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 for tests and observations · (a)(ii) Zn2+: ✗, SO4 2-: ✓, H2O: ✓ · (b)(i) Purple solid jumps, black residue, glowing splint relights · (b)(ii) (dark) green · (c)(i) See table for observations · (c)(ii) manganese (or Mn) · (c)(iii) FA 6/FA 7: +7, FA 8: +2

Chemistry 9701/31 — 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 $(\text{COOH})_2 \cdot \mathbf{x}\text{H}_2\text{O}$ where $\mathbf{x}$ 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 $\mathbf{x}$ in $(\text{COOH})_2 \cdot \mathbf{x}\text{H}_2\text{O}$ by titrating a solution containing ethanedioic acid with manganate(VII) ions.

FA 1 is $6.20 \text{ g dm}^{-3}$ aqueous ethanedioic acid, $(\text{COOH})_2 \cdot \mathbf{x}\text{H}_2\text{O}$ .
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 $2\text{--}3 \text{ cm}^3$ . 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

Rough titration

Initial reading = 0.00 cm³, final reading = 25.10 cm³, rough titre = 25.10 cm³.

Accurate titrations (representative example — your readings will differ)

Titration	Initial reading / cm³	Final reading / cm³	Titre / cm³
1	0.00	24.60	24.60
2	24.60	49.25	24.65
3	0.00	24.60	24.60

All burette readings recorded to the nearest 0.05 cm³; accurate titres concordant within 0.10 cm³ of each other.

Final answer

See working — candidate-dependent readings; representative example: rough titre 25.10 cm³, accurate titres 24.60, 24.65, 24.60 cm³

Detailed explanation

Background Concept

In a redox titration, the analyte (here ethanedioic acid, (COOH)₂) is oxidised and the titrant (manganate(VII) ion, MnO₄⁻) is reduced. The reaction between ethanedioic acid and manganate(VII) ions is slow at room temperature, so the mixture is heated to about 70 °C to increase the rate. The reaction also requires acidic conditions — H⁺ ions are consumed in the reduction of MnO₄⁻ to Mn²⁺ — hence sulfuric acid (FA 3) is added. The end-point is the first permanent pale pink colour, which appears when a tiny excess of MnO₄⁻ remains unreacted.

Understanding the Question

Part (a) asks you to carry out the titration and record your results. You must record: (i) the rough titration — both burette readings and the titre; (ii) at least two accurate titrations — initial and final burette readings for each; and (iii) the volume of FA 2 added (the titre) for each. All readings must be recorded to the nearest 0.05 cm³ (the precision of a burette), and the accurate titres must be concordant (within 0.10 cm³ of each other).

Approach

The marks in (a) are awarded for the quality of your data recording and the accuracy of your titres:

Record the rough titre (two readings + titre).
Perform at least two accurate titrations, recording initial and final readings for each.
Set up a table with correct headings and units (initial reading / cm³, final reading / cm³, titre / cm³).
Record all readings to 0.05 cm³.
Ensure your accurate titres agree within 0.10 cm³.
The accuracy marks depend on how close your mean titre is to the supervisor's value.

Step-by-Step Reasoning

The rough titration gives you an approximate end-point so you know roughly how much FA 2 to expect. For the accurate titrations, you add FA 2 drop-wise near the end-point. The first 2–3 cm³ may take several seconds to decolourise because the reaction is slow even at 70 °C. If the mixture turns brown, this indicates formation of MnO₂ (a brown solid) — this can happen if the mixture cools or if there is insufficient acid. Reheating to 70 °C may restore the colourless solution; if the brown colour persists, the titration must be restarted.

The end-point is the first permanent pale pink colour — this means all the ethanedioic acid has been oxidised and a tiny excess of MnO₄⁻ remains.

For a representative example:

Rough: initial 0.00, final 25.10, titre 25.10 cm³
Accurate 1: initial 0.00, final 24.60, titre 24.60 cm³
Accurate 2: initial 24.60, final 49.25, titre 24.65 cm³
Accurate 3: initial 0.00, final 24.60, titre 24.60 cm³

All readings are to 0.05 cm³. The accurate titres (24.60, 24.65, 24.60) are concordant within 0.10 cm³.

Key Takeaways

Burette readings are recorded to the nearest 0.05 cm³.
A rough titration precedes accurate titrations.
Concordant titres (within 0.10 cm³) are required for reliability.
The end-point of a manganate(VII) titration is the first permanent pale pink colour.

