Chemistry 9701/37 — May/June 2025

Background Concept

Significant figures reflect the precision of a measurement. The given data — $0.0300\text{ mol dm}^{-3}$, 25.0 cm³, 250 cm³ — all carry 3 significant figures, so calculated answers should be quoted to 3 (or 4) significant figures to match that precision.

Understanding the Question

This part is a note to give all answers in (c)(ii)–(c)(iv) to an appropriate number of significant figures. The mark is awarded if all three answers are to 3 or 4 sf.

Approach

Count the significant figures in the least precise given value and match it in the answers.

Step-by-Step Reasoning

• $0.0300$ has 3 significant figures (the leading zeros are not significant).
• 25.0 has 3 significant figures.
• 250 (a volumetric flask marking) is taken as 3 significant figures here.

So the answers should be quoted to 3 significant figures: $7.50 \times 10^{-4}$, $1.88 \times 10^{-3}$ and 0.750.

Key Takeaways

• Match the number of significant figures to the given data (usually 3 sf in these titrations).

Common Mistakes

• Quoting too many decimal places (e.g. $1.875 \times 10^{-3}$ when the data only justify 3 sf).
• Quoting only 1 sf.

Things to Be Careful About

• Leading zeros are not significant; $0.0300$ has 3 significant figures, not 5.

Background Concept

The amount of a substance in moles is found from its concentration and volume: $n = cV$, where $V$ must be in dm³. Since the titre is measured in cm³, it must be divided by 1000 to convert to dm³.

Understanding the Question

You are asked to calculate the amount of $\text{MnO}_4^-$ ions in the mean titre volume of FA2, which has concentration $0.0300\text{ mol dm}^{-3}$. The mean titre from part (b) is 25.00 cm³.

Approach

Use $n = cV$ with $V = 25.00/1000 = 0.02500\text{ dm}^3$ and $c = 0.0300\text{ mol dm}^{-3}$.

Step-by-Step Reasoning

This is the amount of manganate(VII) that reacted with the $\text{H}_2\text{O}_2$ in the 25.0 cm³ sample of FA4.

Key Takeaways

• $n = cV$ with volume in dm³.
• Convert cm³ to dm³ by dividing by 1000.

Common Mistakes

• Forgetting to divide the volume by 1000 (i.e. using $0.0300 \times 25.00 = 0.750$ mol, which is wrong by a factor of 1000).
• Writing the answer to the wrong number of significant figures.

Things to Be Careful About

• Quote the answer to 3 significant figures: $7.50 \times 10^{-4}$ mol.

Background Concept

The balanced equation gives the stoichiometric ratio between the reactants:

So 5 mol of $\text{H}_2\text{O}_2$ react with 2 mol of $\text{MnO}_4^-$, i.e. $n(\text{H}_2\text{O}_2) = n(\text{MnO}_4^-) \times \frac{5}{2} = n(\text{MnO}_4^-) \times 2.5$.

Understanding the Question

Using the moles of manganate(VII) found in (c)(ii), calculate the moles of hydrogen peroxide that reacted in the 25.0 cm³ sample of FA4.

Approach

Multiply the moles of $\text{MnO}_4^-$ by the stoichiometric ratio $\frac{5}{2}$ (or 2.5).

Step-by-Step Reasoning

Rounded to 3 significant figures: $1.88 \times 10^{-3}$ mol. This is the amount of $\text{H}_2\text{O}_2$ in the 25.0 cm³ of FA4 that was titrated.

Key Takeaways

• The stoichiometric ratio comes from the balanced equation: $\text{H}_2\text{O}_2 : \text{MnO}_4^- = 5 : 2$.
• Multiply moles of $\text{MnO}_4^-$ by 2.5 to get moles of $\text{H}_2\text{O}_2$.

Common Mistakes

• Using the inverse ratio ($\frac{2}{5}$ instead of $\frac{5}{2}$) — this would give $3.00 \times 10^{-4}$ mol, which is wrong.
• Forgetting that this amount refers to the 25.0 cm³ FA4 sample, not to FA1.

Things to Be Careful About

• The ratio is $\text{H}_2\text{O}_2 : \text{MnO}_4^- = 5 : 2$, so multiply by 2.5.

