Chemistry 9701/38 — May/June 2025

Background Concept

Significant figures indicate the precision of a measurement. In calculations, the final answer should not imply more precision than the data justify. The mark scheme for this question accepts 3 or 4 significant figures for the final answers in (c)(ii), (c)(iii) and (c)(iv).

Understanding the Question

This is an instruction, not a calculation. You must make sure the numerical answers you give in the next three parts are quoted to an appropriate number of significant figures.

Approach

Use 3 significant figures for all three final answers. This is consistent with the data (0.0100 mol dm-3 has 3 significant figures).

Step-by-Step Reasoning

For example, 1.88 × 10^-4 mol, 0.0376 mol dm-3 and 2.10 g dm-3 all have 3 significant figures. Avoid giving 0.0376000 mol dm-3 or 0.04 mol dm-3.

Key Takeaways

Always check the number of significant figures in the least precise given value and quote final answers to a similar number.

Common Mistakes

• Quoting too many significant figures, e.g. 0.0376000.
• Quoting too few, e.g. 0.04 instead of 0.0376.
• Inconsistent significant figures across related answers.

Things to Be Careful About

• The concentration of FB 2 is given as 0.0100 mol dm-3, which has 3 significant figures.
• The mean titre is quoted to 2 d.p., but that is a burette reading precision, not the number of significant figures for the final concentration.

Background Concept

The amount of a solute in moles is given by amount = concentration × volume, where volume is in dm3. Since burette readings are in cm3, they must be converted by dividing by 1000.

Understanding the Question

You must calculate the moles of manganate(VII) ions in the mean volume of FB 2 found in (b). Use the concentration 0.0100 mol dm-3 and the mean titre, e.g. 18.80 cm3.

Approach

Convert the volume to dm3, then multiply by the concentration.

Step-by-Step Reasoning

Volume = 18.80 cm3 = 18.80/1000 = 0.0188 dm3. Amount = 0.0100 × 0.0188 = 1.88 × 10^-4 mol.

Key Takeaways

Always convert cm3 to dm3 before using concentration in mol dm-3.

Common Mistakes

• Using the volume in cm3 directly: 0.0100 × 18.80 = 0.188 mol, which is wrong.
• Quoting too many significant figures.

Things to Be Careful About

• The answer must be in mol, not mol dm-3.
• Use the mean titre, not the rough titre.

Background Concept

The two half-equations show that one MnO4- ion accepts 5 electrons while each Fe2+ loses 1 electron. Therefore 1 mol MnO4- reacts with 5 mol Fe2+. This 5:1 ratio is essential for the calculation.

Understanding the Question

You now know the moles of MnO4- used. Use the stoichiometric ratio to find the moles of Fe2+ in the 25.0 cm3 sample of FB 1, then divide by the volume in dm3 to get concentration.

Approach

Multiply moles of MnO4- by 5 to get moles of Fe2+. Then divide by 0.0250 dm3.

Step-by-Step Reasoning

Moles Fe2+ = 5 × 1.88 × 10^-4 = 9.40 × 10^-4 mol. Volume = 25.0 cm3 = 0.0250 dm3. Concentration = 9.40 × 10^-4 / 0.0250 = 0.0376 mol dm-3.

Key Takeaways

The stoichiometric ratio from the half-equations is the bridge between the titrant and the analyte.

Common Mistakes

• Using a 1:1 ratio instead of 5:1.
• Forgetting to convert 25.0 cm3 to dm3.
• Using the total volume of the conical flask contents instead of the volume of FB 1.

Things to Be Careful About

• The 25.0 cm3 is the volume of FB 1 pipetted, not the volume after adding FB 3.
• The concentration is of Fe2+ ions, not of FeSO4·7H2O formula units.

Background Concept

Concentration in g dm-3 is obtained by multiplying concentration in mol dm-3 by the molar mass (in g mol-1). For Fe2+, use the atomic mass of iron, 55.8 g mol-1.

Understanding the Question

Convert the molar concentration of Fe2+ from (c)(iii) into a mass concentration.

Approach

Multiply 0.0376 mol dm-3 by 55.8 g mol-1.

Step-by-Step Reasoning

0.0376 × 55.8 = 2.10 g dm-3.

Key Takeaways

Mass concentration = molar concentration × molar mass.