Common Mistakes

Recording readings to only 1 decimal place (e.g. 24.6 instead of 24.60) — the mark scheme requires 0.05 cm³ precision.
Not recording both initial and final readings.
Not including units in table headings.
Using the rough titre as an accurate titre.
Labelling the rough titre as an accurate result.

Things to Be Careful About

The rough titre is not used in the mean calculation.
The mark scheme checks concordance of the accurate titres — if your final accurate titre is more than 0.10 cm³ from another, you lose the mark.
Burette readings must be taken at eye level to avoid parallax error.

Techniques used

record burette readings to the nearest 0.05 cm³perform a rough titrationobtain concordant titres within 0.10 cm³record data in a table with correct headings and 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 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 $(\text{COOH})_2$ react with 2 mol $\text{MnO}_4^-$ .

\text{amount of }(\text{COOH})_2 = 4.924 \times 10^{-4} \times \frac{5}{2} = 1.231 \times 10^{-3} \text{ mol}

\text{concentration} = \frac{1.231 \times 10^{-3}}{25.0/1000} = 0.04924 \text{ mol dm}^{-3}

Answer

Amount of $(\text{COOH})_2$ = $1.23 \times 10^{-3} \text{ mol}$ (3 sf)

Concentration of $(\text{COOH})_2$ = $0.0492 \text{ mol dm}^{-3}$ (3 sf)

Final answer

Amount of (COOH)₂ = 1.23 × 10⁻³ mol; concentration = 0.0492 mol dm⁻³

Detailed explanation

Background Concept

The balanced equation gives the stoichiometric ratio: 5 mol (COOH)₂ react with 2 mol MnO₄⁻. So moles of (COOH)₂ = moles of MnO₄⁻ × 5/2. The concentration of FA 1 is then moles ÷ volume in dm³ (25.0 cm³ = 0.0250 dm³).

Understanding the Question

Two calculations: (1) moles of ethanedioic acid that reacted; (2) its concentration in FA 1.

Approach

Use the 5:2 ratio from the equation.
Divide moles by 0.0250 dm³ to get concentration.

Step-by-Step Reasoning

Moles of (COOH)₂ = 4.924 × 10⁻⁴ × 5/2 = 1.231 × 10⁻³ mol

Concentration = 1.231 × 10⁻³ / 0.0250 = 0.04924 mol dm⁻³ ≈ 0.0492 mol dm⁻³ (3 sf)

The 5/2 factor comes directly from the coefficients in the balanced equation: 5(COOH)₂ + 2MnO₄⁻ → ... The volume 25.0 cm³ is the pipetted volume of FA 1, converted to dm³ by dividing by 1000.

Key Takeaways

The stoichiometric ratio comes from the balanced equation.
concentration = amount / volume (dm³).

Common Mistakes

Using the wrong ratio (e.g. 2/5 instead of 5/2).
Forgetting to convert 25.0 cm³ to 0.0250 dm³.
Not giving the answer to 3–4 sf.

Things to Be Careful About

The mark scheme gives M1 for the 5/2 ratio and M2 for the concentration calculation.
Answer to 3–4 sf.

Techniques used

apply the 5:2 stoichiometric ratio from the balanced equationcalculate concentration from moles and volume in dm³

(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 $(\text{COOH})_2 \cdot \mathbf{x}\text{H}_2\text{O}$ .

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 $\text{ZnSO}_4 \cdot \mathbf{y}\text{H}_2\text{O}$ where $\mathbf{y}$ is an integer.

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

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

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

(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 - all six headings must include a unit (write / g in the heading):

Mass of crucible + lid / g
Mass of crucible + lid + FA4 / g
Mass of crucible + lid + residue after 1st heating / g
Mass of crucible + lid + residue after 2nd heating / g
Mass of FA4 / g
Mass of residue / g

Record all four weighings to the same number of decimal places (2 or 3 dp). The mass after the 2nd heating must be within +0.02 to -0.05 g of the mass after the 1st heating (constant mass).

Representative readings (your own will differ):

Crucible + lid = 20.00 g
Crucible + lid + FA4 = 23.30 g
After 1st heating = 21.86 g
After 2nd heating = 21.85 g

$\text{mass of FA4} = 23.30 - 20.00 = 3.30 \text{ g}$

$\text{mass of residue} = 21.85 - 20.00 = 1.85 \text{ g}$

$\text{mass ratio} = \frac{3.30}{1.85} = 1.78$ (theoretical value for $\text{ZnSO}_4 \cdot 7\text{H}_2\text{O}$ is 1.78)

Final answer

Candidate-dependent. Example: mass FA4 = 3.30 g, mass residue = 1.85 g, mass ratio = 1.78.

