Chemistry 9701/31 — May/June 2025

Background Concept

For gases at room temperature and pressure, 1 mole occupies about 24 dm3 (24000 cm3). The reaction has a 1-to-1 molar ratio between CaCO3 and CO2, so the moles of gas produced equals the moles of CaCO3 in the sample.

Understanding the Question

One mark is given for both values. The answer must use the volume already collected; you are not expected to subtract anything or correct for gas pressure if the apparatus was at laboratory conditions. The moles of CO2 and the moles of CaCO3 are numerically equal.

Approach

Change cm3 to dm3, divide by the molar gas volume, then use the balanced equation to copy the same number of moles to CaCO3.

Step-by-Step Reasoning

1. Using $V=150\:\text{cm}^3$, converting to dimensional units: $0.150\ \text{dm}^3$.
2. $n=\frac{0.150}{24.0}=0.00625\ \text{mol}$.
3. Equation says 1 mol CaCO3 gives 1 mol CO2, so

$n(\text{CaCO}_3)=n(\text{CO}_2)=0.00625\ \text{mol}$

The result should be reported to 2–4 significant figures, matching the scale of the volumes.

Key Takeaways

Molar volume is used only for a gas at ‘room conditions’. Stoichiometry ratios are taken directly from the balanced equation; if the equation showed 1:2, the factor would be different.

Common Mistakes

Using 22.4 instead of 24.0 is a very common error. Mixing cm3 and dm3: dividing cm3 by 24 is also wrong—divide by 24000, or first convert to dm3.

Things to Be Careful About

Write the gas volume in dm3 before dividing so the units cancel. The moles of gas collected are never the same as the moles of acid used; the acid is present in excess.

Background Concept

Percentage purity compares the mass of pure calcium carbonate present in the powder to the total mass of FA1 that was weighed, usually expresses it as a percentage.

Understanding

You are asked to use the moles CaCO3 found in (b)(i) with the mass of FA1 that was actually added. The mass of the sample is already known from part (a).

The required answer carries the worked calculation.

Approach

First determine the mass of pure CaCO3:

mass = moles × molar mass. Then divide by mass of the sample and multiply by 100.

Step-by-Step

Using the illustrative values:

1. moles CaCO3 = 0.00625 mol.
2. $M_r(\text{CaCO}_3)$ = 40.1 + 12.0 + (3×16.0) = 100.1.
3. mass pure CaCO3 = 0.00625 × 100.1 = 0.6256 g.
4. mass of FA1 = 0.80 g.
5. % purity = (0.6256 / 0.80) × 100 = 78.2.

The example values show an impure sample; any larger gas volume and same mass would give a larger purity, and vice versa.

Key Takeaways

The mass of the solid sample is larger than the mass of the pure carbonate; the ratio gives the purity.

Common Mistakes

Using the mass of the total weighed container instead of the mass of FA1 in the purity denominator; forgetting to multiply by 100; using mixture moles from gas but not converting to mass first.

Things to Be Careful About

The molar mass must be one genuine value. Give the final percentage to 2–4 significant figures; the mark scheme does not insist on a particular number of figures, but not huge or tiny numbers. If the candidate’s gas volume gives e.g. 0.0062 mol, use that and help.

Background

In the experiment the acid should be in excess so that all the CaCO3 dissolves. If the acid becomes the limiting reagent, the reaction stops while some CaCO3 is still solid in the flask.

Approach

The only visible sign is the unreacted powder. The stopper must be removed to see inside; after all fizzing has stopped the solid should have vanished if enough acid were present. If it has not, acid (HCl) is not in excess.

Step-by-Step Reasoning

During the reaction CO2 leaves as bubbles, so fizzing. When fizzing stops the calcium carbonate has either all been used up or acid available is exhausted. If solid remains, the amount of acid has become insufficient before the carbonate is fully consumed. This shows the acid is not in excess.

Key Points

The acid being not in excess is not observed directly; it is inferred from the leftover powder. Do not answer “effervescence stops” alone, because that should occur even with an excess acid (after all carbonate reacted).

