PAG 4 and PAG 7 Qualitative Analysis of Unknowns

PAG 4 and PAG 7 asks you to identify several unknown substances, typically colourless organic liquids (here A, B and C) and white inorganic solids (here X, Y and Z), from a fixed menu of test-tube reactions.

Nothing here needs new theory: the tests come from the Group 17 chemistry, the tests for alkenes, alcohols, aldehydes and haloalkanes, and the ion tests you first met at GCSE.

What is new is the discipline: a small portion of the unknown for each test, the correct reagent in the correct order and the right conditions (warming when the test needs it).

Then comes an observation written in words an examiner will credit before any inference is made.

The logic is an evidence trail: unknown → test → observation → inference → identity. Negative results are part of the trail.

A liquid that turns acidified dichromate(VI) green but leaves Fehling’s solution blue is a primary or secondary alcohol; the negative Fehling’s result is what rules out an aldehyde, which dichromate(VI) would also have oxidised.

A solid that fizzes with acid is not shown to be a carbonate until the gas has turned limewater cloudy.

Students lose marks by jumping from one positive test to an identity and by describing a solution with no colour as “clear”.

The written papers (practical skills are examined in every paper) and the practical endorsement.

For this practical that means naming reagents and conditions precisely (warm in a water bath, dilute nitric acid before silver nitrate), giving observations in full (colour before and after, precipitate colour, gas test) and writing ionic equations with state symbols.

It also means explaining why a step is there: why the acid, why the warming, why the solid is dissolved first, why a second reagent is needed to confirm.

The evidence trail for the six unknowns on this page: every test, every observation and the inference each one supports.

The method below follows the sample set on this page: three colourless liquids A, B and C and three white solids X, Y and Z. Each test uses a fresh portion of the unknown, and every “why” is a mark-scheme point.

Four tests separate the organic liquids: bromine water for a C=C bond, warm acidified dichromate(VI) for anything oxidisable, warm Fehling’s solution (Fehling’s solution or Tollens’ reagent are accepted) for an aldehyde, and hydrolysis followed by silver nitrate for a haloalkane.

The table gives the sample observations and the inference each one allows. Notice that the inference column uses the negative results as much as the positive ones.

Three details carry marks. The dichromate(VI) and Fehling’s tubes are warmed in a water bath at about 60 °C; at room temperature both stay unchanged and the negative result is worthless.

Bromine water is decolourised, from orange to colourless, never “goes clear”.

The dichromate(VI) result alone does not identify A: dichromate(VI) would also oxidise an aldehyde, which is exactly why the negative Fehling’s result must be quoted in the inference.

Bromine water added to A, B and C: only B decolourises it, from orange to colourless, which is the evidence for a C=C bond.

After warming in a water bath, only tube A has changed from orange to green: Cr₂O₇²⁻(aq) reduced to Cr³⁺(aq) as A is oxidised.

Fehling’s solution warmed with A, B and C: all three remain blue with no precipitate, so none of them is an aldehyde.

Hydrolysis, acidification with dilute nitric acid and then silver nitrate: only C gives a cream precipitate of silver bromide.

The haloalkane test has five steps and every one of them is examined. First, warm a few drops of C with sodium hydroxide solution and a little ethanol in a water bath.

The C–Br bond is covalent, so there are no bromide ions in the liquid until it has been hydrolysed: CH₃CH₂CH₂Br + OH⁻ → CH₃CH₂CH₂OH + Br⁻ (1-bromopropane is used as the example).

Ethanol is the co-solvent that lets the organic liquid mix with the aqueous alkali, and it is flammable, so the tube is warmed in a water bath and never over a flame.

Second, cool the tube and acidify with dilute nitric acid. The alkali was added in excess, and Ag⁺ reacts with OH⁻ to give a brown precipitate of silver oxide, Ag₂O, which would hide the halide result: 2Ag⁺(aq) + 2OH⁻(aq) → Ag₂O(s) + H₂O(l).

