Chemistry 9701/34 — May/June 2024

Background Concept

Significant figures indicate the precision of a measured or calculated value. A value quoted to 3 significant figures (e.g. 0.00225) claims more precision than one quoted to 2 (0.0023) and less than one quoted to 4 (0.002250). The data in this question — 5.05 g dm⁻³, 5.00 mol dm⁻³, 22.50 g, 25.0 cm³ — carry 3 or 4 significant figures, so the calculated answers should be quoted to the same precision.

Understanding the Question

This one-mark part simply asks you to state how many significant figures you will use for the answers to (c)(ii), (c)(iii) and (c)(iv).

Approach

Quote the answers to 3 or 4 significant figures.

Step-by-Step Reasoning

The mark scheme awards the mark for answers given to 3 or 4 significant figures. This is a convention: the answers must not be over- or under-precise relative to the data. The data in the question (5.05 g dm⁻³, 5.00 mol dm⁻³, 22.50 g, 25.0 cm³) all carry 3 or 4 significant figures, so the calculated answers should match. For example, 0.00225 mol and 0.250 mol dm⁻³ are each 3 significant figures.

Key Takeaways

• Match the significant figures of the data.
• 3 or 4 significant figures is the required convention here.

Common Mistakes

• Giving answers to 2 or 5 significant figures.

Things to Be Careful About

• This is an easy mark to lose — always check significant figures before finalising.

Background Concept

The amount of a substance in moles is given by:

where $c$ is the concentration in mol dm⁻³ and $V$ is the volume in dm³. FB3 is KOH at a concentration of 5.05 g dm⁻³, so its concentration in mol dm⁻³ is found by dividing by its molar mass:

Understanding the Question

Calculate the amount, in mol, of KOH present in 25.0 cm³ of FB3.

Approach

1. Convert the concentration from g dm⁻³ to mol dm⁻³ by dividing by 56.1.
2. Convert 25.0 cm³ to dm³ by dividing by 1000.
3. Multiply: $n = c \times V$.

Step-by-Step Reasoning

This is $2.25 \times 10^{-3}$ mol, quoted to 3 significant figures.

Key Takeaways

• $n = c \times V$ is the fundamental relationship.
• Volumes must be in dm³: divide cm³ by 1000.
• Concentration in g dm⁻³ must be converted to mol dm⁻³ using the molar mass.

Common Mistakes

• Forgetting to convert 25.0 cm³ to 0.0250 dm³.
• Using the wrong molar mass for KOH.

Things to Be Careful About

• Quote the answer to 3 or 4 significant figures: 0.00225 mol or $2.25 \times 10^{-3}$ mol.

Background Concept

The reaction between hydrochloric acid and potassium hydroxide is a neutralisation:

The ionic equation removes the spectator ions (K⁺ and Cl⁻):

At the end-point of the titration, the moles of HCl that have been added exactly equal the moles of KOH originally in the flask. So the moles of HCl in the titre volume equal the moles of KOH calculated in (c)(ii).

Understanding the Question

Write the ionic equation for the neutralisation (with state symbols), then use the mean titre from (b) and the moles from (c)(ii) to calculate the concentration of HCl in FB2.

Approach

1. Write the ionic equation: $\text{H}^+(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{H}_2\text{O}(\text{l})$.
2. $n(\text{HCl}) = n(\text{KOH}) = 0.00225$ mol.
3. Convert the titre to dm³ and divide: $[\text{HCl}] = n/V$.

Step-by-Step Reasoning

The titre is 9.00 cm³ = 0.00900 dm³.

Equivalently, using the formula from the mark scheme:

Key Takeaways

• Neutralisation: $\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}$.
• At the end-point, moles of acid = moles of base.
• $c = n/V$ with volume in dm³.

Common Mistakes

• Forgetting state symbols in the ionic equation — the mark scheme requires (aq) and (l).
• Not converting the titre volume to dm³.