Detailed explanation

Background Concept

Hydrated salts such as $\text{ZnSO}_4 \cdot 7\text{H}_2\text{O}$ contain a fixed number of water molecules - called water of crystallisation - locked into the crystal lattice. On strong heating this water escapes as steam, leaving the anhydrous salt, provided the anhydrous salt is thermally stable. Since the only mass lost is water,

\text{mass of water lost} = \text{mass of hydrated salt} - \text{mass of anhydrous residue}

Heating to constant mass means repeating the heat-cool-weigh cycle until two successive residue masses agree, which confirms all the water has been driven off. Weighing by difference measures a sample indirectly as (container + sample) - (container), avoiding the difficulty of weighing the sample on its own. The theoretical mass ratio for $\text{ZnSO}_4 \cdot 7\text{H}_2\text{O}$ is $M_r(\text{hydrated}) : M_r(\text{anhydrous}) = (161.5 + 126) : 161.5 = 1.78$ .

Understanding the Question

Part (a) is the practical part - the candidate performs the experiment and records their OWN readings. Five marks are awarded for: (I) six correct headings with units; (II) the four weighings recorded to the same number of decimal places, with the reading after the 2nd heating within +0.02 to -0.05 g of the reading after the 1st heating; (III) correct subtractions to give mass of FA4 (between 2.90 and 3.40 g) and mass of residue; and (IV, V) accuracy - how close the candidate's mass ratio (mass FA4 divided by mass residue) is to the theoretical 1.78. Because it is a live practical, there is no single correct set of numbers - the marking rewards careful technique, sensible recording and accuracy.

Approach

Set out the six required rows with units in the header, record the four weighings consistently to 2 or 3 decimal places, subtract the empty-crucible mass to obtain the mass of FA4 and the mass of residue, confirm that the first and second heating residues agree (constant mass), and finally compute the mass ratio to judge accuracy against 1.78.

Step-by-Step Reasoning

Headings and units: the marking rewards six labelled rows, each carrying the unit in the heading (... / g). Writing 'g' once in the column heading covers every entry; the mark is lost if units are scattered or missing.
Record all four weighings to the same number of decimal places - either two or three. Mixing 0.1 g and 0.01 g readings loses the mark.
Both heating stages are needed: the first strong heating removes most of the water rapidly; the second heating (and re-weighing) checks that no further mass is lost - this is the constant-mass check. If the second reading has dropped by more than about 0.05 g, more heating is needed.
Cool before weighing: a hot crucible warms the surrounding air and sets up convection currents that make the balance reading too low. That is why the method says to cool for at least 5 minutes before each weighing.
Subtractions (using the representative readings):

$\text{mass FA4} = 23.30 - 20.00 = 3.30 \text{ g}$

$\text{mass residue} = 21.85 - 20.00 = 1.85 \text{ g}$

Accuracy check: $\text{mass ratio} = 3.30/1.85 = 1.78$ , matching the theoretical value - this would earn both accuracy marks (IV if the ratio is between 1.51 and 2.05, V if between 1.60 and 1.96).

Key Takeaways

Weighing by difference; heating to constant mass; table conventions (quantity with unit in the heading, consistent decimal places); and the accuracy check via a theoretical ratio. These ideas transfer directly to any gravimetric practical.

Common Mistakes

Inconsistent decimal places across the four readings (mixing 2 dp and 3 dp, or writing only 0.1 g precision).
Missing units in column headings.
Weighing the crucible while still hot, giving a falsely low mass.
Only one heating stage, so the residue still contains water.
Forgetting that the ratio compares masses of the hydrated and anhydrous salt, not just any two readings.

Things to Be Careful About

The mark for readings requires the residue masses after the 1st and 2nd heating to agree closely (within +0.02/-0.05 g). Keep the lid on while cooling so the anhydrous salt does not re-absorb moisture. Use the 2nd-heating residue mass for the calculations, since it is the constant-mass value.

