Question

Solution of 0.1N $N{{H}_{4}}OH$ and 0.1N $N{{H}_{4}}Cl$ has pH 9.25, then find out $p{{K}_{b}}$ of $N{{H}_{4}}OH$.
(A) 9.25
(B) 4.75
(C) 3.75
(D) 8.25

Answer 1

Hint: ${{K}_{a}}$ is the acid dissociation constant. $p{{K}_{a}}$ is simply the -log of this constant. Similarly, ${{K}_{b}}$ is the base dissociation constant, while $p{{K}_{b}}$ is the -log of the constant. The acid and base dissociation constants are usually expressed in terms of moles per liter (mol/L).

Complete step by step solution: pH is a measure of hydrogen ion concentration, [${{H}^{+}}$ ], in an aqueous (water) solution. The pH scale ranges from 0 to 14. A low pH value indicates acidity, a pH of 7 is neutral, and a high pH value indicates alkalinity. The degree of ionization of an acid or base and are true indicators of acid or base strength because adding water to a solution will not change the equilibrium constant.
$N{{H}_{4}}OH$ (a base) and $N{{H}_{4}}Cl$ (a salt) undergoes dissociation as follows:-
\[N{{H}_{4}}OH\rightleftharpoons N{{H}^{4+}}+O{{H}^{-}}\]
\[N{{H}_{4}}Cl\to N{{H}_{4}}^{+}+C{{l}^{-}}\]
Due to the common ion effect, the reaction of dissociation of $N{{H}_{4}}OH~$shifts to the left.
Here, the given information is :-
Concentration of salt $N{{H}_{4}}Cl$ $= 0.1N$
Concentration of the base $N{{H}_{4}}OH~$ $=0.1N$
pH of the solution is $9.25$
Since we know that, $pH$$+$$pOH$$=14$
Therefore, $pOH=$ $14 - pH$ $= 14 - 9.25$ $= 4.75$
We know that,
\[pOH=p{{K}_{b}}+log\dfrac{[salt]}{[base]}\]
As given in the question, here concentration of salt = concentration of base and [$lo{{g}_{10}}\left( 1 \right)=0$ ]
 pOH= $p{{K}_{b}}$ $​=4.75$

Hence the correct option is the B option.

Note: The equilibrium between the weak acid and its conjugate base allows the solution to resist changes to pH when small amounts of strong acid or base are added. The buffer pH can be estimated using the Henderson-Hasselbalch equation, which is,
\[pOH=p{{K}_{b}}+log\dfrac{[salt]}{[base]}\]