Calculate the $pH$ of a $0.10 \,M$ ammonia solution. Calculate the pH after $50.0 \,mL$ of this solution is treated with $25.0 \,mL$ of $0.10 \,M$ $HCl$. The dissociation constant of ammonia, $K_{b}=1.77 \times 10^{-5}$

$NH _{3}+ H _{2} O \rightarrow NH _{4}^{+}+ OH ^{-}$

$K_{ b }=\left[ NH _{4}^{+}\right]\left[ OH ^{-}\right] /\left[ NH _{3}\right]=1.77 \times 10^{-5}$

Before neutralization,

$\left[ NH _{4}^{+}\right]=\left[ OH ^{-}\right]= x$

$\left[ {N{H_3}} \right] = 0.10 - x \simeq 0.10$

$x^{2} / 0.10=1.77 \times 10^{-5}$

Thus, $x=1.33 \times 10^{-3}=\left[ OH ^{-}\right]$

Therefore, $\left[ H ^{+}\right]=K_{ w } /\left[ OH ^{-}\right]=10^{-14} /$ $\left(1.33 \times 10^{-3}\right)=7.51 \times 10^{-12}$

$p H=-\log \left(7.5 \times 10^{-12}\right)=11.12$

On addition of $25 \,mL$ of $0.1 \,M$ $HCl$ solution (i.e., $2.5$ $m\,mol$ of $HCl$ ) to $50$ $mL$ of $0.1 \,M$ ammonia solution (i.e., $5 \,m\,mol$ of $NH _{3}$, $2.5 \,m\,mol$ of ammonia molecules are neutralized. The resulting $75\, mL$ solution contains the remaining unneutralized $2.5 \,m\,mol$ of $NH _{3}$ molecules and $2.5 \,m\,mol$ of $NH _{4}^{+}$

$NH _{3}+ HCl \rightarrow NH _{4}^{+}+ Cl ^{-}$

$2.5$              $2.5$           $0$              $0$

At equilibrium

$0$                  $0$             $2.5$           $2.5$

The resulting $75$ $mL$ of solution contains $2.5 \,m\,mol$ of $NH _{4}^{+}$ ions (i.e., $0.033 \,M$ ) and $2.5 \,m\,mol$ (i.e.. $0.033\, M$ ) of uneutralised $NH _{3}$ molecules. This $NH _{3}$ exists in the following equilibrium: 

$NH _{4} OH \quad \rightleftharpoons \quad NH _{4}^{+}+\quad OH ^{-}$

$0.033\,M-y$                        $y$             $y$

where, $y=\left[ OH ^{\top}\right]=\left[ NH _{4}^{+}\right]$

The final $75$ $mL$ solution after neutralisation already contains $2.5 \,m\, mol \,NH _{4}^{+}$ ions (i.e. $0.033 \,M$ ), thus total concentration of $NH _{4}^{+}$ ions is given as:

$\left[ NH _{4}^{+}\right]=0.033+ y$

As y is small, $\left[ {N{H_4}OH} \right] \simeq 0.033\,M$ and $\left[ {NH_4^ + } \right] \simeq 0.033\,M$

$K_{ b }=\left[ NH _{4}^{+}\right][ OH ] /\left[ NH _{4} OH \right]$

$=y(0.033) /(0.033)=1.77 \times 10^{-5} \,M$

Thus, $y=1.77 \times 10^{-5}=\left[O H^{-}\right]$

$\left[H^{+}\right]=10^{-14} / 1.77 \times 10^{-5}=0.56 \times 10^{-9}$

Hence, $p H=9.24$