Common Mistakes

• Using the molar mass of FeSO4·7H2O (278.0) instead of Fe (55.8).
• Forgetting units: g dm-3, not mol dm-3.

Things to Be Careful About

• The question asks for iron(II) ions, so use the atomic mass of Fe.
• Keep 3 significant figures.

Background Concept

FB 1 was made by dissolving 14 tablets in 1.00 dm3 of solution. Therefore the mass of Fe2+ in 1.00 dm3 is the total mass from 14 tablets. Dividing by 14 gives the mass per tablet.

Understanding the Question

Compare the calculated mass per tablet with the manufacturer's claim of a minimum of 150 mg Fe2+ per tablet.

Approach

Convert the concentration from (c)(iv) to mg dm-3, multiply by 1.00 dm3, divide by 14, and compare with 150 mg.

Step-by-Step Reasoning

2.10 g dm-3 = 2100 mg dm-3. In 1.00 dm3 there is 2100 mg Fe2+. Mass per tablet = 2100 / 14 = 150 mg. Since 150 mg is equal to the claimed minimum, the claim is correct.

Key Takeaways

Scaling a solution concentration back to a single tablet requires using the number of tablets and the total volume.

Common Mistakes

• Forgetting to divide by 14.
• Not converting g to mg before comparing with 150 mg.
• Concluding 'incorrect' if the calculated value is slightly different due to rounding; here it is exactly 150 mg.

Things to Be Careful About

• Use the exact calculated value from (c)(iv) before rounding if possible.
• The claim is a minimum, so any value ≥ 150 mg supports it.

Background Concept

Plotting a graph of temperature change against the volume of reactant added is a standard technique in calorimetry to determine the stoichiometric ratio. As the volume of acid increases, more heat is released, and the temperature rises. Once the alkali is completely neutralised, any additional acid adds no more heat but increases the total mass of the solution, which absorbs the heat and lowers the maximum temperature. The graph therefore has a peak. Because experimental data is noisy, we draw two lines of best fit: one for the rising portion (limiting alkali) and one for the falling portion (excess acid). The intersection of these two lines gives the theoretical maximum temperature and the exact volume of acid required for neutralisation.

Understanding the Question

You are asked to plot the data from Table 2.1 on the provided grid. The x-axis is the volume of FB 5 (hydrochloric acid), and the y-axis is the maximum temperature. You must:

1. Choose suitable scales for both axes.
2. Plot all points accurately.
3. Identify and label any anomalous points.
4. Draw two lines of best fit (straight lines are acceptable and often preferred for this type of graph) that account for all non-anomalous points.
5. Extrapolate both lines so they intersect.

Approach

• Axes: x-axis = volume of FB 5 / cm³ (0 to 10), y-axis = maximum temperature / °C (start at initial temp, go up to at least max temp + 2 °C).
• Plotting: Place each (volume, temperature) pair accurately on the grid.
• Anomalous points: If a point is far from the expected trend (e.g., a point that is much lower than the line), label it as anomalous and do not include it in the line of best fit.
• Lines of best fit: Draw a straight line through the rising points (exp 1 to 3/4) and another through the falling points (exp 4 to 5/6/7). Extend both lines until they cross.

Step-by-Step Reasoning

• Scale selection: If temperatures range from 20.0 to 26.5 °C, the y-axis could go from 19.0 to 28.0 °C (giving 2 °C above the max). The x-axis goes from 0.00 to 10.00 cm³.
• Plotting: For example, (1.00, 21.2), (3.00, 23.5), (5.00, 26.5), (7.00, 25.0), (9.00, 23.8).
• Identifying anomalies: Suppose the point (3.00, 23.5) is slightly off the line. Label it 'anomalous'.
• Drawing lines: Draw a line from (0, 20.0) through (1, 21.2) and (5, 26.5). Draw another line from (5, 26.5) through (7, 25.0) and (9, 23.8). Extend both lines to intersect at approximately x = 5.00 cm³, y = 26.5 °C.

Key Takeaways

• The y-axis must have a suitable scale, extending at least 2 °C above the highest maximum temperature.
• Two lines of best fit are required: one for the increase and one for the decrease. They should be straight lines (or smooth curves, but straight lines are standard here).
• Anomalous points must be labelled and excluded from the lines of best fit.
• The intersection gives the volume of acid for complete neutralisation.

