Chemistry 9701/36 — October/November 2025

Background Concept

The amount of substance in moles is calculated by dividing the mass of the substance by its molar mass ($M_r$). For water ($\text{H}_2\text{O}$), the molar mass is approximately $18.02 \text{ g mol}^{-1}$ ($2 \times 1.01 + 16.00$). In this experiment, the mass of water lost during heating corresponds directly to the water of crystallisation in the hydrated salt.

Understanding the Question

You need to calculate the number of moles of water that evaporated from the sample. This is derived from the 'Mass of water lost' calculated in part (a).

Approach

1. Identify the mass of water lost from your results in part (a).
2. Divide this mass by the molar mass of water ($18.0 \text{ g mol}^{-1}$ or $18.02 \text{ g mol}^{-1}$).
3. Report the answer to 2-4 significant figures.

Step-by-Step Reasoning

• Mass of Water: From part (a), the mass of water lost is the difference between the initial mass of FB 1 and the final mass of the anhydrous residue. In the example, this was $0.82$ g.
• Molar Mass: $M_r(\text{H}_2\text{O}) = 18.02 \text{ g mol}^{-1}$.
• Calculation: $$n(\text{H}_2\text{O}) = \frac{0.82 \text{ g}}{18.02 \text{ g mol}^{-1}} \approx 0.045505... \text{ mol}$$
• Significant Figures: The input mass ($0.82$ g) has 2 significant figures, so the answer should ideally be given to 2 or 3 significant figures. $0.0455$ mol is appropriate.

Key Takeaways

• Always use the correct molar mass for the substance being converted to moles.
• The mass lost in this specific experiment is purely water.

Common Mistakes

• Using the mass of the residue instead of the mass of water lost.
• Using the molar mass of the hydrated salt instead of water.
• Incorrect significant figures (e.g., too many or too few).

Things to Be Careful About

• Ensure you are using the mass of water lost, not the mass of the salt.
• Check the molar mass of water used ($18$ vs $18.02$) is consistent with the rest of the calculation.

Background Concept

The anhydrous residue left after heating is aluminium sulfate, $\text{Al}_2(\text{SO}_4)_3$. To find the moles of this substance, we divide its mass by its molar mass. The molar mass is calculated by summing the atomic masses of all atoms in the formula unit.

Understanding the Question

Calculate the moles of anhydrous aluminium sulfate produced. This requires the mass of the residue from part (a) and the molar mass of $\text{Al}_2(\text{SO}_4)_3$.

Approach

1. Identify the mass of the residue from part (a).
2. Calculate the molar mass of $\text{Al}_2(\text{SO}_4)_3$.
3. Divide the mass of the residue by the molar mass.
4. Report the answer to 2-4 significant figures.

Step-by-Step Reasoning

• Mass of Residue: From part (a), the mass of the anhydrous residue is $1.08$ g in the example.
• Molar Mass Calculation:
  • Al: $2 \times 26.98 \approx 54.0$
  • S: $3 \times 32.06 \approx 96.2$
  • O: $12 \times 16.00 = 192.0$
  • Total $M_r \approx 54.0 + 96.2 + 192.0 = 342.2 \text{ g mol}^{-1}$. The mark scheme uses $342.3$.
• Calculation: $$n(\text{Al}_2(\text{SO}_4)_3) = \frac{1.08 \text{ g}}{342.3 \text{ g mol}^{-1}} \approx 0.003155... \text{ mol}$$
• Significant Figures: $0.00316$ mol (3 s.f.) is appropriate.

Key Takeaways

• Accurate calculation of molar mass is essential for stoichiometric problems.
• The residue is the anhydrous salt, not the hydrated form.

Common Mistakes

• Using the initial mass of FB 1 instead of the residue mass.
• Incorrect molar mass calculation (e.g., forgetting there are 12 oxygen atoms).
• Arithmetic errors in division.