Things to Be Careful About

• The ionic equation must show $\text{H}^+(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{H}_2\text{O}(\text{l})$; spectator ions are omitted.
• Quote the concentration to 3 or 4 significant figures.

Background Concept

This is a back-titration. The total amount of HCl added to FB1 is known: 100.0 cm³ of 5.00 mol dm⁻³ HCl. The amount of HCl that remained unreacted is found from the titration: it equals the concentration of HCl in FB2 multiplied by the volume of FB2 (1.00 dm³). The difference between the total added and the amount remaining is the amount of HCl that reacted with FB1. From the balanced equation:

1 mol of MCO₃·M(OH)₂ reacts with 4 mol of HCl. So the amount of FB1 is the reacted HCl divided by 4. Finally, the relative formula mass is:

Understanding the Question

Use the concentration of HCl in FB2 from (c)(iii) together with the preparation details (22.50 g FB1, 100.0 cm³ of 5.00 mol dm⁻³ HCl, made up to 1.00 dm³) to calculate the $M_r$ of MCO₃·M(OH)₂.

Approach

1. Total HCl added = 0.100 × 5.00 = 0.500 mol.
2. HCl remaining in FB2 = $[\text{HCl}] \times 1.00$ dm³.
3. HCl reacted with FB1 = total − remaining.
4. Amount of FB1 = reacted HCl ÷ 4.
5. $M_r$ = 22.50 ÷ amount of FB1.

Step-by-Step Reasoning

Step 1 — total HCl added:

Step 2 — HCl remaining (in the whole 1.00 dm³ of FB2):

Step 3 — HCl that reacted with FB1:

Step 4 — amount of FB1: from the equation, 4 mol HCl react with 1 mol FB1:

Step 5 — relative formula mass:

Key Takeaways

• Back-titration logic: total added − remaining = amount reacted.
• Use the stoichiometric ratio from the balanced equation (4 HCl : 1 FB1).
• $M_r$ = mass ÷ amount.

Common Mistakes

• Forgetting to multiply $[\text{HCl}]$ by 1.00 dm³ (the volume of FB2) — this is the most common error.
• Using the wrong stoichiometric ratio (e.g. dividing by 2 instead of 4).
• Not subtracting the remaining HCl from the total.

Things to Be Careful About

• The balanced equation shows 4 HCl per FB1 — check the stoichiometry carefully.
• Quote the final $M_r$ to 3 or 4 significant figures.

Background Concept

A basic metal carbonate has the formula $\text{MCO}_3\cdot\text{M(OH)}_2$. The dot means the compound contains one carbonate ion and two hydroxide ions combined with two metal ions. On strong heating it decomposes to the metal oxide, carbon dioxide and water vapour.

Understanding the Question

You are asked to write the balanced equation for the decomposition, including state symbols. The products are given, so the task is to balance the equation correctly.

Approach

Write the reactants and products, then balance the atoms. Count the metal atoms, carbon atoms, oxygen atoms and hydrogen atoms on each side.

Step-by-Step Reasoning

The reactant contains:

• 2 M atoms (one from $\text{MCO}_3$ and one from $\text{M(OH)}_2$)
• 1 C atom
• 5 O atoms (3 from carbonate and 2 from hydroxide)
• 2 H atoms

Products:

• $2\text{MO}$ gives 2 M and 2 O
• $\text{CO}_2$ gives 1 C and 2 O
• $\text{H}_2\text{O}$ gives 2 H and 1 O

Total products: 2 M, 1 C, 5 O, 2 H. The equation balances with a coefficient of 2 on $\text{MO}$.

State symbols: the carbonate is a solid, the metal oxide is a solid, carbon dioxide and water vapour are gases.

Key Takeaways

When balancing equations with dot formulae, count the atoms in each part of the formula separately. The dot does not mean multiplication by a number unless shown.

Common Mistakes

• Writing only one $\text{MO}$ instead of $2\text{MO}$.
• Forgetting state symbols.
• Using (aq) for water vapour instead of (g).