Common Mistakes

• Using a non-linear scale or an axis that doesn't start at zero (for the x-axis) or doesn't cover the data range.
• Drawing a single smooth curve through all points instead of two distinct lines of best fit.
• Forgetting to extrapolate the lines so they actually intersect.
• Not labelling axes with quantities and units.

Things to Be Careful About

• The mark scheme requires unambiguous labelled axes with units or correct units. Write 'maximum temperature / °C' and 'volume of FB 5 / cm³'.
• Points must be plotted accurately; even a small plotting error can shift the intersection and cost marks in part (b)(ii).
• Lines of best fit must account for all plotted points ignoring any points labelled anomalous. Do not force a line through an anomalous point.
• Extrapolation must be clearly shown; do not just draw the lines up to the data points.

Background Concept

The intersection of the two lines of best fit on a temperature vs. volume graph represents the theoretical point of complete neutralisation. At this point, the acid and alkali are in exact stoichiometric proportions, releasing the maximum possible heat. The x-coordinate of this intersection gives the volume of the titrant (FB 5) required to neutralise the fixed volume of the analyte (FB 4).

Understanding the Question

You must read the volume of FB 5 at the intersection point of the two lines drawn in part (b)(i). This value is used in subsequent calculations to determine the moles of acid and alkali.

Approach

• Locate the intersection of the two extrapolated lines on the graph.
• Draw a vertical line down to the x-axis.
• Read the value and record it to the appropriate number of decimal places (usually 2 d.p. for burette/pipette volumes, so 5.00 cm³).

Step-by-Step Reasoning

• In the example graph, the lines intersect at x = 5.00 cm³.
• Record this value as the volume of FB 5 required to neutralise 10.0 cm³ of FB 4.

Key Takeaways

• The intersection point gives the stoichiometric volume. Always read this value carefully from the graph.
• Ensure you read the correct axis (x-axis for volume).

Common Mistakes

• Reading the y-value (temperature) instead of the x-value (volume).
• Not reading to the correct precision (e.g., 5 cm³ instead of 5.00 cm³).
• Using a value from a single data point instead of the intersection.

Things to Be Careful About

• The mark scheme awards the mark for a correctly read volume. If your graph is poor, your reading will be wrong, and you may lose marks in later calculation parts (though ecf is often allowed if you state your assumption clearly, the mark scheme here uses the value from (b)(ii) directly).
• Use the value you read from your own graph, even if it differs from the 'true' value, as error carried forward may apply in later parts (check the specific paper guidance, but here the mark scheme uses 'volume from (b)(ii)').

Background Concept

The amount of substance (in moles) is calculated using the formula $n = c \times V$, where $c$ is concentration in mol dm⁻³ and $V$ is volume in dm³. Since volume is usually given in cm³, we divide by 1000. In a neutralisation reaction between a strong acid (HCl) and a strong alkali (NaOH), the equation is $\text{HCl}(\text{aq}) + \text{NaOH}(\text{aq}) \rightarrow \text{NaCl}(\text{aq}) + \text{H}_2\text{O}(\text{l})$. The mole ratio is 1:1, meaning 1 mole of HCl reacts with exactly 1 mole of NaOH.

Understanding the Question

You are asked to calculate the moles of HCl in the volume determined in part (b)(ii), and then deduce the moles of NaOH in 10.0 cm³ of FB 4. The mark scheme allows the use of 5.10 cm³ if you couldn't answer (b)(ii), but you should use your own graph value.

Approach

1. Convert the volume of FB 5 from cm³ to dm³ by dividing by 1000.
2. Multiply by the concentration of FB 5 (2.00 mol dm⁻³) to get moles of HCl.
3. Use the 1:1 stoichiometry to state that moles of NaOH = moles of HCl.
4. Ensure answers are given to 2–4 significant figures.

Step-by-Step Reasoning

• Volume of FB 5 = 5.00 cm³ (from your graph).
• Moles of HCl = $2.00 \times \frac{5.00}{1000} = 0.0100$ mol.
• Since the ratio is 1:1, moles of NaOH = 0.0100 mol.
• Both values are given to 3 significant figures, which satisfies the 2–4 sf requirement.

Key Takeaways

• Always convert volume to dm³ before using $n = c \times V$.
• Strong acid-strong base neutralisation is always 1:1 for monoprotic acids and monobasic alkalis.
• Pay attention to significant figures; the concentration (2.00) has 3 sf, so the answer should have at least 3 sf.