Techniques used

weigh by differenceheat to constant massrecord measurements in a results table

(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

$M_r(\text{ZnSO}_4) = 65.4 + 32.1 + 4(16.0) = 161.5$

\text{amount ZnSO}_4 = \frac{\text{mass of residue}}{161.5} = \frac{1.85}{161.5} = 0.0115 \text{ mol}

$\text{mass of water lost} = 3.30 - 1.85 = 1.45 \text{ g}$

\text{amount H}_2\text{O} = \frac{1.45}{18.0} = 0.0806 \text{ mol}

Answer

amount of ZnSO4 = 0.0115 mol; amount of H2O = 0.0806 mol

Final answer

0.0115 mol; 0.0806 mol

Detailed explanation

Background Concept

The amount of substance in moles is found from mass and molar mass:

n = \frac{m}{M_r}

The anhydrous residue is pure $\text{ZnSO}_4$ ; its molar mass is $M_r(\text{ZnSO}_4) = 65.4 + 32.1 + 4(16.0) = 161.5$ . The water driven off has $M_r = 18.0$ . Because only water is lost, the mass of water lost is the difference between the mass of FA4 and the mass of the residue.

Understanding the Question

Given the mass of FA4 and the mass of anhydrous residue from part (a), convert each to moles: the residue gives the amount of $\text{ZnSO}_4$ ; the difference (FA4 minus residue) gives the mass - and hence amount - of water lost. Two method marks are available, one per mole calculation; the answers must each be quoted to 2-4 significant figures.

Approach

Compute the molar mass of anhydrous zinc sulfate; divide the residue mass by it to obtain the amount of $\text{ZnSO}_4$ ; subtract to find the water mass; then divide the water mass by 18.0.

Step-by-Step Reasoning

Using the representative values (mass FA4 = 3.30 g, mass residue = 1.85 g):

$M_r(\text{ZnSO}_4) = 65.4 + 32.1 + 64.0 = 161.5$ .
$\text{amount ZnSO}_4 = 1.85/161.5 = 0.0115 \text{ mol}$ (3 s.f.).
$\text{mass of water lost} = 3.30 - 1.85 = 1.45 \text{ g}$ .
$\text{amount H}_2\text{O} = 1.45/18.0 = 0.0806 \text{ mol}$ .

Both answers are within 2-4 s.f. as required. If the candidate's readings differ, the method is identical - only the numbers change.

Key Takeaways

Mole conversion from mass; finding a lost mass by difference; quoting answers to an appropriate number of significant figures.

Common Mistakes

Using the molar mass of the hydrated salt (287.5) instead of the anhydrous 161.5 for the zinc sulfate - the residue is anhydrous.
Forgetting to subtract to find the water mass before dividing by 18.
Quoting mole answers to only 1 s.f., losing the significant-figures requirement.
Reversing the two divisions (dividing by FA4 mass instead of residue mass).

Things to Be Careful About

Use the 2nd-heating residue mass (the constant-mass value). Keep units - g and mol - explicit. The mark scheme accepts 2-4 s.f., so both 0.01146 mol and 0.0115 mol would score.

Techniques used

calculate molar massconvert mass to amount in molcalculate mass of water lost by difference

(ii)

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

Show your working.

y = ..............................

2M Similar

DifficultyMedium-Easy

Worked solution

Working

y = \frac{\text{amount H}_2\text{O}}{\text{amount ZnSO}_4} = \frac{0.0806}{0.0115} = 7.0

y is an integer.

Answer

y = 7

( $\text{ZnSO}_4 \cdot 7\text{H}_2\text{O}$ )

Final answer

Detailed explanation

Background Concept

The formula $\text{ZnSO}_4 \cdot y\text{H}_2\text{O}$ means each formula unit of anhydrous salt carries $y$ water molecules. The hydrated substance therefore contains $y$ moles of water for every 1 mole of $\text{ZnSO}_4$ , so

y = \frac{\text{amount of H}_2\text{O}}{\text{amount of ZnSO}_4}

Because a formula ratio is a ratio of whole particles, $y$ must come out as a (small) integer.

Understanding the Question

Divide the two mole values computed in (b)(i). The first mark (M1) is for the correct ratio formula; the second mark (M2) is for evaluating it and quoting $y$ as an integer.

Approach

Divide $n(\text{H}_2\text{O})$ by $n(\text{ZnSO}_4)$ , recognise that the result is very close to a whole number, and quote that integer.

Step-by-Step Reasoning

y = \frac{0.0806}{0.0115} = 7.0

Since $y$ must be an integer, $y = 7$ , i.e. the salt is $\text{ZnSO}_4 \cdot 7\text{H}_2\text{O}$ (zinc sulfate heptahydrate). If the ratio had come out slightly off (e.g. 7.2), the small discrepancy would be experimental error and it should still be rounded to the nearest integer - heating errors always make measured ratios slightly imperfect.

Key Takeaways

The hydration number is a mole ratio rounded to an integer; mole values feed into a formula ratio exactly like $y$ in a molecular formula.

