Question

The metal ion (M²⁺) in the following reaction is:

$M^{2+} + S^{2-} \rightarrow$ black precipitate $\xrightarrow[\Delta]{\text{dil.HNO}_3}$ white precipitate

Answer 1

Pb²⁺ (Lead ion)

Answer 2

When an aqueous solution of a metal ion (M²⁺) is treated with sulfide ions, a metal sulfide precipitate is formed. Lead(II) sulfide (PbS) is black in color. A classical confirmatory test for Pb²⁺ involves forming a black precipitate (PbS) which, upon treatment with dilute nitric acid, converts to a white precipitate (due to formation of a lead salt).

Reaction:

$$\text{Pb}^{2+} + \text{S}^{2-} \rightarrow \text{PbS (black precipitate)}$$

On heating with dilute $\text{HNO}_3$, PbS is converted to a white lead salt, confirming the presence of Pb²⁺.