Derive the equation of relation between weak base ionization constant ${K_b}$ and its conjugate acid ionization constant ${K_a}$

$\mathrm{NH}_{3}$ (ammonia) is a weak base and conjugate acid of $\mathrm{NH}_{3}$ is $\mathrm{NH}_{4}^{+}$. Following equilibrium is established in $\mathrm{NH}_{3}$.

$(i)\mathrm{NH}_{3(\mathrm{aq})}+\mathrm{H}_{2} \mathrm{O}_{(l)} \rightleftharpoons \mathrm{NH}_{4(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-}$

$\mathrm{K}_{b}=\frac{\left[\mathrm{NH}_{4}^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_{3}\right]}\left(\right.$ Suppose $\left.\mathrm{K}_{b}=1.8 \times 10^{-5}\right)$

$\mathrm{NH}_{4}^{+}$is a conjugate acid of $\mathrm{NH}_{3}$. Its equilibrium in aqueous solution act as acid is following.

$(ii)$ $\mathrm{NH}_{4 \text { (aq) }}^{+}+\mathrm{H}_{2} \mathrm{O}_{\text {(l) }} \square \mathrm{H}_{3} \mathrm{O}_{\text {(aq) }}^{+}+\mathrm{NH}_{3 \text { (aq) }}$

Take ionization constant of weak acid $\mathrm{NH}_{4}^{+}$is $\mathrm{K}_{a}$,

$\mathrm{K}_{a}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{NH}_{3}\right]}{\left[\mathrm{NH}_{4}^{+}\right]} \quad\left(\right.$ Suppose $\left.\mathrm{K}_{a}=5.6 \times 10^{-10}\right)$

Addition of reaction $(i)$ and $(ii)$ and net reaction is, $(i)$ $+$ $(ii)$ $=2 \mathrm{H}_{2} \mathrm{O}_{(l)} \square \mathrm{H}_{3} \mathrm{O}_{(\text {aq })}^{+}+\mathrm{OH}_{(\text {aq) }}^{-}$

This reaction is self equilibrium of water,

$\mathrm{K}_{w}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]=1.0 \times 10^{-14}$

$\mathrm{~K}_{b}$ reaction of (i) $\times \mathrm{K}_{a}$ reaction of (ii),

$\therefore \mathrm{K}_{a} \times \mathrm{K}_{b}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{NH}_{3}\right]}{\left[\mathrm{NH}_{4}^{+}\right]} \times \frac{\left[\mathrm{NH}_{4}^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_{3}\right]}$

$\therefore \mathrm{K}_{a} \times \mathrm{K}_{b}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]=\mathrm{K}_{w} \quad \ldots(\mathrm{Eq} . \mathrm{i})$

e.g., $\left(5.8 \times 10^{-10}\right)\left(1.8 \times 10^{-5}\right)=1.0 \times 10^{-14}$

This can be extended to make a generalisation. Thus, one reaction equilibrium constant $=\mathrm{K}_{1}$ and second reaction equilibrium constant $=K_{2}$ and reaction $(1)+$ reaction $(2)=$ reaction $(3)$ So, Reaction $(iii)$ $\mathrm{K}_{3}=\mathrm{K}_{1} \times \mathrm{K}_{2} \quad \ldots$.(Eq.-$ii)$