Common Mistakes

Dividing the MASSES instead of the mole amounts - this gives a completely different, false value.
Quoting 7.0 or 7.03 instead of the integer 7.
Reversing the division ( $n(\text{ZnSO}_4)/n(\text{H}_2\text{O}) = 0.14$ instead of 7).

Things to Be Careful About

The marks are method followed by integer answer: any correct ratio method earns M1 even if arithmetic is slightly off. Only the correctly evaluated expression with the integer quoted earns M2.

Techniques used

calculate the mole ratio of water to anhydrous saltround the ratio to an integer

(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

DifficultyMedium-Easy

Worked solution

Answer

The method is unsuitable because ethanedioic acid (or its anhydrous form) decomposes on heating / is flammable. It does not simply lose water of crystallisation, so the mass loss would not be due to water alone.

Final answer

Ethanedioic acid decomposes on heating (or is flammable), so it does not simply lose water of crystallisation.

Detailed explanation

Background Concept

The thermal method works only when a substance loses water of crystallisation without itself reacting further. Zinc sulfate is an ionic salt that is stable at the temperature needed to drive off its water, so the residue is pure anhydrous salt. Ethanedioic acid (oxalic acid), $\text{HOOC-COOH}$ , is an organic acid. When heated strongly it decomposes - producing carbon monoxide, carbon dioxide and other products (it may char or give off flammable gases) - and it is flammable. So heating it does more than remove water of crystallisation.

Understanding the Question

The teacher rejects the method for hydrated ethanedioic acid. The question asks why. One mark is available; the credited answers are that the (anhydrous) ethanedioic acid would decompose on heating, OR that it is flammable. Give one of these two specific points.

Approach

Ask: does this solid lose only water on heating, or does the solid itself react? Recall that ethanedioic acid decomposes (or catches fire) when heated, so the observed mass change could not be attributed to loss of water of crystallisation alone.

Step-by-Step Reasoning

The entire basis of the zinc sulfate experiment was that the only mass loss is water. For ethanedioic acid, heating would decompose the acid itself (or ignite it), so the residue would not be pure anhydrous ethanedioic acid, and the mass lost could not be equated with water of crystallisation. In short, ethanedioic acid decomposes on heating and is flammable, so strong heating is both invalid and hazardous. Either the decomposition point or the flammability point scores the mark.

Key Takeaways

Thermally decomposing (or flammable) solids cannot be analysed by loss-on-drying methods. The validity of a method depends on the chemistry of the substance, not just the equipment.

Common Mistakes

Vague answers such as 'it melts' or 'it is not a salt' - these are not credited.
Giving only 'it is dangerous' without specifying decomposition or flammability.
Saying 'the water would not be removed' - the point is that the acid itself reacts, not that the water stays behind.

Things to Be Careful About

The mark scheme wants the specific property: decomposes on heating OR is flammable. State clearly that heating does more than drive off water - the acid itself breaks down.

Techniques used

recognise that ethanedioic acid decomposes on heatingevaluate the suitability of a thermal method

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

Test for water of crystallisation:

Heat a sample of solid FA 5 in a boiling tube.
Observation: Condensation / droplets of water / steam form on the cooler parts of the tube.

Test for ions (prepare solution):

Make an aqueous solution of FA 5 by dissolving it in distilled water.

Test for cation:

Add aqueous ammonia ( $\text{NH}_3$ ) to a portion of the solution.
Observation: A white precipitate forms, which is insoluble in excess aqueous ammonia.

Test for sulfate anion:

Add aqueous barium chloride (or barium nitrate) to a portion of the solution.
Observation: A white precipitate forms.
Add dilute hydrochloric acid (or nitric acid) to the precipitate.
Observation: The white precipitate remains insoluble.
(Alternatively: add acidified $\text{KMnO}_4$ to the solution; it remains purple / is not decolourised.)

Final answer

See working for tests and observations

Detailed explanation

Background Concept

Qualitative analysis involves using chemical tests to identify the ions present in an unknown substance. Water of crystallisation is identified by heating the solid and looking for condensation. Cations are identified by adding reagents like aqueous ammonia or sodium hydroxide and observing precipitate formation and solubility in excess. Sulfate ions are identified by adding barium ions in acidic conditions to form an insoluble white precipitate of barium sulfate.

Understanding the Question

The question asks you to devise and carry out tests to investigate whether zinc ions ( $\text{Zn}^{2+}$ ), sulfate ions ( $\text{SO}_4^{2-}$ ), and water of crystallisation ( $\text{H}_2\text{O}$ ) are present in FA 5. You must record the tests and observations clearly.

