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.