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'.…

(a)(i) See working above for the sequence of tests and observations confirming H2O, SO4^2-, and Mg^2+ (not Zn^2+). · (a)(ii) Zn^2+: ✗, SO4^2-: ✓, H2O: ✓ · (b)(i) Purple solid jumps, black residue forms, glowing splint relights. · (b)(ii) dark green · (c)(i) See table above for observations. · (c)(ii) manganese · (c)(iii) FA 6/FA 7: +7, FA 8: +2

Chemistry 9701/37 — October/November 2025

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

Q1Medium-EasyManipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and Evaluation

Hydrated ethanedioic acid is a diprotic acid with the formula $(\text{COOH})_2 \cdot x\text{H}_2\text{O}$ where $x$ is an integer.

Ethanedioic acid reacts with manganate(VII) ions when heated.

5(\text{COOH})_2(\text{aq}) + 2\text{MnO}_4^-(\text{aq}) + 6\text{H}^+(\text{aq}) \rightarrow 10\text{CO}_2(\text{g}) + 2\text{Mn}^{2+}(\text{aq}) + 8\text{H}_2\text{O}(\text{l})

You will determine the value of $x$ in $(\text{COOH})_2 \cdot x\text{H}_2\text{O}$ by titrating a solution containing ethanedioic acid with manganate(VII) ions.

FA 1 is $6.20 \text{ g dm}^{-3}$ aqueous ethanedioic acid, $(\text{COOH})_2 \cdot x\text{H}_2\text{O}$ .
FA 2 is $0.0200 \text{ mol dm}^{-3}$ potassium manganate(VII), $\text{KMnO}_4$ .
FA 3 is $1.0 \text{ mol dm}^{-3}$ sulfuric acid, $\text{H}_2\text{SO}_4$ .

(a)

Method

Fill the burette with FA 2.
Pipette $25.0 \text{ cm}^3$ of FA 1 into a conical flask.
Use the measuring cylinder to add approximately $20 \text{ cm}^3$ of FA 3 to the conical flask.
Place the conical flask on a tripod and gauze and heat carefully until the temperature of the solution is approximately $70 \text{ }^\circ\text{C}$ .
Remove the flame.
Carefully lift the hot conical flask and place it on the white tile under the burette.
Add FA 2 drop-wise for the first $2-3 \text{ cm}^3$ . Any initial pink colouring may take several seconds to disappear.
If the reaction mixture turns brown, reheat it to about $70 \text{ }^\circ\text{C}$ . If the brown colour disappears, continue the titration. If the brown colour remains, discard the contents of the flask and begin a new titration.
The end-point is reached when a permanent pale pink colour is formed.
Perform a rough titration with FA 2. Record your burette readings in the space below.

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

Carry out as many 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 FA 2 added in each accurate titration.

Results

7M Similar

DifficultyMedium-Easy

Worked solution

Answer

Record the rough titre and at least two accurate titrations, with all burette readings to the nearest 0.05 cm3. For example (your readings will differ):

Titration	Initial reading / cm3	Final reading / cm3	Titre / cm3
Rough	0.00	24.85	24.85
1	0.00	24.60	24.60
2	24.60	49.25	24.65

The accurate titres are concordant (within 0.10 cm3 of each other).

Final answer

Example data: titres 24.60 and 24.65 cm3

Detailed explanation

Background Concept

In a titration, the aim is to determine the volume of a solution (the titrant) needed to react completely with a known volume of another solution. For reliable results, the titre must be measured accurately and consistently. Burette readings are typically taken to the nearest 0.05 cm3 because the burette scale allows estimation between the 0.1 cm3 divisions. Concordant titres are those that agree within a small tolerance (here 0.10 cm3), indicating that the reaction has been carried out consistently.

Understanding the Question

This part asks you to perform the titration and record your results properly. You need to record the rough titre and then carry out accurate titrations until you have at least two concordant results. The marks are awarded for the quality of your data recording: correct headings, units, precision, and concordance.

Approach

Carry out a rough titration to get an approximate end-point. Then perform accurate titrations, adding the manganate(VII) drop-wise near the end-point. Record initial and final burette readings for each accurate titration. Calculate the titre as final minus initial. Ensure that your accurate titres agree within 0.10 cm3. Record all readings to 0.05 cm3.

