Question

The presence of $N{H_4}Cl$ in the test solution while precipitating group $III - A$ hydroxides (in qualitative inorganic analysis) helps in
(A) Lowering $[O{H^ - }]$
(B) Lowering $[N{H_4}OH]$
(C) Increasing $[O{H^ - }]$
(D) Increasing $[N{H_4}OH]$

Answer 1

Hint: In qualitative inorganic analysis of cations, cations are grouped on the basis of solubility product and selective precipitation of sparingly soluble salts. Group $III$is precipitated as hydroxides by ammonium hydroxide $(N{H_4}OH)$in presence of ammonium chloride $(N{H_4}Cl)$. The cations precipitated in group $III$are $A{l^{ + 3}},C{r^{ + 3}},M{n^{ + 2}},F{e^{ + 3}}$.

Complete Step by Step Solution:
$(N{H_4}Cl)$is added before $(N{H_4}OH)$while precipitating group $III$analysis.
The reaction representing dissociation of ammonium chloride and ammonium hydroxide are as follow:
$N{H_4}Cl \to N{H_4}^ + + C{l^ - }$

$N{H_4}OH\underset {} \leftrightarrows N{H_4}^ + + O{H^ - }$ (Ammonium hydroxide is a weak base hence this is an equilibrium reaction proceeding in the both the directions)
 It is observed from the above two reactions that the concentration of ammonium ion$N{H_4}^ + $ increases.

Applying Le Chatelier's principle to the equilibrium reaction of dissociation of ammonium hydroxide. The increase in concentration of ammonium ion decreases the concentration of hydroxide ion by suppressing ionisation of $N{H_4}OH$. This is known as the common ion effect.

Due to increase in concentration of ammonium ions, dissociation of $N{H_4}OH$will shift in backward direction decreasing concentration of hydroxide ions sufficiently enough to precipitate group $III$cations.

If $N{H_4}OH$is used alone, then in that case, the concentration of hydroxide ions would be enough to precipitate hydroxides of group $IV,V,VI$cations also. Hence a low concentration of hydroxide is required.
Thus, the correct option is A.

Note: Ammonium sulphate can’t be used in place of ammonium chloride because this would lead to precipitation of group $V$ radicals as their sulphate in group $III$ itself.