Things to Be Careful About

• The water is produced as steam, so use (g).
• The metal oxide is a solid, so use (s).

Background Concept

The amount of a substance in moles is given by $\text{amount} = \frac{\text{mass}}{M_r}$. During heating the mass lost is due to the escape of both carbon dioxide and water vapour, so the relevant molar mass is the sum of $M_r(\text{CO}_2)$ and $M_r(\text{H}_2\text{O})$.

Understanding the Question

You are told the formula for the amount of carbon dioxide produced. You need to use your mass loss from part (a) and calculate the number of moles, giving the answer to an appropriate number of significant figures.

Approach

Find the mass loss during heating, then divide by $44 + 18 = 62$.

Step-by-Step Reasoning

1. Mass loss = mass of FB 1 used − mass of residue.
2. $M_r(\text{CO}_2) = 12 + 2(16) = 44$
3. $M_r(\text{H}_2\text{O}) = 2(1) + 16 = 18$
4. Total = 62
5. $\text{mol CO}_2 = \frac{\text{mass loss}}{62}$

The answer should be given to 2–4 significant figures, matching the precision of the balance readings.

Key Takeaways

Mass loss in a decomposition can be converted to moles using the combined molar mass of the gaseous products.

Common Mistakes

• Using only $M_r(\text{CO}_2) = 44$ and forgetting the water.
• Using the mass of the residue instead of the mass loss.
• Giving too many significant figures, such as 0.032456 mol.

Things to Be Careful About

• The mass loss must be in grams.
• Use the candidate’s own mass loss; there is no single correct numerical answer.

Background Concept

The relative formula mass, $M_r$, is the mass of one mole of the substance in grams. Since one mole of $\text{MCO}_3\cdot\text{M(OH)}_2$ produces one mole of $\text{CO}_2$, the amount of carbonate is equal to the amount of carbon dioxide.

Understanding the Question

You need to use the amount of carbon dioxide from (b)(ii) and the mass of FB 1 used to calculate $M_r$ of the basic carbonate.

Approach

Use $M_r = \frac{\text{mass of FB 1 used}}{\text{mol CO}_2}$.

Step-by-Step Reasoning

1. From (b)(ii), you have the number of moles of $\text{CO}_2$.
2. The same number of moles of $\text{MCO}_3\cdot\text{M(OH)}_2$ was decomposed.
3. Therefore $M_r = \frac{\text{mass of FB 1 used}}{\text{mol CO}_2}$.
4. Give the answer to 2–4 significant figures.

Key Takeaways

The mole ratio between reactant and product is 1:1, so the amount of carbonate equals the amount of carbon dioxide.

Common Mistakes

• Using the mass of the residue instead of the mass of FB 1 used.
• Using the mass loss instead of the original sample mass.
• Forgetting to use the value from (b)(ii).

Things to Be Careful About

• The mass of FB 1 used is the mass before heating, not the mass after.
• Relative formula mass has no units, although it is numerically equal to the molar mass in g mol$^{-1}$.

Background Concept

The formula $\text{MCO}_3\cdot\text{M(OH)}_2$ contains two metal atoms, one carbonate group and two hydroxide groups. Its relative formula mass can be written as $2A_r + 60 + 34$.

Understanding the Question

You are asked to use your value of $M_r$ from (b)(iii) to find the relative atomic mass of metal M.

Approach

Write an expression for $M_r$ in terms of $A_r$, then rearrange to solve for $A_r$.

Step-by-Step Reasoning

1. $M_r(\text{CO}_3) = 12 + 3(16) = 60$
2. $M_r(\text{OH}) = 16 + 1 = 17$, so two hydroxide groups contribute 34.
3. $M_r = 2A_r + 60 + 34 = 2A_r + 94$
4. Rearranging: $A_r = \frac{M_r - 94}{2}$
5. Substitute the value of $M_r$ from (b)(iii).

Key Takeaways

A formula unit containing two metal atoms means the metal contributes twice its relative atomic mass to $M_r$.