Step-by-Step Reasoning

Fill the burette with FA2 (KMnO4). Record the initial reading.
Pipette 25.0 cm3 of FA1 into a conical flask, add about 20 cm3 of FA3 (H2SO4), heat to about 70°C, then titrate.
For the rough titration, add FA2 quickly until the end-point is approached, then drop-wise. Record the rough titre.
For accurate titrations, start with a clean flask and repeat the procedure. Add FA2 drop-wise near the end-point until a permanent pale pink colour appears. Record the final reading.
The titre is the difference between final and initial readings.
Repeat until you have at least two accurate titres within 0.10 cm3 of each other.
Record all readings to the nearest 0.05 cm3.

Key Takeaways

Burette readings must be recorded to 0.05 cm3.
Concordant titres are essential for a reliable mean.
The titre is calculated as final − initial reading.

Common Mistakes

Recording readings to only 0.1 cm3 instead of 0.05 cm3.
Not performing enough accurate titrations to get concordant results.
Forgetting to record units or headings in the results table.

Things to Be Careful About

Ensure the burette is filled correctly and there are no air bubbles.
The end-point is a permanent pale pink colour; a transient pink that disappears is not the end-point.
If the solution turns brown and does not disappear on reheating, discard and start again.

Techniques used

record burette readings to 0.05 cm3perform concordant titrationscalculate titre from initial and final readingsensure consistent results

(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 FA 1 required .............................. $\text{cm}^3$ of FA 2.

1M Similar

(c)

Calculations

(i)

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

1M Similar

(ii)

Calculate the amount, in mol, of manganate(VII) ions, $\text{MnO}_4^-$ , in the volume of FA 2 calculated in (b).

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

1M Similar

(iii)

Calculate the amount, in mol, of ethanedioic acid that reacts with the manganate(VII) ions in (c)(ii).

amount of $(\text{COOH})_2$ = .............................. mol

Hence calculate the concentration, in $\text{mol dm}^{-3}$ , of ethanedioic acid in FA 1.

concentration of $(\text{COOH})_2$ = .............................. $\text{mol dm}^{-3}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

From the equation: 5 mol (COOH)2 react with 2 mol MnO4−.

Amount of (COOH)2 = $4.93 \times 10^{-4} \times \frac{5}{2} = 1.23 \times 10^{-3} \text{ mol}$

This is in 25.0 cm3 = 0.0250 dm3.

Concentration = $\frac{1.23 \times 10^{-3}}{0.0250} = 0.0493 \text{ mol dm}^{-3}$

Answer

Amount of (COOH)2 = $1.23 \times 10^{-3} \text{ mol}$
Concentration = $0.0493 \text{ mol dm}^{-3}$

Final answer

1.23 × 10^-3 mol; 0.0493 mol dm^-3

Detailed explanation

Background Concept

The balanced equation shows the stoichiometric relationship between ethanedioic acid and manganate(VII) ions: 5 mol of acid react with 2 mol of manganate(VII). Therefore, the amount of acid is 5/2 times the amount of manganate(VII). The concentration of the acid in FA1 is then found by dividing the amount by the volume of FA1 used (25.0 cm3 = 0.0250 dm3).

Understanding the Question

You need to calculate the amount of ethanedioic acid that reacted, and then its concentration in FA1. This links the titration result to the unknown concentration.

Approach

Use the mole ratio from the equation to convert moles of MnO4− to moles of (COOH)2. Then divide by the volume of FA1 (in dm3) to get concentration.

Step-by-Step Reasoning

From part (c)(ii), amount of MnO4− = 4.93 × 10^-4 mol.
Mole ratio: 5 (COOH)2 : 2 MnO4−, so amount of (COOH)2 = 4.93 × 10^-4 × 5/2 = 1.2325 × 10^-3 mol.
Round to 3 sf: 1.23 × 10^-3 mol.
Volume of FA1 = 25.0 cm3 = 0.0250 dm3.
Concentration = 1.23 × 10^-3 / 0.0250 = 0.0493 mol dm^-3.

Key Takeaways

Use the stoichiometric ratio correctly.
Concentration = moles / volume (dm3).

Common Mistakes

Using the ratio upside down (2/5 instead of 5/2).
Forgetting to convert volume to dm3.

Things to Be Careful About

The amount of acid is calculated from the reaction with the manganate(VII), not from the volume of FA1 directly.

Techniques used

apply stoichiometric ratiocalculate concentration

(iv)

Calculate the relative molecular mass, $M_r$ , of the ethanedioic acid in FA 1.

$M_r$ = ..............................

1M Similar

(v)

Calculate the value of $x$ in $(\text{COOH})_2 \cdot x\text{H}_2\text{O}$ .

Show your working.

$x$ = ..............................

1M Similar

(d)

Explain why it is necessary to add FA 3 in each titration.

1M Similar

Q2Medium-EasyManipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and Evaluation

Hydrated zinc sulfate has the formula $\text{ZnSO}_4 \cdot y\text{H}_2\text{O}$ where $y$ is an integer.