Nitric acid is chosen because hydrochloric acid would add chloride ions and give a white precipitate of AgCl whatever the unknown, and sulfuric acid would add sulfate ions.

Third, add silver nitrate solution: Ag⁺(aq) + Br⁻(aq) → AgBr(s), a cream precipitate.

Fourth and fifth, add dilute ammonia solution (the cream precipitate stays) and then concentrated ammonia solution (it dissolves): AgBr(s) + 2NH₃(aq) → [Ag(NH₃)₂]⁺(aq) + Br⁻(aq).

The five steps for testing an organic liquid for a halogen: hydrolyse with NaOH(aq) and ethanol in a water bath, cool, acidify with dilute nitric acid, add silver nitrate, confirm with dilute then concentrated ammonia.

The same silver nitrate chemistry is used to compare the rates of hydrolysis of chloro-, bromo- and iodoalkanes in Practical Skills: Rates of Hydrolysis (hydrolysis of halogenoalkanes), where the time taken for the precipitate to appear is the measurement.

Here the precipitate is only the evidence for which halogen is present.

The three white solids are tested for their anions in solution, so the first step is to dissolve a spatula tip of each in a little distilled water.

A dry solid cannot react with silver nitrate or barium chloride solution, and tap water would add chloride ions.

On this board the cation is identified separately, by the scheme in the cation card below; in this sample set X is a calcium salt, Y a sodium salt and Z a potassium salt.

The table gives the anion evidence with the inference beside each observation.

Identities: X is calcium bromide, CaBr₂; Y is sodium sulfate, Na₂SO₄; Z is potassium carbonate, K₂CO₃.

Each anion has one primary test and one confirmation: silver nitrate then ammonia for the halide, acid then barium chloride for the sulfate, acid then limewater for the carbonate.

The chlorine water result for X is a second, independent line of evidence for bromide, which is what makes the identification secure.

The acid before each precipitation test is not decoration. Carbonate ions give a white precipitate with silver ions (Ag₂CO₃) and with barium ions (BaCO₃), so a carbonate would give a false positive in both tests.

Dilute acid converts the carbonate to carbon dioxide and removes the problem.

The acid must add nothing that is being tested for: nitric acid before silver nitrate (hydrochloric acid would add Cl⁻), hydrochloric or nitric acid before barium chloride (sulfuric acid would add SO₄²⁻).

The silver halide test: acidify with dilute nitric acid, add silver nitrate, then use dilute and concentrated ammonia to tell white AgCl, cream AgBr and yellow AgI apart.

The sulfate test: acidify with dilute hydrochloric acid, then add barium chloride solution. A white precipitate of BaSO₄ shows sulfate; a carbonate fizzes with the acid instead.

The carbonate test done properly: acid added to the solid in a boiling tube, bung fitted at once, and the gas bubbled through limewater, which turns cloudy.

The anion tests on X, Y and Z side by side, with the orange bromine in the upper cyclohexane layer of the chlorine water tube for X.

Every precipitation and gas equation in this practical is expected as an ionic equation with state symbols; an equation without them loses the mark. The full set for the sample unknowns is below, together with the equations for the confirmatory and organic steps.

Three reagent roles are asked about again and again. Dilute nitric acid before silver nitrate has two jobs.

In the inorganic test it removes carbonate ions, which would give a white precipitate of Ag₂CO₃ and mask the halide. In the organic test it neutralises the excess sodium hydroxide from the hydrolysis, which would otherwise give brown Ag₂O.

Why nitric acid and not another acid: hydrochloric acid adds chloride ions (false white AgCl) and sulfuric acid adds sulfate ions (false white BaSO₄ in the sulfate test).

Dilute hydrochloric acid before barium chloride removes carbonate ions, which would give white BaCO₃; sulfuric acid can never be used here.

