Cation Group I — Pb²⁺ (Dilute HCl Group)

Where Group I sits in the scheme

Qualitative analysis of cations is built on a single idea: do not test every ion at once, but split the cations into groups using group reagents that precipitate one set while leaving the rest in solution. The NCERT Lab Manual (Unit 7, Systematic Qualitative Analysis) arranges the syllabus cations into six analytical groups plus the zero group. Group I is the very first separation you carry out on the original solution, immediately after the zero-group test for the ammonium ion.

In the classical analytical scheme, Group I — the dilute-HCl or "insoluble chloride" group — contains three cations: silver $\ce{Ag+}$, mercurous $\ce{Hg2^2+}$, and lead $\ce{Pb^2+}$. All three form chlorides that are insoluble enough to drop out when dilute hydrochloric acid is added. The NCERT Class XII syllabus, however, restricts the cations under study, and the only Group I ion you must analyse is lead, $\ce{Pb^2+}$. So for NEET you should know the classical trio by name, but every reaction, separation and confirmatory test you actually perform is for lead.

Notice that lead appears in both Group I and Group II. This is not a printing error — it is a direct consequence of the solubility of lead chloride, and we return to it below because it is a favourite NEET trap.

The group reagent: dilute HCl

To test for Group I, a small portion of the original solution is taken and dilute hydrochloric acid is added in the cold. If lead is present, a white precipitate of lead chloride forms at once:

$\ce{Pb^2+ + 2Cl- -> PbCl2 v}$   (white precipitate)

If the original solution was prepared in hot concentrated HCl, the NCERT procedure has a neat variant: add cold water and cool the tube under the tap. The chloride ion is already supplied by the HCl, and cooling drops the white $\ce{PbCl2}$ out of solution. Either way, a white precipitate at this stage is the signal for Group I.

The first SVG below shows where this single step sits in the larger separation flow chart.

Why dilute HCl works: the Ksp logic

The NCERT theory section names the two principles that drive every group separation: the solubility product ($K_{sp}$) and the common ion effect. A sparingly soluble salt $\ce{PbCl2}$ sits in equilibrium with its ions:

$\ce{PbCl2 <=> Pb^2+ + 2Cl-}$,   $K_{sp} = [\ce{Pb^2+}][\ce{Cl-}]^2$

Precipitation occurs only when the ionic product exceeds $K_{sp}$. By adding dilute HCl we flood the solution with chloride ions; the term $[\ce{Cl-}]^2$ climbs steeply, the ionic product overshoots $K_{sp}$, and lead chloride is forced out as a solid. This is the common ion effect doing the precipitating.

The reason dilute (not concentrated) HCl is specified is just as important. The chlorides of Group II–VI cations are far more soluble, so a moderate chloride concentration is not enough to exceed their solubility products — they stay dissolved. Only the genuinely insoluble Group I chloride drops out. This selectivity is the whole point of a group reagent.

It is worth being precise about what "exceeds the solubility product" means in practice. Before any acid is added, the original solution holds dissolved lead with only the trace of chloride supplied by the salt itself, so the ionic product $[\ce{Pb^2+}][\ce{Cl-}]^2$ sits below $K_{sp}$ and everything stays in solution. The instant dilute HCl floods the system with chloride, the squared chloride term dominates the product; it shoots past $K_{sp}$ and the system relieves the supersaturation by depositing solid lead chloride. The common ion effect — adding an ion that the salt already contains — is therefore not a separate principle from the solubility product but the lever used to push the ionic product above it.

Partial solubility and the Group II overlap

Lead chloride is described as sparingly soluble, not completely insoluble. Even after dilute HCl is added, a measurable fraction of lead remains in solution as free $\ce{Pb^2+}$. That residual lead is carried forward, and when $\ce{H2S}$ is later passed through the filtrate in Group II it precipitates as black lead sulphide:

$\ce{Pb^2+ + H2S -> PbS v + 2H+}$   (black precipitate)

This is exactly why the NCERT flow chart lists $\ce{Pb^2+}$ under both Group I and Group II. A salt of lead will give a white chloride in Group I and, because precipitation there is incomplete, also give a black sulphide in Group II. Lead is the one cation that legitimately appears twice in the scheme.

For the exam this has a practical reading. If a question describes a salt that gives a white precipitate with dilute HCl and a black precipitate with H₂S, you should not conclude that two different cations are present — a single lead salt accounts for both observations. Equally, the absence of any precipitate with dilute HCl does not rule out lead entirely; a dilute lead solution may carry too little chloride to overshoot the solubility product in the cold, and the lead will only declare itself as black PbS once H₂S is passed. Reading the scheme this way prevents the most common over-counting error in salt analysis.

Separating PbCl₂ with hot water

Having obtained a white precipitate, the analyst must confirm it really is lead chloride and not, say, a carbonate or sulphate of some other ion. The key physical property is that lead chloride is insoluble in cold water but readily soluble in hot water. The precipitate is therefore treated with hot water; if it dissolves, that alone points strongly to Group I.

The hot lead chloride solution is then divided into three parts for the confirmatory tests. The dependence on temperature is also the basis of a striking observation: when the yellow lead iodide formed in the first test (below) is boiled, it dissolves, and on cooling it re-deposits as glittering crystals — the famous "golden spangles." The same hot-soluble / cold-insoluble behaviour reappears throughout the lead chemistry.

Three confirmatory tests for Pb²⁺

From the hot lead chloride solution, three independent confirmations are drawn. Each gives a characteristic precipitate with its own solubility behaviour, and together they leave no doubt that the cation is lead.

1. Potassium iodide test — yellow PbI₂ ("golden spangles")

Adding potassium iodide solution to the first part gives a yellow precipitate of lead iodide:

$\ce{PbCl2 + 2KI -> PbI2 v + 2KCl}$   (yellow precipitate)

This precipitate is soluble in boiling water and re-appears on cooling as shining, needle-like crystals — the classic golden-spangle effect that makes the iodide test memorable and unambiguous.

2. Potassium chromate test — yellow PbCrO₄

The second part is treated with potassium chromate solution, giving a yellow precipitate of lead chromate:

$\ce{PbCl2 + K2CrO4 -> PbCrO4 v + 2KCl}$   (yellow precipitate)

Lead chromate is soluble in hot NaOH (forming the soluble tetrahydroxoplumbate(II) ion) but insoluble in ammonium acetate. This solubility pattern distinguishes it cleanly:

$\ce{PbCrO4 + 4NaOH -> Na2[Pb(OH)4] + Na2CrO4}$

3. Dilute sulphuric acid test — white PbSO₄

To the third part, a few drops of alcohol followed by dilute sulphuric acid give a white precipitate of lead sulphate:

$\ce{PbCl2 + H2SO4 -> PbSO4 v + 2HCl}$   (white precipitate)

Lead sulphate is soluble in ammonium acetate through the formation of the ammonium tetraacetoplumbate(II) complex, a reaction promoted by adding a little acetic acid:

$\ce{PbSO4 + 4CH3COONH4 -> (NH4)2[Pb(CH3COO)4] + (NH4)2SO4}$

Worked confirmation sequence

Observation → inference table