Things to Be Careful About

• Ensure the molar mass used matches the atomic masses provided in the question data booklet or standard values ($342.3 \text{ g mol}^{-1}$ is the standard value here).
• Keep sufficient precision in intermediate steps to avoid rounding errors in the final ratio calculation.

Background Concept

The formula of a hydrated salt $\text{Salt} \cdot x\text{H}_2\text{O}$ indicates the ratio of moles of water to moles of anhydrous salt. By calculating the moles of water lost and the moles of anhydrous salt remaining, we can determine this ratio. The value of $x$ must be an integer, so we round the calculated ratio to the nearest whole number.

Understanding the Question

Determine the value of $x$ in $\text{Al}_2(\text{SO}_4)_3 \cdot x\text{H}_2\text{O}$. This is done by dividing the moles of water (from b.i) by the moles of anhydrous salt (from b.ii).

Approach

1. Take the mole values calculated in parts (b)(i) and (b)(ii).
2. Divide the moles of water by the moles of anhydrous salt.
3. Round the result to the nearest integer.

Step-by-Step Reasoning

• Moles of Water: $0.0455$ mol (from b.i).
• Moles of Anhydrous Salt: $0.00316$ mol (from b.ii).
• Ratio Calculation: $$x = \frac{0.0455}{0.00316} \approx 14.398$$
• Rounding: The question states $x$ is an integer. $14.398$ is closest to $14$. (Note: Actual hydrated aluminium sulfate is often $\text{Al}_2(\text{SO}_4)_3 \cdot 18\text{H}_2\text{O}$ or $\text{Al}_2(\text{SO}_4)_3 \cdot 16\text{H}_2\text{O}$, but experimental error or specific hydrate forms can lead to other values. Based on the example data provided, 14 is the result. In a real exam, your value depends on your specific measurements. The mark scheme expects the calculation from your numbers).

Key Takeaways

• The coefficient in a chemical formula represents a mole ratio.
• Experimental data often yields a non-integer ratio due to errors; rounding to the nearest integer is the standard procedure.

Common Mistakes

• Dividing moles of salt by moles of water (inverting the ratio).
• Failing to round to the nearest integer.
• Using masses directly instead of converting to moles first.

Things to Be Careful About

• Ensure you use the unrounded mole values from the previous steps to minimize rounding errors.
• Check if the result is reasonably close to a known hydrate (e.g., 16, 18) to validate your experimental technique, though you must report the calculated value rounded to the nearest integer.

Background Concept

Hydrated salts typically form well-defined crystals or fine powders due to the ordered lattice structure incorporating water molecules. When heated, the loss of water can disrupt this structure, causing the remaining anhydrous salt to sinter or fuse into a harder, less defined mass, often described as lumpy or having a crust.

Understanding the Question

Compare the visual appearance of the starting material (FB 1) with the final residue after heating.

Approach

Describe the texture/form of the initial solid and contrast it with the final solid.

Step-by-Step Reasoning

• FB 1 (Hydrated): Usually appears as a white, crystalline solid or a fine powder.
• Residue (Anhydrous): After strong heating, the structure collapses or fuses. It often appears as a white, lumpy solid, sometimes with a 'crusty' top layer or 'skin' formed by the melting and resolidifying of surface particles.

Key Takeaways

• Physical appearance changes can indicate chemical or structural changes during heating.

Common Mistakes

• Stating only that the color changes (both are typically white).
• Failing to mention the texture difference (powder vs. lump/crust).

Things to Be Careful About

• Use descriptive terms like 'crystalline', 'powder', 'lumpy', 'crusty', or 'skin' as these are the key observation points.

Background Concept

When hydrated salts are heated, they often dissolve in their own water of crystallisation before the water evaporates. This can cause vigorous boiling or 'spitting', where droplets of the solution are ejected from the crucible. This leads to a loss of sample mass that is not due to water evaporation, causing experimental error.

Understanding the Question

Explain why the crucible is heated with the lid on for the first two minutes.

