Chemistry 9701/33 — May/June 2025

Background Concept

Significant figures reflect the precision of the measurements. The concentration $0.0300\text{ mol dm}^{-3}$ has 3 significant figures (the trailing zeros after the decimal point count). The answers to the calculations should therefore be given to 3 (or 4) significant figures.

Understanding the Question

This is a quality mark for expressing the answers to (c)(ii), (c)(iii) and (c)(iv) to an appropriate number of significant figures — 3 or 4 s.f. is required.

Approach

Give every calculated answer to 3 significant figures, matching the precision of the $0.0300\text{ mol dm}^{-3}$ concentration.

Step-by-Step Reasoning

$0.0300$ has 3 s.f. So the answers should be quoted as $7.50 \times 10^{-4}$, $1.88 \times 10^{-3}$ (or $1.875 \times 10^{-3}$, which is 4 s.f.) and $0.750$ (or $0.7500$), all of which are 3 or 4 s.f.

Key Takeaways

Match the number of significant figures to the least precise given value. Trailing zeros after a decimal point are significant.

Common Mistakes

• Giving answers to 1 or 2 s.f. (e.g. 0.8 instead of 0.750).
• Truncating instead of rounding (e.g. 1.87 instead of 1.88 for 1.875).

Things to Be Careful About

• $7.50 \times 10^{-4}$ has 3 s.f.; $7.5 \times 10^{-4}$ has only 2 s.f. — keep the trailing zero.
• Round correctly: 1.875 rounds to 1.88 (3 s.f.).

Background Concept

Moles = concentration × volume, with volume in dm³. The titre is measured in cm³, so it must be divided by 1000 to convert to dm³.

Understanding the Question

Calculate the amount of manganate(VII) ions in the mean titre volume of FA2 ($0.0300\text{ mol dm}^{-3}$ $\text{KMnO}_4$).

Approach

$n(\text{MnO}_4^-) = c \times V = 0.0300 \times (\text{mean titre} / 1000)$.

Step-by-Step Reasoning

With mean titre 25.00 cm³:

This is the amount of $\text{MnO}_4^-$ that reacted with the $\text{H}_2\text{O}_2$ in the 25.0 cm³ sample of FA4.

Key Takeaways

Always convert cm³ to dm³ (divide by 1000) before using $c = n/V$.

Common Mistakes

• Forgetting to divide the titre by 1000, giving an answer 1000 times too large.
• Using the wrong concentration (e.g. FA1 or FA4 instead of FA2).

Things to Be Careful About

• The unit is mol.
• Quote to 3 s.f.: $7.50 \times 10^{-4}$ mol.

Background Concept

The balanced equation gives the stoichiometric ratio between the reactants:

So the amount of $\text{H}_2\text{O}_2$ is $5/2 = 2.5$ times the amount of $\text{MnO}_4^-$.

Understanding the Question

Use the moles of $\text{MnO}_4^-$ from (c)(ii) to find the moles of $\text{H}_2\text{O}_2$ in the 25.0 cm³ sample of FA4 that was titrated.

Approach

$n(\text{H}_2\text{O}_2) = n(\text{MnO}_4^-) \times (5/2) = n(\text{MnO}_4^-) \times 2.5$.

Step-by-Step Reasoning

This is the amount of $\text{H}_2\text{O}_2$ in 25.0 cm³ of FA4. Note the ratio is 5/2 ($\text{H}_2\text{O}_2 : \text{MnO}_4^-$), not 2/5.

Key Takeaways

The stoichiometric ratio comes from the coefficients in the balanced equation. Here 5 mol $\text{H}_2\text{O}_2$ react with 2 mol $\text{MnO}_4^-$.

Common Mistakes

• Using the inverted ratio (2/5 instead of 5/2).
• Not realising this is the amount in the 25.0 cm³ sample, not in the whole 250 cm³ flask.

