Question

\[N{{H}_{4}}OH\] is a weak base but it becomes still weaker in the aqueous solution of:
(A) 0.1 M\[HCl\]
(B) 0.1 M\[N{{H}_{4}}Cl\]
(C) 0.1 M\[{{H}_{2}}S{{O}_{4}}\]
(D) 0.1 M\[C{{H}_{3}}COOH\]

Answer 1

Hint: By using the common ion effect we can answer the question in a better way. “Common-ion effect describes the suppressing effect on degree of ionization ionization of an electrolyte when another electrolyte is added that shares a common ion”.

Complete step by step answer:
* The given chemical is\[N{{H}_{4}}OH\], a weak base.
* We know that whenever a weak base is going to add into aqueous solution it will show the following reaction.
\[N{{H}_{4}}OH\underset{{}}{\overset{{}}{\longleftrightarrow}}N{{H}_{4}}^{+}+O{{H}^{-}}\]
* Now coming to the concept of common ion, by adding a common ion prevents the weak acid or weak base from degree of ionizing as much as it would without the added common ion.
* The common ion effect decreases the ionization of a weak acid by adding more of an ion that is a product of this equilibrium.
* In the given options,
Option A, is 0.1 M \[HCl\], it won’t form any common ion to \[N{{H}_{4}}OH\]after dissociation in aqueous solution.
Option C is 0.1 M \[{{H}_{2}}S{{O}_{4}}\], it won’t form any common ion to \[N{{H}_{4}}OH\]after dissociation in aqueous solution.
Option D is 0.1 M \[C{{H}_{3}}COOH\], it won’t form any common ion to \[N{{H}_{4}}OH\]after dissociation in aqueous solution.
Option B is 0.1 M \[N{{H}_{4}}Cl\], it will form a common ion to \[N{{H}_{4}}OH\]after dissociation in aqueous solution.

\[N{{H}_{4}}Cl\underset{{}}{\overset{{}}{\longleftrightarrow}}N{{H}_{4}}^{+}+C{{l}^{-}}\]

* \[N{{H}_{4}}^{+}\]is the common ion produce by the \[N{{H}_{4}}Cl\]when it is added to aqueous solution
So, the correct option is B.

Note: Don’t be confused about the degree of ionization and common ion effect.
Degree of ionization of \[N{{H}_{4}}OH\] is suppressed by \[N{{H}_{4}}Cl\] due to common ion effects.