Common Mistakes

Stating “a gas is given off” is too vague. The required observation is specifically that unreacted solid remains, or that all the solid has not disappeared by the end.

Things to Be Careful About

Observations should be objective and physical. Do not claim to see HCl molecules; write about the solid’s appearance or absence.

Background

With insufficient HCl, CaCO3 is no longer the only limiting reagent; the reaction stops when all the acid is used. The stoichiometry 1 CaCO3 : 2 HCl fixes the maximum amount of carbonate that can react.

Understanding

In this part, the acid is deliberately not in excess, so the carbon dioxide volume is limited by the acid, not approximated by the amount of solid weighed. The question asks how this changes the purity calculated from (b)(ii).

Approach

Compare the volume that should have been collected with the true value: acid limited → CaCO3 is left unreacted → less CO2 → the calculated amount of CaCO3 is too low → purity appears too low.

Step-by-Step Reasoning

Because acid is not in excess, some calcium carbonate bleibt unreacted. Consequently the measured gas volume collected is lower. In (b)(i) we use that volume to calculate n(CO2), then set n(CaCO3) = n(CO2). Hence the deduced amount of CaCO3 is too low. The percentage purity = mass pure CaCO3 / mass sample, and the numerator is smaller, so the computed purity decreases.

This is not a “no effect” case, because the mass of CaCO3 used up to the calculation is no longer the true mass initially weighed.

Key Takeaways

Lit. results only reflect reaction that actually occurred; a limiting reactant makes the laboratory result appear lower than real purity.

Common Mistakes

Stating “% purity stays the same” because the sample mass did not change; the calculated amount is now limited deep. Saying “more acid would make purity higher” is also incorrect.

Things to Be Careful About

The marks are two: (1) say the gas volume is less, and (2) say the smaller gas volume lowers calculated moles/mass of CaCO3. Both ideas score. Present them clearly as two separate sentences.

Background Concept

Significant figures reflect the precision of the data. The concentrations (0.0900 mol dm⁻³ and 2.00 mol dm⁻³) and volumes (25.0, 250, 19.0 cm³) are given to 3 significant figures, so calculated answers should carry 3 (or 4) significant figures to avoid false precision.

Understanding the Question

This instruction tells you how to present the numerical answers in the following parts: each must be to 3 or 4 significant figures. It is a presentation requirement, not a calculation.

Approach

After each calculation, count the significant figures in the result and adjust the rounding so that 3 or 4 significant figures are shown.

Step-by-Step Reasoning

For example, 0.0024012 mol has 5 significant figures and must be rounded to 0.00240 mol (3 sf). Leading zeros do not count as significant figures. A value such as 0.240 mol has 3 significant figures (the trailing zero after the decimal point counts).

Key Takeaways

Match the precision of your answers to the precision of the given data. Leading zeros are placeholders; trailing zeros after a decimal point are significant.

Common Mistakes

• Quoting calculator output with 5 or more significant figures.
• Quoting too few significant figures (e.g. 0.002 mol instead of 0.00240 mol).

Things to Be Careful About

Leading zeros (e.g. in 0.00240) never count as significant figures. A trailing zero after the decimal point (e.g. in 0.240) does count.

Background Concept

The amount of a dissolved substance is given by $n = cV$, where $c$ is concentration in mol dm⁻³ and $V$ is volume in dm³. Since burette volumes are measured in cm³, they must be divided by 1000 to convert to dm³.

Understanding the Question

You know the concentration of FA 5 (0.0900 mol dm⁻³) and the volume used in the titration from part (b) (26.68 cm³). Calculate the amount of NaOH in that volume.

Approach

Substitute the concentration and the converted volume into $n = cV$.

Step-by-Step Reasoning

Volume = 26.68 cm³ = 26.68/1000 = 0.02668 dm³. Amount = 0.0900 × 0.02668 = 0.0024012 mol. To 3 significant figures, this is 0.00240 mol. The leading zeros are placeholders and do not count as significant figures.

Key Takeaways

Always convert cm³ to dm³ before using $n = cV$. The result must be to 3 or 4 significant figures.

