Derive ${K_w} = {K_a} \times {K_b}$ and ${K_w} = p{K_a} \times p{K_b}$ for weak base $B$ and its conjugate acid ${B{H^ + }}$.
Derive ${K_w} = {K_a} \times {K_b}$ and ${K_w} = p{K_a} \times p{K_b}$ for weak base $B$ and its conjugate acid ${B{H^ + }}$.
If weak base $\mathrm{B}$, so equilibrium in its solution is,
$\mathrm{B}_{(\mathrm{aq})}+\mathrm{H}_{2} \mathrm{O}_{(l)}+\mathrm{BH}_{(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-} \quad \ldots \text { (i) }$
In above ionic equilibrium of base, the constant is $\mathrm{K}_{b}$ and $\mathrm{H}_{2} \mathrm{O}_{(t)}$ is taken as constant...
$\mathrm{K}_{b}=\frac{\left[\mathrm{BH}^{+}\right]\left[\mathrm{OH}^{-}\right]}{[\mathrm{B}]}$
If this expression is multiplied \& divided by $\left[\mathrm{H}^{+}\right]$,
$\mathrm{K}_{b}=\frac{\left[\mathrm{BH}^{+}\right]\left[\mathrm{OH}^{-}\right]\left[\mathrm{H}^{+}\right]}{[\mathrm{B}] \quad\left[\mathrm{H}^{+}\right]}=\frac{\left[\mathrm{OH}^{-}\right]\left[\mathrm{H}^{+}\right]\left[\mathrm{BH}^{+}\right]}{[\mathrm{B}]\left[\mathrm{H}^{+}\right]}$
In it $\left[\mathrm{OH}^{-}\right]\left[\mathrm{H}^{+}\right]=\mathrm{K}_{w}$ and $\frac{\left[\mathrm{BH}^{+}\right]}{[\mathrm{B}]\left[\mathrm{H}^{+}\right]}=\frac{1}{\mathrm{~K}_{a}}$
Because, $\mathrm{BH}^{+}$(acid) $+\mathrm{B}_{\text {(aq) }}+\mathrm{H}_{\text {(aq) }}^{+}$
So, $\mathrm{K}_{b}=\frac{\mathrm{K}_{w}}{\mathrm{~K}_{a}}$ and $\mathrm{K}_{w}=\left(\mathrm{K}_{a}\right)\left(\mathrm{K}_{b}\right)$
According to above,
$\mathrm{K}_{b} \times \mathrm{K}_{a}=\mathrm{K}_{w}=1 \times 10^{-14}$
taking log both the side
$\therefore\left(-\log \mathrm{K}_{b}\right)+\left(-\log \mathrm{K}_{a}\right)=-\log \mathrm{K}_{w}=\log \left(1 \times 10^{-14}\right)$
$\therefore \mathrm{pK}_{b}+\mathrm{pK}_{a}=\mathrm{pK}_{w}=+14 \quad \ldots \text { (Eq.-iii) }$
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