Question:
Upon treatment with ammoniacal $H_2 S$, the metal ion that precipitates as a sulphide is
Upon treatment with ammoniacal $H_2 S$, the metal ion that precipitates as a sulphide is
Updated On: Jun 14, 2022
- Fe (III)
- Al(III)
- Mg (II)
- Zn(II)
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The Correct Option is D
Solution and Explanation
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.
$\hspace20mm H_2 S \rightleftharpoons Zn^+ + S^{2-} \, (I)$
$\hspace10mm H^+ + OH^- \rightleftharpoons H_2 O \, (II)$
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.
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