Approach

Identify the physical phenomenon occurring during gentle heating (dissolution/spitting) and the consequence of not using a lid (loss of solid).

Step-by-Step Reasoning

• Initial Heating: Gentle heating causes the hydrated salt to melt/dissolve in its water of crystallisation.
• Risk: This liquid can bubble or spit, throwing solid material out of the crucible.
• Function of Lid: The lid acts as a barrier, keeping the solid inside the crucible even if spitting occurs.
• Subsequent Heating: Once the water has largely evaporated and the solid is dry, the lid is removed to allow remaining water vapor to escape freely during strong heating.

Key Takeaways

• Experimental steps are designed to minimize systematic errors (like loss of sample).
• 'Spitting' is a common issue when heating hydrated salts.

Common Mistakes

• Saying it prevents heat loss (the lid is removed later, so this isn't the primary reason for the initial gentle heating phase).
• Saying it prevents contamination from the air (less relevant than sample loss).

Things to Be Careful About

• Focus on the loss of solid material, not just water vapor.

Background Concept

The value of $x$ is calculated as $\frac{\text{moles of water}}{\text{moles of anhydrous salt}}$. Contamination affects the numerator and denominator differently depending on the nature of the contaminant. Anhydrous salt contributes to the 'residue' mass but adds no 'water' mass.

Understanding the Question

A student claims that contamination with anhydrous salt causes an overestimation of $x$. We must verify if this logic holds.

Approach

1. Assume the sample is a mix of hydrated salt and anhydrous salt.
2. Determine how this affects the measured 'Mass of water lost'.
3. Determine how this affects the measured 'Mass of residue'.
4. Analyze the effect on the ratio $x = \frac{n(\text{water})}{n(\text{salt})}$.

Step-by-Step Reasoning

• Effect on Water Mass: Anhydrous salt contains no water. Therefore, a contaminated sample will release less water per gram of sample than pure hydrated salt. The measured 'Mass of water lost' will be lower than expected for the mass of FB 1 taken.
• Effect on Residue Mass: The anhydrous contaminant remains as residue. This means the 'Mass of residue' will be higher than expected for the amount of water lost.
• Effect on Moles:
  • $n(\text{water})$ will be lower.
  • $n(\text{salt})$ (calculated from residue mass) will be higher.
• Effect on Ratio $x$: $$x = \frac{n(\text{water})}{n(\text{salt})}$$ Since the numerator decreases and the denominator increases, the value of $x$ will decrease.
• Conclusion: The student obtained a higher $x$. Contamination with anhydrous salt causes a lower $x$. Therefore, the student's suggestion is incorrect. (A higher $x$ would be caused by contamination with a more hydrated salt or incomplete drying of the residue, or loss of salt during heating).

Key Takeaways

• Always trace the effect of an error through the calculation steps.
• Contamination with the anhydrous form reduces the water content of the sample.

Common Mistakes

• Assuming any contamination increases the residue mass and therefore increases $x$ (forgetting the water mass also changes).
• Confusing the direction of the error (thinking anhydrous contamination leads to higher $x$).

Things to Be Careful About

• Clearly state whether the suggestion is correct or incorrect.
• Provide the reasoning linking the contamination to the change in measured masses and the final ratio.

Background Concept

The concentration of FB 3 is given as a mass concentration in $\text{g dm}^{-3}$. To find the amount in mol, first divide by the molar mass to get a concentration in $\text{mol dm}^{-3}$, then multiply by the volume used in dm3.

Understanding the Question

Part (c)(i) asks for the amount of $\text{KMnO}_4$ in the volume found in part (b). The only extra information needed is the molar mass of $\text{KMnO}_4$, which can be calculated from $\text{K} + \text{Mn} + 4\text{O}$.

Approach

1. Calculate $M_r(\text{KMnO}_4)$.
2. Convert $3.48\ \text{g dm}^{-3}$ to $\text{mol dm}^{-3}$ by dividing by $M_r$.
3. Convert the mean titre from cm3 to dm3 by dividing by 1000.
4. Multiply the molar concentration by the volume in dm3.