Things to Be Careful About

• Keep the answer to 3 or 4 s.f.: $1.88 \times 10^{-3}$ mol (3 s.f.) or $1.875 \times 10^{-3}$ mol (4 s.f.).

Background Concept

Concentration = moles / volume (in dm³). The moles from (c)(iii) are in 25.0 cm³ = 0.0250 dm³ of FA4. FA4 was made by diluting FA1 tenfold (25.0 → 250 cm³), so FA1 is 10 times more concentrated than FA4.

Understanding the Question

Find the concentration of $\text{H}_2\text{O}_2$ in the original solution FA1, accounting for both the volume of the titrated sample and the dilution factor.

Approach

First find $[\text{H}_2\text{O}_2]$ in FA4: divide the moles by 0.0250 dm³. Then multiply by the dilution factor 10 to get $[\text{H}_2\text{O}_2]$ in FA1.

Step-by-Step Reasoning

Dilution factor $= 250/25 = 10$.

The mark scheme route is (c)(iii) × 40 × 10: the ×40 converts from per 25.0 cm³ to per dm³, and the ×10 is the dilution factor.

Key Takeaways

When a solution is diluted, its concentration decreases by the dilution factor; multiply by the factor to recover the original concentration.

Common Mistakes

• Forgetting the ×10 dilution factor, giving 0.0750 instead of 0.750 mol dm⁻³.
• Using 25 cm³ instead of 0.025 dm³ in the concentration formula.

Things to Be Careful About

• The answer is the concentration in FA1 (the original solution), not FA4.
• Quote to 3 s.f.: 0.750 mol dm⁻³.

Background Concept

The energy change in a calorimetry experiment is given by $q = mc\Delta T$, where $m$ is the mass of the solution being heated, $c$ is its specific heat capacity, and $\Delta T$ is the temperature rise. For dilute aqueous solutions, the density is taken as 1 g cm⁻³, so the mass in grams equals the volume in cm³. The specific heat capacity of the solution is taken as that of water, 4.18 J g⁻¹ °C⁻¹.

Understanding the Question

In Experiment 2 you used 40.0 cm³ of FA 1, so the mass of solution is taken as 40 g. You measured the temperature rise $\Delta T_2$. Part (b)(i) asks for the energy change in joules: $q = 40 \times 4.18 \times \Delta T_2$.

Approach

Identify $m$ = 40 g, $c$ = 4.18 J g⁻¹ °C⁻¹, $\Delta T$ = $\Delta T_2$ (your measured value). Substitute into $q = mc\Delta T$ and give the answer to 2–4 significant figures.

Step-by-Step Reasoning

Using a representative $\Delta T_2$ = 10.0 °C:

The mark scheme accepts the answer to 2–4 significant figures, so 1672 J or $1.67 \times 10^3$ J both score. Substitute your own $\Delta T_2$ to get your own value.

Key Takeaways

• $q = mc\Delta T$ is the fundamental calorimetry equation.
• For aqueous solutions, mass in g = volume in cm³.
• $c$ = 4.18 J g⁻¹ °C⁻¹ for water/dilute aqueous solutions.

Common Mistakes

• Using 25.0 cm³ (Experiment 1's volume) instead of 40.0 cm³ for Experiment 2.
• Forgetting to use the temperature rise — using the maximum temperature instead.
• Giving the answer in kJ instead of J, or with the wrong number of significant figures.

Things to Be Careful About

• The answer must be in joules (J), not kJ.
• Give the answer to 2–4 significant figures as the mark scheme requires.
• The density assumption (1 g cm⁻³) is implicit — the mass is taken as 40 g.

Background Concept

The enthalpy change of the decomposition is $\Delta H$ = −98.2 kJ mol⁻¹, meaning 1 mol of $\text{H}_2\text{O}_2$ releases 98.2 kJ (98 200 J) of heat. So the number of moles of $\text{H}_2\text{O}_2$ that decomposed equals the energy change (in J) divided by 98 200 J mol⁻¹. The concentration is then the number of moles divided by the volume in dm³.

