Visual walkthrough — Common anions — Cl⁻, Br⁻, I⁻, SO₄²⁻, NO₃⁻, CO₃²⁻ — confirmatory tests

Before anything, the notation we will use constantly:

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Step 1 — The three handles: what can an invisible ion be forced to do?

WHAT. We cannot see ions. So we must make them act. There are only three kinds of action a dissolved ion can be forced into:

1. Stick to a partner and fall out as a solid (solubility handle).
2. Give up or grab an electron and change colour (redox handle).
3. Break apart when attacked by acid and release a gas (acid handle).

WHY these three and not others. These are the only changes that produce something a human eye can register directly: a solid appears, a colour appears, or a gas bubbles out. Every confirmatory test in the parent note is one of these three — nothing more exotic is needed.

PICTURE. Three doors leading out of the "unknown ion" box, each door labelled with the visible clue it produces.

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Step 2 — The solubility handle: turning ${\text{Cl}}^{-}$, ${\text{Br}}^{-}$, ${\text{I}}^{-}$ into coloured solids

WHAT. We add silver ions, written ${\text{Ag}}^{+}$ (a silver atom missing one electron, so $+1$). Silver loves halides (the family ${\text{Cl}}^{-},{\text{Br}}^{-},{\text{I}}^{-}$) and locks onto them as a solid salt $\text{AgX}$, where ${\text{X}}^{-}$ stands for any one halide.

${\text{Ag}}^{+}+{\text{X}}^{-}\to \text{AgX}\downarrow$

• ${\text{Ag}}^{+}$ — the positive silver we added.
• ${\text{X}}^{-}$ — whichever halide was hiding in the unknown.
• $\text{AgX}\downarrow$ — the coloured solid that drops out. The $\downarrow$ is our clue.

WHY silver and not something else. We need a partner whose halide salts are very insoluble so they actually fall out, and whose three salts happen to differ in colour so they tell the three halides apart. Silver does both.

PICTURE. Three test tubes side by side: white, cream, yellow — the colour deepens as the halide gets bigger.

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Step 3 — The confirmation twist: ammonia sorts the three silvers

WHAT. All three gave a pale precipitate — so far ${\text{Br}}^{-}$ and ${\text{I}}^{-}$ could still be confused. We now add ammonia, ${\text{NH}}_3$. Ammonia can peel silver off the solid by wrapping around it as a complex:

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

• $2{\text{NH}}_3$ — two ammonia molecules grab the silver.
• $[{\text{Ag(NH}}_3{)}_2{]}^{+}$ — the soluble silver-ammonia complex (see Coordination Complexes); because it is soluble, the solid dissolves.
• ${\text{Cl}}^{-}$ — the chloride, released back into solution.

WHY it separates them. Ammonia only wins the tug-of-war if the salt was willing to release some ${\text{Ag}}^{+}$ in the first place.

So "willingness to release ${\text{Ag}}^{+}$" is exactly what $K_{sp}$ measures: bigger $K_{sp}$ = easier for ammonia to pull silver out.

Salt	$K_{sp}\approx$	Ammonia verdict
$\text{AgCl}$	$1.8\times 10^{-10}$	fully dissolves
$\text{AgBr}$	$5.4\times 10^{-13}$	barely dissolves
$\text{AgI}$	$8.5\times 10^{-17}$	stays solid

PICTURE. A see-saw: ammonia's pull on one side, the salt's grip ($K_{sp}$) on the other; chloride tips toward "dissolve", iodide toward "hold".