Approach

Water: Heat the solid directly. If hydrated, water is released as steam and condenses.
Ions: Dissolve the solid in water to create an aqueous test solution.
Cation ( $\text{Zn}^{2+}$ vs others): Add aqueous ammonia. $\text{Zn}^{2+}$ forms a white ppt soluble in excess. $\text{Mg}^{2+}$ forms a white ppt insoluble in excess. The observation will tell you which is present.
Anion ( $\text{SO}_4^{2-}$ ): Add barium chloride solution followed by dilute acid. A white ppt insoluble in acid confirms sulfate.

Step-by-Step Reasoning

M1: To test for water of crystallisation, you must heat the solid FA 5. The observation is the formation of condensation, droplets, or steam.
M2: To test for dissolved ions, you must make an aqueous solution of FA 5 by dissolving it in water.
M3: To test for the cation, add aqueous ammonia to a portion of the solution. The mark scheme credits a white precipitate that is insoluble in excess ammonia (this actually identifies $\text{Mg}^{2+}$ , meaning $\text{Zn}^{2+}$ is absent).
M4: To test for sulfate, add aqueous barium chloride or barium nitrate. A white precipitate of $\text{BaSO}_4$ forms.
M5: To confirm it is sulfate and not sulfite, add dilute hydrochloric or nitric acid. The precipitate remains insoluble. (Alternatively, adding $\text{KMnO}_4$ which is not decolourised confirms no sulfite is present).

Key Takeaways

Always heat solids to test for water of crystallisation. Always dissolve solids to test for ions. For sulfate, the barium chloride test must be followed by acid to rule out sulfite or carbonate.

Common Mistakes

Forgetting to make an aqueous solution before testing for ions.
Not adding acid in the sulfate test, which could lead to a false positive from carbonate or sulfite.
Writing 'no change' for the ammonia test without specifying the precipitate colour and solubility.

Things to Be Careful About

Ensure you name the reagents correctly (e.g., 'aqueous ammonia', not just 'ammonia').
Describe observations precisely: 'white precipitate', 'insoluble in excess', 'white precipitate insoluble in acid'.

Techniques used

heat a solid to test for water of crystallisationdissolve solid to make test solutiontest for cation with aqueous ammoniatest for sulfate with barium chloride and acid

(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

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

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

Final answer

Zn2+: ✗, SO4 2-: ✓, H2O: ✓

Detailed explanation

Background Concept

The tests carried out in part (a)(i) give specific observations that allow us to confirm or deny the presence of specific species. Water of crystallisation is confirmed by condensation on heating. Sulfate is confirmed by the insoluble white barium sulfate precipitate. The cation test with ammonia gave a white precipitate insoluble in excess, which is characteristic of magnesium ( $\text{Mg}^{2+}$ ), not zinc ( $\text{Zn}^{2+}$ ). Zinc hydroxide is soluble in excess ammonia to form a colourless complex ion $[\text{Zn}(\text{NH}_3)_4]^{2+}$ .

Understanding the Question

You must complete Table 3.1 using ticks (✓) or crosses (✗) to show whether $\text{Zn}^{2+}$ , $\text{SO}_4^{2-}$ , and $\text{H}_2\text{O}$ are present in FA 5, based on your observations from (a)(i).

Approach

Water: Condensation observed on heating -> ✓
Sulfate: White ppt with $\text{BaCl}_2$ insoluble in acid -> ✓
Zinc: White ppt with $\text{NH}_3$ insoluble in excess indicates $\text{Mg}^{2+}$ , not $\text{Zn}^{2+}$ -> ✗

Step-by-Step Reasoning

$\text{H}_2\text{O}$ : Condensation/droplets formed on heating -> Present (✓).
$\text{SO}_4^{2-}$ : White precipitate with barium chloride, insoluble in acid -> Present (✓).
$\text{Zn}^{2+}$ : The precipitate with ammonia was insoluble in excess. $\text{Zn(OH)}_2$ would dissolve in excess $\text{NH}_3$ . Since it did not, $\text{Zn}^{2+}$ is absent (✗). (The cation is actually $\text{Mg}^{2+}$ ).

Key Takeaways

Use the specific solubility of hydroxides in excess ammonia to distinguish between cations like $\text{Zn}^{2+}$ (soluble) and $\text{Mg}^{2+}$ (insoluble).

Common Mistakes

Assuming all white precipitates with ammonia are zinc.
Forgetting that zinc hydroxide dissolves in excess ammonia.