Common Mistakes

• Forgetting to divide the volume by 1000.
• Quoting the unrounded value 0.0024012 (5 sf) instead of 0.00240 (3 sf).

Things to Be Careful About

The concentration 0.0900 has 3 significant figures. The volume 26.68 has 4. The answer is quoted to 3 sf to match the least precise given value.

Background Concept

Hydrochloric acid and sodium hydroxide neutralise in a 1:1 ratio: HCl + NaOH → NaCl + H₂O. Therefore the amount of HCl in the 25.0 cm³ portion of FA 4 titrated equals the amount of NaOH in the titre. To find the amount in the whole 250 cm³ flask, multiply by 10. Then, because FA 4 was prepared by diluting 25.0 cm³ of FA 3 to 250 cm³ (a 10× dilution), the amount of HCl in 250 cm³ of FA 3 is 10× the amount in 250 cm³ of FA 4.

Understanding the Question

This part asks for two amounts: the HCl in the full 250 cm³ of FA 4, and the HCl in the original 250 cm³ of FA 3. Both require scaling up from the titre result.

Approach

Step 1: amount HCl in 25.0 cm³ FA 4 = amount NaOH (from c(ii)). Step 2: multiply by 10 to get the amount in 250 cm³ FA 4. Step 3: multiply by 10 again because FA 4 is a 10× dilution of FA 3.

Step-by-Step Reasoning

The titre used 25.0 cm³ of FA 4, so the amount of HCl in that portion is 0.00240 mol. The volumetric flask holds 250 cm³, which is 10× the titre volume, so the amount in the flask is 0.00240 × 10 = 0.0240 mol. FA 4 was made by diluting 25.0 cm³ of FA 3 to 250 cm³, so FA 4 is 10× more dilute than FA 3; therefore the amount of HCl in 250 cm³ of FA 3 is 0.0240 × 10 = 0.240 mol.

Key Takeaways

Dilution factors and volume ratios both multiply the amount. Keep track of which solution is being scaled and in which direction.

Common Mistakes

• Applying only one ×10 instead of two.
• Confusing the direction of the dilution factor (FA 3 is more concentrated than FA 4).

Things to Be Careful About

The two ×10 factors are different in origin: one is the volume ratio (25 → 250 cm³), the other is the dilution factor (FA 3 → FA 4). Both must be applied.

Background Concept

FA 3 was prepared by adding 19.0 g of FA 1 to 250 cm³ of 2.00 mol dm⁻³ hydrochloric acid. The amount of HCl initially present is found from $n = cV$: concentration 2.00 mol dm⁻³ and volume 250 cm³ = 0.250 dm³.

Understanding the Question

Calculate the total amount of HCl that was used to prepare FA 3, before any reaction with the calcium carbonate occurred.

Approach

Convert the volume to dm³ and multiply by the concentration.

Step-by-Step Reasoning

Volume = 250 cm³ = 0.250 dm³. Amount = 0.250 × 2.00 = 0.500 mol. This is the initial amount of acid before any reacted with the calcium carbonate in FA 1.

Key Takeaways

The initial amount of HCl is a fixed reference value; the amount that reacted is found by subtracting the amount remaining (from c(iii)).

Common Mistakes

• Forgetting to convert cm³ to dm³.
• Using the wrong concentration (e.g. 0.0900 instead of 2.00).

Things to Be Careful About

The value 0.500 mol has 3 significant figures. This is the acid added in step 1, not the acid in FA 4.

Background Concept

In step 1, FA 1 (impure calcium carbonate) was added to an excess of hydrochloric acid. The acid that reacted with the carbonate was consumed; the acid that remained is what was titrated in step 3. Therefore: acid reacted = initial acid − acid remaining.

Understanding the Question

Use the initial amount of HCl (c(iv) = 0.500 mol) and the amount remaining in FA 3 (c(iii) = 0.240 mol) to find how much acid reacted with the calcium carbonate.

Approach

Subtract the remaining amount from the initial amount.

Step-by-Step Reasoning

Initial HCl = 0.500 mol. HCl remaining after reaction = 0.240 mol. HCl that reacted = 0.500 − 0.240 = 0.260 mol.

