Question

State True or False.
\[{\rm{N}}{{\rm{H}}_{\rm{4}}}{\rm{Cl}}\] When dissolved in water undergoes anionic hydrolysis
(A) True
(B) False

Answer 1

Hint:
 As we know that the \[{\rm{N}}{{\rm{H}}_{\rm{4}}}{\rm{Cl}}\] is a salt. So, it is commonly observed that different salts, on dissolution in water do not always form neutral solutions such as copper sulphate solution (aqueous solution) form acidic solution whereas sodium acetate solution form basic solution. This is due to the fact that ions can interact with water thereby producing either acidic or an alkaline solution.

Complete step by step solution
Ammonium chloride is a salt and It is prepared from a weak base ( \[{\rm{N}}{{\rm{H}}_{\rm{4}}}{\rm{OH}}\])and a strong acid (\[{\rm{HCl}}\]). When it is dissolved in water it produces aqueous cation and aqueous anion.
Hydrolysis is the interaction of cation or anion of the salt with water molecules hence by bonding with water, salts produce either an acidic or basic solution. The \[{\rm{pH}}\]of the solution gets affected by this interaction.
The anions of strong acid and cations of strong bases do not hydrolyse and thus, the solution formed is neutral. On the other hand, salts of weak acid and strong base are basic in nature and salts of weak base and strong acid are acidic in nature.
Let’s see the interaction of its cation and anion on hydrolysis-
\[{\rm{N}}{{\rm{H}}_{\rm{4}}}{\rm{Cl + }}{{\rm{H}}_{\rm{2}}}{\rm{O}} \to {\rm{NH}}_{\rm{4}}^{\rm{ + }}{\rm{(aq) + C}}{{\rm{l}}^{\rm{ - }}}{\rm{(aq)}}\]
According to polarisation effect, the ammonium cation polarises most effectively to water molecules and hence it undergoes as
\[{\rm{NH}}_{\rm{4}}^{\rm{ + }}{\rm{ + }}{{\rm{H}}_{\rm{2}}}{\rm{O}} \to {\rm{N}}{{\rm{H}}_{\rm{4}}}{\rm{OH + }}{{\rm{H}}^{\rm{ + }}}\]

Here, ammonium hydroxide is a weak base and neutralises with proton ion and the solution becomes acidic.

Therefore, cation of salt is hydrolysed and the given statement is false.

Note:
The calculation of \[{\rm{pH}}\]of hydrolysed weak salt solution can be followed as
\[{\rm{pH = }}\dfrac{{\rm{1}}}{{\rm{2}}}{\rm{p}}{{\rm{K}}_{\rm{W}}}{\rm{ - }}\dfrac{{\rm{1}}}{{\rm{2}}}{\rm{p}}{{\rm{K}}_{\rm{b}}}{\rm{ - }}\dfrac{{\rm{1}}}{{\rm{2}}}{\rm{log \, C}}\]