0.1 mol of CH₃NH₂ (K_b = 5 x 10^-4) is mixed with 0.08 mol of HCI and diluted to 1 L. What will be the H⁺ concentration in the solution?

$$\begin{gathered} {\mathrm{CH}}_3{\mathrm{NH}}_2+{\mathrm{H}}^{+}{\mathrm{Cl}}^{-}\rightleftharpoons {\mathrm{CH}}_3{\mathrm{NH}}_3^{+}{\mathrm{Cl}}^{-} \\ \mathrm{Initial} 0.1 0.08 - \\ \mathrm{Final} 0.02 - 0.08 \mathrm{M} \end{gathered}$$
$$\begin{gathered} {\mathrm{CH}}_3{\mathrm{NH}}_2+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{CH}}_3{\mathrm{NH}}_3^{+}+{\mathrm{OH}}^{-} \\ {\mathrm{K}}_{\mathrm{b}}=\frac{\left[{\mathrm{CH}}_3{\mathrm{NH}}_3^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{CH}}_3{\mathrm{NH}}_2\right]} \\ 5\times {10}^{-4}=\frac{(0.08)\left[{\mathrm{OH}}^{-}\right]}{(0.02)} \\ \left[{\mathrm{OH}}^{-}\right]=1.25\times {10}^{-4}\mathrm{M} \\ \left[{\mathrm{H}}^{+}\right]=\frac{{\mathrm{K}}_{\mathrm{w}}}{\left[{\mathrm{OH}}^{-}\right]}=\frac{1\times {10}^{-14}}{1.25\times {10}^{-4}} \\ =8\times {10}^{-11}\mathrm{M} \end{gathered}$$