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:

  1. 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.



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