Common Mistakes

• Forgetting to divide the volume by 1000 (calculating moles as $2.00 \times 5.00 = 10.0$ mol).
• Using the wrong volume (e.g., using 10.0 cm³ instead of the intersection volume).
• Not stating the 1:1 ratio explicitly, though the mark scheme just requires the correct final value for NaOH.

Things to Be Careful About

• The mark scheme requires both answers to 2–4 sf. 0.0100 is 3 sf. Writing 0.01 might be marked as 1 sf and could lose a mark if the examiner is strict, though often 2 sf (0.010) is accepted. Better to write 0.0100.
• If you used the alternative volume 5.10 cm³, moles HCl = $2.00 \times 5.10 / 1000 = 0.0102$ mol. Use your own value consistently.

Background Concept

The energy change (heat released) in a solution is calculated using the equation $q = mc\Delta T$, where:

• $m$ is the mass of the solution in grams.
• $c$ is the specific heat capacity of the solution, typically assumed to be that of water ($4.18 \text{ J g}^{-1} \text{ K}^{-1}$).
• $\Delta T$ is the temperature change in °C or K.

For dilute aqueous solutions, we assume the density is $1.00 \text{ g cm}^{-3}$, so the mass in grams is equal to the volume in cm³. The total volume here is $10.0 \text{ cm}^3 \text{ (NaOH)} + 10.0 \text{ cm}^3 \text{ (water + HCl)} = 20.0 \text{ cm}^3$, so $m = 20.0 \text{ g}$.

Understanding the Question

You must calculate the energy released when the amounts of HCl and NaOH calculated in (c)(i) neutralise each other. This is the energy corresponding to the maximum temperature rise in the experiment where complete neutralisation occurred (at the intersection point).

Approach

1. Calculate the total volume of the solution: $10.0 \text{ cm}^3 + 10.0 \text{ cm}^3 = 20.0 \text{ cm}^3$.
2. Assume mass $m = 20.0 \text{ g}$ and $c = 4.18 \text{ J g}^{-1} \text{ K}^{-1}$.
3. Calculate $\Delta T = T_{\text{max}} - T_{\text{initial}}$.
4. Calculate $q = 20.0 \times 4.18 \times \Delta T$.
5. Round to 2–4 significant figures.

Step-by-Step Reasoning

• Total volume = 20.0 cm³, so mass = 20.0 g.
• $\Delta T = 26.5 - 20.0 = 6.5$ °C.
• Energy = $20.0 \times 4.18 \times 6.5 = 543.4$ J.
• Round to 3 sf: 543 J.

Key Takeaways

• The mass $m$ is the total mass of the mixed solution, not just the acid or alkali.
• The specific heat capacity is not given in the question; you must assume $4.18 \text{ J g}^{-1} \text{ K}^{-1}$ (value for water).
• $\Delta T$ is the difference between the maximum temperature and the initial temperature of the alkali (FB 4). Do not use the initial temperature of the acid if it differs (though in this experiment, both are assumed to be at room temperature, so $T_{\text{initial}}$ is the same).

Common Mistakes

• Using the wrong mass (e.g., using 10.0 g instead of 20.0 g).
• Forgetting to use $\Delta T$ and just using the maximum temperature.
• Not showing working; the mark scheme requires the formula and substitution.

Things to Be Careful About

• The mark scheme explicitly states: Energy = $20 \times 4.18 \times \Delta T$. You must use 20 for the mass.
• $\Delta T$ must be calculated as max T - initial T of FB 4. If your initial temp was 20.0 and max was 26.5, $\Delta T = 6.5$.
• The answer must be in Joules (J), not kJ. The next part will convert it.

Background Concept

The enthalpy change of neutralisation ($\Delta H$) is the energy released when one mole of water is formed from the reaction of an acid and an alkali. It is calculated by dividing the total energy change (in kJ) by the number of moles of water formed (which equals the moles of acid or alkali reacted, since the ratio is 1:1).

The negative sign is used because neutralisation is an exothermic reaction (heat is released to the surroundings, so the system loses energy).

Understanding the Question

You must calculate the enthalpy change in kJ mol⁻¹ when one mole of FB 4 is neutralised by one mole of FB 5. This is the standard enthalpy change of neutralisation for a strong acid and strong alkali.

