Topic 12 is where examiners separate students who have done chemistry from students who have only read it. The traps are all practical and all predictable: baselines drawn in ink that dissolve and wreck the chromatogram, solvent poured above the baseline so the spots wash away, Rf values greater than 1 reported without a flicker of doubt, solutions that "turn clear" when the examiner will only pay for colourless, aluminium and zinc declared identical because nobody reached for the excess ammonia, iron's two ions swapping colours, silver halides in the wrong order, sulfate answers given to sulfite questions, lighted and glowing splints exchanged, copper sulfate boiled to a grey crust instead of crystallised, rough titres averaged into the mean, and impure samples said to melt higher. Every one of these is on the list below — hunted down, explained, and drilled until Paper 6 feels like open-book.
Twelve traps that cost students marks on Topic 12 questions. Every one of them appears on challenge papers regularly.
Six challenging questions broken down step by step. Try each step yourself before revealing the next.
Use a volumetric pipette (with safety filler) to transfer exactly 25.0 cm³ of the NaOH into a conical flask standing on a white tile. Add a few drops of thymolphthalein — the solution turns blue (alkaline). Fill the burette with the hydrochloric acid (after rinsing it with a little of the acid) and record the initial reading at the bottom of the meniscus, at eye level.
Run acid in from the burette, swirling the flask constantly. Near the expected end-point, add drop by drop. Stop at the end-point: the moment the indicator just changes from blue to colourless permanently. Read the burette again; titre = final − initial reading. Repeat until concordant results are obtained. (Say "colourless" — never "clear".)
24.00 and 24.10 agree within 0.10 cm³ — concordant. 24.90 is the rough titre, run quickly to find the end-point's neighbourhood; it is excluded. Average = (24.00 + 24.10) ÷ 2 = 24.05 cm³. Note the shape of every burette value: two decimal places, ending in 0 or 5.
You know the NaOH's volume and concentration, so start there. Moles NaOH = concentration × volume in dm³ = 0.100 × (25.0 ÷ 1000) = 2.50 × 10⁻³ mol. The unit conversion cm³ → dm³ (divide by 1000) is where most calculation errors are born.
Moles HCl = moles NaOH = 2.50 × 10⁻³ mol, delivered in 24.05 cm³ = 0.02405 dm³. Concentration = moles ÷ volume = 2.50 × 10⁻³ ÷ 0.02405 = 0.104 mol/dm³ (3 s.f.). Always finish with units — mol/dm³.
Draw a baseline in pencil near the bottom of the paper (reason: graphite is insoluble, so the line will not dissolve and run). Place small spots of G, T, X and Y on the line, labelled in pencil. Stand the paper in solvent with the level below the baseline (reason: otherwise the spots dissolve off the paper into the solvent). Cover, let the solvent rise, and mark the solvent front immediately on removal.
G separates into two spots, so the "single" green colouring is a mixture of at least two substances. (A blue and a yellow dye blended to look green is the classic reality behind such results.) One spot would have meant pure; two or more means mixture — the fastest purity verdict in chemistry.
G's spot at 6.8 cm sits at exactly the height of T (tartrazine-B) — the banned dye is present. G's spot at 3.0 cm matches X, a permitted dye. Nothing in G lines up with Y at 5.1 cm, so Y is absent. Three conclusions, each anchored to a height comparison on the same paper in the same solvent.
Rf = distance moved by substance ÷ distance moved by solvent = 6.8 ÷ 8.5 = 0.80. Sanity check before writing anything: is it less than 1? Yes. No units, two decimal places. (8.5 ÷ 6.8 = 1.25 is the upside-down trap — impossible, therefore instantly detectable.)
Examiners love bolting this on: if the substances were colourless (amino acids, sugars), the separated spots would be invisible. Spray the dried chromatogram with a locating agent, which reacts with the substances to form coloured spots, then measure positions and Rf values exactly as before.
Most common cations give colourless solutions. A pale green solution points at iron(II) before a single reagent is added (copper(II) would be blue, iron(III) yellow-brown). Treat solution colour as evidence — it is often a printed observation the examiner expects you to use.
A green precipitate insoluble in excess NaOH is iron(II) hydroxide, Fe(OH)₂ — the fingerprint of Fe²⁺. Not red-brown (that would be Fe³⁺), not blue (Cu²⁺), not white (Al³⁺, Zn²⁺, Ca²⁺). One colour word carries the identification.
With aqueous ammonia iron(II) behaves identically: green precipitate, insoluble in excess. This rules out any cation whose precipitate dissolves in excess ammonia (zinc, copper — copper's would go deep blue). Consistency between the two reagents strengthens the identification — exactly how real analysis works.