Background Concept

Concentration is amount divided by volume: $c = n/V$. The amount found in (c)(iii) is in 25.0 cm³ (0.0250 dm³) of FA4, so the concentration of FA4 is $n/0.0250$. FA1 was diluted 10× (25 cm³ made up to 250 cm³), so FA1 is 10× more concentrated than FA4.

Understanding the Question

Find the concentration of $\text{H}_2\text{O}_2$ in the original FA1, accounting for the 10× dilution performed at the start of the experiment.

Approach

First find the concentration of FA4 from the moles in (c)(iii) and the 25.0 cm³ sample volume, then multiply by the dilution factor (10).

Step-by-Step Reasoning

1. Concentration in FA4:

1. Dilution factor: 25.0 cm³ was diluted to 250 cm³, so $250/25 = 10$.
2. FA1 is 10× more concentrated than FA4:

Key Takeaways

• Dilution factor = final volume / aliquot volume = $250/25 = 10$.
• FA1 is 10× more concentrated than FA4.

Common Mistakes

• Forgetting the dilution factor and giving 0.0750 mol dm⁻³ instead of 0.750 mol dm⁻³.
• Dividing instead of multiplying by the dilution factor.
• Using the full 250 cm³ volume instead of the 25.0 cm³ sample volume when finding the concentration of FA4.

Things to Be Careful About

• The amount in (c)(iii) is for the 25.0 cm³ FA4 sample, so divide by 0.0250 dm³.
• The answer is quoted to 3 significant figures: 0.750 mol dm⁻³.

Background Concept

The energy change when a substance is heated or cooled is given by $q = mc\Delta T$, where $m$ is the mass (in g), $c$ is the specific heat capacity (4.18 J g⁻¹ K⁻¹ for water and dilute aqueous solutions), and $\Delta T$ is the temperature change. For dilute aqueous solutions, 1 cm³ ≈ 1 g, so the volume in cm³ is numerically equal to the mass in g.

Understanding the Question

You need to calculate the energy change in Experiment 2, which used 40.0 cm³ of FA 1. You use your measured $\Delta T$ from Experiment 2. The formula is: energy = 40 × 4.18 × $\Delta T$.

Approach

Substitute your measured $\Delta T$ into the formula $q = mc\Delta T$, with $m$ = 40 g (from 40 cm³), $c$ = 4.18 J g⁻¹ K⁻¹.

Step-by-Step Reasoning

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

The answer should be given to 2–4 significant figures.

Key Takeaways

• $q = mc\Delta T$ is the fundamental equation for energy change.
• For dilute aqueous solutions, 1 cm³ ≈ 1 g, so volume in cm³ = mass in g.

Common Mistakes

• Using the wrong volume (25 cm³ instead of 40 cm³).
• Forgetting to multiply by 4.18.
• Giving the answer to too few significant figures.

Things to Be Careful About

• The answer must be in J (not kJ).
• Use the $\Delta T$ from Experiment 2 specifically.

Background Concept

The energy released comes from the decomposition of $\text{H}_2\text{O}_2$. The enthalpy change of the reaction is $\Delta H = -98.2\text{ kJ mol}^{-1} = -98200\text{ J mol}^{-1}$. The number of moles of $\text{H}_2\text{O}_2$ decomposed is $n = \text{energy} / 98200$. The concentration is then $n / V$, where $V$ is the volume in dm³. Here $V$ = 40 cm³ = 0.040 dm³. So:

The mark scheme writes this as (b)(i) / (98.2 × 40), which is numerically equivalent.

Understanding the Question

Using the energy change from (b)(i), calculate the concentration of $\text{H}_2\text{O}_2$ in FA 1. The mark scheme gives the formula: $[\text{H}_2\text{O}_2]$ = (b)(i) / (98.2 × 40) = (b)(i) / 3928.

Approach

Divide the energy change (in J) by 3928 to get the concentration in mol dm⁻³.

Step-by-Step Reasoning

With energy = 1672 J:

The answer should be given to 2–4 significant figures.

Key Takeaways

• The relationship between energy released, moles, and concentration.
• Unit conversion: cm³ to dm³ (divide by 1000).

Common Mistakes

• Using the wrong volume.
• Forgetting to convert kJ to J.
• Not showing working (M1 requires some working shown).

Things to Be Careful About

• The mark scheme allows (b)(i) / 98.2, (b)(i) / 98200, or (b)(i) / 3928 as correct working.
• Answer to 2–4 sf.

