IIT-JAM Inorganic Qualitative Analysis — Salt Analysis and Its Chemistry

By Aniket Bhardwaj · 22 September 2026 · IIT-JAM Chemistry

Salt analysis is taught in the laboratory as a sequence of steps to follow, and that is exactly how not to prepare it for IIT-JAM. A written paper cannot ask you to smell a gas; it asks why ammonium chloride is added before ammonia, why hydrogen sulphide separates copper from zinc, and which reagent will fail if an interfering radical is present. Every one of those answers is solubility-product and common-ion equilibrium in disguise. This guide walks through the group scheme with the equilibrium arithmetic actually computed, then lists the confirmatory tests and the equations worth writing out in full.

The one principle behind the whole scheme

A precipitate forms when the ionic product Q exceeds K_sp:
For M_xA_y(s) ⇌ xM^y+ + yA^x−, Q = [M^y+]^x[A^x−]^y

Q > K_sp → precipitation · Q = K_sp → saturated · Q < K_sp → stays in solution

The whole group scheme is one idea used five times over: the analyst controls the concentration of the precipitating anion — Cl⁻, S²⁻, OH⁻ or CO₃²⁻ — so that Q crosses K_sp for one set of cations and not for the next. The control is exerted through pH and the common-ion effect. Nothing else is going on.

The group scheme in one table

Group	Reagent	Cations	Precipitate and colour
0	NaOH, warm	NH₄⁺	NH₃ gas, no precipitate
I	dilute HCl	Pb²⁺, Ag⁺, Hg₂²⁺	PbCl₂, AgCl, Hg₂Cl₂ — all white
II A	H₂S in dilute HCl	Cu²⁺, Cd²⁺, Bi³⁺, Pb²⁺, Hg²⁺	CuS black, CdS yellow, Bi₂S₃ brown, PbS black, HgS black
II B	same, then yellow (NH₄)₂S_x	As³⁺, Sb³⁺, Sn^2+/4+	As₂S₃ yellow, Sb₂S₃ orange, SnS₂ yellow — these redissolve as thio-anions
III	NH₄Cl then NH₄OH	Fe³⁺, Al³⁺, Cr³⁺	Fe(OH)₃ reddish-brown, Al(OH)₃ white gelatinous, Cr(OH)₃ green
IV	H₂S in ammoniacal solution	Zn²⁺, Mn²⁺, Ni²⁺, Co²⁺	ZnS white, MnS buff, NiS black, CoS black
V	(NH₄)₂CO₃ in ammoniacal solution	Ba²⁺, Sr²⁺, Ca²⁺	BaCO₃, SrCO₃, CaCO₃ — all white
VI	none (soluble group)	Mg²⁺, Na⁺, K⁺	identified by flame test and specific reagents

Group II B is separated from II A by the amphoteric, thiophilic character of arsenic, antimony and tin: their sulphides dissolve in yellow ammonium sulphide as thio-anions such as AsS₄³⁻, while the II A sulphides do not.

Worked example 1 — why ammonium chloride goes in before ammonia

Show numerically that adding NH₄Cl before NH₄OH stops magnesium from precipitating in group III, while still precipitating aluminium. Take K_b(NH₃) = 1.8 × 10⁻⁵, both NH₃ and NH₄⁺ at 0.10 M, and each metal ion at 0.010 M.

Step 1 — hydroxide concentration from 0.10 M ammonia alone.
[OH⁻] = √(K_b × C) = √(1.8 × 10⁻⁵ × 0.10) = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ M
(check: (1.34 × 10⁻³)² = 1.80 × 10⁻⁶)

Step 2 — hydroxide concentration in the NH₃/NH₄⁺ buffer.
[OH⁻] = K_b × [NH₃]/[NH₄⁺] = 1.8 × 10⁻⁵ × (0.10 / 0.10) = 1.8 × 10⁻⁵ M
That is a fall by a factor of 1.34 × 10⁻³ ÷ 1.8 × 10⁻⁵ ≈ 75.

Step 3 — test magnesium. Using K_sp[Mg(OH)₂] ≈ 1.8 × 10⁻¹¹:
Without the buffer: Q = 0.010 × (1.34 × 10⁻³)² = 0.010 × 1.80 × 10⁻⁶ = 1.8 × 10⁻⁸, which is about a thousand times larger than K_sp → Mg(OH)₂ would precipitate, wrongly putting magnesium in group III.
With the buffer: Q = 0.010 × (1.8 × 10⁻⁵)² = 0.010 × 3.24 × 10⁻¹⁰ = 3.2 × 10⁻¹², which is below K_sp → magnesium stays in solution and reaches group VI where it belongs.