Key Takeaways

The "back-titration" idea: measure what is left over after a reaction to deduce how much was consumed.

Common Mistakes

• Subtracting in the wrong order (0.240 − 0.500 gives a negative value).
• Using the amount in FA 4 instead of FA 3.

Things to Be Careful About

Both values must refer to the same volume (250 cm³ of FA 3). The result 0.260 mol has 3 significant figures.

Background Concept

The balanced equation is CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g). The stoichiometric ratio is 1 mol CaCO₃ : 2 mol HCl. To find the amount of calcium carbonate that reacted, divide the amount of HCl that reacted by 2.

Understanding the Question

Use the amount of HCl that reacted with FA 1 (c(v) = 0.260 mol) and the balanced equation to find the amount of CaCO₃ in the sample.

Approach

Divide the amount of HCl by the stoichiometric coefficient 2.

Step-by-Step Reasoning

0.260 mol of HCl reacted. Since 2 mol HCl react with 1 mol CaCO₃, the amount of CaCO₃ = 0.260/2 = 0.130 mol. This is the amount of pure calcium carbonate in the 19.0 g sample of FA 1.

Key Takeaways

Always read the stoichiometric ratio from the balanced equation before converting between amounts of different substances.

Common Mistakes

• Multiplying by 2 instead of dividing (0.260 × 2 = 0.520, which is wrong).
• Using the wrong equation or forgetting the ratio.

Things to Be Careful About

The coefficient of HCl is 2, so CaCO₃ is half the amount of HCl. The answer 0.130 mol has 3 significant figures.

Background Concept

Percentage purity is the mass of pure substance divided by the mass of the impure sample, multiplied by 100. The mass of pure CaCO₃ is found from moles × molar mass, where $M_r(\text{CaCO}_3) = 40.1 + 12.0 + 3 \times 16.0 = 100.1$.

Understanding the Question

From the amount of CaCO₃ found in (c)(vi) = 0.130 mol, calculate the mass of pure calcium carbonate, then express it as a percentage of the 19.0 g sample of FA 1 that was used.

Approach

Convert moles of CaCO₃ to mass using the molar mass, then divide by the sample mass and multiply by 100.

Step-by-Step Reasoning

Mass of pure CaCO₃ = 0.130 × 100.1 = 13.01 g. The sample of FA 1 weighed 19.0 g. Percentage purity = (13.01/19.0) × 100 = 68.5%. This means 68.5% of the impure sample is calcium carbonate; the rest is the impurity.

Key Takeaways

Percentage purity links the amount of a pure component to the mass of the impure sample. The molar mass must be calculated correctly (100.1 for CaCO₃).

Common Mistakes

• Using the wrong molar mass (e.g. forgetting the oxygen atoms).
• Dividing by the wrong mass (e.g. 13.01/100.1 instead of 13.01/19.0).
• Quoting the answer to too many significant figures.

Things to Be Careful About

The sample mass is 19.0 g (3 sf), so the percentage should be quoted to 3 significant figures (68.5%). Show your working, as the mark scheme requires it.

Background Concept

In the gas collection method, the volume of carbon dioxide produced by the reaction CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g) is measured (typically by collecting gas over water in a measuring cylinder). The percentage purity is then calculated from the gas volume. Several practical issues make this method less accurate than titration.

Understanding the Question

The question asks for two reasons why the gas collection method gives a less accurate percentage purity than the titration method. You need to identify specific, credible sources of error in collecting and measuring the gas.

Approach

Think about where gas can be lost or the measured volume distorted during the collection process: dissolution in the collecting water, escape before sealing, and interference from the impurity.

Step-by-Step Reasoning

Reason 1: Carbon dioxide is soluble in water. When the gas is collected over water, some CO₂ dissolves, so the measured volume is less than the volume actually produced. This makes the calculated purity too low.

Reason 2: When the acid is added to the flask containing the calcium carbonate, gas is produced immediately. If the bung is not replaced instantly, some gas escapes before the apparatus is sealed, again reducing the measured volume.

Reason 3 (alternative): The impurity in FA 1 might itself react with the acid to produce gas, which would make the measured volume too high and inflate the calculated purity.