Step-by-Step Reasoning

$M_r(\text{KMnO}_4) = 39.1 + 54.9 + 4 \times 16.0 = 158.0$.

Concentration of $\text{KMnO}_4$:

Volume used: $24.50\ \text{cm}^3 = 0.02450\ \text{dm}^3$.

Amount:

A suitable final answer is $5.40 \times 10^{-4}$ mol to 3 s.f. or $5.396 \times 10^{-4}$ mol to 4 s.f.

Key Takeaways

Whenever a solute concentration is given in $\text{g dm}^{-3}$, convert it to $\text{mol dm}^{-3}$ using molar mass before finding $n$. Volumes in cm3 must be divided by 1000 before using $n = cV$.

Common Mistakes

• Using $3.48$ as the concentration in $\text{mol dm}^{-3}$ without dividing by $M_r$.
• Leaving the titre in cm3 instead of converting to dm3.
• Using the rough titre rather than the mean titre from part (b).
• Quoting too many or too few significant figures.

Things to Be Careful About

Use $M_r = 158.0$ consistently. Keep more significant figures during the calculation and only round at the final answer. The mark scheme accepts 3 or 4 significant figures for the final value.

Background Concept

In acidified manganate(VII), the oxidising agent in this titration is $\text{MnO}_4^-$. It is reduced to $\text{Mn}^{2+}$, while iron(II) is oxidised to iron(III). The equation must balance for atoms and charge.

Understanding the Question

The mole ratio of $\text{Fe}^{2+}$ to $\text{MnO}_4^-$ is already given as $5:1$. The task is to complete the coefficients in front of $\text{H}^+$, $\text{Fe}^{3+}$, and $\text{H}_2\text{O}$.

Approach

Write the two half-equations, balance electrons, then combine them. The oxidation half-equation releases one electron per $\text{Fe}^{2+}$; the reduction half-equation needs five electrons, so five iron(II) ions are oxidised.

Step-by-Step Reasoning

Oxidation:

Reduction:

Multiply the oxidation half-equation by 5:

Add the two half-equations, cancelling 5 electrons:

Check: on the left, total charge is $5(+2) + (-1) + 8(+1) = +17$. On the right, $5(+3) + (+2) = +17$. Atoms are also balanced.

Key Takeaways

The redox equation is the basis of the mole ratio used in (c)(iii): 5 mol $\text{Fe}^{2+}$ react with 1 mol $\text{MnO}_4^-$. Balancing redox equations always requires equal numbers of electrons transferred.

Common Mistakes

• Writing $5\text{H}^+$ instead of $8\text{H}^+$.
• Forgetting $4\text{H}_2\text{O}$.
• Balancing atoms but not charge.
• Confusing the oxidation and reduction half-equations.

Things to Be Careful About

Make sure the coefficient of $\text{Fe}^{3+}$ is 5, not 1. The equation needs both atom balance and charge balance.

Background Concept

The balanced equation shows that 5 mol of $\text{Fe}^{2+}$ react with 1 mol of $\text{MnO}_4^-$. Therefore the amount of iron(II) in the 25.0 cm3 sample is five times the amount of manganate(VII) that reacted. Concentration is then moles divided by volume in dm3.

Understanding the Question

Part (c)(iii) asks for the concentration of iron(II) sulfate in FB 2 in $\text{mol dm}^{-3}$. The volume used in the titration was $25.0$ cm3, and the amount of $\text{KMnO}_4$ was found in (c)(i).

Approach

1. Multiply the amount of $\text{KMnO}_4$ by 5.
2. Divide by $0.0250$ dm3.

Step-by-Step Reasoning

From (c)(i), the amount of $\text{KMnO}_4$ used is $5.396 \times 10^{-4}$ mol.

Amount of $\text{Fe}^{2+}$:

This amount was present in $25.0$ cm3, i.e. $0.0250$ dm3.