Background Concept

In this method, the concentration is calculated from the temperature rise. If heat is lost to the surroundings, the measured maximum temperature (and hence $\Delta T$) is lower than the true value. This means the calculated energy released is lower, and therefore the calculated moles (and concentration) of $\text{H}_2\text{O}_2$ is lower than the true value.

Understanding the Question

You need to describe and explain the effect of heat loss on the calculated concentration. The mark scheme requires: concentration will be lower, AND two of: $\Delta T$ lower; energy released lower; moles of $\text{H}_2\text{O}_2$ lower.

Approach

Trace the chain of reasoning: heat loss → lower $\Delta T$ → lower energy → lower moles → lower 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.
2. Therefore $\Delta T$ is lower.
3. The calculated energy released ($q = mc\Delta T$) is lower.
4. The calculated amount (moles) of $\text{H}_2\text{O}_2$ is lower.
5. Therefore the calculated concentration is lower than the true value.

Key Takeaways

• Heat loss systematically lowers the calculated concentration.
• This is a systematic error, not a random one.

Common Mistakes

• Saying the concentration is higher (wrong direction).
• Only stating "concentration is lower" without explaining why.

Things to Be Careful About

• The mark scheme requires the direction (lower) plus two supporting reasons.

Background Concept

At lower concentration, the decomposition of $\text{H}_2\text{O}_2$ is slower. This has two effects: (1) the reaction takes longer, so more heat escapes to the surroundings during the reaction; (2) the temperature rise is smaller, so the percentage error in measuring $\Delta T$ is larger (a fixed reading error of ±0.5 °C is a larger fraction of a smaller $\Delta T$).

Understanding the Question

The student suggests the method is less accurate at lower concentration. You need to agree or disagree and explain. The mark scheme accepts several lines of reasoning.

Approach

Consider the two effects of lower concentration: slower reaction (more heat loss) and smaller $\Delta T$ (larger percentage error). Both support the student's claim.

Step-by-Step Reasoning

The student is correct. At lower concentration:

• The reaction is slower, so more time is available for heat to escape to the surroundings, increasing heat loss.
• The temperature rise is smaller, so the percentage error in measuring $\Delta T$ is greater (a ±0.5 °C reading error is a larger fraction of a smaller $\Delta T$).

Key Takeaways

• Lower concentration → slower reaction → more heat loss.
• Smaller $\Delta T$ → larger percentage error.

Common Mistakes

• Saying the student is wrong without justification.
• Not linking lower concentration to slower reaction.

Things to Be Careful About

• The mark scheme accepts several alternative lines of reasoning, including the argument that the student is not correct (if $\Delta T$ is lower, heat loss is less, or the lower $\Delta T$ is cancelled by the slower reaction). Any consistent argument with two supporting points scores.

Background Concept

In alkaline conditions, manganate(VII) ions ($\text{MnO}_4^-$) can be reduced to manganate(VI) ions ($\text{MnO}_4^{2-}$). The $\text{MnO}_4^{2-}$ ion is characteristically dark green. When an alkaline solution containing $\text{MnO}_4^{2-}$ is acidified, the manganate(VI) ion is unstable and undergoes disproportionation, reverting to manganate(VII) ($\text{MnO}_4^-$, which is pink/purple) and manganese(IV) oxide ($\text{MnO}_2$, a brown/black solid). The reaction is:

Understanding the Question

The candidate mixes $\text{KMnO}_4$ (purple), aqueous $\text{NaOH}$, and solid $\text{MnO}_2$. The $\text{MnO}_2$ acts as a reducing agent in the alkaline medium, reducing some $\text{MnO}_4^-$ to $\text{MnO}_4^{2-}$ while being oxidised itself to $\text{MnO}_4^{2-}$. The mixture is filtered to remove unreacted solid $\text{MnO}_2$. The candidate must record the colour of the filtrate, and then the observation when dilute sulfuric acid is added to it.

Approach

1. Identify the product of the reaction between $\text{KMnO}_4$, $\text{NaOH}$, and $\text{MnO}_2$: manganate(VI), $\text{MnO}_4^{2-}$, which is green.
2. Filter removes the solid $\text{MnO}_2$, leaving a green solution of $\text{MnO}_4^{2-}$.
3. Acidifying the green $\text{MnO}_4^{2-}$ solution causes disproportionation back to pink/purple $\text{MnO}_4^-$.