Understanding the Question

You know the energy change from (b)(i). You need the concentration of $\text{H}_2\text{O}_2$ in FA 1 in mol dm⁻³. The volume used in Experiment 2 was 40.0 cm³ = 0.0400 dm³.

Approach

Two steps:

1. $n(\text{H}_2\text{O}_2)$ = energy change (J) / 98 200 (J mol⁻¹)
2. $[\text{H}_2\text{O}_2]$ = $n$ / 0.0400 (dm³)

Step-by-Step Reasoning

Using the representative energy change from (b)(i), 1672 J:

The mark scheme also accepts the combined formula $[\text{H}_2\text{O}_2]$ = (b)(i) / (98.2 × 40) = 1672 / 3928 = 0.426 mol dm⁻³. This works because dividing the energy in J by 98 200 (J mol⁻¹) gives moles, then dividing by 0.040 dm³ gives concentration; 98 200 × 0.040 = 3928 = 98.2 × 40. So the shortcut is mathematically identical to the two-step route.

Key Takeaways

• Energy change links to moles via $\Delta H$.
• Concentration = moles / volume (in dm³).
• The combined formula $[\text{H}_2\text{O}_2]$ = E / (98.2 × 40) is a valid shortcut.

Common Mistakes

• Forgetting to convert cm³ to dm³ (40 cm³ = 0.040 dm³).
• Using $\Delta H$ in kJ without converting to J.
• Using the wrong volume (25.0 instead of 40.0 cm³).
• Not showing working — the accuracy mark requires "some working shown".

Things to Be Careful About

• Answer to 2–4 significant figures.
• Show your working to secure both method marks.
• The mark scheme's M1 is for the correct formula; M2 for the correct answer with working.

Background Concept

In an uninsulated cup, heat generated by the reaction is lost to the surroundings (the beaker, the air, the thermometer). The thermometer therefore records a maximum temperature lower than the true adiabatic value, so the measured temperature rise $\Delta T$ is an underestimate.

Understanding the Question

You must state the effect of heat loss on the calculated concentration and explain why. The command "describe and explain" requires both the direction of the effect and the reasoning chain.

Approach

Trace the chain of reasoning: heat loss → lower measured $\Delta T$ → lower calculated energy change → lower calculated moles of $\text{H}_2\text{O}_2$ → lower calculated concentration.

Step-by-Step Reasoning

1. Heat is lost to the surroundings, so the maximum temperature recorded is lower than it would be without heat loss. Hence $\Delta T$ is lower.
2. The energy change is calculated from $q = mc\Delta T$, so a lower $\Delta T$ gives a lower (apparent) energy change.
3. The moles of $\text{H}_2\text{O}_2$ are found from energy change ÷ 98.2 kJ mol⁻¹, so lower energy gives lower moles.
4. Concentration = moles ÷ volume, so the calculated concentration of $\text{H}_2\text{O}_2$ is lower than the true value.

Key Takeaways

• Heat loss is a systematic error that biases the result in one direction (here, downward).
• Trace the effect through each step of the calculation.

Common Mistakes

• Saying the concentration would be higher (wrong direction).
• Stopping at "$\Delta T$ is lower" without explaining the effect on the calculated concentration.
• Not mentioning that the energy change (and hence moles) is also affected.

Things to Be Careful About

• The mark scheme requires "concentration will be lower" plus two supporting points ($\Delta T$ lower; energy change lower; moles lower). Give all three for safety.

Background Concept

Percentage error matters more than absolute error. If a fixed amount of heat is lost, that loss is a larger fraction of a small total energy change than of a large one. So when $\Delta T$ is small (low concentration), the percentage error is larger. Additionally, at lower concentration the reaction is slower, giving more time for heat to escape.