The ammonia steps are also reagent roles. Dilute ammonia dissolves AgCl by forming the soluble complex [Ag(NH₃)₂]⁺; AgBr is less soluble and needs concentrated ammonia; AgI is too insoluble to dissolve in either.

So the ammonia result is not a second test for the same thing, it is what separates the three halides once a precipitate has been seen.

A qualitative practical has no numerical result to compare with an accepted value, so the evaluation questions ask about false positives, false negatives and the steps that prevent them. The table lists the errors students actually make and what each does to the conclusion.

Where a quantity is measured, its percentage uncertainty still follows the usual rule.

If a method asks for “about 0.5 g” of solid and a 2 d.p. balance reads 0.50 g, the uncertainty is ±0.005 g, so the percentage uncertainty is (0.005 ÷ 0.50) × 100 = 1.0%.

Measuring 5 cm³ of water with a 10 cm³ measuring cylinder (±0.5 cm³) gives (0.5 ÷ 5) × 100 = 10%.

Neither matters here, because a precipitate forms or does not form whatever the exact mass or volume, and that is the answer to “why is ‘approximately’ acceptable in this method”: the result is qualitative, so the measurements need only be roughly right.

The composition data in the exam question are different: 38.7% quoted to 1 decimal place has an uncertainty of ±0.05%, about 0.1% of the value.

That is why the mole ratio comes out as 1.00 : 3.00 : 1.00 and not as something that needs rounding.

• Writing “clear” for a solution with no colour. Bromine water is decolourised; a solution with no colour is colourless; “clear” only means not cloudy.
• Not warming the dichromate(VI) and Fehling’s tubes, then reporting “no reaction”. A negative result at room temperature proves nothing.
• Calling A “an oxidisable alcohol”. The inference is “a primary or secondary alcohol”; the negative Fehling’s result is what rules out an aldehyde.
• Adding silver nitrate straight to the organic liquid. The halogen is covalently bonded; hydrolyse first, then acidify with nitric acid, then test.
• Testing a solid with silver nitrate or barium chloride solution without dissolving it in distilled water first.
• Giving “fizzes with acid” as proof of a carbonate. The gas must turn limewater cloudy.
• Stopping at “cream precipitate”. Without the dilute and concentrated ammonia results, AgBr has not been distinguished from AgI.
• Writing “chlorine displaces bromide ions”. Chlorine oxidises bromide ions to bromine; the element bromine is what is displaced.
• Ionic equations without state symbols, and minus signs missing from Br⁻ and SO₄²⁻.
• Rounding the mole ratio carelessly (CH₂O instead of CH₃O) and quoting a fragment whose mass does not add up (CHO⁺ is 29, CH₂OH⁺ is 31).

Say

“Warm with acidified potassium dichromate(VI) in a water bath: orange to green, so the liquid is oxidised.”

“Fehling’s solution remains blue on warming, so it is not an aldehyde; A is a primary or secondary alcohol.”

“Dissolve the solid in distilled water, add dilute nitric acid, then silver nitrate solution: cream precipitate, insoluble in dilute ammonia, soluble in concentrated ammonia, so bromide ions.”

“Effervescence; the gas turns limewater cloudy, so carbonate ions.”

“Chlorine oxidises bromide ions to bromine, which gives an orange upper layer in cyclohexane.”

Do not say

“Goes clear.” “Positive result.” “Oxidisable alcohol.” “Chlorine displaces bromide.” “Add acid” without saying which acid and why. “Fizzes, so carbonate” without the limewater step. Equations without state symbols. “No reaction” when a soluble product has formed.

Watch for

The order of the tests and the reason for it (acid before silver nitrate or barium chloride; carbonate removed first).

Which acid: nitric before silver nitrate, hydrochloric or nitric before barium chloride, never sulfuric. Water bath, not a naked flame, for every warming step.

Percentages in a composition question must add to 100%, and every fragment mass must add up.