Question

On additon of KI to the solution gives a yellow precipitate which soluble in boiling water and reappears on cooling as shining crystals. The cation present in the solution is

• A. $B{i}^{3+}$
• B. $A{l}^{3+}$
• C. $C{u}^{2+}$
• D. $P{b}^{2+}$

Answer 1

The correct option is D $P{b}^{2+}$

On adding of potassium iodide (KI) to the $P{b}^{2+}$ ion solution, a yellow precipitate of lead iodide is obtained. This confirms the presence of $P{b}^{2+}$ ion.

$PbC{l}_2+2KI\to \underset{\text{Yellow precipitae}}{Pb{I}_2}\downarrow +2KCl$​

The precipitate soluble in boiling water and reappears on cooling as shining crystals​.

Hence, option (A) is correct.