Understanding the Question

The student claims the method is less accurate at lower concentration. You must say whether this is correct and explain. Note the mark scheme accepts several valid lines of reasoning — the key is to give a clear, chemically sound argument.

Approach

Consider the student's claim. The most direct argument: at lower concentration, $\Delta T$ is smaller, so the same absolute heat loss is a larger percentage of the total, making the percentage error greater. A supporting argument: lower concentration → slower reaction → more time for heat loss.

Step-by-Step Reasoning

The student is correct. At lower concentration, less $\text{H}_2\text{O}_2$ decomposes, so less heat is released and $\Delta T$ is smaller. Heat loss is roughly a fixed absolute amount, so it represents a larger percentage of the smaller energy change — the percentage error is greater. Also, the decomposition is slower at lower concentration, giving more time for heat to escape before the maximum temperature is reached.

(Alternative acceptable arguments per the mark scheme: the student is correct because the reaction is slower giving more time for heat loss; or the student is not correct because although $\Delta T$ is lower, the heat loss is also less, or the lower $\Delta T$ is cancelled out by the slower reaction. Any one clear, correct argument scores the mark.)

Key Takeaways

• Distinguish absolute error from percentage error.
• Rate of reaction affects the time available for heat loss in calorimetry.

Common Mistakes

• Giving a vague answer without a mechanism (e.g. just "less accurate" without explaining why).
• Confusing the direction: saying the concentration would be overestimated.
• Not addressing the student's specific claim.

Things to Be Careful About

• The mark scheme accepts "correct" or "not correct" with appropriate reasoning — but the most defensible answer is that the student is correct, with the percentage-error argument.
• Give both the $\Delta T$ argument and the rate argument for a robust answer.

Background Concept

Manganate(VII) ions, $\text{MnO}_4^-$, are deep purple. In alkaline conditions, they can react with manganese(IV) oxide, $\text{MnO}_2$, to form manganate(VI) ions, $\text{MnO}_4^{2-}$, which are dark green. When a green manganate(VI) solution is acidified, it disproportionates back into purple manganate(VII) and brown manganese(IV) oxide precipitate.

Understanding the Question

You are mixing $\text{KMnO}_4$ (purple) with $\text{NaOH}$ and $\text{MnO}_2$ (solid), then filtering. The first observation is the colour of the filtrate. Then, acid is added to the filtrate, and you must observe the colour change.

Approach

1. Recognise that $\text{MnO}_4^-$ and $\text{MnO}_2$ react in $\text{OH}^-$ to form $\text{MnO}_4^{2-}$ (green). The filtrate will be green.
2. Recognise that acidifying $\text{MnO}_4^{2-}$ causes disproportionation: $3\text{MnO}_4^{2-} + 4\text{H}^+ \rightarrow 2\text{MnO}_4^- + \text{MnO}_2 + 2\text{H}_2\text{O}$. The solution turns purple/pink (and a brown precipitate may form, though the mark scheme focuses on the solution colour turning pink/purple).

Step-by-Step Reasoning

First observation: When $\text{KMnO}_4(\text{aq})$ is mixed with $\text{NaOH}(\text{aq})$ and $\text{MnO}_2(\text{s})$, the purple $\text{MnO}_4^-$ is reduced by $\text{MnO}_2$ in alkaline conditions to form green $\text{MnO}_4^{2-}$. After filtering, the filtrate is a dark green solution.

Second observation: Adding dilute $\text{H}_2\text{SO}_4$ acidifies the green $\text{MnO}_4^{2-}$ solution. The manganate(VI) ion disproportionates into manganate(VII) (purple/pink) and manganese(IV) oxide (brown solid). The solution turns pink or purple.

Key Takeaways

• $\text{MnO}_4^-$ (purple) + $\text{MnO}_2$ + $\text{OH}^-$ $\rightarrow$ $\text{MnO}_4^{2-}$ (green).
• Acidifying $\text{MnO}_4^{2-}$ (green) causes disproportionation to $\text{MnO}_4^-$ (purple) and $\text{MnO}_2$ (brown ppt).

