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

The silver halide $\text{AgX}$ forms: ${\text{Ag}}^{+}+{\text{X}}^{-}\to \text{AgX}\downarrow$.

• (a) $\text{AgCl}$ = white
• (b) $\text{AgBr}$ = pale yellow (cream)
• (c) $\text{AgI}$ = yellow

What it looks like: as we move Cl → Br → I the anion gets bigger and softer (more polarisable — see Polarisation and Fajans' Rules), the bond gets more covalent, and the colour deepens from white toward yellow.

Add dilute HCl. You see brisk effervescence of a colourless, odourless gas: ${\text{CO}}_3^{2-}+2{\text{H}}^{+}\to {\text{H}}_2\text{O}+{\text{CO}}_2\uparrow$ Pass the gas into lime water $\left({\text{Ca(OH)}}_2\right)$ and it turns milky: ${\text{CO}}_2+{\text{Ca(OH)}}_2\to {\text{CaCO}}_3\downarrow +{\text{H}}_2\text{O}$

Reagents: freshly prepared ${\text{FeSO}}_4$ solution, then concentrated ${\text{H}}_2{\text{SO}}_4$ poured gently down the side of the tube. A brown ring appears at the acid–solution interface, confirming ==nitrate (${\text{NO}}_3^{-}$)==.

• Cream colour sits between white (Cl) and yellow (I) → suspect ${\text{Br}}^{-}$.
• Solubility in ${\text{NH}}_3$ follows Ksp: $\text{AgCl}$ (Ksp $\approx 1.8\times 10^{-10}$) fully dissolves, $\text{AgBr}$ (Ksp $\approx 5.4\times 10^{-13}$) barely/partially, $\text{AgI}$ (Ksp $\approx 8.5\times 10^{-17}$) not at all. "Partially soluble" fits AgBr exactly.
• Answer: Br⁻. Confirm further with chlorine water + ${\text{CCl}}_4$ → orange organic layer.

Chlorine is a stronger oxidiser than bromine, so it steals electrons from ${\text{Br}}^{-}$, liberating ${\text{Br}}_2$: ${\text{Cl}}_2+2{\text{Br}}^{-}\to 2{\text{Cl}}^{-}+{\text{Br}}_2$ ${\text{Br}}_2$ dissolves in ${\text{CCl}}_4$/${\text{CHCl}}_3$ giving an orange/brown organic layer. (See ease-of-oxidation order in Redox Series and Standard Potentials.)

${\text{Ba}}^{2+}+{\text{SO}}_4^{2-}\to {\text{BaSO}}_4\downarrow (\text{white})$ Extra step: acidify with dilute HCl first (or check the ppt is acid-insoluble). ${\text{BaCO}}_3$ and ${\text{BaSO}}_3$ are also white but dissolve in acid; only ${\text{BaSO}}_4$ survives. Acid is the filter that removes impostors.

Ammonia works by pulling ${\text{Ag}}^{+}$ away into a complex (see Coordination Complexes): $\text{AgX}+2{\text{NH}}_3\rightleftharpoons [{\text{Ag(NH}}_3{)}_2{]}^{+}+{\text{X}}^{-}$ This only proceeds if the salt is willing to release some ${\text{Ag}}^{+}$ in the first place — i.e. if Ksp is large enough. $\text{AgCl}$ has the largest Ksp ($1.8\times 10^{-10}$), so enough ${\text{Ag}}^{+}$ is available for ${\text{NH}}_3$ to grab and shift the equilibrium right. $\text{AgI}$ has a Ksp roughly a million million times smaller ($8.5\times 10^{-17}$); it clings to ${\text{Ag}}^{+}$ so tightly that ${\text{NH}}_3$ cannot win. Numerically the ratio $\frac{K_{sp}(\text{AgCl})}{K_{sp}(\text{AgI})}\approx 2.1\times 10^6$ shows AgCl is over two million times more willing to give up silver.

Both can be reduced to NO, which forms the brown complex $[{\text{Fe(H}}_2{\text{O)}}_5\text{NO}{]}^{2+}$. The difference is how much acid strength each needs:

• ${\text{NO}}_2^{-}$ gives a brown ring even with dilute acid (it's already easy to reduce).
• ${\text{NO}}_3^{-}$ needs the harsh oxidising power of concentrated ${\text{H}}_2{\text{SO}}_4$. Distinguish: first run the test with dilute acid. If a ring forms → nitrite present. Destroy/remove nitrite (e.g. with urea/sulphamic acid), then do the concentrated-${\text{H}}_2{\text{SO}}_4$ ring test — a ring now confirms nitrate.

