Derive ${K_a} \times {K_b} = {K_w}$ equation.

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

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

$\mathrm{K}_{b}=\frac{\left[\mathrm{NH}_{4}^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_{3}\right]} \quad\left(\right.$ 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.

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

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

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

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),

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

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

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 $=\mathrm{K}_{2}$ and reaction $(1)$ $+$ reaction $(2)$ $=$ reaction $(3)$

So, Reaction ($iii$) $\mathrm{K}_{3}=\mathrm{K}_{1} \times \mathrm{K}_{2} \quad \ldots .$ (Eq.-$ii$)