Common Mistakes

• Forgetting the factor of 2 for the two metal atoms.
• Using $M_r(\text{OH}) = 17$ but forgetting there are two hydroxide groups.
• Arithmetic errors when subtracting 94 and dividing by 2.

Things to Be Careful About

• The value of $M_r$ must come from (b)(iii).
• Check that your final $A_r$ is a sensible value for a metal.

Background Concept

When a substance is heated and decomposes, it changes into different substances. The original compound is no longer present.

Understanding the Question

The results table labels the third and fourth readings as “mass of crucible, lid and FB 1 (after first/second heating)”. These headings are unsuitable because after heating the solid is not FB 1.

Approach

Identify what the solid actually is after heating: the metal oxide residue.

Step-by-Step Reasoning

1. FB 1 is $\text{MCO}_3\cdot\text{M(OH)}_2$.
2. On heating it decomposes to $\text{MO}$, $\text{CO}_2$ and $\text{H}_2\text{O}$.
3. The $\text{CO}_2$ and $\text{H}_2\text{O}$ escape as gases, leaving only $\text{MO}$ in the crucible.
4. Therefore the headings should refer to the residue or metal oxide, not FB 1.

Key Takeaways

Labels in a results table must describe what is actually being weighed.

Common Mistakes

• Saying the mass changed without identifying the residue.
• Continuing to call the solid FB 1 after decomposition.

Things to Be Careful About

• Use the term “residue” or “metal oxide” in the heading.

Background Concept

Heating to constant mass is a way to check that decomposition is complete. If two consecutive weighings are the same, no more gas is being lost.

Understanding the Question

You must decide, using your own results, whether a third heating would improve accuracy and explain why.

Approach

Compare the third and fourth readings. If they are equal within 0.02 g, decomposition is complete and a third heating would not help. If the fourth is lower, decomposition was not complete and another heating would help.

Step-by-Step Reasoning

1. Look at the third and fourth readings in your results.
2. If they are the same within 0.02 g, the mass is constant, so all the $\text{CO}_2$ and $\text{H}_2\text{O}$ have been driven off. A third heating would not change the mass and would not improve accuracy.
3. If the fourth reading is lower than the third, decomposition may still be occurring. A third heating would make the result more accurate because it would ensure constant mass.

Key Takeaways

The criterion for complete decomposition is constant mass on reheating.

Common Mistakes

• Giving an answer without referring to the actual readings.
• Saying “yes” or “no” without explaining why.

Things to Be Careful About

• The mark scheme accepts either conclusion as long as it matches your results and is explained.

Background Concept

The accuracy of an experiment depends on the number of measurements and the ease of judging the end-point. Fewer steps generally mean fewer opportunities for error.

Understanding the Question

You need to give one reason why the thermal decomposition method in Question 2 is expected to be more accurate than the titration in Question 1.

Approach

Compare the two methods in terms of number of readings, ease of end-point, and equipment precision.

Step-by-Step Reasoning

1. The thermal decomposition method involves fewer readings and fewer steps than the titration, so there are fewer sources of error.
2. The end-point of the decomposition is easy to judge: it is when the mass becomes constant. A titration end-point depends on judging a colour change, which is more subjective.
3. A balance generally has a smaller percentage error than a burette or pipette.

Any one of these reasons is sufficient.

Key Takeaways

Accuracy is improved by reducing the number of steps and by using an end-point that is easy to judge.

Common Mistakes

• Giving a vague answer such as “human error”.
• Saying the experiment is more accurate without giving a reason.

Things to Be Careful About

• Give a specific, scientific reason, not a general statement.

Background Concept

FB 5 is ammonium carbonate, $(\text{NH}_4)_2\text{CO}_3$. On strong heating it undergoes thermal decomposition:

All three products are gases at the temperature of a Bunsen flame, so nothing solid remains in the tube. The water vapour condenses to droplets (condensation) on the cooler upper part of the tube. Ammonia is an alkaline gas and turns moist red litmus paper blue. This is a Paper 3 qualitative-analysis task: the candidate heats a solid and records every observation at the stage it occurs.

