Worked examples — Common anions — Cl⁻, Br⁻, I⁻, SO₄²⁻, NO₃⁻, CO₃²⁻ — confirmatory tests

You have read the parent note the topic note. Now we drill it. This page is a courtroom: an unknown salt is the suspect, and each test is a clue. We will walk through every kind of trap the exam can set — clean single anions, mixtures that lie to you, "nothing happens" negatives, and one numbers question that uses Solubility Product (Ksp).

Before anything, one promise: we will not use a single symbol you have not seen decoded. Let us decode the three that appear most.

The scenario matrix

Every question this topic can throw at you falls into one of these cells. Our examples below are labelled by cell so you can see the whole board is covered.

Nine cells. Eight examples, because Ex 2 and Ex 3 each hit two cells. Nothing is left unshown.

Example 1 — Cell A: the cleanest halide

Steps.

1. Colour is white → shortlist is Cl⁻ (white), not Br⁻ (cream) or I⁻ (yellow). Why this step? Colour is the first, cheapest discriminator — larger, softer anions (Br⁻, I⁻) make more covalent, more coloured salts (this trend is Polarisation and Fajans' Rules).

2. Ammonia fully dissolves the ppt → confirms Cl⁻. Why this step? Ammonia grabs Ag⁺ into a complex $[{\text{Ag(NH}}_3{)}_2{]}^{+}$ (see Coordination Complexes). It only succeeds if the salt lets go of Ag⁺ easily — i.e. the salt with the largest $K_{sp}$, which is AgCl.

$\text{AgCl}+2{\text{NH}}_3\to [{\text{Ag(NH}}_3{)}_2{]}^{+}+{\text{Cl}}^{-}$

Answer: Cl⁻.

Example 2 — Cells A + B: cream that we double-confirm

Steps.

1. Cream + only partial NH₃ solubility → Br⁻. Why? Br⁻ is middle-sized: AgBr $K_{sp}\approx 5.4\times 10^{-13}$ is between AgCl and AgI, so ammonia barely touches it. Colour + partial solubility together pin Br⁻.

2. Add chlorine water. Chlorine is a stronger oxidiser than bromine, so it steals electrons from Br⁻: ${\text{Cl}}_2+2{\text{Br}}^{-}\to 2{\text{Cl}}^{-}+{\text{Br}}_2$ Why this step? Ease of oxidation runs ${\text{I}}^{-}>{\text{Br}}^{-}>{\text{Cl}}^{-}$ (Redox Series and Standard Potentials). Cl₂ can oxidise anything below it — Br⁻ and I⁻ — but not Cl⁻ itself.

3. Shake with ${\text{CCl}}_4$. The freed ${\text{Br}}_2$ prefers the oily layer and colours it orange/brown. Why the organic solvent? ${\text{Br}}_2$ and ${\text{I}}_2$ are non-polar; they migrate into ${\text{CCl}}_4$, concentrating the colour where it is easy to read. See figure.

Answer: Br⁻, confirmed by orange ${\text{CCl}}_4$ layer.

Example 3 — Cells B + F: iodide, and a built-in negative case

Steps.

1. Tube 1 (I⁻): ${\text{Cl}}_2+2{\text{I}}^{-}\to 2{\text{Cl}}^{-}+{\text{I}}_2$. The ${\text{CCl}}_4$ layer turns violet/purple; adding starch gives blue-black. Why? I⁻ is the easiest halide to oxidise (biggest, softest, loosest electron), so chlorine oxidises it readily.

2. Tube 2 (Cl⁻): no colour change — this is the degenerate/negative case (Cell F). Why nothing happens? Cl₂ cannot oxidise Cl⁻; you would be asking chlorine to oxidise itself. There is no reaction, so no ${\text{Cl}}_2$/coloured product forms in the layer.

Answer: Tube 1 → violet layer (I⁻ confirmed). Tube 2 → no change (this very inertness fingerprints Cl⁻).