Approach

1. Convert the energy change from (c)(ii) from Joules to kilojoules by dividing by 1000.
2. Divide by the number of moles of NaOH (or HCl) from (c)(i).
3. Apply a negative sign because the reaction is exothermic.
4. Round to at least 2 significant figures.

Step-by-Step Reasoning

• Energy change = 543.4 J = 0.5434 kJ.
• Moles = 0.0100 mol.
• $\Delta H = -\frac{0.5434}{0.0100} = -54.34 \text{ kJ mol}^{-1}$.
• Round to 3 sf: -54.3 kJ mol⁻¹.
• The mark scheme requires the sign and the value. The value is typically around -57.1 kJ mol⁻¹ for strong acid/strong base; experimental values are lower due to heat loss.

Key Takeaways

• Always convert J to kJ before calculating enthalpy change in kJ mol⁻¹.
• The sign MUST be negative for exothermic reactions. Forgetting the sign costs a mark.
• The value should be around -57 kJ mol⁻¹; if you get a positive value or a value > -60, check your working.

Common Mistakes

• Forgetting the negative sign.
• Forgetting to divide the energy by 1000 to convert J to kJ (getting -5434 kJ mol⁻¹).
• Dividing by the wrong number of moles (e.g., using 10.0 instead of 0.0100).
• Not rounding to at least 2 sf.

Things to Be Careful About

• The mark scheme explicitly asks for the sign and the value: 'enthalpy change = (sign) (value) kJ mol⁻¹'. Write '-54.3'.
• Ensure you use the moles from (c)(i), not the concentration or volume.
• Experimental enthalpy changes of neutralisation are typically less exothermic than the theoretical -57.1 kJ mol⁻¹ due to heat loss to the surroundings, heat absorbed by the cup, and incomplete mixing. A value of -50 to -55 kJ mol⁻¹ is common and acceptable.

Background Concept

Qualitative analysis identifies substances by observing how they react with selected reagents. A key skill is linking an observation to a specific chemical reaction. Here three aqueous unknowns are tested with three reagents: manganese(IV) oxide (a catalyst for the decomposition of hydrogen peroxide), magnesium metal (a reducing metal that reacts with acids), and iron(II) sulfate (a source of $\text{Fe}^{2+}$ ions).

Understanding the Question

You are given three aqueous solutions FB 6, FB 7 and FB 8, each containing at least one oxygen atom. You must carry out three tests on each and record observations. The mark scheme rewards specific observations: effervescence, gas identification, colour changes, and precipitate formation.

Approach

Think about what each reagent does:

• $\text{MnO}_2$ catalyses the decomposition of hydrogen peroxide: $2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2$. Only $\text{H}_2\text{O}_2$ gives a gas.
• Mg reacts with acids: $\text{Mg} + 2\text{H}^+ \rightarrow \text{Mg}^{2+} + \text{H}_2$. Only an acid gives a gas.
• $\text{FeSO}_4$: with alkali (NaOH) gives green $\text{Fe}(\text{OH})_2$ precipitate which turns brown on oxidation; with $\text{H}_2\text{O}_2$, $\text{Fe}^{2+}$ is oxidised to $\text{Fe}^{3+}$ giving a yellow/brown solution.

Step-by-Step Reasoning

FB 6 = $\text{CH}_3\text{COOH}$ (ethanoic acid):

• Test 1 ($\text{MnO}_2$): no change — no $\text{H}_2\text{O}_2$ present to decompose.
• Test 2 (Mg): fizzing — acid reacts with Mg: $\text{Mg} + 2\text{CH}_3\text{COOH} \rightarrow (\text{CH}_3\text{COO})_2\text{Mg} + \text{H}_2$. The gas gives a pop with a lit splint ($\text{H}_2$). Heat is released (exothermic).
• Test 3 ($\text{FeSO}_4$): no change — $\text{CH}_3\text{COOH}$ does not react with $\text{Fe}^{2+}$ to give a visible change.

FB 7 = $\text{NaOH}$:

• Test 1 ($\text{MnO}_2$): no change.
• Test 2 (Mg): no change — no acid present.
• Test 3 ($\text{FeSO}_4$): green precipitate of $\text{Fe}(\text{OH})_2$ forms: $\text{Fe}^{2+} + 2\text{OH}^- \rightarrow \text{Fe}(\text{OH})_2$. On standing, air oxidises $\text{Fe}(\text{OH})_2$ to $\text{Fe}(\text{OH})_3$, which is brown.