Understanding the Question

Part (a)(i) asks you to heat a small spatula measure of FB 5 in a hard-glass test-tube until no further change occurs and to record all observations. Two marks are available, and the mark scheme requires two distinct observations per mark — so four creditable points are needed. The observations must include the appearance at the start, what happens during heating (condensation, gas), and what remains at the end.

Approach

Work through the heating in stages: note the initial appearance; watch for condensation as water vapour reaches the cooler part of the tube; test any gas evolved with moist red litmus paper; and finally check whether any solid residue remains. Record each observation at the stage it is made.

Step-by-Step Reasoning

1. Initial appearance: FB 5 is a white solid/powder (or white/colourless crystals). This is the first observation.
2. On gentle heating, water vapour is produced immediately and condenses as droplets (condensation/steam) on the cooler part of the tube. This is the second observation.
3. A gas is evolved. Test it by holding moist red litmus paper at the mouth of the tube: it turns blue, identifying ammonia. This is the third observation.
4. At the end, no residue remains — the solid has disappeared because it has decomposed completely into gases. This is the fourth observation.
   Each of these is a separate creditable point; the mark scheme needs two points per mark.

Key Takeaways

• Thermal decomposition of ammonium carbonate gives only gases, so no residue remains.
• Ammonia is identified by turning moist red litmus paper blue.
• In Paper 3, record observations at the stage they occur and include the initial appearance.

Common Mistakes

• Writing only "gas formed" without testing or identifying it — the mark requires the litmus result.
• Not recording the condensation (water vapour) — an easy mark missed.
• Not noting that no residue remains.
• Using blue litmus paper — it is already blue, so the ammonia test shows no change; red litmus must be used.

Things to Be Careful About

• Use a hard-glass test-tube when heating a solid (as instructed in the paper's preamble).
• The mark scheme links "condensation/steam" to gentle heating — record it as water vapour being produced immediately on heating.
• Do not write "sublimes" unless you mean the solid disappears; the mark scheme allows "solid disappears / evaporates / sublimes".

Background Concept

The ammonium ion, $\text{NH}_4^+$, is identified by warming the compound with aqueous sodium hydroxide. Hydroxide ions deprotonate the ammonium ion:

Ammonia is a soluble, alkaline gas that turns moist red litmus paper blue. This is the standard, specific test for the ammonium cation.

Understanding the Question

Part (a)(ii) asks for another test — separate from the heating in (a)(i) — to positively identify the cation in FB 5. One mark is available. The cation is ammonium, $\text{NH}_4^+$.

Approach

Use the classic ammonium test: warm a little FB 5 with aqueous NaOH and test the gas with moist red litmus paper.

Step-by-Step Reasoning

1. Place a small amount of FB 5 in a test-tube and add aqueous sodium hydroxide.
2. Warm the mixture gently (use a boiling tube as instructed when warming).
3. Ammonia gas is evolved; hold a piece of moist red litmus paper at the mouth of the tube.
4. The paper turns blue, confirming ammonia and therefore the ammonium ion.

Key Takeaways

• The ammonium test: warm with NaOH(aq); ammonia turns moist red litmus blue.
• This is a positive identification of $\text{NH}_4^+$, not just an observation.

Common Mistakes

• Not warming the mixture — the reaction is slow at room temperature.
• Using blue litmus (no visible change).
• Testing the gas with limewater instead of litmus (that is for $\text{CO}_2$).

Things to Be Careful About

• The mark scheme requires both the reagent (aqueous NaOH) and the test result (gas turns red litmus blue).
• Warming must be gentle; use a boiling tube.