Step 4 — check aluminium still comes down. K_sp[Al(OH)₃] is of the order of 10⁻³³. In the buffer, Q = 0.010 × (1.8 × 10⁻⁵)³ = 0.010 × 5.83 × 10⁻¹⁵ = 5.8 × 10⁻¹⁷, which is roughly sixteen orders of magnitude above K_sp. Aluminium precipitates comfortably.

That is the whole answer to "why NH₄Cl first". The common ion NH₄⁺ pushes the ammonia equilibrium back and drops [OH⁻] into the narrow window that is high enough for the group III trivalent hydroxides and too low for the group VI divalent ones. Note that this margin is comfortable, so the conclusion survives the real spread in published K_sp values.

Worked example 2 — why H₂S separates group II from group IV

Compute the sulphide-ion concentration in 0.30 M HCl and at pH 9, for a solution saturated with H₂S at about 0.10 M.

Combining the two dissociation steps of H₂S gives

[S²⁻] = K_a1K_a2 × [H₂S] ÷ [H⁺]²

Take the commonly quoted representative values K_a1 ≈ 1.0 × 10⁻⁷ and K_a2 ≈ 1.0 × 10⁻¹⁴, so K_a1K_a2 ≈ 1.0 × 10⁻²¹. (Books differ noticeably on K_a2; use the value your own textbook gives and state it — the ratio below is what matters and it is unaffected.)

In 0.30 M HCl: [H⁺] = 0.30, so [H⁺]² = 0.090.
[S²⁻] = (1.0 × 10⁻²¹ × 0.10) ÷ 0.090 = 1.0 × 10⁻²² ÷ 0.090 = 1.1 × 10⁻²¹ M

At pH 9 (ammoniacal): [H⁺] = 1.0 × 10⁻⁹, so [H⁺]² = 1.0 × 10⁻¹⁸.
[S²⁻] = 1.0 × 10⁻²² ÷ 1.0 × 10⁻¹⁸ = 1.0 × 10⁻⁴ M

Ratio: 1.0 × 10⁻⁴ ÷ 1.1 × 10⁻²¹ ≈ 9 × 10¹⁶ — the sulphide concentration rises by roughly seventeen orders of magnitude simply by changing the acidity.

What that buys: with 0.010 M metal ion, Q = 0.010 × 1.1 × 10⁻²¹ = 1.1 × 10⁻²³ in acid. CuS, whose K_sp is quoted around 10⁻³⁶, is far below that and comes down at once. ZnS, whose quoted K_sp scatters between about 10⁻²¹ and 10⁻²⁴ depending on the source and on which crystalline form is meant, sits close to the line and does not precipitate reliably in acid. Move to pH 9 and Q = 0.010 × 1.0 × 10⁻⁴ = 1.0 × 10⁻⁶, which is enormously above any quoted ZnS value, so zinc precipitates without question. That is group II and group IV, separated by nothing but pH.

Anion (acid radical) tests worth knowing by equation

These four turn up as "write the reaction" questions rather than as lab procedure.

Brown ring (NO₃⁻): nitrate is reduced by Fe²⁺ in concentrated H₂SO₄, and the NO produced is captured as the brown complex
[Fe(H₂O)₆]²⁺ + NO → [Fe(H₂O)₅(NO)]²⁺ + H₂O

Chromyl chloride (Cl⁻), red vapours:
4NaCl + K₂Cr₂O₇ + 6H₂SO₄ → 2CrO₂Cl₂ + 2KHSO₄ + 4NaHSO₄ + 3H₂O

Nessler's reagent (NH₄⁺): K₂[HgI₄] in KOH gives a brown precipitate

Carbonate: CO₃²⁻ + 2H⁺ → H₂O + CO₂, and CO₂ + Ca(OH)₂ → CaCO₃↓ + H₂O (limewater turns milky)

The chromyl chloride equation is the one most often written unbalanced. Count sodium, potassium, chlorine, chromium, sulphur, hydrogen and oxygen separately: 4 Na, 2 K, 4 Cl, 2 Cr, 6 S, 12 H and 31 O on each side.