FB 8 = $\text{H}_2\text{O}_2$:

• Test 1 ($\text{MnO}_2$): $\text{MnO}_2$ catalyses decomposition: $2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2$. Fizzing; the gas relights a glowing splint (oxygen).
• Test 2 (Mg): no change — $\text{H}_2\text{O}_2$ is not acidic enough to react with Mg.
• Test 3 ($\text{FeSO}_4$): $\text{H}_2\text{O}_2$ oxidises $\text{Fe}^{2+}$ to $\text{Fe}^{3+}$: $2\text{Fe}^{2+} + \text{H}_2\text{O}_2 + 2\text{H}^+ \rightarrow 2\text{Fe}^{3+} + 2\text{H}_2\text{O}$. The solution turns yellow/orange-brown ($\text{Fe}^{3+}$). $\text{H}_2\text{O}_2$ also decomposes, giving fizzing and oxygen that relights a glowing splint.

Key Takeaways

• Link each observation to a specific reaction.
• Gas tests: $\text{O}_2$ relights a glowing splint; $\text{H}_2$ gives a pop with a lit splint; $\text{CO}_2$ turns limewater milky.
• $\text{Fe}(\text{OH})_2$ is green and turns brown on oxidation to $\text{Fe}(\text{OH})_3$.
• $\text{Fe}^{3+}$ solutions are yellow/brown.

Common Mistakes

• Writing "gas produced" without identifying it — the mark requires the gas test.
• Not recording "no change" — this is a valid and credited observation.
• Confusing the green $\text{Fe}(\text{OH})_2$ precipitate with a brown one; the colour change on standing is a separate mark.
• Forgetting the heat released observation where the mark scheme awards it.

Things to Be Careful About

• Record at what stage an observation is made (e.g. "on warming", "on standing").
• Use the correct formula or name of reagents.
• If a solid is heated use a hard-glass tube; if a solution is warmed use a boiling tube.
• The mark scheme rounds down: 2 asterisks = 1 mark, so partial observations still score.

Background Concept

The observations in (a)(i) are the evidence. Each observation is characteristic of a functional group or compound type: an acid reacts with Mg to give $\text{H}_2$; an alkali gives a green precipitate with $\text{Fe}^{2+}$ that turns brown; $\text{H}_2\text{O}_2$ decomposes catalytically to $\text{O}_2$ and oxidises $\text{Fe}^{2+}$ to $\text{Fe}^{3+}$.

Understanding the Question

You must suggest a possible formula for each of FB 6, FB 7 and FB 8 based on your observations. The mark scheme accepts $\text{CH}_3\text{COOH}$ (or any acid containing oxygen), $\text{NaOH}$ (or any Group 1 hydroxide), and $\text{H}_2\text{O}_2$.

Approach

Match the observations to the chemistry:

• Fizzes with Mg, $\text{H}_2$ gas → acid. Contains O → $\text{CH}_3\text{COOH}$.
• Green precipitate with $\text{FeSO}_4$ → alkali ($\text{OH}^-$). Contains O → $\text{NaOH}$.
• Decomposes with $\text{MnO}_2$ giving $\text{O}_2$, oxidises $\text{Fe}^{2+}$ → $\text{H}_2\text{O}_2$.

Step-by-Step Reasoning

FB 6: only the Mg test gave a gas ($\text{H}_2$). This identifies an acid. Since it must contain oxygen, $\text{CH}_3\text{COOH}$ (ethanoic acid) is a valid formula. Any acid containing O would score.

FB 7: only the $\text{FeSO}_4$ test gave a green precipitate turning brown. This is $\text{Fe}(\text{OH})_2 \rightarrow \text{Fe}(\text{OH})_3$, requiring $\text{OH}^-$ ions. $\text{NaOH}$ (or any Group 1 hydroxide) is the answer.

FB 8: only the $\text{MnO}_2$ test gave oxygen (relights glowing splint), and $\text{FeSO}_4$ turned yellow-brown ($\text{Fe}^{3+}$ formed). Both point to $\text{H}_2\text{O}_2$, which decomposes catalytically and is an oxidising agent.

Key Takeaways

• Observations are evidence for functional groups.
• A single distinctive test can identify a compound.

Common Mistakes

• Writing "$\text{H}_2\text{O}$" instead of $\text{H}_2\text{O}_2$ — water does not decompose with $\text{MnO}_2$.
• Giving a formula that does not contain oxygen, contradicting the stem.