Electron bookkeeping: N goes from $+5$ (in ${\text{NO}}_3^{-}$) to $+2$ (in NO) → gains 3 electrons. Each Fe goes $+2\to +3$ → loses 1 electron. So 3 Fe²⁺ are needed per nitrate: $a=c=3$. Balance O: left has 3 O (from ${\text{NO}}_3^{-}$), right has 1 O in NO + $d$ O in water → $d=2$. Balance H: right has $2d=4$ H → $b=4$. $3{\text{Fe}}^{2+}+4{\text{H}}^{+}+{\text{NO}}_3^{-}\to 3{\text{Fe}}^{3+}+\text{NO}+2{\text{H}}_2\text{O}$ Check charge: left $=3(+2)+4(+1)+(-1)=+9$; right $=3(+3)=+9$. ✓

A mixture is hard because AgX colours overlap. Use oxidation selectivity instead, exploiting order of ease of oxidation ${\text{I}}^{-}>{\text{Br}}^{-}>{\text{Cl}}^{-}$:

1. Add chlorine water drop by drop and shake with ${\text{CCl}}_4$.
2. ${\text{I}}^{-}$ oxidises first (easiest) → violet layer. Confirm with starch → blue-black.
3. Keep adding ${\text{Cl}}_2$ until violet fades (all ${\text{I}}_2$ over-oxidised to colourless iodate); now ${\text{Br}}^{-}$ oxidises → orange layer.
4. For ${\text{Cl}}^{-}$: ${\text{Cl}}_2$ cannot oxidise it, so use a separate portion — acidify with dil. ${\text{HNO}}_3$, add ${\text{AgNO}}_3$; a white ppt fully soluble in ${\text{NH}}_3$ confirms ${\text{Cl}}^{-}$ (bromide/iodide would give cream/yellow, not fully NH₃-soluble).

Reading the leaves top to bottom:

• fizz + milky lime water → Carbonate (${\text{CO}}_3^{2-}$)
• white ppt, ${\text{NH}}_3$-soluble → Chloride
• cream ppt, partially ${\text{NH}}_3$-soluble → Bromide
• yellow ppt, ${\text{NH}}_3$-insoluble → Iodide
• white ppt insoluble in acid (with ${\text{BaCl}}_2$) → Sulphate
• brown ring (conc. ${\text{H}}_2{\text{SO}}_4$) → Nitrate

The order matters: carbonate is destroyed by the very acid used to prep the halide test, so we test it first.

Precipitation begins when the ion product equals Ksp: $[{\text{Ag}}^{+}][{\text{Cl}}^{-}]=K_{sp}⟹[{\text{Ag}}^{+}]=\frac{K_{sp}}{[{\text{Cl}}^{-}]}=\frac{1.8\times 10^{-10}}{0.010}$ $[{\text{Ag}}^{+}{]}_{\min }=1.8\times 10^{-8}{\text{mol L}}^{-1}$ Interpretation: an astonishingly tiny silver concentration already triggers the white ppt — that's why the halide test is so sensitive.

Minimum $[{\text{Ag}}^{+}]$ for each: $\frac{K_{sp}}{0.010}$. $\frac{[{\text{Ag}}^{+}{]}_{\text{AgCl}}}{[{\text{Ag}}^{+}{]}_{\text{AgI}}}=\frac{K_{sp}(\text{AgCl})}{K_{sp}(\text{AgI})}=\frac{1.8\times 10^{-10}}{8.5\times 10^{-17}}\approx 2.1\times 10^6$ So AgI precipitates at a silver concentration about 2.1 million times smaller. Prediction: as ${\text{Ag}}^{+}$ is added drop by drop to a mixed solution, $\text{AgI}$ (yellow) forms first, then AgBr, then AgCl last — a real "selective/fractional precipitation" effect.

Each Cl in ${\text{Cl}}_2$ goes $0\to -1$, gaining 1 electron → 2 electrons gained per ${\text{Cl}}_2$. Each I in ${\text{I}}^{-}$ goes $-1\to 0$, losing 1 electron → to supply 2 electrons we need 2 ${\text{I}}^{-}$, producing 1 ${\text{I}}_2$. Electrons balance ($2=2$), charge balances (left $-2$, right $-2$), atoms balance. Hence: ${\text{Cl}}_2+2{\text{I}}^{-}\to 2{\text{Cl}}^{-}+{\text{I}}_2$ The liberated ${\text{I}}_2$ gives the violet organic layer and blue-black with starch.

Carbonate first (fizz), then AgNO₃ for halides (White-Cream-Yellow = Cl-Br-I), Barium for sulphate (white, acid-proof), Fe+conc.H₂SO₄ for nitrate (brown ring). Order: fizz → silver → barium → ring.