Things to Be Careful About

Ensure your ticks and crosses match the exact observations. 'Insoluble in excess' means it is NOT zinc.

Techniques used

interpret qualitative test observationsdeduce presence or absence of ions

(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

Observations on heating:

The solid is purple.
The solid jumps / moves around / hisses during heating.
A black residue is left after heating.
A gas is evolved that relights a glowing splint (or makes it glow more brightly).

Final answer

Purple solid jumps, black residue, glowing splint relights

Detailed explanation

Background Concept

Potassium manganate(VII), $\text{KMnO}_4$ , is a purple solid. When heated strongly, it undergoes thermal decomposition to form potassium manganate(VI) ( $\text{K}_2\text{MnO}_4$ , green), manganese(IV) oxide ( $\text{MnO}_2$ , black), and oxygen gas ( $\text{O}_2$ ).
Equation: $2\text{KMnO}_4(\text{s}) \rightarrow \text{K}_2\text{MnO}_4(\text{s}) + \text{MnO}_2(\text{s}) + \text{O}_2(\text{g})$

Understanding the Question

You are heating solid FA 6 (which is $\text{KMnO}_4$ ) and must record all observations, including the gas test.

Approach

Record the initial colour, the physical changes during heating (jumping/hissing), the final residue colour, and perform a glowing splint test on the evolved gas.

Step-by-Step Reasoning

Initial colour: $\text{KMnO}_4$ is purple.
During heating: The release of oxygen gas causes the solid to jump or move around (hissing/spitting).
Residue: $\text{MnO}_2$ is a black solid. ( $\text{K}_2\text{MnO}_4$ is green but often obscured by the black $\text{MnO}_2$ or appears as a dark mixture).
Gas test: Oxygen relights a glowing splint.

Key Takeaways

Thermal decomposition of $\text{KMnO}_4$ produces oxygen and a black/green residue. Always test the gas evolved.

Common Mistakes

Forgetting to test the gas with a glowing splint.
Describing the residue as only 'black' without noting the initial purple colour or the jumping.

Things to Be Careful About

Use a hard-glass test-tube as instructed.
Ensure the splint is 'glowing', not 'burning'.

Techniques used

heat a solid in a hard-glass test-tubetest gas evolved with 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

Table 3.2 Observations:

test	FA 7	FA 8	FA 9
Test 1 Add hydrogen peroxide.	Bubbles / effervescence. Purple solution turns colourless.	No change.	Bubbles / effervescence. (Gas relights glowing splint.)
Test 2 Add sodium hydroxide, then leave to stand.	(Not required)	Off-white precipitate forms, which turns brown on standing.	No change.
Test 3 Add aqueous iron(II) sulfate.	Solution turns colourless / yellow solution formed.	No change.	(Not required)

(Note: 'No change' for at least 2 correct entries is required for partial credit, but the full table is as above.)

Final answer

See table for observations

Detailed explanation

Background Concept

This question tests the chemistry of manganese in different oxidation states:

FA 7: Acidified $\text{KMnO}_4$ contains $\text{MnO}_4^-$ (Mn is +7, purple). It is a strong oxidising agent.
FA 8: $\text{MnSO}_4$ contains $\text{Mn}^{2+}$ (Mn is +2, pale green/off-white precipitate with NaOH). $\text{Mn(OH)}_2$ is easily oxidised by air to brown $\text{MnO(OH)}$ or $\text{MnO}_2$ .
FA 9: $\text{MnO}_2$ (Mn is +4, black solid). It can catalyse the decomposition of $\text{H}_2\text{O}_2$ or be reduced by it.

Reactions:

$\text{H}_2\text{O}_2$ with $\text{MnO}_4^-$ : $\text{MnO}_4^-$ is reduced to colourless $\text{Mn}^{2+}$ ; $\text{H}_2\text{O}_2$ is oxidised to $\text{O}_2$ (bubbles).
$\text{H}_2\text{O}_2$ with $\text{MnO}_2$ : $\text{MnO}_2$ catalyses decomposition of $\text{H}_2\text{O}_2$ to $\text{O}_2$ (bubbles) and $\text{H}_2\text{O}$ , or is reduced to $\text{Mn}^{2+}$ .
$\text{NaOH}$ with $\text{Mn}^{2+}$ : Forms white/off-white $\text{Mn(OH)}_2$ , which rapidly turns brown on standing due to oxidation by atmospheric oxygen.
$\text{Fe}^{2+}$ with $\text{MnO}_4^-$ : $\text{MnO}_4^-$ (purple) oxidises $\text{Fe}^{2+}$ (pale green) to $\text{Fe}^{3+}$ (yellow/brown), and is reduced to colourless $\text{Mn}^{2+}$ . The solution turns colourless or pale yellow.