Hydrated zinc sulfate decomposes when heated, losing only its water of crystallisation and becoming anhydrous.

You will determine the value of $y$ in $\text{ZnSO}_4 \cdot y\text{H}_2\text{O}$ by heating the hydrated salt until it becomes anhydrous.

FA 4 is hydrated zinc sulfate, $\text{ZnSO}_4 \cdot y\text{H}_2\text{O}$ .

(a)

Method

Weigh the crucible with its lid. Record the mass in the space for Results.
Add between $3.20 \text{ g}$ and $3.40 \text{ g}$ of FA 4 to the crucible.
Weigh the crucible with its lid and FA 4. Record the mass.
Place the crucible on the pipeclay triangle. Gently heat the crucible and contents for approximately 2 minutes with the lid on.
Remove the lid. Heat the crucible and contents strongly for approximately 4 minutes.
Replace the lid and leave the crucible and residue to cool for at least 5 minutes.

You may wish to begin work on Question 3 while the crucible is cooling.

Weigh the crucible with its lid and its contents. Record the mass.
Remove the lid. Heat the crucible strongly for approximately 3 minutes.
Replace the lid and leave the crucible and residue to cool for at least 5 minutes.
Weigh the crucible with its lid and its contents. Record the mass.
Calculate the mass of FA 4 used and the mass of residue obtained. Record the masses.

Results

5M Similar

DifficultyMedium-Easy

Worked solution

Answer

Results table (example readings — record your own):

Heading	Mass / g
(Mass) crucible + lid	20.00
(Mass) crucible + lid + FA 4	23.25
(Mass) crucible + lid + residue after first heating	21.85
(Mass) crucible + lid + residue after second heating	21.83
(Mass) FA 4 used	3.25
(Mass) residue	1.83

All four weighings are recorded to the same number of decimal places (2 dp). The reading after the second heating (21.83 g) is within +0.02 / −0.05 of the reading after the first heating (21.85 g), confirming constant mass.

Mass of FA 4 = 23.25 − 20.00 = 3.25 g
Mass of residue = 21.83 − 20.00 = 1.83 g

Final answer

Example readings: FA 4 = 3.25 g, residue = 1.83 g (candidate-dependent)

Detailed explanation

Background Concept

Water of crystallisation is water chemically bound within the crystal lattice of a hydrated salt. When a hydrated salt such as zinc sulfate is heated, this water is driven off as steam, leaving the anhydrous salt. The mass lost on heating equals the mass of water of crystallisation. By comparing the mass of the anhydrous salt with the mass of water lost, the mole ratio can be found and hence the formula determined. For reliable results, heating must continue until constant mass is reached — two consecutive weighings that agree within a small tolerance show that all water has been removed.

Understanding the Question

This part asks you to carry out the practical procedure and record the data correctly. The five marks are for: (1) a properly headed results table with units, (2) four weighings recorded to a consistent number of decimal places, (3) correct subtraction to find the mass of FA 4 and the residue, and (4–5) the accuracy of the result judged against the theoretical mass ratio. The readings themselves are your own — there is no single correct value.

Approach

Set up the apparatus: crucible with lid on a pipeclay triangle over a tripod, heated with a Bunsen burner. Weigh the empty crucible with its lid. Add 3.20–3.40 g of FA 4 and reweigh. Heat gently with the lid on, then strongly with the lid off, allow to cool, and reweigh. Repeat the strong heating and reweighing until the mass is constant. Record all four weighings to the same number of decimal places (2 or 3 dp). Subtract to find the mass of FA 4 and the mass of residue.

Step-by-Step Reasoning

The six headings required, each with a unit:

(Mass) crucible + lid / g
(Mass) crucible + lid + FA 4 / g
(Mass) crucible + lid + residue after first heating / g
(Mass) crucible + lid + residue after second heating / g
(Mass) FA 4 used / g
(Mass) residue / g

All four weighings must be recorded to the same number of decimal places — either all to 2 dp or all to 3 dp. The reading after the second heating must lie within +0.02 and −0.05 of the reading after the first heating; this is the constant-mass check that proves all water of crystallisation has been driven off. The mass of FA 4 must be between 2.90 and 3.40 g.

Using the example readings:

crucible + lid = 20.00 g
crucible + lid + FA 4 = 23.25 g
crucible + lid + residue after first heating = 21.85 g
crucible + lid + residue after second heating = 21.83 g

Mass of FA 4 = 23.25 − 20.00 = 3.25 g
Mass of residue = 21.83 − 20.00 = 1.83 g

The mass ratio = mass FA4 / mass residue = 3.25 / 1.83 = 1.78, which matches the theoretical value exactly. This earns both accuracy marks (IV and V).