Step-by-Step Reasoning

• First observation: After stirring $\text{KMnO}_4$, $\text{NaOH}$, and $\text{MnO}_2$ and filtering, the filtrate contains $\text{MnO}_4^{2-}$ ions. The mark scheme requires stating that the filtrate/solution is (dark) green.
• Second observation: Adding dilute sulfuric acid provides $\text{H}^+$ ions, which trigger the disproportionation of $\text{MnO}_4^{2-}$ into $\text{MnO}_4^-$ and $\text{MnO}_2$. The solution will turn pink / purple (or remain green and turn pink/purple). The mark scheme accepts 'green solution OR turns AND pink / purple'.

Key Takeaways

Manganate(VII) is reduced to manganate(VI) (green) in alkaline conditions. Acidification causes manganate(VI) to disproportionate back to manganate(VII) (pink/purple).

Common Mistakes

• Describing the filtrate as colourless or purple (forgetting that $\text{MnO}_4^{2-}$ is green).
• Not mentioning the colour change to pink/purple upon acidification.
• Writing 'brown precipitate' for the second observation (while $\text{MnO}_2$ does form, the primary observable change in the solution colour is to pink/purple, and the mark scheme specifically looks for the solution turning pink/purple).

Things to Be Careful About

• Must mention both the green filtrate and the pink/purple colour on acidification to earn both marks.
• Use precise colour terminology: 'dark green' for $\text{MnO}_4^{2-}$ and 'pink' or 'purple' for $\text{MnO}_4^-$.

Background Concept

Iron(II) ions ($\text{Fe}^{2+}$) are colourless or pale green in dilute solution. In the presence of an oxidising agent and acid, $\text{Fe}^{2+}$ is oxidised to iron(III) ions ($\text{Fe}^{3+}$), which form a yellow or pale brown solution. When aqueous sodium hydroxide ($\text{NaOH}$) is added to $\text{Fe}^{3+}$, a characteristic red-brown (or brown) precipitate of iron(III) hydroxide, $\text{Fe(OH)}_3$, forms. This precipitate is insoluble in excess $\text{NaOH}$.

Understanding the Question

The candidate mixes aqueous iron(II) sulfate, dilute sulfuric acid, and solid $\text{MnO}_2$, then warms the mixture. $\text{MnO}_2$ acts as an oxidising agent in acidic conditions, oxidising $\text{Fe}^{2+}$ to $\text{Fe}^{3+}$. The mixture is filtered to remove unreacted $\text{MnO}_2$. The candidate must record the colour of the filtrate, and then the observation when aqueous $\text{NaOH}$ is added dropwise.

Approach

1. Identify the product of the oxidation: $\text{Fe}^{3+}$, which gives a yellow/pale brown solution.
2. Filter removes the solid $\text{MnO}_2$, leaving a yellow/pale brown solution containing $\text{Fe}^{3+}$ and $\text{Mn}^{2+}$.
3. Adding $\text{NaOH}$ to $\text{Fe}^{3+}$ produces a red-brown precipitate of $\text{Fe(OH)}_3$, which does not dissolve in excess $\text{NaOH}$.

Step-by-Step Reasoning

• First observation: The reaction is $\text{MnO}_2(\text{s}) + 4\text{H}^+(\text{aq}) + 2\text{Fe}^{2+}(\text{aq}) \rightarrow \text{Mn}^{2+}(\text{aq}) + 2\text{Fe}^{3+}(\text{aq}) + 2\text{H}_2\text{O}(\text{l})$. The filtrate contains $\text{Fe}^{3+}$ ions. The mark scheme requires stating that the filtrate/solution is yellow / (pale) brown.
• Second observation: Adding aqueous $\text{NaOH}$ to $\text{Fe}^{3+}$ gives: $\text{Fe}^{3+}(\text{aq}) + 3\text{OH}^-(\text{aq}) \rightarrow \text{Fe(OH)}_3(\text{s})$. The observation is a red-brown / brown precipitate. The mark scheme also requires stating that it is insoluble in excess $\text{NaOH}$ to distinguish it from amphoteric hydroxides like $\text{Al(OH)}_3$ or $\text{Zn(OH)}_2$.

Key Takeaways

$\text{MnO}_2$ oxidises $\text{Fe}^{2+}$ to $\text{Fe}^{3+}$ in acid. $\text{Fe}^{3+}$ gives a yellow/brown solution and a red-brown precipitate with $\text{NaOH}$ that is insoluble in excess.

