0.1 mole of ${\mathrm{CH}}_3{\mathrm{NH}}_2 \left({\mathrm{K}}_{\mathrm{b}} = 5 \times {10}^{-4}\right)$ is mixed with 0.08 mole of HCl and diluted to one litre. The $\left[{\mathrm{H}}^{+}\right]$ in solution is

${\mathrm{CH}}_3{\mathrm{NH}}_2 + \mathrm{HCl} \to \mathrm{C}{\mathrm{H}}_3\stackrel{+}{\mathrm{N}}{\mathrm{H}}_3{\mathrm{Cl}}^{-}$

0.1                 0.08                0         

0.02               0                  0.08    

[ 0.08mole HCl neutralizes 0.08 mole methylamine. Concentration of base left is (0.1-0.08) = 0.02]                                                 

This is basic buffer solution.           

$\left[{\mathrm{OH}}^{-}\right] = {\mathrm{K}}_{\mathrm{b}} \times \frac{\left[\mathrm{Base}\right]}{\left[\mathrm{Salt}\right]}$

$=5 \times {10}^{-4} \times \frac{0.02}{0.08} = 1.25 \times {10}^{-4}$

$$\begin{gathered} \therefore \left[{\mathrm{H}}^{+}\right] = \frac{{10}^{-14}}{\left[{\mathrm{OH}}^{-}\right]} = \frac{{10}^{-14}}{1.25 \times {10}^{-4}} \\ = 8 \times {10}^{-11} \mathrm{M} \end{gathered}$$