Confirmatory tests for the common cations

Cation	Reagent	Observation
Pb²⁺	KI, or K₂CrO₄	Yellow PbI₂ (golden spangles on cooling); yellow PbCrO₄
Ag⁺	excess NH₃	AgCl dissolves as [Ag(NH₃)₂]⁺, reappears on acidifying
Cu²⁺	excess NH₃; or K₄[Fe(CN)₆]	Deep blue [Cu(NH₃)₄]²⁺; chocolate-brown Cu₂[Fe(CN)₆]
Fe³⁺	K₄[Fe(CN)₆]; or KSCN	Prussian blue; blood-red [Fe(SCN)]²⁺
Al³⁺	NaOH then blue litmus solution	Hydroxide dissolves in excess NaOH (amphoteric); blue lake
Cr³⁺	NaOH + H₂O₂	Yellow chromate, then orange dichromate on acidifying
Zn²⁺	NaOH	White hydroxide dissolving in excess as zincate (amphoteric)
Mn²⁺	NaOH in air	White hydroxide darkening to brown on oxidation
Ni²⁺	dimethylglyoxime in ammonia	Rose-red chelate precipitate
Co²⁺	NH₄SCN in amyl alcohol	Blue [Co(SCN)₄]²⁻ in the organic layer
Ba²⁺, Sr²⁺, Ca²⁺	flame test	Apple-green, crimson and brick-red respectively
Mg²⁺	Na₂HPO₄ in ammonia	White crystalline Mg(NH₄)PO₄

Flame colours are a genuine spectroscopic observation, not a trick: the heat excites a valence electron, and the light emitted as it falls back has a wavelength set by the energy gap, which is characteristic of the element.

Interfering radicals — the standard exam question

Certain anions bind cations so tightly that the group reagent never gets a chance, or they carry a cation forward into the wrong group. The textbook set is oxalate, borate, fluoride, phosphate and tartrate. Phosphate is the classic: it precipitates the group V and VI cations as phosphates along with the group III hydroxides, so a correct scheme removes phosphate first (as ferric phosphate in the presence of excess Fe³⁺, in acetic acid) before proceeding. If a question tells you the salt contains phosphate, the expected answer names the interference and the removal step — not a colour.

Common mistakes that cost marks

Explaining group order by "solubility" rather than by K_sp and Q. The reagent concentration is what is being manipulated; say so explicitly.
Adding NH₄OH before NH₄Cl. The order is the whole point, and worked example 1 is the reason.
Saying H₂S in acid "does not ionise". It does; its ionisation is simply suppressed, and the suppression is quantitative — a factor of about 10¹⁷ in [S²⁻].
Confusing PbCl₂ with AgCl. Both are white, but PbCl₂ dissolves in hot water and AgCl dissolves in ammonia. Lead therefore appears in both group I and group II, which is expected, not an error.
Writing chromyl chloride from memory without balancing. Check every element count.
Quoting a K_sp as if it were exact. Published sulphide values scatter by orders of magnitude; state the source value you used and reason from the ratio.
Forgetting that the brown ring must be done with cold, concentrated H₂SO₄ layered underneath. Mixing destroys the ring.

How to prepare this unit

Theme	What you must be able to do without hesitation
Q versus K_sp	Decide precipitation for any cation at a stated ion concentration
Common-ion effect	Compute [OH⁻] with and without a buffer and compare the two
Sulphide equilibrium	Use [S²⁻] = K_a1K_a2[H₂S]/[H⁺]² at any pH
Group scheme	Name reagent, cations, precipitate and colour for all six groups
Amphoterism	Explain Al, Zn, Sn and Cr hydroxides dissolving in excess alkali
Confirmatory reactions	Write balanced equations, not just colours
Interfering radicals	Name them and state the removal step for phosphate

Treat that as a revision checklist, not as a prediction of the paper. For the syllabus and the current pattern, read the official IIT-JAM notification for your year.

Practise the precipitation decision itself. Every step of the group scheme is a K_sp comparison, and the fastest way to build a feel for it is to run the numbers on real salts — solubility to K_sp and back, and the effect of a common ion. The solubility product calculator does exactly that, so you can check the working in examples 1 and 2 and then try your own.

Open the Solubility Product (K_sp) Calculator →

Preparing for IIT-JAM, GATE, CSIR-NET or CUET-PG? ABC Chemistry runs dedicated competitive-exam batches — classroom at the Gurugram coaching centre and live online classes for students anywhere in India. Details at abcchemistry.in.