JEE Challenger
More from p-Block Elements
ChemMedAdv 2013 • Paper 2

Identification of Cation in Hot Water Soluble Precipitate P

Comprehension Passage

An aqueous solution containing a mixture of two inorganic salts yields a precipitate (P)(P) and a filtrate (Q)(Q) upon treatment with dilute HClHCl. The precipitate PP is soluble in hot water. The filtrate (Q)(Q) remains unchanged when treated with H2SH_2S in dilute mineral acid medium, but gives a precipitate (R)(R) with H2SH_2S in ammoniacal medium. Precipitate RR gives a coloured solution (S)(S) upon treatment with H2O2H_2O_2 in aqueous NaOHNaOH medium.

Select the ion present in precipitate PP:

Options

A

Pb2+Pb^{2+}

Correct
B

Hg22+Hg_2^{2+}

C

Ag+Ag^+

D

Hg2+Hg^{2+}

Topics & Concepts

p-Block Elementsp-Block ElementsQualitative Analysis

Step-by-Step Solution

To determine the ion present in precipitate PP, let us analyze the qualitative analysis scheme provided in the passage:

  1. Precipitation with dilute HCl\mathrm{HCl} (Group I Analysis): When dilute HCl\mathrm{HCl} is added to an aqueous solution containing cations, Group I cations (Pb2+\mathrm{Pb}^{2+}, Ag+\mathrm{Ag}^+, and Hg22+\mathrm{Hg}_2^{2+}) precipitate as insoluble chlorides: Pb2++2Cl−⟶PbCl2↓(white precipitate)\mathrm{Pb}^{2+} + 2\mathrm{Cl}^- \longrightarrow \mathrm{PbCl}_2 \downarrow \quad (\text{white precipitate}) Ag++Cl−⟶AgCl↓(white precipitate)\mathrm{Ag}^+ + \mathrm{Cl}^- \longrightarrow \mathrm{AgCl} \downarrow \quad (\text{white precipitate}) Hg22++2Cl−⟶Hg2Cl2↓(white precipitate)\mathrm{Hg}_2^{2+} + 2\mathrm{Cl}^- \longrightarrow \mathrm{Hg}_2\mathrm{Cl}_2 \downarrow \quad (\text{white precipitate})

  2. Solubility in Hot Water: Among these insoluble chlorides:

    • AgCl\mathrm{AgCl} and Hg2Cl2\mathrm{Hg}_2\mathrm{Cl}_2 are insoluble in hot water.
    • Lead(II) chloride (PbCl2\mathrm{PbCl}_2) has a relatively high temperature coefficient of solubility; it dissolves readily in hot water and recrystallizes upon cooling.

Since the precipitate PP is specified to be soluble in hot water, the precipitate PP is PbCl2\mathrm{PbCl}_2, and the cation present in PP must be Pb2+\mathrm{Pb}^{2+}.

Therefore, the correct option is (A).

PreviousNext