Chemistry - Ionic Equilibrium Question with Solution | TestHub

${\mathrm{NH}}_4\mathrm{OH}\rightleftharpoons {\mathrm{NH}}_4^{+}+{\mathrm{OH}}^{-}$ $K_b=\frac{\left[{\mathrm{NH}}_4^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{NH}}_4\mathrm{OH}\right]}$ $1.8\times 10^{-5}=\frac{\left[{\mathrm{NH}}_4^{+}\right][0.01]}{[0.02]}$ $\therefore \left[{\mathrm{NH}}_4^{+}\right]=3.6\times 10^{-5}\mathrm{M}$