Question

Upon treatment with ammoniacal $H_2 S$, the metal ion that precipitates as a sulphide is

• A. Fe (III)
• B. Al(III)
• C. Mg (II)
• D. Zn(II)

Answer 1

The Correct Option is D

PLAN $K_{sp} (ZnS)$ is very high and $Zn^{2+}$ is precipitated as ZnS by high concentration of $S^{2-}$ formed when $H_2 S$ is passed in ammoniacal solution.
$\hspace20mm H_2 S \rightleftharpoons Zn^+ + S^{2-} \, (I)$
$\hspace10mm H^+ + OH^- \rightleftharpoons H_2 O \, (II)$
Reaction (I) is favoured in forward side if $H^+$ is removed immediately by $OH^- (NH_4 OH)$.
$\hspace10mm Zm^{2+} + S^{2-} \longrightarrow \, \underset{White \, ppt}{ \, ZnS\downarrow}$
$Fe^{3+} \, \, and \, \, AI^{3+}$ are precipitated as hydroxide.