Background Concept

Carbonate ions react with acids to give carbon dioxide:

The effervescence (fizzing) is the visible sign. Carbon dioxide is identified by bubbling it through limewater (aqueous calcium hydroxide): it forms a white precipitate of calcium carbonate, turning the limewater milky:

Understanding the Question

Part (a)(iii) asks you to add dilute hydrochloric acid to a small spatula measure of FB 5 and record observations. Two marks are available, needing four creditable points: the effervescence, the colourless solution, the limewater test, and the white precipitate.

Approach

Add the acid, observe the effervescence and the solution colour, then test the gas with limewater.

Step-by-Step Reasoning

1. On adding dilute HCl, vigorous effervescence/fizzing occurs — carbon dioxide is being released.
2. The solid dissolves, giving a colourless solution (ammonium chloride forms).
3. Collect or bubble the gas into limewater.
4. The limewater turns milky — a white precipitate of $\text{CaCO}_3$ forms — confirming $\text{CO}_2$, hence carbonate.

Key Takeaways

• Carbonate + acid $\rightarrow$ $\text{CO}_2$; test with limewater (white ppt/milky).
• Effervescence is the observation; the limewater test identifies the gas.

Common Mistakes

• Writing "gas formed" instead of "effervescence/fizzing" — the mark scheme rejects the vague phrase.
• Writing "milky" alone — the mark scheme requires "white precipitate/solid with limewater".
• Not testing the gas at all.

Things to Be Careful About

• The mark scheme explicitly says the limewater result must be "white ppt/solid", not just "milky".
• Record the stage: effervescence on addition; limewater test afterwards.

Background Concept

The observations in (a)(i)–(a)(iii) identify the cation and anion of FB 5. The gas that turns red litmus blue (ammonia) on heating and with NaOH shows the cation is ammonium, $\text{NH}_4^+$. The effervescence with acid and the white precipitate with limewater show the anion is carbonate, $\text{CO}_3^{2-}$. The formula must balance charges: $\text{NH}_4^+$ is +1 and $\text{CO}_3^{2-}$ is −2, so two ammonium ions are needed per carbonate ion.

Understanding the Question

Part (a)(iv) asks you to deduce the formula of FB 5 from the observations. One mark.

Approach

Identify the cation and anion from the earlier tests, then combine them in the charge-balanced ratio.

Step-by-Step Reasoning

1. From (a)(i) and (a)(ii): ammonia is evolved $\rightarrow$ $\text{NH}_4^+$.
2. From (a)(iii): $\text{CO}_2$ is evolved with acid $\rightarrow$ $\text{CO}_3^{2-}$.
3. Charge balance: $2 \times (+1) + (-2) = 0$, so the formula is $(\text{NH}_4)_2\text{CO}_3$, ammonium carbonate.

Key Takeaways

• Deduce ions from specific tests, then write a charge-balanced formula.
• Ammonium is written in parentheses when there is more than one: $(\text{NH}_4)_2$.

Common Mistakes

• Writing $\text{NH}_4\text{CO}_3$ — charges do not balance.
• Forgetting the parentheses around $\text{NH}_4$.

Things to Be Careful About

• The formula must be $(\text{NH}_4)_2\text{CO}_3$, not $\text{NH}_4\text{CO}_3$ or $(\text{NH}_4)_2(\text{CO}_3)_2$.

Background Concept

Chromium(III) and iron(II) can be distinguished by their reactions with aqueous sodium hydroxide. $\text{Cr}^{3+}$ gives a grey-green precipitate of $\text{Cr(OH)}_3$ which is amphoteric — it dissolves in excess NaOH to give a dark green solution containing $[\text{Cr(OH)}_6]^{3-}$. $\text{Fe}^{2+}$ gives a dirty green precipitate of $\text{Fe(OH)}_2$ which is not soluble in excess NaOH. Alternatively, acidified potassium manganate(VII) is decolourised by $\text{Fe}^{2+}$ (a reducing agent) but not by $\text{Cr}^{3+}$.

Understanding the Question

FB 6 is either aqueous chromium(III) sulfate or aqueous iron(II) sulfate. Devise a test, record observations, and state the conclusion. Two marks: M1 for the reagent, M2 for the observation and conclusion.