Key Takeaways

Record all readings to the same number of decimal places.
Heat to constant mass — two consecutive weighings agreeing within tolerance proves all water is gone.
Label every row with a quantity and a unit.
The accuracy of the result is judged by the mass ratio, not by any single reading.

Common Mistakes

Forgetting units in the headings — the unit (g) must appear for every row.
Recording weighings to different decimal places (e.g. 20.0 and 23.25).
Not heating to constant mass (only one heating), so the residue still contains water and y comes out too high.
Using the wrong subtraction, e.g. subtracting the residue mass from the FA 4 mass instead of subtracting the empty crucible mass.
Weighing the crucible without its lid.

Things to Be Careful About

The lid is included in every weighing — never weigh the crucible without its lid.
Wait for the crucible to cool before weighing; hot crucibles give false low readings because of convection currents in the balance.
The mass after the second heating must be within +0.02/−0.05 of the first heating reading — this is the constant-mass check.

Techniques used

set up crucible and pipeclay triangle apparatusrecord weighings to consistent decimal placescalculate masses by subtractionheat to constant mass

(b)

Calculations

(i)

Calculate the amount, in mol, of anhydrous zinc sulfate residue formed in the decomposition of FA 4.

amount of $\text{ZnSO}_4$ = .............................. mol

Calculate the amount, in mol, of water of crystallisation lost.

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

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Molar mass of ZnSO4 = 65.4 + 32.1 + 4(16.0) = 161.5 $\text{g mol}^{-1}$

n(\text{ZnSO}_4) = \frac{m(\text{residue})}{161.5} = \frac{1.83}{161.5} = 0.0113 \text{ mol}

Mass of water lost = 3.25 − 1.83 = 1.42 g

n(\text{H}_2\text{O}) = \frac{1.42}{18} = 0.0789 \text{ mol}

Answer

amount of ZnSO4 = 0.0113 mol
amount of H2O = 0.0789 mol

Final answer

ZnSO4 = 0.0113 mol; H2O = 0.0789 mol (using example masses)

Detailed explanation

Background Concept

The amount of a substance in moles is found from the relationship $n = m / M$ , where $m$ is the mass in grams and $M$ is the molar mass in $\text{g mol}^{-1}$ . The molar mass of anhydrous zinc sulfate ZnSO4 is the sum of the relative atomic masses of its atoms: Zn (65.4) + S (32.1) + 4 × O (16.0) = 161.5 $\text{g mol}^{-1}$ . The molar mass of water is 18 $\text{g mol}^{-1}$ (2 × 1.0 + 16.0).

Understanding the Question

You are given the mass of the residue (anhydrous ZnSO4) and the mass of FA 4 (the hydrated salt). The mass of water lost during heating equals the mass of FA 4 minus the mass of the residue. You must convert both masses to amounts in mol.

Approach

Use $n = m / M$ for each substance. For ZnSO4, use the residue mass and $M = 161.5$ $\text{g mol}^{-1}$ . For water, first find the mass lost by subtracting the residue mass from the FA 4 mass, then divide by 18 $\text{g mol}^{-1}$ .

Step-by-Step Reasoning

Using the example masses: residue = 1.83 g, FA 4 = 3.25 g.

Amount of ZnSO4 = 1.83 / 161.5 = 0.01133... = 0.0113 mol (3 sf)

Mass of water lost = 3.25 − 1.83 = 1.42 g
Amount of H2O = 1.42 / 18 = 0.0789 mol (3 sf)

The mark scheme requires answers to 2–4 significant figures. Both values are quoted to 3 sf here.

Key Takeaways

$n = m / M$ is the fundamental conversion between mass and amount.
The mass of water lost is found by difference: hydrated mass − anhydrous mass.

Common Mistakes

Using the mass of FA 4 (3.25 g) instead of the residue mass when calculating the amount of ZnSO4.
Forgetting to subtract to find the mass of water lost.
Using the wrong molar mass, e.g. including water in the molar mass of ZnSO4.

Things to Be Careful About

Molar mass of ZnSO4 = 161.5 $\text{g mol}^{-1}$ — use the precise value, not a rounded 161.
Give answers to 2–4 significant figures as the mark scheme requires.
Include the unit mol.

Techniques used

calculate moles from mass and molar massuse molar mass of anhydrous zinc sulfateuse molar mass of water

(ii)

Calculate the value of $y$ in the formula $\text{ZnSO}_4 \cdot y\text{H}_2\text{O}$ .

