AQA A-Level Chemistry 7405 · 3.2.3 Halogens

Part 2: Halide tests and practical identification

All 4 parts available · worked answers and exam guidance included. Reviewed 2 October 2026.

Plan reagent choice, observations and confirmatory tests. Explain why the acid matters and distinguish a precipitate from a coloured solution.

A complete halide test

Use a fresh portion of the aqueous unknown. Acidify with dilute nitric acid, then add aqueous silver nitrate. A silver-halide precipitate can form. Record its colour, then use ammonia to distinguish similar-looking precipitates.

Nitric acid removes carbonate interference: otherwise silver carbonate could precipitate. Do not use hydrochloric acid, because it introduces Cl⁻ and can produce a false chloride result. Keep the sulfate test on a separate portion; its hydrochloric acid and barium chloride both introduce chloride.

AgF is soluble, so fluoride does not give the usual silver-halide precipitate. No precipitate alone is not proof of fluoride: it may mean that no detectable halide is present.

Ag⁺(aq) + Cl⁻(aq) → AgCl(s)
Ag⁺(aq) + Br⁻(aq) → AgBr(s)
Ag⁺(aq) + I⁻(aq) → AgI(s)
CO₃²⁻(aq) + 2H⁺(aq) → CO₂(g) + H₂O(l)
Observe a solid, then check its solubility
IonSilver nitrate resultDilute NH₃(aq)Concentrated NH₃(aq)
Cl⁻White AgCl precipitateDissolvesDissolves
Br⁻Cream AgBr precipitateDoes not appreciably dissolveDissolves
I⁻Yellow AgI precipitateDoes not dissolveDoes not dissolve

Why ammonia is a useful second test

White, cream and pale yellow can be difficult to distinguish in a small sample. Ammonia adds an independent observation: whether the solid dissolves. When AgCl dissolves, the silver is present mainly as the soluble [Ag(NH₃)₂]⁺ complex.

“The solution goes colourless” is not enough if the liquid was already colourless: say that the precipitate dissolves to form a colourless solution. If two possible ions are specified, use the smallest set of tests that distinguishes that pair.

AgCl(s) + 2NH₃(aq) ⇌ [Ag(NH₃)₂]⁺(aq) + Cl⁻(aq)

Diagram placeholder

Halide-test decision diagram

Labels to include:

  • Fresh unknown → dilute nitric acid → silver nitrate
  • Three solid colours: white, cream, yellow
  • Dilute ammonia: AgCl dissolves
  • Concentrated ammonia: AgBr dissolves; AgI remains
  • Use text labels as well as colours; do not depict a precipitate as a clear coloured solution

The colour narrows the identity; ammonia confirms it. Nitric acid is added before silver nitrate to remove interfering carbonate.

Required Practical 4: use separate portions

Label fresh portions of the unknown before adding reagents. A reagent used in one test can introduce ions or neutralise a species needed in another. Record reagent, observation and inference separately, with full names or formulae for reagents.

The related Group 2 pages explain hydroxide and sulfate precipitation. The table below joins these tests to the other RP4 ions. In an unknown mixture, results require interpretation; an alkaline indicator result alone does not uniquely identify hydroxide because other ions can also make solutions alkaline.

NH₄⁺(aq) + OH⁻(aq) → NH₃(g) + H₂O(l)
Practical identification overview
TargetReagent / methodPositive observation
HalideDilute HNO₃, then AgNO₃; ammonia confirmationCharacteristic precipitate and ammonia solubility
SO₄²⁻Dilute HCl, then BaCl₂ solutionWhite BaSO₄ precipitate
CO₃²⁻Dilute acid; test the gas with limewaterEffervescence; gas makes limewater cloudy
NH₄⁺Add aqueous NaOH and warm gentlyGas turns damp red litmus blue (NH₃)
OH⁻ in a simple unknownIndicator / pH testAlkaline response; confirm against the possible identities
Group 2 cationsCompare hydroxide/sulfate precipitation under specified conditionsUse solubility pattern, not a single white precipitate as unique proof

Do not confuse three different halogen experiments

With silver nitrate, you are looking for a solid silver halide. With an aqueous halogen displacement, you are observing the colour of a newly formed free halogen. With concentrated sulfuric acid and solid halide, you may see fumes, vapours or solid reduction products.

The 2023 Q06.2 report highlights confusion between the silver nitrate and concentrated sulfuric acid experiments. Start every answer by reading the reagent and physical state of the starting halide.

Use goggles and the prescribed small-scale method. Ammonia vapour is irritating; concentrated ammonia needs suitable ventilation. Collect silver-containing waste as instructed. Do not smell gases directly.

Using precipitation for an amount calculation

If all chloride in a sample is precipitated, one mole Cl⁻ gives one mole AgCl. For example, 0.287 g dry, pure AgCl with Mr = 143.5 corresponds to 0.00200 mol AgCl and therefore 0.00200 mol Cl⁻. In an original 25.0 cm³ aliquot, [Cl⁻] = 0.00200 ÷ 0.0250 = 0.0800 mol dm⁻³.

Excess AgNO₃ helps collect all the halide. Filter, wash and dry before weighing. Do not use the total mass of a mixed AgCl/AgBr/AgI precipitate as if it had one molar mass unless the composition is known or another measurement resolves it.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official AQA mark allocations.

Q1. Why are nitric acid and hydrochloric acid not interchangeable in the silver nitrate test?Show answer

Nitric acid removes carbonate without supplying a halide. HCl supplies Cl⁻, which forms AgCl and may give a false positive.

Q2. A cream precipitate remains in dilute ammonia but dissolves in concentrated ammonia. Identify the ion.Show answer

Br⁻. The precipitate is AgBr. The concentration-dependent ammonia result distinguishes it from AgCl and AgI.

Q3. Write the ionic equation for the iodide test and state the observation.Show answer

Ag⁺(aq) + I⁻(aq) → AgI(s). A yellow precipitate forms; it remains in concentrated ammonia.

Q4. Why should a sample already used for the acidified BaCl₂ sulfate test not be reused for a chloride test?Show answer

The added BaCl₂ and usually HCl introduce chloride ions. Silver nitrate could then produce AgCl from those reagents rather than from the original unknown.

Q5. 0.4305 g dry AgCl is obtained from 20.0 cm³ of a chloride solution. Calculate [Cl⁻], given Mr(AgCl) = 143.5 and complete precipitation.Show answer

n(AgCl) = 0.4305 ÷ 143.5 = 0.003000 mol. n(Cl⁻) is the same. [Cl⁻] = 0.003000 ÷ 0.0200 = 0.150 mol dm⁻³.

Sources

Sources and examiner guidance (reviewed 2 October 2026)

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