Question

Which mixture is not separated by conc. Aqueous solution of ammonium hydroxide?
(A) $A{l^{ + 3}},S{n^{ + 3}}$
(B) $A{l^{ + 3}},F{e^{ + 3}}$
(C) $A{l^{ + 3}},Z{n^{ + 2}}$
(D) $Z{n^{ + 2}},P{b^{ + 2}}$

Answer 1

Hint: In group\[3\], ferrous oxide is not precipitated because of the very high value of the solubility product. Nitric acid is added to convert ferrous to ferric before adding ammonium chloride group \[3\] cations are precipitated in the form of hydroxides by ammonium hydroxide in presence of ammonium chloride by common ion effect.

Complete Step by Step Solution:
Option A: \[A{l^{ + 3}}\]is precipitated in group \[3\] while $S{n^{ + 3}}$will remain soluble. This is because to cause precipitation of tin, the ionic product has to exceed the solubility product. Hence, the two can be separated as one will precipitate and other will not leading to separation.

Option B: \[A{l^{ + 3}}\] and \[F{e^{ + 3}}\] both belong to group \[3\] and get precipitated because the concentration of hydroxide will be sufficiently enough to exceed the solubility product by common ion effect.
Aluminium hydroxide is white in colour and ferric oxide is red in colour. The two although being different in colour will remain mixed and separation of two by colour might not be possible.

Option C: aluminium will convert to aluminium hydroxide by conc. aqueous solution of ammonium hydroxide and will get precipitated hence will separate from zinc easily as zinc hydroxide will not get precipitated.

Option D: zinc hydroxide and lead hydroxide both have a very high solubility product value. So both are not precipitated hence remain soluble in the solution.
This, correct answer is B.

Note: In group \[3\], ferrous hydroxide is not precipitated because of very high value of the solubility product. For this reason, concentrated nitric acid is added to convert ferrous to ferric and only ferric hydroxide is precipitated. Concentrated nitric acid is added before ammonium chloride and ammonium hydroxide. ${H_2}S$ is boiled before precipitation of group \[3\]. If ${H_2}S$ is not boiled then it would react with concentrated nitric acid that would oxidise it to colloidal sulphur that would interfere in the further analysis.