Common Mistakes

• Describing the $\text{Fe(OH)}_3$ precipitate as 'green' (which is for $\text{Fe}^{2+}$).
• Forgetting to state that the precipitate is 'insoluble in excess'. This is a critical mark point to distinguish from amphoteric hydroxides.
• Writing 'brown' without 'red-brown' is acceptable, but 'red-brown' is more precise.

Things to Be Careful About

• Must mention both the yellow/pale brown filtrate and the red-brown precipitate insoluble in excess.
• The precipitate is $\text{Fe(OH)}_3$, not $\text{Fe}_2(\text{SO}_4)_3$.

Background Concept

In a redox reaction, the species that is reduced (gains electrons) causes the other species to be oxidised. Therefore, the species that is reduced is called the oxidising agent (or oxidant). In part (a)(ii), $\text{MnO}_2$ (where Mn is +4) was reduced to $\text{Mn}^{2+}$ (where Mn is +2), while $\text{Fe}^{2+}$ was oxidised to $\text{Fe}^{3+}$. Since $\text{MnO}_2$ caused the oxidation of $\text{Fe}^{2+}$, it acted as the oxidising agent.

Understanding the Question

The question asks for a conclusion about the chemical behaviour of $\text{FA 5}$ ($\text{MnO}_2$) based on the observations in (a)(ii). In (a)(ii), $\text{MnO}_2$ was added to $\text{Fe}^{2+}$ in acid, and the resulting filtrate turned yellow/brown and gave a red-brown precipitate with $\text{NaOH}$, indicating the formation of $\text{Fe}^{3+}$. This means $\text{Fe}^{2+}$ was oxidised to $\text{Fe}^{3+}$.

Approach

Since $\text{MnO}_2$ caused $\text{Fe}^{2+}$ to be oxidised, $\text{MnO}_2$ itself must have been reduced. Therefore, $\text{MnO}_2$ acted as an oxidising agent.

Step-by-Step Reasoning

• Observation in (a)(ii): Filtrate is yellow/brown, and gives a red-brown ppt with $\text{NaOH}$ insoluble in excess. This confirms $\text{Fe}^{3+}$ is present.
• Conclusion: $\text{Fe}^{2+}$ was oxidised to $\text{Fe}^{3+}$ by $\text{MnO}_2$.
• Therefore, $\text{MnO}_2$ acted as an oxidising agent / oxidant.

Key Takeaways

If a reagent causes another species to be oxidised, the reagent itself is reduced and acts as an oxidising agent.

Common Mistakes

• Writing 'reducing agent' (this would be the case if $\text{MnO}_2$ was oxidised, e.g., in part (a)(i) where it formed $\text{MnO}_4^{2-}$, but in (a)(ii) it is reduced).
• Writing 'catalyst' ($\text{MnO}_2$ is consumed in the reaction, as seen by the need to filter it out, and its oxidation state changes).

Things to Be Careful About

• Use the exact term 'oxidising agent' or 'oxidant'. 'Oxidiser' is also acceptable.

Background Concept

When aqueous sodium hydroxide is added to a solution containing $\text{Fe}^{3+}$ ions, a precipitation reaction occurs, forming insoluble iron(III) hydroxide. The ionic equation shows only the species that change during the reaction: the aqueous ions combining to form the solid precipitate. State symbols are mandatory in CIE A-Level Chemistry for ionic equations involving precipitation or neutralisation.

Additionally, the filtrate from (a)(ii) was made by adding dilute sulfuric acid, so it contains excess $\text{H}^+$ ions. Adding $\text{NaOH}$ will first neutralise the acid: $\text{H}^+(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{H}_2\text{O}(\text{l})$. The mark scheme accepts this as an alternative correct ionic equation for 'the reaction between aqueous sodium hydroxide and the filtrate'.

Understanding the Question

The question asks for an ionic equation for the reaction between aqueous $\text{NaOH}$ and the filtrate from (a)(ii), including state symbols. The filtrate contains $\text{Fe}^{3+}$ (and excess $\text{H}^+$ from the dilute sulfuric acid added earlier).

Approach

Write the ionic equation for the precipitation of $\text{Fe(OH)}_3$, or the neutralisation of excess acid. Both are acceptable according to the mark scheme.

Step-by-Step Reasoning

• Primary equation: The main observable reaction is the precipitation of $\text{Fe}^{3+}$ by $\text{OH}^-$.