Example 4 — Cell C: sulphate with a carbonate impostor

Steps.

1. The naive white ppt is ambiguous — it could be ${\text{BaSO}}_4$ or ${\text{BaCO}}_3$. Why? Both barium salts are insoluble and white; colour cannot separate them.

2. First acidify with dilute $\text{HCl}$. Carbonate reacts with acid and leaves as gas: ${\text{CO}}_3^{2-}+2{\text{H}}^{+}\to {\text{H}}_2\text{O}+{\text{CO}}_2\uparrow$ Why this step? Acid is a filter: it destroys the acid-soluble impostor (${\text{CO}}_3^{2-}$) before we ever form a barium ppt. (Acid–Base Reactions of Salts explains why carbonate is basic enough to be eaten by acid.)

3. Now add ${\text{BaCl}}_2$. A white ppt that stays in the acid confirms sulphate: ${\text{Ba}}^{2+}+{\text{SO}}_4^{2-}\to {\text{BaSO}}_4\downarrow$ Why? ${\text{BaSO}}_4$ has a tiny $K_{sp}$ and does not dissolve in dilute acid — it is the only survivor.

Answer: No, the direct test is invalid. Acidify with dil. HCl first; the acid-insoluble white BaSO₄ then confirms ${\text{SO}}_4^{2-}$.

Example 5 — Cell D: the brown ring

Steps.

1. Conc. ${\text{H}}_2{\text{SO}}_4$ supplies ${\text{H}}^{+}$ and oxidising conditions; nitrate is reduced to nitric oxide (NO): $3{\text{Fe}}^{2+}+4{\text{H}}^{+}+{\text{NO}}_3^{-}\to 3{\text{Fe}}^{3+}+\text{NO}+2{\text{H}}_2\text{O}$ Why this step? Nitrate is a mild oxidiser; only in strong acid does it accept electrons from ${\text{Fe}}^{2+}$. This is a redox handle (Redox Series and Standard Potentials).

2. The freed NO binds to remaining ${\text{Fe}}^{2+}$, forming a brown complex: $\text{NO}+[{\text{Fe(H}}_2{\text{O)}}_6{]}^{2+}\to [{\text{Fe(H}}_2{\text{O)}}_5\text{NO}{]}^{2+}+{\text{H}}_2\text{O}$ Why a complex? NO replaces one water ligand on iron (Coordination Complexes); the new complex $[{\text{Fe(H}}_2{\text{O)}}_5\text{NO}{]}^{2+}$ absorbs light to look brown.

3. The ring forms only at the interface where dense acid (bottom) meets the iron/nitrate solution (top). Why a thin ring? Reaction needs both zones' concentrations to overlap; that overlap is a thin boundary layer, so brown appears as a band. See figure.

Answer: ${\text{NO}}_3^{-}$ confirmed by the brown ring of $[{\text{Fe(H}}_2{\text{O)}}_5\text{NO}{]}^{2+}$.

Example 6 — Cells E + I: the fizzy real-world puzzle

Steps.

1. Brisk fizz with acid → carbonate: ${\text{CO}}_3^{2-}+2{\text{H}}^{+}\to {\text{H}}_2\text{O}+{\text{CO}}_2\uparrow$ Why? Only carbonate (and bicarbonate) release ${\text{CO}}_2$ this readily; the gas is colourless and odourless.

2. Milky lime water confirms ${\text{CO}}_2$: ${\text{CO}}_2+{\text{Ca(OH)}}_2\to {\text{CaCO}}_3\downarrow +{\text{H}}_2\text{O}$ Why milky? ${\text{CaCO}}_3$ is an insoluble white solid suspended in the water → looks like milk.

3. Excess ${\text{CO}}_2$ clears the milk by dissolving the solid as soluble bicarbonate: ${\text{CaCO}}_3+{\text{CO}}_2+{\text{H}}_2\text{O}\to {\text{Ca(HCO}}_3{)}_2$ Why does more gas reverse it? ${\text{Ca(HCO}}_3{)}_2$ is water-soluble, so the suspended white redissolves — the same chemistry that makes hard-water caves.