Common Mistakes

• Writing 'brown' for the first observation instead of 'green'. The solid $\text{MnO}_2$ is filtered off; the filtrate is green.
• Forgetting that the second observation is about the solution colour changing to pink/purple, not just 'brown precipitate forms'.

Things to Be Careful About

• The mark scheme specifically awards marks for 'filtrate / solution' and 'dark green', and 'solution' and 'pink / purple'. Be precise about what is being observed (the liquid, not the solid).

Background Concept

Iron(II) ions, $\text{Fe}^{2+}$, are pale green in solution. Iron(III) ions, $\text{Fe}^{3+}$, are yellow or pale brown. When aqueous $\text{NaOH}$ is added to $\text{Fe}^{3+}$, a red-brown (or brown) precipitate of iron(III) hydroxide, $\text{Fe(OH)}_3$, forms. This precipitate is insoluble in excess $\text{NaOH}$.

Manganese(IV) oxide, $\text{MnO}_2$, is an oxidising agent. In acidic conditions, it can oxidise $\text{Fe}^{2+}$ to $\text{Fe}^{3+}$.

Understanding the Question

You warm $\text{FeSO}_4$ with dilute $\text{H}_2\text{SO}_4$ and $\text{MnO}_2$. The $\text{MnO}_2$ oxidises $\text{Fe}^{2+}$ to $\text{Fe}^{3+}$. You filter and observe the filtrate colour. Then you add $\text{NaOH}$ and observe the precipitate.

Approach

1. Identify that $\text{MnO}_2$ oxidises $\text{Fe}^{2+}$ to $\text{Fe}^{3+}$ in acid. The filtrate will contain $\text{Fe}^{3+}$, which is yellow/pale brown.
2. Recall the test for $\text{Fe}^{3+}$ with $\text{NaOH}$: red-brown precipitate, insoluble in excess.

Step-by-Step Reasoning

First observation: The mixture is warmed in dilute acid. $\text{MnO}_2$ oxidises $\text{Fe}^{2+}$ to $\text{Fe}^{3+}$. The filtrate contains $\text{Fe}^{3+}(\text{aq})$, which is yellow or pale brown.

Second observation: Adding $\text{NaOH}(\text{aq})$ to the $\text{Fe}^{3+}$ solution produces a red-brown (or brown) precipitate of $\text{Fe(OH)}_3$. This precipitate does not dissolve in excess $\text{NaOH}$.

Key Takeaways

• $\text{MnO}_2$ oxidises $\text{Fe}^{2+}$ to $\text{Fe}^{3+}$ in acidic solution.
• $\text{Fe}^{3+}$ gives a red-brown ppt with $\text{NaOH}$, insoluble in excess.

Common Mistakes

• Describing the precipitate as 'green' (that's for $\text{Fe}^{2+}$).
• Saying the precipitate is 'soluble in excess' (only amphoteric hydroxides like $\text{Al(OH)}_3$, $\text{Zn(OH)}_2$, $\text{Pb(OH)}_2$ are soluble in excess $\text{NaOH}$).

Things to Be Careful About

• The mark scheme accepts 'yellow', 'pale brown', 'red-brown', or 'brown'. Use precise terminology.
• Must state 'insoluble in excess' for the second mark.

Background Concept

In a redox reaction, the substance that is reduced (gains electrons) is the oxidising agent (or oxidant). It causes another substance to be oxidised.

Understanding the Question

In part (a)(ii), $\text{MnO}_2$ caused $\text{Fe}^{2+}$ to be oxidised to $\text{Fe}^{3+}$. Therefore, $\text{MnO}_2$ itself was reduced. You must state the role $\text{MnO}_2$ played.

Approach

Since $\text{MnO}_2$ oxidised $\text{Fe}^{2+}$, it acted as an oxidising agent.

