Qualitative Analysis - Result Question 5
5. Upon treatment with ammoniacal $H _2 S$, the metal ion that precipitates as a sulphide is
(2013 Adv.)
(a) $Fe$ (III)
(b) $Al$ (III)
(c) $Mg$ (II)
(d) $Zn$ (II)
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Answer:
Correct Answer: 5. (d)
Solution:
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.
$\begin{aligned} H _2 S & \rightleftharpoons Zn^{+}+S^{2-}(I) \\ H^{+}+OH^{-} & \rightleftharpoons H _2 O(II) \end{aligned}$
Reaction (I) is favoured in forward side if $H^{+}$is removed immediately by $OH^{-}\left(NH _4 OH\right)$.
$Zn^{2+}+S^{2-} \longrightarrow \underset{\text { White ppt }}{ZnS \downarrow}$
$Fe^{3+}$ and $Al^{3+}$ are precipitated as hydroxide.
BETA