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Step 4 — The redox handle: catching ${\text{Br}}^{-}$ and ${\text{I}}^{-}$ by stealing an electron

WHAT. Instead of precipitating, we now oxidise the halide — meaning we take one electron away, turning the ion ${\text{X}}^{-}$ into the neutral coloured molecule ${\text{X}}_2$. Our electron-thief is chlorine water, ${\text{Cl}}_2$:

${\text{Cl}}_2+2{\text{Br}}^{-}\to 2{\text{Cl}}^{-}+{\text{Br}}_2$ ${\text{Cl}}_2+2{\text{I}}^{-}\to 2{\text{Cl}}^{-}+{\text{I}}_2$

• ${\text{Cl}}_2$ — takes electrons (it is reduced to ${\text{Cl}}^{-}$).
• ${\text{Br}}_2/{\text{I}}_2$ — the freed coloured element; this is the new clue.

WHY chlorine, and why it can't test chloride. Ease of losing an electron runs ${\text{I}}^{-}>{\text{Br}}^{-}>{\text{Cl}}^{-}$ (bigger ion → looser electron). Chlorine sits above bromide and iodide in oxidising power, so it can rob them — but it cannot rob ${\text{Cl}}^{-}$ (nothing left below it to give). That "no reaction with chloride" is itself informative. This ladder is the Redox Series and Standard Potentials.

PICTURE. A layered tube: watery top, dense organic solvent bottom that pulls the freed ${\text{Br}}_2$ (orange) or ${\text{I}}_2$ (violet) into itself.

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Step 5 — The sulphate case: acid as a bouncer

WHAT. For sulphate, ${\text{SO}}_4^{2-}$ (a sulphur-oxygen group carrying $-2$), we add barium ions ${\text{Ba}}^{2+}$:

${\text{Ba}}^{2+}+{\text{SO}}_4^{2-}\to {\text{BaSO}}_4\downarrow (\text{white})$

WHY acidify first, and why HCl is acceptable here (unlike the halide test). The impostors ${\text{CO}}_3^{2-}$ and ${\text{SO}}_3^{2-}$ also make white barium solids. The trick: acid destroys those impostor solids (they react with ${\text{H}}^{+}$ and dissolve away as gas), while ${\text{BaSO}}_4$ shrugs acid off — see Acid–Base Reactions of Salts.

PICTURE. A doorway: acid at the door; ${\text{BaCO}}_3$/${\text{BaSO}}_3$ turned away (fizz off as gas), ${\text{BaSO}}_4$ standing firm inside.

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Step 6 — The nitrate case: the brown ring at a boundary

WHAT. Add fresh iron(II) sulphate (${\text{FeSO}}_4$, giving ${\text{Fe}}^{2+}$), then pour dense concentrated ${\text{H}}_2{\text{SO}}_4$ gently down the side so it sinks without mixing. Two reactions happen at the meeting layer:

$3{\text{Fe}}^{2+}+4{\text{H}}^{+}+{\text{NO}}_3^{-}\to 3{\text{Fe}}^{3+}+\text{NO}+2{\text{H}}_2\text{O}$ $\text{NO}+[{\text{Fe(H}}_2{\text{O)}}_6{]}^{2+}\to [{\text{Fe(H}}_2{\text{O)}}_5\text{NO}{]}^{2+}+{\text{H}}_2\text{O}$

• Nitrate ${\text{NO}}_3^{-}$ is reduced to the gas $\text{NO}$ using the acid's oxidising strength.
• $\text{NO}$ then slots into an iron complex, giving the brown $[{\text{Fe(H}}_2{\text{O)}}_5\text{NO}{]}^{2+}$.

WHY a ring and not a whole colour. Dense acid sinks and stays below; the reaction only occurs in the thin boundary where nitrate solution meets acid. The brown colour therefore appears as a band at the interface — a ring.

PICTURE. Vertical tube: light solution on top, dense acid at bottom, a glowing amber ring exactly at their contact line.