Show your working.

$y$ = ..............................

2M Similar

(c)

A student suggests using this thermal decomposition method to investigate the number of moles of water of crystallisation in hydrated ethanedioic acid. The teacher says that this method is unsuitable.

Suggest why this method is unsuitable.

1M Similar

Q3Medium-HardQualitative Analysis

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)

A bottle labelled FA 5 is thought to contain hydrated zinc sulfate. It would therefore contain zinc ions and sulfate ions as well as water of crystallisation.

(i)

Devise and carry out tests to investigate whether zinc ions, sulfate ions and water of crystallisation are present.

Record the tests you carry out and the observations you see in the space provided.

5M Similar

DifficultyMedium

Worked solution

Answer

Heat a few crystals of solid FA 5 in a boiling tube. Observe condensation of water droplets / steam / water vapour on the cooler parts of the tube.
Make an aqueous solution of FA 5 by dissolving the solid in distilled water.
To a portion of the solution, add aqueous ammonia dropwise, then add excess. Observe a white precipitate that is insoluble in excess aqueous ammonia.
To a separate portion of the solution, add aqueous barium chloride (or barium nitrate). Observe a white precipitate.
Add dilute hydrochloric acid (or nitric acid) to the white precipitate; it remains insoluble. (Alternatively, add acidified potassium manganate(VII) to the solution; the purple colour is not decolourised.)

Final answer

See working above for the sequence of tests and observations confirming H2O, SO4^2-, and Mg^2+ (not Zn^2+).

Detailed explanation

Background Concept

Qualitative analysis of unknown substances requires a systematic approach: first testing for water of crystallisation, then preparing an aqueous solution for ion tests. Cations are identified by their reactions with aqueous ammonia, while anions like sulfate are identified by precipitation with barium salts. It is crucial to confirm the identity of the anion (e.g., distinguishing sulfate from sulfite) and to observe the behaviour of precipitates in excess reagent to differentiate between similar cations like Mg^2+ and Zn^2+.

Understanding the Question

The student is given FA 5, which is labelled as hydrated zinc sulfate, but the actual contents must be determined. The task is to devise and carry out tests to investigate the presence of zinc ions, sulfate ions, and water of crystallisation. The student must record observations for each test and then use these to complete a table indicating which species are actually present.

Approach

Water of crystallisation: Heat the solid and look for condensation.
Solution preparation: Dissolve the solid in water to test for ions.
Cation test (Zn^2+ vs Mg^2+): Add aqueous ammonia. Zn^2+ forms a white ppt soluble in excess; Mg^2+ forms a white ppt insoluble in excess. The observation will rule out Zn^2+ and confirm Mg^2+.
Anion test (SO4^2-): Add barium chloride to form a white ppt of BaSO4.
Confirmation: Distinguish sulfate from sulfite by adding acid (sulfite ppt dissolves with effervescence, sulfate does not) or acidified KMnO4 (sulfite decolourises it, sulfate does not).

Step-by-Step Reasoning

M1: Heating the solid FA 5 releases water of crystallisation. The observation is condensation (water droplets, steam, or water vapour) on the cooler upper parts of the boiling tube.
M2: To test for ions, the solid must be dissolved in distilled water to form an aqueous solution.
M3: Adding aqueous NH3 to a portion of the solution produces a white precipitate of Mg(OH)2. Unlike Zn(OH)2, Mg(OH)2 is insoluble in excess NH3. This observation rules out Zn^2+ (which would form a soluble complex with excess NH3) and confirms Mg^2+.
M4: Adding aqueous BaCl2 (or Ba(NO3)2) to another portion produces a white precipitate of BaSO4, indicating the presence of SO4^2-.
M5: To ensure the anion is sulfate and not sulfite (which also forms a white BaSO3 ppt), add dilute HCl or HNO3. BaSO3 would dissolve with effervescence (SO2 gas), while BaSO4 remains insoluble. Alternatively, acidified KMnO4(aq) would be decolourised by sulfite but remains purple with sulfate.

Key Takeaways

Always test for water of crystallisation by heating before dissolving the solid.
Ammonia test is a classic method to distinguish Mg^2+ (insoluble white ppt) from Zn^2+ (soluble white ppt in excess).
Sulfate confirmation requires showing the barium precipitate is insoluble in acid, or that it does not reduce acidified permanganate.

Common Mistakes

Writing 'no change' for the ammonia test when a white precipitate forms.
Claiming the white precipitate is soluble in excess ammonia (this would indicate Zn^2+, which is incorrect here).
Forgetting to add acid to the barium precipitate to rule out sulfite or carbonate.
Not specifying that the precipitate is white.