Approach

Choose the NaOH test: add aqueous NaOH, observe the precipitate, then add excess and test solubility. (The KMnO4 route is an alternative.)

Step-by-Step Reasoning

1. Add aqueous NaOH to FB 6.
2. A grey-green precipitate forms — this is $\text{Cr(OH)}_3$ ($\text{Fe}^{2+}$ would give a dirty green precipitate, similar at first glance).
3. Add excess NaOH: the precipitate dissolves, giving a dark green solution. This is the key observation — $\text{Cr(OH)}_3$ is amphoteric.
4. Conclusion: FB 6 is chromium(III) sulfate, $\text{Cr}_2(\text{SO}_4)_3$.
   Alternative: add acidified KMnO4 — it remains purple (not decolourised) because there is no $\text{Fe}^{2+}$ to reduce it, so FB 6 is $\text{Cr}_2(\text{SO}_4)_3$.

Key Takeaways

• $\text{Cr(OH)}_3$ is amphoteric: soluble in excess NaOH; $\text{Fe(OH)}_2$ is not.
• $\text{Fe}^{2+}$ decolourises acidified KMnO4; $\text{Cr}^{3+}$ does not.

Common Mistakes

• Stopping after the first precipitate without testing solubility in excess — the solubility is the discriminating observation.
• Confusing the colours: $\text{Fe(OH)}_2$ is dirty green and insoluble; $\text{Cr(OH)}_3$ is grey-green and soluble in excess.

Things to Be Careful About

• The mark scheme accepts either the NaOH route or the KMnO4 route; give the observation AND the conclusion for the mark.

Background Concept

Nitrate ions are identified by reduction to ammonia. Heating the solution with aqueous sodium hydroxide and aluminium (or Devarda's alloy) reduces $\text{NO}_3^-$ to $\text{NH}_3$:

The ammonia is detected by its alkaline smell and by turning moist red litmus paper blue. Hydrobromic acid contains bromide, which gives no such gas.

Understanding the Question

FB 7 is either dilute hydrobromic acid or dilute nitric acid. Devise a test, record observations, conclude. Two marks. The question notes that if the chosen test gives a negative result, a further positive test must be carried out.

Approach

Use the nitrate test: heat with excess NaOH and Al; test the gas with moist red litmus.

Step-by-Step Reasoning

1. Add excess aqueous NaOH and a piece of aluminium to FB 7 in a boiling tube.
2. Warm the mixture.
3. If nitrate is present, ammonia is evolved; moist red litmus turns blue.
4. Positive result $\rightarrow$ nitrate present $\rightarrow$ FB 7 is nitric acid, $\text{HNO}_3$.
   (If no ammonia were evolved, FB 7 would be HBr, and a further positive test — e.g. adding $\text{AgNO}_3$ to give a cream precipitate of AgBr — would be needed.)

Key Takeaways

• Nitrate test: NaOH + Al, heat $\rightarrow$ $\text{NH}_3$ $\rightarrow$ red litmus blue.
• The question's note about negative results means you must confirm with a positive test.

Common Mistakes

• Testing for a halide instead of nitrate (e.g. with $\text{AgNO}_3$) — that would give a cream ppt for HBr and no ppt for $\text{HNO}_3$, but the mark scheme wants the nitrate test.
• Not using aluminium — NaOH alone will not reduce nitrate to ammonia.

Things to Be Careful About

• Use excess NaOH and Al; heat the mixture (boiling tube).
• The gas must be identified by the litmus result.

Background Concept

Zinc and magnesium both form white hydroxides with aqueous ammonia or NaOH, but $\text{Zn(OH)}_2$ is amphoteric and dissolves in excess reagent (forming $[\text{Zn(NH}_3)_4]^{2+}$ with ammonia or $[\text{Zn(OH)}_4]^{2-}$ with NaOH), whereas $\text{Mg(OH)}_2$ is insoluble in excess. Since the carbonates are insoluble in water, they must first be dissolved in a mineral acid. Alternatively, heating the carbonate alone distinguishes them: ZnO is yellow when hot and white when cold, while MgO stays white.