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Step 7 — The carbonate case: fizz, then a reversible cloud

WHAT. Add dilute HCl to carbonate ${\text{CO}}_3^{2-}$; it breaks apart and releases carbon dioxide gas (recall $\uparrow$ = "escapes as bubbles"):

${\text{CO}}_3^{2-}+2{\text{H}}^{+}\to {\text{H}}_2\text{O}+{\text{CO}}_2\uparrow$

Pass that gas through lime water ${\text{Ca(OH)}}_2$:

${\text{CO}}_2+{\text{Ca(OH)}}_2\to {\text{CaCO}}_3\downarrow +{\text{H}}_2\text{O}(\text{milky})$

WHY it re-clears with excess gas (the degenerate/limiting case). Keep bubbling and the white ${\text{CaCO}}_3$ re-dissolves:

${\text{CaCO}}_3+{\text{CO}}_2+{\text{H}}_2\text{O}\to {\text{Ca(HCO}}_3{)}_2(\text{soluble})$

So "milky then clear on excess" is the fingerprint separating ${\text{CO}}_2$ from the smelly ${\text{SO}}_2$ that sulphite would give.

PICTURE. A U-shaped path: fizz from the tube → milky lime water → (keep bubbling) → clear again.

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The one-picture summary

Everything above is one decision tree: which handle → which reagent → which visible clue → which ion.

Recall Feynman retelling of the whole walkthrough

We can't see the ions, so we set traps that make each one show itself in one of three ways. Trap 1 – silver: the halides fall out as coloured dust — chloride white, bromide cream, iodide yellow. Then we offer them ammonia: only chloride is loose enough to dissolve back in, so ammonia sorts the family. Trap 2 – chlorine: we steal an electron; iodine and bromine break free and dye an oily layer violet or orange, but chloride refuses to be robbed, which itself gives it away. Trap 3 – barium plus an acid bouncer: many white solids form, but only true sulphate stays solid when acid tries to wash the fakes away — and here chloride from the acid is harmless because barium chloride never falls out. Trap 4 – iron and heavy acid: nitrate turns into a gas that paints a brown ring exactly where the two liquids meet, so long as we keep the tube cool and let the acid slide gently underneath without stirring — but sneaky nitrite paints a ring too, so we first blow the nitrite off with urea, and any ring that survives is truly nitrate. Trap 5 – acid fizz: carbonate bubbles out carbon dioxide that clouds lime water, and then, if you overdo it, the cloud clears again — proof it was carbon dioxide and not its smelly cousin sulphur dioxide. Five traps, five confessions.

Recall

Why deepen colour Cl→Br→I in silver salts ::: bigger, softer halide → more covalent bond → colour shifts into visible Which handle catches nitrate ::: redox (nitrate reduced to NO, forms brown iron complex) One-line role of acid in the sulphate test ::: bouncer — dissolves BaCO₃/BaSO₃ impostors, BaSO₄ survives Why is dilute HCl safe in the sulphate test but not the halide test ::: BaCl₂ is soluble so Cl⁻ is a harmless spectator; but a silver-halide clue would be faked by stray Cl⁻ Sign that lime-water cloud is really CO₂ ::: excess gas re-dissolves the cloud (forms soluble Ca(HCO₃)₂) How to distinguish nitrate from nitrite in the brown-ring test ::: ring with dilute acid = nitrite; ring only after urea clean-up + conc. H₂SO₄ = nitrate Three operational rules for a clean brown ring ::: keep the tube cool, trickle acid gently down the side without stirring, use fresh FeSO₄ What does the ↑ symbol mean ::: the substance leaves the liquid as a gas (bubbles rising out)

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Connections

• Solubility Product (Ksp) — the $K_{sp}$ see-saw of Step 3
• Polarisation and Fajans' Rules — colour deepening in Step 2
• Redox Series and Standard Potentials — the electron ladder of Step 4
• Coordination Complexes — $[{\text{Ag(NH}}_3{)}_2{]}^{+}$ and $[{\text{Fe(H}}_2{\text{O)}}_5\text{NO}{]}^{2+}$
• Acid–Base Reactions of Salts — the acid bouncer in Steps 5 and 7
• Group Analysis of Cations — same precipitation logic, mirror image