Step-by-Step Reasoning

$\text{MnO}_2$ accepted electrons from $\text{Fe}^{2+}$, reducing the iron(II) to iron(III) (wait, no, it caused iron to be oxidised, so $\text{MnO}_2$ was reduced). By oxidising $\text{Fe}^{2+}$ to $\text{Fe}^{3+}$, $\text{MnO}_2$ acted as an oxidising agent.

Key Takeaways

• If a reagent causes another species to be oxidised, it is the oxidising agent.

Common Mistakes

• Writing 'reducing agent' (that would be if it caused another species to be reduced).
• Writing 'catalyst' (it is consumed in the reaction, as seen by the filtration step removing unreacted solid).

Things to Be Careful About

• The mark scheme accepts 'oxidising agent' or 'oxidant'.

Background Concept

When aqueous sodium hydroxide is added to a solution containing metal ions, insoluble metal hydroxides may precipitate. The ionic equation shows only the species that actually change.

Understanding the Question

The filtrate from (a)(ii) contains $\text{Fe}^{3+}$ ions (and excess $\text{H}^+$ from the dilute sulfuric acid). You must write the ionic equation for the reaction with $\text{NaOH}$ that produced the red-brown precipitate observed.

Approach

The primary observation was the formation of a red-brown precipitate. This is $\text{Fe(OH)}_3$. Write the ionic equation for its formation. State symbols are required.

Step-by-Step Reasoning

$\text{Fe}^{3+}(\text{aq})$ reacts with $\text{OH}^-(\text{aq})$ to form solid $\text{Fe(OH)}_3$.

Note: The mark scheme also accepts the neutralisation of excess acid: $\text{H}^+(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{H}_2\text{O}(\text{l})$, but the precipitation equation is the primary one linked to the observation in part (ii).

Key Takeaways

• Always include state symbols in ionic equations for precipitation reactions.
• Balance the charges and atoms.

Common Mistakes

• Forgetting state symbols.
• Writing the full molecular equation instead of the ionic equation.
• Incorrect balancing (e.g., missing the coefficient 3 for $\text{OH}^-$).

Things to Be Careful About

• The mark scheme requires state symbols: (aq) for ions, (s) for the precipitate.

Background Concept

FA 6 is $\text{FeCl}_3$: Contains $\text{Fe}^{3+}$ (yellow/brown solution) and $\text{Cl}^-$ (chloride ion). $\text{Fe}^{3+}$ gives a red-brown ppt with $\text{NaOH}$ and $\text{Na}_2\text{CO}_3$. $\text{Fe}^{3+}$ is not oxidised by $\text{KMnO}_4$ or $\text{H}_2\text{O}_2$.

FA 7 is KI: Contains $\text{K}^+$ (no characteristic tests) and $\text{I}^-$ (iodide ion). $\text{I}^-$ is oxidised by $\text{KMnO}_4$ (purple to brown $\text{I}_2$) and by $\text{H}_2\text{O}_2$ (to $\text{I}_2$, yellow/brown, with $\text{O}_2$ evolution). $\text{I}_2$ turns starch blue-black.

Understanding the Question

You must perform four tests on both FA 6 and FA 7 and record observations. Test 3 is shaded out for FA 6.

Approach

Work through each test systematically, predicting the observation for each solution based on the ions present.

Step-by-Step Reasoning

Test 1: + NaOH (aq), then warm

• FA 6 ($\text{Fe}^{3+}$): Red-brown / brown precipitate of $\text{Fe(OH)}_3$, insoluble in excess $\text{NaOH}$. Warming produces no change (no ammonia gas, so red litmus does not turn blue).
• FA 7 ($\text{K}^+$, $\text{I}^-$): No precipitate forms (both $\text{KOH}$ and $\text{NaI}$ are soluble). Warming produces no change.