Understanding the Question

FB 8 is either magnesium carbonate or zinc carbonate. Devise a test, record observations, conclude. Two marks.

Approach

Route 1: dissolve the carbonate in dilute acid, then add aqueous ammonia/NaOH and test solubility of the precipitate in excess. Route 2: heat the solid and observe the colour of the hot residue.

Step-by-Step Reasoning

Route 1:

1. Dissolve FB 8 in a mineral acid (e.g. dilute $\text{H}_2\text{SO}_4$) — effervescence as $\text{CO}_2$ is released.
2. Add aqueous ammonia (or NaOH): a white precipitate forms.
3. Add excess: the precipitate dissolves $\rightarrow$ $\text{Zn}^{2+}$ $\rightarrow$ FB 8 is $\text{ZnCO}_3$. ($\text{Mg(OH)}_2$ would not dissolve in excess.)
   Route 2:
4. Heat FB 8 alone in a hard-glass test-tube.
5. Observe the hot residue: yellow $\rightarrow$ ZnO $\rightarrow$ FB 8 is $\text{ZnCO}_3$. (MgO is white when hot.)

Key Takeaways

• $\text{Zn(OH)}_2$ is amphoteric (soluble in excess NH3/NaOH); $\text{Mg(OH)}_2$ is not.
• ZnO is yellow when hot, white when cold — a quick distinguishing test.

Common Mistakes

• Adding NaOH directly to the solid carbonate without dissolving first — carbonates are insoluble.
• Confusing which hydroxide dissolves in excess.

Things to Be Careful About

• The mark scheme accepts either route; give the observation and the conclusion.
• For route 2, the residue must be observed while hot (yellow).

Background Concept

The iodoform test distinguishes ethanol from methanol. Ethanol, $\text{CH}_3\text{CH}_2\text{OH}$, contains the $\text{CH}_3\text{CH(OH)}-$ group. With iodine in alkaline solution (NaOH), it is oxidised to acetaldehyde and then iodinated to triiodoacetaldehyde, which is hydrolysed to iodoform, $\text{CHI}_3$ — a pale yellow/cream solid with a characteristic antiseptic smell. Methanol, $\text{CH}_3\text{OH}$, has no $\text{CH}_3\text{CH(OH)}-$ group and gives no iodoform. The test often needs time to stand for the precipitate to form.

Understanding the Question

FB 9 is either aqueous methanol or aqueous ethanol. Devise a test, record observations, conclude. Two marks. The question notes that FB 9 is flammable and must not be heated with a flame; the mixture may need to stand.

Approach

Use the iodoform test: add iodine solution and aqueous NaOH, leave to stand, look for a yellow precipitate.

Step-by-Step Reasoning

1. Add iodine solution (or iodine in KI) and aqueous NaOH to FB 9.
2. Leave the mixture to stand (do not heat with a flame — flammable).
3. A pale yellow/cream precipitate of iodoform forms $\rightarrow$ ethanol present $\rightarrow$ FB 9 is ethanol, $\text{C}_2\text{H}_5\text{OH}$.
4. Methanol would give no precipitate (no $\text{CH}_3\text{CH(OH)}-$ group).

Key Takeaways

• Iodoform test: $\text{I}_2$ + NaOH $\rightarrow$ yellow ppt ($\text{CHI}_3$) for ethanol ($\text{CH}_3\text{CH(OH)}-$ group), not methanol.
• The test may take time; leave to stand.

Common Mistakes

• Heating with a flame — the solution is flammable.
• Expecting an immediate result — the precipitate forms on standing.
• Confusing methanol and ethanol (methanol gives no iodoform).

Things to Be Careful About

• Must not heat with a flame (explicit instruction).
• The precipitate is pale yellow/cream, formed on standing.