Things to Be Careful About

• The stem states each compound contains at least one oxygen atom — your formula must satisfy this.
• The mark scheme allows any acid containing O and any Group 1 hydroxide.

Background Concept

Carbonates react with acids to give $\text{CO}_2$: $\text{CO}_3^{2-} + 2\text{H}^+ \rightarrow \text{CO}_2 + \text{H}_2\text{O}$. $\text{CO}_2$ is confirmed by bubbling it through limewater ($\text{Ca}(\text{OH})_2$), which turns milky due to $\text{CaCO}_3$ precipitate.

Understanding the Question

FB 9 contains two anions and two cations. Adding dilute nitric acid to the solid tests for carbonate (or other acid-reactive anions). You record observations and keep the solution for (b)(ii).

Approach

Add dilute nitric acid to the solid. If carbonate is present, effervescence occurs. Test the gas with limewater to confirm $\text{CO}_2$.

Step-by-Step Reasoning

• Fizzing on adding $\text{HNO}_3$: carbonate present, $\text{CO}_2$ released.
• Colourless solution: the solid dissolves.
• Limewater test: gas bubbled through limewater gives white precipitate ($\text{CaCO}_3$), confirming $\text{CO}_2$.

Note: nitric acid is used rather than HCl or $\text{H}_2\text{SO}_4$ because it does not interfere with later halide tests (no $\text{Cl}^-$ or $\text{SO}_4^{2-}$ introduced).

Key Takeaways

• Carbonate test: acid + limewater.
• Nitric acid is the acid of choice before silver nitrate tests.

Common Mistakes

• Using HCl — introduces $\text{Cl}^-$ which would interfere with the silver nitrate test in (b)(ii).
• Not testing the gas with limewater.

Things to Be Careful About

• Record the limewater result as "milky" or "white precipitate".
• Keep the solution for the next test.

Background Concept

Silver nitrate + halide → silver halide precipitate. $\text{AgCl}$ is white and soluble in dilute ammonia; $\text{AgBr}$ is cream and soluble in concentrated ammonia; $\text{AgI}$ is yellow and insoluble in ammonia. The colour and ammonia solubility identify the halide.

Understanding the Question

To the solution from (b)(i) (which contains the anions), add $\text{AgNO}_3$ then excess $\text{NH}_3$. The observation identifies the halide present.

Approach

Add $\text{AgNO}_3$: a precipitate forms. Its colour (cream) suggests $\text{AgBr}$. Test solubility in ammonia to confirm.

Step-by-Step Reasoning

• $\text{Ag}^+ + \text{Br}^- \rightarrow \text{AgBr}(\text{s})$: cream precipitate.
• $\text{AgBr}$ is only partially soluble / insoluble in aqueous ammonia (unlike $\text{AgCl}$ which dissolves in dilute $\text{NH}_3$).

Since the precipitate is cream and insoluble in ammonia, the halide is bromide.

Key Takeaways

• $\text{AgCl}$ white (soluble in dilute $\text{NH}_3$); $\text{AgBr}$ cream (soluble in concentrated $\text{NH}_3$); $\text{AgI}$ yellow (insoluble).
• Ammonia solubility distinguishes halides.

Common Mistakes

• Writing "white precipitate" — that would suggest chloride.
• Saying the precipitate dissolves in ammonia — that would suggest chloride.

Things to Be Careful About

• Use "cream" not "white".
• The solution from (b)(i) is acidic — this is fine for the silver nitrate test.

Background Concept

• Ammonium salts + NaOH on warming → $\text{NH}_3$ gas: $\text{NH}_4^+ + \text{OH}^- \rightarrow \text{NH}_3 + \text{H}_2\text{O}$. $\text{NH}_3$ is alkaline, turning damp red litmus blue.
• Halide + chlorine water: $\text{Cl}_2$ displaces the halide: $2\text{Br}^- + \text{Cl}_2 \rightarrow \text{Br}_2 + 2\text{Cl}^-$. $\text{Br}_2$ in solution is brown/orange.

Understanding the Question

Two tests on the aqueous solution of FB 9: (1) NaOH + warm → tests for $\text{NH}_4^+$; (2) HCl + chlorine water → tests for halide ($\text{Br}^-$ gives brown $\text{Br}_2$).