Things to Be Careful About

State the reagent clearly (e.g., 'aqueous ammonia', not just 'ammonia').
Describe the precipitate colour (white) and its solubility in excess reagent.
For the sulfate confirmation, explicitly state that the precipitate is 'insoluble' in acid, or that KMnO4 'remains purple'.

Techniques used

test for water of crystallisation by heatingdissolve solid to make aqueous solutionidentify cation using ammonia testidentify sulfate anion using barium chloride testdistinguish sulfate from sulfite using acid or permanganate

(ii)

Use your observations in (a)(i) to complete Table 3.1 to show whether each species is present in FA 5.

Use a tick (✓) if the species is present.
Use a cross (✗) if the species is not present.

Table 3.1

Species	Present
$\text{Zn}^{2+}$
$\text{SO}_4^{2-}$
$\text{H}_2\text{O}$

1M Similar

DifficultyEasy

Worked solution

Answer

Species	Present
Zn^2+	✗
SO4^2-	✓
H2O	✓

Final answer

Zn^2+: ✗, SO4^2-: ✓, H2O: ✓

Detailed explanation

Background Concept

Based on the observations from part (a)(i), we can deduce the actual composition of FA 5. The heating test confirmed water. The ammonia test ruled out Zn^2+ and confirmed Mg^2+. The barium chloride test confirmed SO4^2-. Therefore, FA 5 is magnesium sulfate heptahydrate (MgSO4•7H2O), not zinc sulfate.

Understanding the Question

The student must use the observations from part (a)(i) to complete Table 3.1, indicating whether Zn^2+, SO4^2-, and H2O are present in FA 5 using a tick (✓) or a cross (✗).

Approach

Zn^2+: The ammonia test showed a white ppt insoluble in excess, which is characteristic of Mg^2+, not Zn^2+. Therefore, Zn^2+ is absent (✗).
SO4^2-: The barium chloride test gave a white ppt insoluble in acid, confirming SO4^2- is present (✓).
H2O: Heating the solid produced condensation, confirming water of crystallisation is present (✓).

Step-by-Step Reasoning

Zn^2+: If Zn^2+ were present, the white Zn(OH)2 precipitate would dissolve in excess aqueous ammonia to form a colourless [Zn(NH3)4]2+ complex. Since it did not, Zn^2+ is not present.
SO4^2-: The formation of a white precipitate with Ba^2+ that is insoluble in dilute acid is the definitive test for sulfate ions.
H2O: Condensation on the cooler parts of the test tube upon heating is the standard test for water of crystallisation in hydrated salts.

Key Takeaways

Qualitative analysis results must be directly linked to the specific tests performed.
A negative result for a suspected ion (like Zn^2+) is just as important as a positive result and must be recorded correctly.

Common Mistakes

Assuming the label on the bottle is correct and ticking Zn^2+.
Forgetting that water of crystallisation is 'H2O' and not 'OH-'.

Things to Be Careful About

Ensure the tick and cross symbols are clear and unambiguous.
Only include the species asked for in the table.

Techniques used

interpret test observations to identify speciesuse tick and cross notation for presence/absence

(b)

You are provided with solid FA 6.

(i)

Heat a few crystals of FA 6 in a hard-glass test-tube until no further gas is evolved. Record all your observations.

Leave the test-tube until it is cool.

Keep the cooled residue for use in (b)(ii).

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

The solid is purple and jumps / moves around as it is heated.
A black residue is left in the test tube.
A glowing splint is inserted into the test tube and relights / glows more brightly.

Final answer

Purple solid jumps, black residue forms, glowing splint relights.

Detailed explanation

Background Concept

Potassium manganate(VII), KMnO4, is a purple solid that undergoes thermal decomposition when heated strongly. It breaks down into potassium manganate(VI) (K2MnO4, green), manganese(IV) oxide (MnO2, black), and oxygen gas. The oxygen can be tested with a glowing splint, which will relight.

Understanding the Question

The student is given solid FA 6 (KMnO4) and must heat it until no further gas is evolved, record all observations, and leave the residue for the next part.

Approach

Note the initial colour of the solid.
Observe physical changes during heating (jumping/melting).
Note the colour of the final solid residue.
Test the evolved gas with a glowing splint to identify it as oxygen.

Step-by-Step Reasoning

Initial colour: KMnO4 is a deep purple solid.
During heating: The solid may jump or move around in the tube as oxygen gas is rapidly evolved. It may also appear to melt or decompose vigorously.
Residue: The solid residue is a mixture of K2MnO4 (green) and MnO2 (black). The overall appearance is often described as a black or dark residue (the black MnO2 masks the green K2MnO4 in the solid state).
Gas test: Oxygen supports combustion. Inserting a glowing splint into the test tube will cause it to relight or glow more brightly.