Test 2: + acidified $\text{KMnO}_4$(aq)

• FA 6 ($\text{Fe}^{3+}$, $\text{Cl}^-$): $\text{Fe}^{3+}$ cannot be oxidised further. Purple colour of $\text{KMnO}_4$ remains (no change). Note: dilute $\text{Cl}^-$ does not decolourise $\text{KMnO}_4$ at room temperature.
• FA 7 ($\text{I}^-$): $\text{I}^-$ is oxidised to $\text{I}_2$. The solution turns brown / red-brown.

Test 3: + $\text{H}_2\text{O}_2$, then starch

• FA 6: Shaded out, do not perform.
• FA 7 ($\text{I}^-$): $\text{H}_2\text{O}_2$ oxidises $\text{I}^-$ to $\text{I}_2$ (solution turns yellow / red-brown). $\text{H}_2\text{O}_2$ also decomposes to give $\text{O}_2$ gas (effervescence / fizzing; gas relights a glowing splint). Adding starch turns the solution dark blue / blue-black.

Test 4: + $\text{Na}_2\text{CO}_3$(aq)

• FA 6 ($\text{Fe}^{3+}$): $\text{Fe}^{3+}$ hydrolyses water, producing $\text{H}^+$ which reacts with $\text{CO}_3^{2-}$ to give $\text{CO}_2$ (effervescence; gas turns limewater milky). A red-brown / brown precipitate of $\text{Fe(OH)}_3$ also forms (not $\text{Fe}_2(\text{CO}_3)_3$ which is unstable in water).
• FA 7 ($\text{K}^+$, $\text{I}^-$): No reaction. No change.

Key Takeaways

• $\text{Fe}^{3+}$ gives red-brown ppt with $\text{NaOH}$ and $\text{Na}_2\text{CO}_3$ (with effervescence).
• $\text{I}^-$ is oxidised by $\text{KMnO}_4$ and $\text{H}_2\text{O}_2$ to give brown $\text{I}_2$ and blue-black with starch.
• $\text{K}^+$ and $\text{Cl}^-$ give no characteristic observations in these tests.

Common Mistakes

• Saying $\text{FeCl}_3$ gives a blue-black with starch (only $\text{I}_2$ does).
• Saying $\text{Fe}^{3+}$ is oxidised by $\text{KMnO}_4$ (it is already in its highest common oxidation state).
• Forgetting the effervescence in Test 4 for $\text{Fe}^{3+}$ (hydrolysis of $\text{Fe}^{3+}$ with carbonate produces $\text{CO}_2$).

Things to Be Careful About

• Test 3 is shaded out for FA 6; do not record observations for it.
• In Test 1, you must record observations for both adding $\text{NaOH}$ and warming.
• Use precise colour descriptions: 'red-brown', 'brown', 'purple', 'blue-black'.

Background Concept

The anion in FA 6 is not stated, but we know FA 6 is $\text{FeCl}_3$ (from the mark scheme and the fact that the anion doesn't contain nitrogen or sulfur). The halide test uses acidified silver nitrate, $\text{AgNO}_3(\text{aq})$.

• $\text{Cl}^-$ gives a white precipitate of $\text{AgCl}$, soluble in dilute ammonia.
• $\text{Br}^-$ gives a cream precipitate of $\text{AgBr}$, sparingly soluble in dilute ammonia.
• $\text{I}^-$ gives a yellow precipitate of $\text{AgI}$, insoluble in dilute ammonia.

Understanding the Question

You must select a reagent to identify the anion in FA 6. Since FA 6 is $\text{FeCl}_3$, the anion is $\text{Cl}^-$. You need the test for chloride.

Approach

Use aqueous silver nitrate (acidified with dilute nitric acid, though the mark scheme just requires $\text{AgNO}_3$). The observation is a white precipitate.

Step-by-Step Reasoning

Add aqueous silver nitrate, $\text{AgNO}_3(\text{aq})$, to FA 6. $\text{Cl}^-$ reacts with $\text{Ag}^+$ to form a white precipitate of silver chloride, $\text{AgCl}$.