Key Takeaways

Gas collection methods lose accuracy through gas dissolving, gas escaping before sealing, and side reactions of impurities. Each error either reduces or inflates the measured volume and hence distorts the calculated purity.

Common Mistakes

• Giving vague answers such as "human error" or "not accurate" without a specific mechanism.
• Saying gas "evaporates" instead of "dissolves" — CO₂ dissolves in water; it does not evaporate from it.
• Repeating the same idea twice in different words (both marks require distinct reasons).

Things to Be Careful About

Each reason must be a distinct, specific source of error. The mark scheme accepts "some gas dissolves in the water", "some gas escapes before the bung is replaced", and "the impurity may react to produce gas".

Background Concept

The main source of error in the gas collection method is gas escaping before the apparatus is sealed, or gas dissolving in the collecting water. Improvements target these specific losses.

Understanding the Question

Describe one change to the gas collection method that would make the percentage purity more accurate. The change must address one of the error sources identified in (d)(i).

Approach

Choose one error from part (d)(i) and describe a concrete modification that eliminates or reduces it.

Step-by-Step Reasoning

If the problem is gas escaping before the bung is replaced, the fix is to have both reactants inside the sealed flask from the start, separated so they only mix when you are ready. For example, put the CaCO₃ in a small container (or a paper package held by a thread) inside the flask, add the acid, seal the flask, then tip the container over to start the reaction. No gas can escape because the flask was sealed before the reaction began.

Alternatively, to reduce CO₂ dissolving, saturate the water with CO₂ before the experiment, or use warmer water (less gas dissolves at higher temperature). Another option is to use a gas syringe instead of collecting over water, which avoids dissolution entirely.

Key Takeaways

A good improvement is specific and directly addresses a named source of error. "Be more careful" is not an improvement; describing how the apparatus is changed to prevent gas loss is.

Common Mistakes

• Giving a vague answer such as "repeat the experiment" or "be more careful".
• Suggesting an improvement that does not address a specific error (e.g. "use a bigger flask").

Things to Be Careful About

The improvement must be a change to the method or apparatus, not a general statement about accuracy. The mark scheme accepts: separating CaCO₃ and acid within the stoppered flask until reaction starts, saturating the water with CO₂ or using warmer water, or using a gas syringe.

Background Concept

Qualitative analysis of aqueous solutions relies on characteristic reactions of ions with specific reagents. Transition metal cations like $\text{Cu}^{2+}$ form coloured precipitates with $\text{OH}^-$ and $\text{NH}_3$, and can form soluble complex ions with excess ammonia. Ammonium ions ($\text{NH}_4^+$) release ammonia gas when warmed with strong alkali. Halide anions form characteristic precipitates with acidified silver nitrate. Oxidising cations like $\text{Cu}^{2+}$ can oxidise iodide ions to iodine, which forms a blue-black complex with starch.

Understanding the Question

You are given FA 7, a solution containing two cations and one anion. You must perform six tests (Table 3.1 and 3.2) and record the exact observations. The tests cover cation identification (ammonia, sodium hydroxide, potassium iodide), anion identification (silver nitrate, barium salts, nitric acid), and gas testing (warming with NaOH).

Approach

Systematically apply each test to FA 7 and record what is seen. Compare your observations against known reactions:

• $\text{Cu}^{2+}$ gives a pale blue ppt with $\text{NH}_3$(aq) and $\text{NaOH}$(aq), soluble in excess $\text{NH}_3$ to give a deep blue complex.
• $\text{NH}_4^+$ gives $\text{NH}_3$ gas with hot $\text{NaOH}$, turning damp red litmus blue.
• $\text{Cl}^-$ gives a white ppt with $\text{AgNO}_3$.
• $\text{Cu}^{2+}$ oxidises $\text{I}^-$ to $\text{I}_2$, turning the solution brown and starch blue-black.
• Absence of $\text{SO}_4^{2-}$ or $\text{CO}_3^{2-}$ is shown by no reaction with $\text{Ba}^{2+}$ or dilute $\text{HNO}_3$.