Key Takeaways

Thermal decomposition of KMnO4 produces O2, which is identified by the glowing splint test.
Always record observations at different stages (initial, during heating, final residue, gas test).

Common Mistakes

Saying the residue is 'white' or 'colourless'.
Forgetting to describe the initial colour of the solid.
Not performing or describing the glowing splint test.

Things to Be Careful About

The question asks for 'all your observations', so include the initial colour, physical changes, residue colour, and gas test.
Use precise language: 'glowing splint relights', not 'splint burns'.

Techniques used

thermal decomposition of a solidtest for oxygen using a glowing splint

(ii)

To the cooled residue from (b)(i), add approximately $3 \text{ cm}$ depth of distilled water and stir. Filter the solution formed into a test-tube.

The colour of the solution is .............................. .

1M Similar

(c)

You are provided with aqueous solutions FA 7 and FA 8 and with solid FA 9.

FA 7 is an aqueous solution of FA 6.
FA 7, FA 8 and FA 9 contain compounds which all have one metal that is the same but which may be in different oxidation states.

(i)

Carry out the following tests on FA 7, FA 8 and FA 9 and record your observations in Table 3.2. For each test use a $1 \text{ cm}$ depth of a solution or a spatula measure of solid.

Table 3.2

4M Similar

DifficultyMedium-Hard

Worked solution

Answer

Test 1: Add hydrogen peroxide

FA 7: Bubbles / effervescence; purple solution turns colourless; gas relights a glowing splint.
FA 8: No change.
FA 9: Bubbles / effervescence; gas relights a glowing splint.

Test 2: Add aqueous sodium hydroxide, then leave to stand

FA 8: Off-white precipitate forms; turns brown on standing.
FA 9: No change.

Test 3: Add aqueous iron(II) sulfate

FA 7: Purple solution turns colourless / yellow solution formed.
FA 8: No change.

Final answer

See table above for observations.

Detailed explanation

Background Concept

This question explores the chemistry of manganese in different oxidation states: +7 (MnO4- in FA 7), +2 (Mn^2+ in FA 8), and +4 (MnO2 in FA 9).

MnO4- (+7): Strong oxidising agent, purple solution. Reduced to Mn^2+ (colourless) or Mn^3+ (yellow) by reducing agents like H2O2 or Fe^2+.
Mn^2+ (+2): Pale pink / colourless solution. Forms a white/off-white precipitate of Mn(OH)2 with NaOH, which is rapidly oxidised by air to brown MnO(OH)2 or MnO2.
MnO2 (+4): Black insoluble solid. Acts as a catalyst for the decomposition of H2O2 into O2 and H2O.

Understanding the Question

The student must carry out three tests on FA 7 (acidified KMnO4), FA 8 (MnSO4/MnCl2), and FA 9 (MnO2) and record the observations in Table 3.2. Some cells are crossed out (FA 7 for Test 2, FA 9 for Test 3) because the tests are not applicable or would not give a meaningful observation.

Approach

Test 1 (H2O2): MnO4- oxidises H2O2 to O2 (effervescence, glowing splint relights, purple to colourless). Mn^2+ does not react. MnO2 catalyses H2O2 decomposition (effervescence, glowing splint relights).
Test 2 (NaOH): Mn^2+ forms Mn(OH)2 (white/off-white ppt), which oxidises in air to brown. MnO2 does not react with NaOH.
Test 3 (FeSO4): MnO4- oxidises Fe^2+ to Fe^3+ (yellow/brown) and is reduced to Mn^2+ (colourless), so the purple colour disappears. Mn^2+ does not react.