Key Takeaways

• Chloride ions give a white precipitate with acidified silver nitrate.

Common Mistakes

• Using barium chloride or barium nitrate (that's for sulfate).
• Using lead(II) nitrate (that's for halides but less common in CIE; silver nitrate is standard).
• Not specifying 'white' precipitate.

Things to Be Careful About

• The mark scheme accepts 'silver nitrate' or '$\text{AgNO}_3$' and 'white ppt'.

Background Concept

FA 6 contains two cations and one anion. FA 7 contains one cation and one anion. Neither contains sulfur. The anion in FA 6 does not contain nitrogen.

Understanding the Question

Use your observations from (b)(i) and (b)(ii) to identify the ions.

Approach

FA 6:

• Test 1 (NaOH): Red-brown ppt $\rightarrow$ $\text{Fe}^{3+}$.
• Test 4 ($\text{Na}_2\text{CO}_3$): Effervescence $\rightarrow$ acidic solution $\rightarrow$ $\text{H}^+$ (or hydrolysis of $\text{Fe}^{3+}$, but the mark scheme explicitly includes $\text{H}^+$ as one of the cations, likely from the preparation of the solution or excess acid).
• Test 2 ($\text{KMnO}_4$): No change $\rightarrow$ not $\text{Fe}^{2+}$, not halide that reduces $\text{MnO}_4^-$ easily (dilute $\text{Cl}^-$ doesn't).
• Test (ii) ($\text{AgNO}_3$): White ppt $\rightarrow$ $\text{Cl}^-$.
• Ions: $\text{Fe}^{3+}$, $\text{H}^+$, $\text{Cl}^-$.

FA 7:

• Test 1 (NaOH): No change $\rightarrow$ $\text{K}^+$ (or other alkali metal, but 'unknown' is accepted if not tested).
• Test 2 ($\text{KMnO}_4$): Turns brown $\rightarrow$ halide, specifically $\text{I}^-$ (or $\text{Br}^-$, but Test 3 confirms $\text{I}^-$).
• Test 3 ($\text{H}_2\text{O}_2$ + starch): Blue-black $\rightarrow$ $\text{I}^-$.
• Cation: $\text{K}^+$ is likely, but since no flame test or specific test was done, 'unknown' is acceptable per the mark scheme.
• Anion: $\text{I}^-$.
• Ions: $\text{I}^-$, unknown.

Step-by-Step Reasoning

FA 6:

• Red-brown ppt with $\text{NaOH}$ identifies $\text{Fe}^{3+}$.
• Effervescence with $\text{Na}_2\text{CO}_3$ indicates acidity ($\text{H}^+$).
• White ppt with $\text{AgNO}_3$ identifies $\text{Cl}^-$.
• Formulae: $\text{Fe}^{3+}$, $\text{H}^+$, $\text{Cl}^-$.

FA 7:

• Brown colour with $\text{KMnO}_4$ and blue-black with starch/$\text{H}_2\text{O}_2$ identifies $\text{I}^-$.
• No test performed for the cation (no $\text{NaOH}$ ppt, no flame test), so it is 'unknown'.
• Formulae: $\text{I}^-$, unknown.

Key Takeaways

• Synthesise multiple tests to confirm ion identity.
• If an ion cannot be identified from the tests performed, write 'unknown'.

Common Mistakes

• Guessing $\text{K}^+$ for FA 7 without a flame test (the mark scheme accepts 'unknown').
• Forgetting $\text{H}^+$ in FA 6 (the effervescence with carbonate indicates acid).
• Writing molecular formulae instead of ion formulae.

Things to Be Careful About

• The mark scheme awards marks for correct ions: 2-3 correct = 1 mark, 4 correct = 2 marks, 5 correct = 3 marks.
• FA 6 has 3 ions, FA 7 has 2 ions. Total 5 possible marks.