Concentration:

A suitable final answer is $0.108$ mol dm3 to 3 s.f. or $0.1079$ mol dm3 to 4 s.f. The unrounded value is used in the next part.

Key Takeaways

The mole ratio comes directly from the balanced redox equation. Always convert a pipette volume in cm3 to dm3 before dividing to find concentration.

Common Mistakes

• Forgetting to multiply the $\text{KMnO}_4$ amount by 5.
• Dividing by $25$ instead of $0.025$.
• Quoting the volume as dm3 without conversion.
• Rounding to 1 or 6 significant figures.

Things to Be Careful About

Use the unrounded amount from (c)(i) in the calculation. If the rounded value $5.40 \times 10^{-4}$ is used, the concentration becomes $0.108$ mol dm3, which is still acceptable for part (c)(iv).

Background Concept

FB 2 contains $30.00\ \text{g dm}^{-3}$ of $\text{FeSO}_4\cdot y\text{H}_2\text{O}$. Dividing a mass concentration in $\text{g dm}^{-3}$ by a molar concentration in $\text{mol dm}^{-3}$ gives the molar mass in $\text{g mol}^{-1}$. The hydrated salt has molar mass $151.9 + 18y$, where $151.9$ is the $M_r$ of anhydrous $\text{FeSO}_4$ and $18$ is the $M_r$ of one water molecule.

Understanding the Question

The aim of the experiment is to determine the integer value of $y$. We now know the molar concentration of hydrated iron(II) sulfate from part (c)(iii) and the mass concentration from the original data. This gives $M_r$ of the hydrate, from which $y$ can be solved.

Approach

1. Use $M_r = \dfrac{\text{mass concentration}}{\text{molar concentration}}$.
2. Subtract $151.9$ from this value to find the total molar mass of water in one mole of hydrate.
3. Divide by 18.0.
4. Give the answer as the nearest integer.

Step-by-Step Reasoning

Using the unrounded concentration from (c)(iii), $0.1079$ mol dm3:

Mass of water per mole of hydrate:

Moles of water per mole of hydrate:

So $y = 7$. This is consistent because $\text{FeSO}_4\cdot 7\text{H}_2\text{O}$ has $M_r = 151.9 + 126 = 277.9$.

Key Takeaways

The value of $y$ is found indirectly: titration gives the molar concentration, and the known mass concentration gives the molar mass. The final answer must be reported as an integer.

Common Mistakes

• Using the amount of $\text{KMnO}_4$ instead of the concentration of $\text{FeSO}_4$ when finding $M_r$.
• Substituting $151.9$ as the hydrated $M_r$ instead of the anhydrous $M_r$.
• Rounding the concentration to only 1 or 2 significant figures before calculating $y$.
• Leaving $y$ as $7.0$ or $7.005$ instead of giving the integer 7.

Things to Be Careful About

Use the most precise value available from part (c)(iii). The mark scheme allows up to 4 significant figures in intermediate answers, and requires the final $y$ to be an integer.

Background Concept

This is a qualitative analysis exercise. The unknown solid FB 5 is zinc metal ($\text{Zn}$), though you must deduce this from your observations. The key chemistry is the reactivity series: zinc sits above both iron and copper, so it displaces them from their salt solutions. Zinc also reacts with dilute acids to liberate hydrogen gas. The three tests probe these properties through visible changes — colour changes in solution, formation of precipitates, evolution of gas, and temperature changes.

Understanding the Question

You are asked to perform three tests on a small quantity of FB 5 and record all observations in Table 3.1. Test 1 uses iron(III) chloride solution, Test 2 uses copper(II) sulfate solution, and Test 3 uses dilute sulfuric acid with two drops of copper(II) sulfate. The general instructions emphasise recording the stage at which each observation is made — e.g. "on adding FB 5" versus "on adding $\text{NaOH}$" — and to write "no change" where nothing happens.