Step-by-Step Reasoning

1. Test 1 ($\text{NH}_3$(aq)): $\text{Cu}^{2+}$ reacts to form $\text{Cu(OH)}_2$(pale blue ppt). Excess $\text{NH}_3$ forms the tetraamminecopper(II) complex $[\text{Cu(NH}_3)_4\text{(H}_2\text{O})_2]^{2+}$, which is deep blue and soluble. Observation: (pale) blue ppt soluble in excess to give a dark blue solution.
2. Test 2 ($\text{NaOH}$(aq) then warm): $\text{Cu}^{2+}$ forms $\text{Cu(OH)}_2$ (pale blue ppt), insoluble in excess. Warming with $\text{NaOH}$ releases $\text{NH}_3$ gas from $\text{NH}_4^+$ (turns red litmus blue) and $\text{Cu(OH)}_2$ decomposes to black $\text{CuO}$ residue.
3. Test 3 ($\text{AgNO}_3$): $\text{Cl}^-$ forms $\text{AgCl}$, a white precipitate.
4. Test 4 ($\text{Ba}^{2+}$): No $\text{SO}_4^{2-}$ or $\text{CO}_3^{2-}$ present, so no precipitate. Solution remains pale blue.
5. Test 5 (dilute $\text{HNO}_3$ then stand): Confirms no carbonate/sulfite (no effervescence). No change.
6. Test 6 ($\text{KI}$ then starch): $2\text{Cu}^{2+} + 4\text{I}^- \rightarrow 2\text{CuI}(s) + \text{I}_2(aq)$. Iodine turns solution brown; starch turns it blue-black.

Key Takeaways

Precise observation recording is critical in practical papers. Note colour changes, precipitate formation and solubility, and gas evolution with appropriate tests. Complexation with excess ammonia is a key differentiator for copper(II).

Common Mistakes

• Writing 'blue solution' instead of 'blue precipitate soluble in excess' for Test 1.
• Forgetting to mention the black solid ($\text{CuO}$) formed upon warming in Test 2.
• Stating 'white ppt' for Test 3 without specifying it is insoluble in excess (though not strictly required here, precision matters).
• Claiming effervescence in Test 4 or 5.

Things to Be Careful About

• Record observations at the correct stage (e.g., before and after adding excess reagent, before and after warming).
• Use precise colour terms: 'pale blue' for $\text{Cu(OH)}_2$, 'dark/deep blue' for the complex, 'white' for $\text{AgCl}$, 'brown' for iodine water, 'blue-black' for starch-iodine.
• State symbols are not required in the observation tables, but must be in ionic equations.

Background Concept

Ionic equations represent the actual chemical species involved in a reaction, omitting spectator ions. State symbols ((s), (l), (g), (aq)) are mandatory in Cambridge practical papers to show understanding of the physical state of each species. Precipitation reactions occur when soluble ions combine to form an insoluble solid.

Understanding the Question

You must write an ionic equation for any reaction in (a)(i) that produced a precipitate. The mark scheme accepts three valid options based on the tests performed.

Approach

Identify the tests that formed precipitates: Test 2 ($\text{Cu(OH)}_2$), Test 3 ($\text{AgCl}$), and Test 6 ($\text{CuI}$ and $\text{I}_2$, though $\text{I}_2$ is aqueous, $\text{CuI}$ is solid). Write the net ionic equation for one of them, ensuring charges and atoms are balanced and state symbols are included.

Step-by-Step Reasoning

1. Option 1: Reaction of copper(II) with hydroxide ions. $\text{Cu}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq}) \rightarrow \text{Cu(OH)}_2(\text{s})$. Balanced, correct states.
2. Option 2: Reaction of silver ions with chloride ions. $\text{Ag}^+(\text{aq}) + \text{Cl}^-(\text{aq}) \rightarrow \text{AgCl}(\text{s})$. Balanced, correct states.
3. Option 3: Redox reaction between copper(II) and iodide. $2\text{Cu}^{2+}(\text{aq}) + 4\text{I}^-(\text{aq}) \rightarrow 2\text{CuI}(\text{s}) + \text{I}_2(\text{aq})$. Note that $\text{CuI}$ is the precipitate here.