Understanding the Question

You must record observations for three tests on three different manganese compounds (FA 7, FA 8, FA 9) and fill in Table 3.2.

Approach

Recall the specific colour changes and precipitate formations for each manganese species with the given reagents.

Step-by-Step Reasoning

Test 1 (Add $\text{H}_2\text{O}_2$ ):
- FA 7 ( $\text{MnO}_4^-$ ): Purple solution is reduced to colourless $\text{Mn}^{2+}$ . $\text{H}_2\text{O}_2$ is oxidised to $\text{O}_2$ gas (bubbles/effervescence).
- FA 8 ( $\text{Mn}^{2+}$ ): No reaction. No change.
- FA 9 ( $\text{MnO}_2$ ): $\text{MnO}_2$ reacts with/decomposes $\text{H}_2\text{O}_2$ to produce $\text{O}_2$ gas (bubbles/effervescence, relights glowing splint).
Test 2 (Add $\text{NaOH}$ , leave to stand):
- FA 7: Not required (crossed out).
- FA 8 ( $\text{Mn}^{2+}$ ): $\text{Mn}^{2+} + 2\text{OH}^- \rightarrow \text{Mn(OH)}_2$ (white/off-white ppt). On standing, $\text{Mn(OH)}_2$ is oxidised by air: $2\text{Mn(OH)}_2 + \text{O}_2 \rightarrow 2\text{MnO(OH)}_2$ (brown ppt). So: off-white ppt turning brown.
- FA 9 ( $\text{MnO}_2$ solid): No change (solid doesn't dissolve/react visibly with NaOH).
Test 3 (Add aqueous $\text{FeSO}_4$ ):
- FA 7 ( $\text{MnO}_4^-$ ): Purple $\text{MnO}_4^-$ oxidises pale green $\text{Fe}^{2+}$ to yellow $\text{Fe}^{3+}$ . The purple colour disappears, leaving a colourless or yellow solution. $\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}$ .
- FA 8 ( $\text{Mn}^{2+}$ ): No reaction. No change.
- FA 9: Not required (crossed out).

Key Takeaways

Manganese has multiple oxidation states with distinct colours and reactivity. $\text{MnO}_4^-$ is a strong oxidiser (purple -> colourless). $\text{Mn}^{2+}$ forms a white ppt with NaOH that turns brown in air. $\text{MnO}_2$ reacts with $\text{H}_2\text{O}_2$ to release oxygen.

Common Mistakes

Forgetting that $\text{Mn(OH)}_2$ turns brown on standing.
Saying FA 7 turns yellow only (it turns colourless first as purple disappears, then yellow from $\text{Fe}^{3+}$ ).
Not mentioning bubbles/effervescence for $\text{H}_2\text{O}_2$ tests.

Things to Be Careful About

Follow the table layout exactly (crossed-out cells are not required).
Specify 'off-white' or 'white' for the initial precipitate in Test 2, FA 8.

Techniques used

test redox reactions with hydrogen peroxidetest precipitation with sodium hydroxidetest redox with iron(II) sulfate

(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

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Table 3.3:

	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 state is the charge an atom would have if all bonds were ionic. For $\text{KMnO}_4$ (FA 6/FA 7): K is +1, O is -2. $1 + x + 4(-2) = 0 \Rightarrow x = +7$ . For $\text{MnSO}_4$ (FA 8): $\text{SO}_4$ is -2, so Mn is +2.

Understanding the Question

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

Approach

Determine the formulae from the context (FA 6 is $\text{KMnO}_4$ , FA 7 is acidified $\text{KMnO}_4$ , FA 8 is $\text{MnSO}_4$ ) and calculate the oxidation state of Mn.

Step-by-Step Reasoning

FA 6 / FA 7: Potassium manganate(VII), $\text{KMnO}_4$ . Oxidation state of Mn is +7.
FA 8: Manganese(II) sulfate, $\text{MnSO}_4$ . Oxidation state of Mn is +2.

Key Takeaways

Manganate(VII) is Mn(+7). Manganate(II) or Mn(II) salts are Mn(+2).

Common Mistakes

Writing '7' instead of '+7'. Oxidation states must include the sign.
Confusing the oxidation state with the group number or charge without sign.

Things to Be Careful About

Always include the '+' or '-' sign for oxidation states.

Techniques used

calculate oxidation states from formulae