• Alternative equation: The filtrate contains excess $\text{H}^+$ from the dilute sulfuric acid. $\text{NaOH}$ will neutralise this:

• Both equations must have correct state symbols: (aq) for ions, (s) for precipitate, (l) for water.

Key Takeaways

Always include state symbols in ionic equations. The primary reaction is precipitation, but neutralisation of excess acid is also a valid reaction occurring in the filtrate.

Common Mistakes

• Forgetting state symbols (e.g., writing $\text{Fe}^{3+} + 3\text{OH}^- \rightarrow \text{Fe(OH)}_3$).
• Writing the full molecular equation instead of the ionic equation (e.g., $\text{FeCl}_3 + 3\text{NaOH} \rightarrow \text{Fe(OH)}_3 + 3\text{NaCl}$).
• Balancing errors (e.g., $\text{Fe}^{3+} + \text{OH}^- \rightarrow \text{FeOH}_3$).

Things to Be Careful About

• The charge on iron(III) is 3+, not 2+.
• The precipitate is $\text{Fe(OH)}_3$, not $\text{FeOH}_3$ or $\text{Fe}_2(\text{OH})_6$ (though the latter is stoichiometrically equivalent, $\text{Fe(OH)}_3$ is the standard form).
• State symbols must be exact: (aq), (s), (l).

Background Concept

This question requires systematic qualitative analysis of two unknown salt solutions, FA 6 and FA 7.

• FA 6 is identified as $\text{FeCl}_3$ (containing $\text{Fe}^{3+}$, $\text{H}^+$, and $\text{Cl}^-$). $\text{Fe}^{3+}$ gives a red-brown precipitate with $\text{NaOH}$ (insoluble in excess). $\text{Fe}^{3+}$ solutions are acidic due to hydrolysis, so they contain excess $\text{H}^+$. $\text{Cl}^-$ gives no reaction with acidified $\text{KMnO}_4$ or $\text{Na}_2\text{CO}_3$, but reacts with $\text{Na}_2\text{CO}_3$ to produce $\text{CO}_2$ (effervescence) due to the acidity of the $\text{Fe}^{3+}$ solution.
• FA 7 is identified as $\text{KI}$ (containing $\text{I}^-$ and $\text{K}^+$). $\text{I}^-$ is a reducing agent. It is oxidised by acidified $\text{KMnO}_4$ to $\text{I}_2$ (brown solution). It is oxidised by $\text{H}_2\text{O}_2$ to $\text{I}_2$ (brown solution, turns blue-black with starch) and $\text{O}_2$ gas (effervescence, relights glowing splint). $\text{I}^-$ does not react with $\text{NaOH}$ or $\text{Na}_2\text{CO}_3$.

Understanding the Question

The candidate must perform four tests on both FA 6 and FA 7 and record observations. The tests are:

• Test 1: $\text{NaOH}$ (aq) then warm gently. (Tests for cations like $\text{NH}_4^+$, $\text{Fe}^{2+}$, $\text{Fe}^{3+}$, $\text{Cu}^{2+}$, etc.)
• Test 2: Acidified $\text{KMnO}_4$. (Tests for reducing agents like $\text{Fe}^{2+}$, $\text{I}^-$, $\text{SO}_3^{2-}$)
• Test 3: $\text{H}_2\text{O}_2$ then starch. (Tests for reducing agents like $\text{I}^-$; FA 6 is crossed out)
• Test 4: $\text{Na}_2\text{CO}_3$. (Tests for cations that form insoluble carbonates or react with carbonate to produce $\text{CO}_2$)

Approach

1. Recall the expected observations for $\text{Fe}^{3+}$ and $\text{Cl}^-$ (FA 6) in each test.
2. Recall the expected observations for $\text{I}^-$ (FA 7) in each test.
3. Record them systematically in the table.

Step-by-Step Reasoning

Test 1: + NaOH (aq), then warm gently

• FA 6 ($\text{FeCl}_3$): $\text{Fe}^{3+}$ reacts with $\text{OH}^-$ to form $\text{Fe(OH)}_3$, a red-brown / brown / rust precipitate. This is insoluble in excess $\text{NaOH}$. Warming does not produce $\text{NH}_3$, so no change / red litmus does not turn blue.
• FA 7 ($\text{KI}$): $\text{K}^+$ and $\text{I}^-$ do not react with $\text{NaOH}$. No change / no precipitate. Warming produces no gas, so no change / red litmus does not turn blue.

