them.

An alternative test uses acidified potassium manganate(VII): sulfite reduces $\text{KMnO}_4$ (it is decolourised), whereas sulfate does not (the purple colour remains).

Understanding the Question

The anion in FB 7 is either sulfite or sulfate. You must carry out a test to identify which is present, recording the reagents used and the results.

Approach

Choose a test that gives a different result for the two anions. The barium chloride + acid test is the standard one; acidified $\text{KMnO}_4$ is a valid alternative.

Step-by-Step Reasoning

Since FB 7 is $\text{CuSO}_4$ , the sulfate ion is present. Adding barium chloride gives a white precipitate of $\text{BaSO}_4$ , which remains insoluble when dilute acid is added. (If sulfite were present, the white $\text{BaSO}_3$ precipitate would dissolve in the acid.)

If using acidified $\text{KMnO}_4$ instead: the purple colour remains because sulfate cannot reduce manganate(VII); a sulfite would decolourise it.

Key Takeaways

$\text{BaSO}_4$ is insoluble in acid; $\text{BaSO}_3$ is soluble — this is the discriminating test.
Sulfite reduces acidified $\text{KMnO}_4$ (decolourises); sulfate does not.

Common Mistakes

Using silver nitrate — it does not discriminate sulfate from sulfite here.
Not adding acid — both anions give a white precipitate with barium ions, so acid is essential.

Things to Be Careful About

Record both the reagent(s) used and the result clearly.
If using $\text{KMnO}_4$ , it must be acidified for the redox test to work.

Techniques used

select a reagent to distinguish sulfate from sulfiteinterpret precipitate solubility in acid

(iii)

Give the formula of the salt in FB 7.

1M Similar