Approach

Work through each test using the reactivity series. Predict what zinc does in each solution: displacement of iron, displacement of copper, and reaction with acid. Then translate each chemical change into observable signs: colour changes, precipitates, gas bubbles, temperature rise. For Test 1, remember the second stage — pouring off some solution and adding $\text{NaOH}$ — tests for the cation left in solution.

Step-by-Step Reasoning

Test 1: Zinc is more reactive than iron, so $\text{Zn}(\text{s}) + \text{FeCl}_3(\text{aq}) \rightarrow \text{ZnCl}_2(\text{aq}) + \text{Fe}(\text{s})$. The yellow $\text{Fe}^{3+}$ ions are removed from solution, so the solution becomes paler or colourless. Some fizzing may occur as zinc reacts with any trace acid in the solution. When $\text{NaOH}$ is added to the poured-off solution (which contains $\text{Zn}^{2+}$ and possibly $\text{Fe}^{2+}$), a pale green or green-white precipitate of $\text{Fe(OH)}_2$ forms; this is insoluble in excess $\text{NaOH}$. The precipitate may turn brown at the surface as $\text{Fe}^{2+}$ is oxidised to $\text{Fe}^{3+}$ by air.

Test 2: Zinc displaces copper: $\text{Zn}(\text{s}) + \text{CuSO}_4(\text{aq}) \rightarrow \text{ZnSO}_4(\text{aq}) + \text{Cu}(\text{s})$. A pink-brown solid (copper metal) forms. The blue $\text{Cu}^{2+}$ ions are removed, so the solution pales or becomes colourless. The reaction is exothermic, so the tube gets hotter.

Test 3: Zinc reacts with dilute sulfuric acid: $\text{Zn}(\text{s}) + \text{H}_2\text{SO}_4(\text{aq}) \rightarrow \text{ZnSO}_4(\text{aq}) + \text{H}_2(\text{g})$. The copper(II) sulfate acts as a catalyst for this reaction, but the key observation is fizzing/effervescence as hydrogen is evolved. The gas pops with a lighted splint — the classic test for hydrogen.

Key Takeaways

The reactivity series predicts displacement reactions. Hydrogen gas is identified by the "pop" with a lighted splint. Observations must be recorded precisely, at the correct stage, with colours and physical states described accurately.

Common Mistakes

• Writing "brown" instead of "pink-brown" for copper metal.
• Forgetting to record the gas test in Test 3.
• Confusing the precipitate colours (green $\text{Fe(OH)}_2$ vs white $\text{Zn(OH)}_2$).
• Not recording that the solution pales in Test 2.

Things to Be Careful About

• Record the exact colour ("pale green", not just "green").
• Note the stage: "on adding $\text{NaOH}$" vs "on adding FB 5".
• The precipitate in Test 1 comes from the solution after pouring off — make clear which stage you are describing.
• "No change" must be written if nothing happens — but here changes do occur.

Background Concept

Aqueous ammonia is a weak base and a source of $\text{OH}^-$ ions. Many metal cations form insoluble hydroxides when ammonia is added. Zinc(II) is amphoteric: its hydroxide dissolves in excess alkali (both $\text{NaOH}$ and $\text{NH}_3$), forming a soluble complex ion, $[\text{Zn(NH}_3)_4]^{2+}$, with ammonia.

Understanding the Question

FB 6 is the filtrate from Test 3 — the solution left after zinc reacted with dilute sulfuric acid and copper(II) sulfate. It contains $\text{ZnSO}_4(\text{aq})$. You add aqueous ammonia and record what happens: first a white precipitate of $\text{Zn(OH)}_2$, then dissolution in excess ammonia.

Approach

Add ammonia dropwise. First observation: white precipitate. Continue adding excess ammonia: precipitate dissolves. Record both stages.