Key Takeaways

Always include state symbols in ionic equations for practical questions. Check that the total charge is balanced on both sides.

Common Mistakes

• Forgetting state symbols (especially (aq) for ions and (s) for precipitates).
• Writing molecular equations instead of ionic equations (e.g., $\text{CuSO}_4 + 2\text{NaOH} \rightarrow \text{Cu(OH)}_2 + \text{Na}_2\text{SO}_4$).
• Incorrectly balancing charges or atoms.

Things to Be Careful About

The mark scheme explicitly allows any of the three equations. Ensure you only write one if asked for 'an' equation, though writing multiple correct ones is usually not penalised unless specified. The redox equation for iodide is often overlooked but is fully creditable.

Background Concept

Deduction of ions in a mixture requires cross-referencing positive and negative tests. A cation test that is positive confirms the ion's presence; a negative test (no reaction) helps rule out others. The combination of tests must be consistent with the observed colours and precipitates.

Understanding the Question

You must give the formula of each ion in FA 7. Two cations and one anion are present. You have 2 marks for this (2 correct = 1 mark, all 3 = 2 marks).

Approach

• Cations: Test 1 and 2 confirm $\text{Cu}^{2+}$ (blue ppt, deep blue complex). Warming with NaOH confirms $\text{NH}_4^+$ (gas turning litmus blue). No other cations are indicated.
• Anion: Test 3 confirms $\text{Cl}^-$ (white ppt with $\text{AgNO}_3$). Tests 4 and 5 rule out $\text{SO}_4^{2-}$ and $\text{CO}_3^{2-}$.

Step-by-Step Reasoning

1. Pale blue ppt with $\text{NH}_3$ and $\text{NaOH}$, soluble in excess $\text{NH}_3$ to deep blue $ightarrow$ $\text{Cu}^{2+}$.
2. Gas with hot $\text{NaOH}$ turning red litmus blue $ightarrow$ $\text{NH}_4^+$.
3. White ppt with $\text{AgNO}_3$, insoluble in dilute $\text{HNO}_3$ (implied by no change in Test 5) $ightarrow$ $\text{Cl}^-$.
4. No ppt with $\text{Ba}^{2+}$ $ightarrow$ no $\text{SO}_4^{2-}$.

Key Takeaways

Use a process of elimination. Positive tests identify ions; negative tests (no reaction) rule out common anions like sulfate and carbonate.

Common Mistakes

• Writing $\text{Cu}^+$ instead of $\text{Cu}^{2+}$ (copper(I) is not stable in aqueous solution with these reagents).
• Forgetting the charge on ions (e.g., writing $\text{Cl}$ instead of $\text{Cl}^-$).
• Missing the ammonium ion and assuming the gas was from something else.

Things to Be Careful About

The question asks for 'formula', so write $\text{Cu}^{2+}$, not 'copper(II) ion'. Include charges explicitly.

Background Concept

Thermal decomposition of hydrated salts or bicarbonates often releases water of crystallisation or water from the decomposition reaction. When heated in a test-tube, water vapour travels to the cooler upper parts and condenses into visible droplets.

Understanding the Question

You are heating FA 8, an anhydrous sodium compound. You must record observations during gentle then strong heating.

Approach

Sodium bicarbonate ($\text{NaHCO}_3$) decomposes on heating to sodium carbonate, water, and carbon dioxide. The water will condense on the cooler parts of the tube.

Step-by-Step Reasoning

1. Heat FA 8 ($\text{NaHCO}_3$).
2. Reaction: $2\text{NaHCO}_3(\text{s}) \rightarrow \text{Na}_2\text{CO}_3(\text{s}) + \text{H}_2\text{O}(\text{g}) + \text{CO}_2(\text{g})$.
3. Water vapour condenses on the cooler upper walls of the test-tube.
4. Observation: condensation, colourless droplets, or steam.

Key Takeaways

When heating solids that release water, look for condensation on the cooler parts of the apparatus.