Test 2: + acidified KMnO4(aq)

• FA 6 ($\text{FeCl}_3$): $\text{Fe}^{3+}$ and $\text{Cl}^-$ are not reducing agents strong enough to reduce $\text{MnO}_4^-$ under these conditions (or $\text{Cl}^-$ requires concentrated acid and heat). No change / purple colour remains.
• FA 7 ($\text{KI}$): $\text{I}^-$ is oxidised to $\text{I}_2$ by $\text{MnO}_4^-$. The solution turns brown / red-brown.

Test 3: + H2O2, then starch solution

• FA 6: Crossed out, no answer required.
• FA 7 ($\text{KI}$): $\text{H}_2\text{O}_2$ oxidises $\text{I}^-$ to $\text{I}_2$ and is reduced to $\text{H}_2\text{O}$, but excess $\text{H}_2\text{O}_2$ can also decompose to give $\text{O}_2$. Observations:
  • Solution turns yellow / red-brown / brown (due to $\text{I}_2$).
  • Effervescence / fizzing / bubbling (due to $\text{O}_2$ from decomposition of excess $\text{H}_2\text{O}_2$).
  • Gas relights a glowing splint (confirms $\text{O}_2$).
  • Starch turns dark blue / blue-black / black (confirms $\text{I}_2$).

Test 4: + Na2CO3(aq)

• FA 6 ($\text{FeCl}_3$): $\text{Fe}^{3+}$ is acidic (hydrolysis produces $\text{H}^+$). $\text{H}^+$ reacts with $\text{CO}_3^{2-}$ to produce $\text{CO}_2$ gas: effervescence. The gas turns limewater milky (though not asked). $\text{Fe}^{3+}$ also reacts with $\text{CO}_3^{2-}$ / $\text{OH}^-$ (from carbonate hydrolysis) to form a red-brown / orange-brown / brown precipitate of $\text{Fe(OH)}_3$ or basic iron carbonate.
• FA 7 ($\text{KI}$): $\text{K}^+$ and $\text{I}^-$ do not react with $\text{Na}_2\text{CO}_3$. No change.

Key Takeaways

• $\text{Fe}^{3+}$ gives a red-brown ppt with $\text{NaOH}$ (insoluble in excess) and reacts with $\text{Na}_2\text{CO}_3$ to give effervescence and a brown ppt.
• $\text{I}^-$ is a reducing agent: turns acidified $\text{KMnO}_4$ brown, oxidised by $\text{H}_2\text{O}_2$ to give brown $\text{I}_2$ (blue-black with starch) and $\text{O}_2$ gas.
• Group 1 cations ($\text{K}^+$) and halide anions ($\text{I}^-$, $\text{Cl}^-$) generally give no reaction with $\text{NaOH}$ or $\text{Na}_2\text{CO}_3$ (except when the solution is acidic).

Common Mistakes

• Describing the $\text{Fe(OH)}_3$ precipitate as 'green' (that's for $\text{Fe}^{2+}$).
• Forgetting to state 'insoluble in excess' for the $\text{Fe(OH)}_3$ precipitate.
• Writing 'no change' for Test 2 on FA 7 (forgetting that $\text{I}^-$ reduces $\text{MnO}_4^-$).
• Forgetting the starch colour change (blue-black) in Test 3 for FA 7.
• Writing 'brown ppt' for Test 4 on FA 7 (no reaction occurs).

Things to Be Careful About

• Test 3 for FA 6 is crossed out in the table; do not write an answer for it.
• In Test 1, must record observations for BOTH adding $\text{NaOH}$ AND warming. For $\text{Fe}^{3+}$, the ppt forms immediately; warming does nothing special (no $\text{NH}_3$). For $\text{KI}$, no ppt forms, and warming does nothing.
• In Test 4 for FA 6, the effervescence is due to the acidity of the $\text{Fe}^{3+}$ solution (hydrolysis), not because $\text{Cl}^-$ is a carbonate. The brown ppt is due to $\text{Fe}^{3+}$ reacting with the basic carbonate / hydroxide.