Step-by-Step Reasoning

Test 1, FA 7: H2O2 is oxidised by MnO4- in acidic solution: 2MnO4- + 5H2O2 + 6H+ -> 2Mn^2+ + 5O2 + 8H2O. Observation: effervescence (O2 gas), purple solution turns colourless (Mn^2+ formed), O2 relights glowing splint.
Test 1, FA 8: Mn^2+ is already in its lowest common oxidation state and does not react with H2O2 under these conditions. Observation: no change.
Test 1, FA 9: MnO2 catalyses the decomposition of H2O2: 2H2O2 -> 2H2O + O2. Observation: effervescence, O2 relights glowing splint. (No colour change as MnO2 is a solid catalyst and remains black, but the solution may bubble vigorously).
Test 2, FA 8: Mn^2+ + 2OH- -> Mn(OH)2 (s). Mn(OH)2 is a white or off-white precipitate. On standing in air, it is oxidised: 2Mn(OH)2 + O2 -> 2MnO(OH)2 (brown). Observation: off-white ppt, turns brown on standing.
Test 2, FA 9: MnO2 is a basic oxide but does not react with dilute NaOH. Observation: no change (black solid remains).
Test 3, FA 7: MnO4- oxidises Fe^2+ to Fe^3+: MnO4- + 5Fe^2+ + 8H+ -> Mn^2+ + 5Fe^3+ + 4H2O. Fe^3+ is yellow/brown in solution, and Mn^2+ is colourless. Observation: purple solution turns colourless or yellow (depending on concentration of Fe^3+ formed).
Test 3, FA 8: Mn^2+ does not react with Fe^2+. Observation: no change.

Key Takeaways

Manganese exhibits rich redox chemistry across oxidation states +2, +4, and +7.
MnO4- is a powerful oxidising agent that is decolourised when reduced to Mn^2+.
Mn^2+ precipitates as a white hydroxide that is easily oxidised in air to a brown product.
MnO2 acts as a heterogeneous catalyst for H2O2 decomposition.

Common Mistakes

Saying FA 7 turns 'brown' in Test 1 (it turns colourless as Mn^2+ is formed; Fe^3+ in Test 3 may be yellow/brown, but Mn^2+ is colourless).
Forgetting to mention the glowing splint test for the gas in Test 1.
Describing the Mn(OH)2 precipitate as 'white' without noting it turns 'brown on standing'.
Saying FA 9 reacts with NaOH or FeSO4 (it does not; it is unreactive towards these reagents in this context).

Things to Be Careful About

Follow the table structure exactly; do not write observations in crossed-out cells.
For Test 2, separate the immediate observation (off-white ppt) from the observation on standing (brown ppt).
Use precise colour descriptions: 'purple to colourless', 'off-white', 'brown', 'yellow'.

Techniques used

observe reactions of permanganate with hydrogen peroxideobserve reactions of manganese(II) with sodium hydroxideobserve catalytic decomposition of hydrogen peroxide by manganese(IV) oxideobserve redox reaction between permanganate and iron(II)

(ii)

Suggest the identity of the metal in FA 6/FA 7, FA 8 and FA 9.

The metal is .............................. .

1M Similar

(iii)

Complete Table 3.3 to suggest the oxidation state of the metal in FA 6/FA 7 and FA 8.

Table 3.3

	FA 6/FA 7	FA 8
oxidation state

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

	FA 6 / FA 7	FA 8
oxidation state	+7	+2

Final answer

FA 6/FA 7: +7, FA 8: +2

Detailed explanation

Background Concept

Oxidation states can be determined from the chemical formulae of the compounds. For manganese:

KMnO4 (potassium manganate(VII)): K is +1, O is -2. 1 + x + 4(-2) = 0 => x = +7.
MnSO4 or MnCl2 (manganese(II) sulfate/chloride): SO4 is -2, Cl is -1. Mn must be +2 to balance.

Understanding the Question

The student must complete Table 3.3 to state the oxidation state of manganese in FA 6/FA 7 and in FA 8.

Approach

FA 6 and FA 7 contain the MnO4- ion (from KMnO4). The oxidation state of Mn in MnO4- is +7.
FA 8 contains Mn^2+ ions (from MnSO4 or MnCl2). The oxidation state is +2.

Step-by-Step Reasoning

FA 6 / FA 7: The purple colour and oxidising behaviour identify this as potassium manganate(VII), KMnO4. In MnO4-, oxygen is -2, so Mn + 4(-2) = -1 => Mn = +7. The oxidation state is +7.
FA 8: The formation of Mn(OH)2 with NaOH and lack of reaction with H2O2 or Fe^2+ identifies this as a manganese(II) salt, e.g., MnSO4. The ion is Mn^2+, so the oxidation state is +2.

Key Takeaways

Oxidation states can be calculated from formulae using known oxidation states of common ions (O = -2, K = +1, SO4 = -2, Cl = -1).
The Roman numeral in the name of a compound (e.g., manganate(VII)) directly gives the oxidation state of the central metal.

Common Mistakes

Writing '7' instead of '+7'.
Confusing the oxidation state of Mn in MnO4- (+7) with Mn in MnO2 (+4).
Not including the '+' sign.

Things to Be Careful About

Always include the sign (+ or -) for oxidation states.
Ensure the oxidation state matches the correct substance (FA 6/FA 7 is +7, FA 8 is +2).

Techniques used

determine oxidation states from chemical formulae and reactions