Approach

• NaOH + heat: if $\text{NH}_4^+$ present, $\text{NH}_3$ evolved.
• Chlorine water: if $\text{Br}^-$ present, $\text{Br}_2$ formed → brown/yellow solution.

Step-by-Step Reasoning

Test 1: No precipitate with NaOH (no metal cation that precipitates), but on warming $\text{NH}_3$ is evolved — confirms $\text{NH}_4^+$.

Test 2: HCl acidifies; chlorine water oxidises $\text{Br}^-$ to $\text{Br}_2$. The brown/yellow colour confirms bromide. (If it were $\text{I}^-$, a brown/black or violet colour would form; if $\text{Cl}^-$, no change.)

Key Takeaways

• $\text{NH}_4^+$ test: NaOH + heat, $\text{NH}_3$ turns red litmus blue.
• Halide test: chlorine water displaces $\text{Br}_2$ (brown) or $\text{I}_2$ (brown/violet).

Common Mistakes

• Not warming the NaOH — $\text{NH}_3$ may not be evolved.
• Using the wrong indicator (must be damp red litmus).
• Saying "colourless" for the chlorine water test — $\text{Br}_2$ gives a colour.

Things to Be Careful About

• Rinse the test-tube immediately after the chlorine test (chlorine is hazardous).
• Record the colour change precisely.

Background Concept

Each test identifies a specific ion:

• (b)(i) $\text{CO}_2$ → carbonate.
• (b)(ii) cream precipitate insoluble in $\text{NH}_3$ → bromide.
• (b)(iii) $\text{NH}_3$ evolved → ammonium.
  The second cation is not identified by the tests — it is "unknown" ($\text{Na}^+$).

Understanding the Question

Deduce the cations and anions in FB 9 from all observations. FB 9 = $\text{Na}_2\text{CO}_3 + \text{NH}_4\text{Br}$.

Approach

Collect the evidence:

• $\text{CO}_2$ with acid → $\text{CO}_3^{2-}$
• Cream precipitate, insoluble in $\text{NH}_3$ → $\text{Br}^-$
• $\text{NH}_3$ with NaOH + heat → $\text{NH}_4^+$
• Second cation: not tested → unknown

Step-by-Step Reasoning

• (b)(i): fizzing + limewater milky → $\text{CO}_3^{2-}$.
• (b)(ii): cream precipitate insoluble in ammonia → $\text{Br}^-$.
• (b)(iii) Test 1: $\text{NH}_3$ evolved → $\text{NH}_4^+$.
• (b)(iii) Test 2: brown solution → $\text{Br}^-$ confirmed.
• The cation that balances $\text{CO}_3^{2-}$ is not identified by these tests; it is "unknown" ($\text{Na}^+$).

Key Takeaways

• A systematic series of tests identifies ions.
• "Unknown" is a valid answer when the test does not identify an ion.

Common Mistakes

• Guessing the second cation without evidence.
• Missing the carbonate deduction from the limewater test.

Things to Be Careful About

• The mark scheme accepts "unknown" for the second cation.
• Both anions must be stated.

Background Concept

When $\text{AgNO}_3$ is added to a solution containing $\text{Br}^-$, the $\text{Ag}^+$ and $\text{Br}^-$ combine to form insoluble $\text{AgBr}$. The nitrate and any other spectator ions are omitted.

Understanding the Question

Write the ionic equation for the reaction in (b)(ii) — the formation of the cream precipitate.

Approach

Identify the reacting ions: $\text{Ag}^+$ and $\text{Br}^-$. Combine them into the insoluble product $\text{AgBr}$. Add state symbols.

Step-by-Step Reasoning

$\text{AgNO}_3$ dissociates to $\text{Ag}^+(\text{aq}) + \text{NO}_3^-(\text{aq})$. $\text{Br}^-$ comes from the bromide salt. $\text{Ag}^+ + \text{Br}^- \rightarrow \text{AgBr}(\text{s})$. $\text{NO}_3^-$ is a spectator ion and is omitted.

Key Takeaways

• Ionic equations omit spectator ions.
• State symbols are required.

Common Mistakes

• Including $\text{NO}_3^-$ in the equation.
• Missing state symbols — (aq) and (s) are required.
• Writing $\text{AgBr}(\text{aq})$ instead of $(\text{s})$.

Things to Be Careful About

• The mark scheme requires state symbols.
• Balance the charges and atoms.