Common Mistakes

• Saying 'the solid melts' (sodium compounds generally have high melting points and won't melt in a test-tube).
• Not specifying where the condensation forms (upper/cooler part of the tube).

Things to Be Careful About

Use precise language: 'condensation' or 'colourless droplets'. Do not just say 'water' without describing its appearance as droplets/steam.

Background Concept

Carbonates react with dilute acids to produce carbon dioxide gas, water, and a salt. This is a standard test for the carbonate ion ($\text{CO}_3^{2-}$). The residue from heating sodium bicarbonate is sodium carbonate.

Understanding the Question

FA 9 is the residue from heating FA 8. You add it to dilute nitric acid. You must record observations.

Approach

FA 8 is $\text{NaHCO}_3$. Heating gives FA 9 ($\text{Na}_2\text{CO}_3$). Adding dilute $\text{HNO}_3$ to $\text{Na}_2\text{CO}_3$ produces $\text{CO}_2$ gas.

Step-by-Step Reasoning

1. FA 9 is $\text{Na}_2\text{CO}_3$ (white solid).
2. Add dilute $\text{HNO}_3$.
3. Reaction: $\text{Na}_2\text{CO}_3(\text{s}) + 2\text{HNO}_3(\text{aq}) \rightarrow 2\text{NaNO}_3(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})$.
4. Observation: effervescence (bubbling/fizzing) as $\text{CO}_2$ is released. The solid dissolves to form a colourless solution.
5. Confirmatory test (if mentioned): $\text{CO}_2$ turns limewater milky (white ppt of $\text{CaCO}_3$).

Key Takeaways

Carbonates fizz with dilute acids. The gas is $\text{CO}_2$, confirmed by limewater.

Common Mistakes

• Saying 'no reaction' (carbonates definitely react with acids).
• Not mentioning the solid dissolving or the solution becoming colourless.
• Confusing the gas with hydrogen (hydrogen is colourless and odourless, but $\text{CO}_2$ is the product here).

Things to Be Careful About

The mark scheme awards M1 for effervescence and M2 for either the gas identification (limewater test) or the formation of a colourless solution. Include both if possible for full marks.

Background Concept

Sodium hydrogen carbonate (bicarbonate) decomposes on heating to sodium carbonate, water, and carbon dioxide. This is distinct from sodium carbonate, which is stable to heat (under normal test-tube conditions).

Understanding the Question

Deduce the formula of FA 8 based on the observations in (b)(i) and (b)(ii).

Approach

• (b)(i) shows water is released on heating $ightarrow$ contains hydrogen or is a hydrate. Since it's anhydrous, it must contain hydrogen in a decomposable group like $\text{HCO}_3^-$.
• (b)(ii) shows $\text{CO}_2$ is released with acid $ightarrow$ contains carbonate or hydrogen carbonate.
• Combined: the compound is a hydrogen carbonate. Since it's a sodium compound, it is $\text{NaHCO}_3$.

Step-by-Step Reasoning

1. Heating FA 8 releases water (condensation) $ightarrow$ $\text{NaHCO}_3$ decomposes: $2\text{NaHCO}_3 \rightarrow \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2$.
2. Residue FA 9 ($\text{Na}_2\text{CO}_3$) reacts with acid to give $\text{CO}_2$ (effervescence).
3. Therefore, FA 8 is sodium hydrogen carbonate, $\text{NaHCO}_3$.

Key Takeaways

Thermal decomposition of bicarbonates produces carbonates, water, and $\text{CO}_2$. This is a key distinguishing test from carbonates.

Common Mistakes

• Writing $\text{Na}_2\text{CO}_3$ (this is the residue, not the original compound; it doesn't release water on heating).
• Writing $\text{NaOH}$ (doesn't decompose to give $\text{CO}_2$).
• Forgetting to balance the formula (e.g., $\text{NaHCO}_3$ is correct, $\text{Na}_2\text{HCO}_3$ is wrong).

Things to Be Careful About

Ensure the formula is written correctly: $\text{NaHCO}_3$, not $\text{Na}_2\text{CO}_3$ or $\text{NaHCO}$. The question states FA 8 is anhydrous, so don't add water of crystallisation.