Step-by-Step Reasoning

$\text{Zn}^{2+}(\text{aq}) + 2\text{NH}_3(\text{aq}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow \text{Zn(OH)}_2(\text{s}) + 2\text{NH}_4^+(\text{aq})$. The white gelatinous precipitate is zinc hydroxide. On adding excess ammonia, $\text{Zn(OH)}_2(\text{s}) + 4\text{NH}_3(\text{aq}) \rightarrow [\text{Zn(NH}_3)_4]^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq})$ — the precipitate dissolves to give a colourless solution.

Key Takeaways

$\text{Zn}^{2+}$ gives a white precipitate with $\text{NH}_3$, soluble in excess — a key distinguishing test for $\text{Zn}^{2+}$ (compare $\text{Al}^{3+}$, which gives a white precipitate soluble in excess $\text{NaOH}$ but NOT in excess $\text{NH}_3$).

Common Mistakes

• Forgetting to record the solubility in excess.
• Writing "no change" — the precipitate does form.
• Confusing with $\text{Al}^{3+}$ behaviour.

Things to Be Careful About

• Record both stages: precipitate forms, then dissolves in excess.
• The precipitate is white and gelatinous.

Background Concept

The observations from (a)(i) — displacement of iron and copper, and reaction with acid to give hydrogen — are characteristic of a metal above iron and copper in the reactivity series. Zinc fits perfectly.

Understanding the Question

Identify FB 5 from the accumulated observations in the three tests.

Approach

Match the observed behaviour to a known metal. Zinc displaces Fe and Cu and reacts with dilute acid to give $\text{H}_2$.

Step-by-Step Reasoning

The pink-brown copper deposit in Test 2 and hydrogen gas in Test 3 identify FB 5 as a metal more reactive than copper and able to reduce $\text{H}^+$ to $\text{H}_2$. Zinc is the metal that fits all observations.

Key Takeaways

Qualitative tests build a profile; the profile identifies the substance.

Common Mistakes

• Guessing without linking observations.
• Confusing with other reactive metals (e.g. Mg would also give $\text{H}_2$, but the displacement colours differ).

Things to Be Careful About

• The answer must be the element or correct formula: zinc / $\text{Zn}$.

Background Concept

A redox reaction involves simultaneous oxidation and reduction. Oxidation is loss of electrons (or increase in oxidation state); reduction is gain of electrons (or decrease in oxidation state). In a displacement reaction, the more reactive metal is oxidised and the less reactive metal's ions are reduced.

Understanding the Question

Explain why Test 2 ($\text{Zn}$ + $\text{CuSO}_4$) is a redox reaction, using your observations.

Approach

Identify the species that change oxidation state and state what happens to each.

Step-by-Step Reasoning

In Test 2: $\text{Zn}(\text{s}) + \text{Cu}^{2+}(\text{aq}) \rightarrow \text{Zn}^{2+}(\text{aq}) + \text{Cu}(\text{s})$. Zinc loses electrons (oxidised, oxidation state 0 $\rightarrow$ +2). Copper ions gain electrons (reduced, oxidation state +2 $\rightarrow$ 0). Both processes occur, so it is a redox reaction. The pink-brown copper solid is the visible evidence of reduction.

Key Takeaways

A displacement reaction is always redox: the more reactive metal is oxidised, the less reactive metal's ion is reduced.

Common Mistakes

• Saying only "copper is formed" without mentioning electron transfer or oxidation-state change.
• Confusing which species is oxidised and which is reduced.

Things to Be Careful About

• State both the species and the electron transfer (or oxidation-state change) for full marks.

Background Concept

The first reaction in (a)(ii) is the formation of zinc hydroxide: $\text{Zn}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Zn(OH)}_2(\text{s})$. This is the ionic equation for the precipitation.

Understanding the Question

Write the ionic equation, with state symbols, for the first reaction observed when ammonia is added to FB 6 ($\text{ZnSO}_4$ solution).

Approach

Identify the ions involved: $\text{Zn}^{2+}$ and $\text{OH}^-$ (from ammonia). Product: $\text{Zn(OH)}_2(\text{s})$. Balance charges and atoms.