Background Concept

To identify the halide anion in a solution, the standard test is to add dilute nitric acid (to remove interfering ions like carbonate or hydroxide) followed by aqueous silver nitrate ($\text{AgNO}_3$). The colour of the precipitate identifies the halide:

• $\text{Cl}^-$: white precipitate of $\text{AgCl}$ (soluble in dilute ammonia)
• $\text{Br}^-$: cream precipitate of $\text{AgBr}$ (soluble in concentrated ammonia)
• $\text{I}^-$: yellow precipitate of $\text{AgI}$ (insoluble in ammonia)

The question states that FA 6 does not contain nitrogen, and from the previous tests, we know it contains $\text{Fe}^{3+}$ and is acidic. The anion must be identified. Since $\text{FA 6}$ is $\text{FeCl}_3$, the anion is $\text{Cl}^-$.

Understanding the Question

The candidate must select one further reagent to identify the anion in FA 6 (which is $\text{Cl}^-$), carry out the test, and record observations in Table 3.2.

Approach

1. Identify the anion in FA 6 as chloride ($\text{Cl}^-$) based on the formula $\text{FeCl}_3$ (or deduce it from the context that it's a halide test).
2. Select the reagent for testing halides: silver nitrate ($\text{AgNO}_3$) (and dilute nitric acid, though the mark scheme primarily rewards 'silver nitrate').
3. Record the observation for chloride: white precipitate.

Step-by-Step Reasoning

• Reagent: silver nitrate (or $\text{AgNO}_3$). (Ideally, dilute nitric acid is added first, but 'silver nitrate' is the key creditable reagent in the mark scheme).
• Observation: Addition of $\text{AgNO}_3$ to $\text{Cl}^-$ produces a white precipitate of $\text{AgCl}$.

Key Takeaways

The test for chloride ions uses silver nitrate (and dilute nitric acid) to produce a white precipitate of silver chloride.

Common Mistakes

• Selecting barium chloride or barium nitrate (these test for sulfate, not chloride).
• Writing 'precipitate' without specifying the colour 'white'.
• Forgetting to mention dilute nitric acid (though the mark scheme is lenient, it is good practice).

Things to Be Careful About

• The reagent must be 'silver nitrate' or '$\text{AgNO}_3$'.
• The observation must be 'white precipitate'. 'White solid' or 'white浑浊' (turbid) may not be accepted; 'precipitate' is the key term.

Background Concept

The question states:

• FA 6 contains two cations and one anion. Two of the ions are listed in the Qualitative analysis notes.
• FA 7 contains one cation and one anion. One of the ions is listed in the Qualitative analysis notes.
• Neither solution includes an ion that contains sulfur.

From the tests:

• FA 6 gave a red-brown ppt with $\text{NaOH}$ (insoluble in excess) -> $\text{Fe}^{3+}$ is present.

• FA 6 gave effervescence with $\text{Na}_2\text{CO}_3$ -> acidic solution, so $\text{H}^+$ is present (due to hydrolysis of $\text{Fe}^{3+}$ or added acid).

• FA 6 gave no change with acidified $\text{KMnO}_4$ -> no reducing anion like $\text{I}^-$ or $\text{SO}_3^{2-}$.

• FA 6 test (ii) identified the anion as chloride ($\text{Cl}^-$) via silver nitrate test (white ppt).

• So FA 6 contains $\text{Fe}^{3+}$, $\text{H}^+$, and $\text{Cl}^-$.

• FA 7 gave no change with $\text{NaOH}$ -> cation is not $\text{Fe}^{2+}$, $\text{Fe}^{3+}$, $\text{Cu}^{2+}$, $\text{Al}^{3+}$, $\text{Zn}^{2+}$, etc. Likely a Group 1 cation like $\text{K}^+$ or $\text{Na}^+$.

• FA 7 turned brown with acidified $\text{KMnO}_4$ -> reducing anion, likely $\text{I}^-$ (or $\text{Fe}^{2+}$, but that's a cation, and we already ruled out cations that react with $\text{NaOH}$).

• FA 7 turned brown with $\text{H}_2\text{O}_2$ and blue-black with starch -> confirms $\text{I}^-$.

• FA 7 gave no change with $\text{Na}_2\text{CO}_3$ -> consistent with $\text{K}^+$ or $\text{Na}^+$ and $\text{I}^-$.

• So FA 7 contains $\text{I}^-$ and a cation that is not in the qualitative analysis notes (like $\text{K}^+$ or $\text{Na}^+$). The mark scheme accepts 'unknown' for this cation.