Answer: Carbonate. Excess ${\text{CO}}_2$ forms soluble calcium bicarbonate, clearing the milkiness.

Example 7 — Cells F + G: a mixture that ambushes you

Steps.

1. Without acid, ${\text{AgNO}}_3$ precipitates both silver chloride and silver carbonate: $2{\text{Ag}}^{+}+{\text{CO}}_3^{2-}\to {\text{Ag}}_2{\text{CO}}_3\downarrow$ Why the trap? ${\text{Ag}}_2{\text{CO}}_3$ is pale/whitish — it mimics AgCl, so the white ppt no longer uniquely means chloride. This is the mixture ambush (Cell G).

2. Correct order: first add dilute ${\text{HNO}}_3$ to destroy carbonate: ${\text{CO}}_3^{2-}+2{\text{H}}^{+}\to {\text{H}}_2\text{O}+{\text{CO}}_2\uparrow$ Why ${\text{HNO}}_3$ and never $\text{HCl}$? $\text{HCl}$ would add chloride ions and create false AgCl — the ultimate own-goal. ${\text{HNO}}_3$ removes the impostor without contributing any halide.

3. Now add ${\text{AgNO}}_3$. Any white ppt is genuinely AgCl → chloride confirmed. If the sample had been carbonate only, step 2 removes it and step 3 gives no ppt — the honest negative case (Cell F).

Answer: The unacidified test is fooled by ${\text{Ag}}_2{\text{CO}}_3$. Acidify with dil. ${\text{HNO}}_3$ first (not HCl), then ${\text{AgNO}}_3$.

Example 8 — Cell H: does a precipitate actually form? (numbers)

Steps.

1. Mixing equal volumes halves each concentration: $[{\text{Ag}}^{+}]=[{\text{Cl}}^{-}]=\frac{2.0\times 10^{-5}}{2}=1.0\times 10^{-5}{\text{mol L}}^{-1}.$ Why halve? Each solution is diluted into twice its original volume, so its concentration drops by a factor of 2 (dilution law: concentration $\times$ volume is conserved).

2. Compute the ionic product $Q=[{\text{Ag}}^{+}][{\text{Cl}}^{-}]$: $Q=(1.0\times 10^{-5})(1.0\times 10^{-5})=1.0\times 10^{-10}.$ Why this quantity? $K_{sp}$ is defined for AgCl as $K_{sp}=[{\text{Ag}}^{+}][{\text{Cl}}^{-}]$ at saturation. The same product using actual concentrations is called $Q$. Comparing $Q$ to $K_{sp}$ tells us which side of saturation we are on (Solubility Product (Ksp)).

3. Compare: $Q=1.0\times 10^{-10}$ vs $K_{sp}=1.8\times 10^{-10}$. Since $Q<K_{sp}$, the solution is unsaturated → no precipitate. Why this rule? $Q<K_{sp}$ means we have not yet packed enough ions to force a solid out; the salt can still hold all of it in solution.

Answer: $Q=1.0\times 10^{-10}<1.8\times 10^{-10}=K_{sp}$, so no AgCl precipitates — the AgNO₃ test would (barely) fail at these dilutions.

Active-Recall

Connections

• Parent topic note
• Solubility Product (Ksp) — the Q vs Ksp maths behind Example 8
• Redox Series and Standard Potentials — why I⁻ > Br⁻ > Cl⁻ in oxidation
• Coordination Complexes — $[{\text{Ag(NH}}_3{)}_2{]}^{+}$ and $[{\text{Fe(H}}_2{\text{O)}}_5\text{NO}{]}^{2+}$
• Acid–Base Reactions of Salts — carbonate eaten by acid
• Polarisation and Fajans' Rules — colour/insolubility trend down the halides
• Group Analysis of Cations — the mirror-image precipitation logic