Step-by-Step Reasoning

$\text{Zn}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Zn(OH)}_2(\text{s})$. The 2+ charge on $\text{Zn}^{2+}$ is balanced by two $\text{OH}^-$ ions. The product is a solid precipitate. State symbols: (aq) for ions in solution, (s) for the precipitate.

Key Takeaways

Ionic equations show only the species that change. Charge and atom balance are essential.

Common Mistakes

• Omitting state symbols.
• Writing the full equation with $\text{NH}_3$ instead of the ionic form.
• Unbalanced charges.

Things to Be Careful About

• State symbols are required by the question.
• The equation must be balanced in both atoms and charge.

Background Concept

Many metal carbonates decompose on heating to give the metal oxide and carbon dioxide: $\text{MCO}_3(\text{s}) \rightarrow \text{MO}(\text{s}) + \text{CO}_2(\text{g})$. Zinc carbonate is white; zinc oxide is white when cold but yellow when hot — a characteristic colour change. Carbon dioxide is identified by turning limewater milky (white precipitate of $\text{CaCO}_3$).

Understanding the Question

Heat FB 7 ($\text{ZnCO}_3$) gently then strongly in a hard-glass test-tube, and record all observations including any gas identification. The anion contains oxygen but not nitrogen, and both ions are in the Qualitative analysis notes — this hints at a carbonate.

Approach

Predict the thermal decomposition of a carbonate. Watch for colour change of the residue, condensation, and test any gas with limewater.

Step-by-Step Reasoning

$\text{ZnCO}_3(\text{s}) \rightarrow \text{ZnO}(\text{s}) + \text{CO}_2(\text{g})$. FB 7 is a white powder. On heating, condensation (water droplets) forms on the cooler parts of the tube (from moisture). The solid turns yellow when hot — characteristic of $\text{ZnO}$. On cooling, it returns to white. The gas evolved is $\text{CO}_2$: it turns limewater milky (white precipitate of $\text{CaCO}_3$).

Key Takeaways

Carbonates decompose to oxide + $\text{CO}_2$. $\text{ZnO}$ is yellow when hot, white when cold. $\text{CO}_2$ is confirmed with limewater.

Common Mistakes

• Not testing the gas with limewater.
• Writing "turns brown" instead of "turns yellow".
• Not recording the colour change on cooling.

Things to Be Careful About

• Record the colour at each temperature stage (hot vs cold).
• Condensation is a valid observation.
• The gas test must be described for the $\text{CO}_2$ identification mark.

Background Concept

The confirmatory test for carbonate is to add a dilute mineral acid: $\text{CO}_3^{2-}(\text{s}) + 2\text{H}^+(\text{aq}) \rightarrow \text{CO}_2(\text{g}) + \text{H}_2\text{O}(\text{l})$. Effervescence (fizzing) indicates $\text{CO}_2$ evolution, confirming carbonate.

Understanding the Question

Carry out one further positive test to confirm the anion in FB 7. The anion is carbonate. You must describe the test and the positive observation.

Approach

Add a dilute mineral acid (e.g. $\text{HCl}$, $\text{HNO}_3$, or $\text{H}_2\text{SO}_4$) to FB 7 and look for effervescence.

Step-by-Step Reasoning

Add dilute hydrochloric acid (or nitric/sulfuric acid) to FB 7. Effervescence/fizzing confirms carbonate. The gas is $\text{CO}_2$ (optionally test with limewater for extra confirmation).

Key Takeaways

Carbonate + acid $\rightarrow$ $\text{CO}_2$ + water. Effervescence is the positive test.

Common Mistakes

• Using a non-mineral acid or no acid at all.
• Not naming the acid or giving its formula.
• Confusing with other anion tests.

Things to Be Careful About

• The acid must be a mineral acid ($\text{HCl}$, $\text{HNO}_3$, $\text{H}_2\text{SO}_4$).
• Record the positive observation (effervescence) only.