Acidified barium nitrate giving a white precipitate (BaSO₄) identifies sulfate, SO₄²⁻. The prior acidification matters: it destroyed any carbonate that could fake the white precipitate. Since the mixture was acidified and still precipitated, sulfate is certain.
Warming with NaOH gives no gas, so there are no ammonium ions in Q (NH₄⁺ would have released ammonia, turning damp red litmus blue). Negative results are evidence: they eliminate. Q's ions are Fe²⁺ and SO₄²⁻ only.
Sand: insoluble — filtration territory. Sodium chloride: dissolved in the water — crystallisation territory. Water: the solvent itself — distillation territory. Every separation question starts with this inventory; the properties choose the methods for you.
Filter the mixture: the sand stays on the filter paper as the residue; the salt solution passes through as the filtrate. Wash the sand on the paper with a little distilled water (to rinse off salt solution), then dry it in a warm oven. Pure dry sand: done.
Crystallisation recovers the salt but lets the water escape as vapour; simple distillation collects the water but leaves damp salt behind. Needing both products from one filtrate, use simple distillation and keep both ends: the water distils over and is collected; concentrated salt solution remains in the flask for crystallising. One apparatus, two products.
Stop distilling while solution remains; the flask now holds a concentrated (near-saturated) salt solution. Cool it slowly so crystals form, filter them off and dry between filter papers. (Sodium chloride is not hydrated, so evaporating further would not destroy it — but crystallisation is the answer examiners want for "pure crystals", and it is essential for hydrated salts.)
Measure the boiling point: pure water boils sharply at exactly 100 °C (at standard pressure). Dissolved impurities would raise it and spread it over a range. (Anhydrous copper(II) sulfate turning blue only proves water is present — it cannot prove purity. Choosing the wrong test here is Topic 12's favourite cross-examination.)
Add a little NaOH to each: all three give white precipitates. Add a little ammonia instead: aluminium and zinc again give white precipitates. The initial precipitate is almost useless here — the information is entirely in what happens in excess reagent. That is the insight the question is built on.
Continue adding NaOH to excess in each tube. Two precipitates dissolve (aluminium and zinc — colourless solutions form); one remains: that tube is calcium nitrate. One reagent, one ion nailed. Record it as an observation pair: "white precipitate, insoluble in excess NaOH".
Both remaining ions gave white precipitates soluble in excess NaOH — indistinguishable. Stating this explicitly often carries a mark: the examiner wants you to know why a second reagent is needed, not just to reach for it by luck.
Take fresh samples of the two unknowns and add aqueous ammonia to excess. Both first give white precipitates; then the difference: zinc's dissolves in excess ammonia (colourless solution), aluminium's does not. The tube whose precipitate vanishes is zinc nitrate; the stubborn one is aluminium nitrate.
Zinc: dissolves in excess of both. Aluminium: excess NaOH only. Calcium: neither. Every white-precipitate puzzle in IGCSE unlocks with this line — it converts a six-mark planning question into a fill-in-the-blanks exercise.
Hydrogen: hold a lighted splint at the jar's mouth — it burns with a squeaky 'pop'. Oxygen: insert a glowing splint — it relights. Each answer is apparatus + observation, welded together. Swap the splints and both tests fail: that is the exact trap this question sets.
Bubble the gas through limewater (aqueous calcium hydroxide): it turns milky (cloudy white) as insoluble calcium carbonate forms. "Milky" is the accepted observation word. Limewater identifies CO₂ and nothing else on the list.
Hold damp red litmus paper in the gas: ammonia turns it blue. It is the only common gas that turns litmus blue, so this test is definitive. State the paper's starting colour — "litmus turns blue" without "red" and "damp" is a weakened answer.
Litmus responds to ions in aqueous solution. The gas must first dissolve in the film of water on the damp paper — ammonia forming the alkaline NH₃(aq)/NH₄OH — before any colour change can happen. Dry paper, dry gas, no solution, no response. The same logic applies to chlorine's test.
The gas is chlorine. First it dissolves to give an acidic solution, turning the blue litmus red; then chlorine bleaches the dye, so the paper turns white. The bleaching is the identifying observation — no other common gas whitens litmus. Candidates who stop at "turns red" have identified an acid, not chlorine.
Pairs of questions that look nearly identical but have different answers. Spot the key distinction.
Click each node to see how the subtopics connect.
Spot the error in each student's answer. Think before revealing.
Ten Cambridge-style challenge questions. Write your answer, then reveal the model answer with mark scheme and examiner's notes.