Derive the equation of ionization constant $({K_b})$ of weak base.

The general equation of weak base is $MOH$. The ionization of base $MOH$ can be represented by equation.

$Equilibrium : \mathrm{MOH}_{(\mathrm{aq})}+(\mathrm{aq})+\mathrm{M}_{\mathrm{aq}}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-}$

Concentration ${M}: CM  \ \ 0 \ \  0$ Change in conce. : -$\mathrm{C} \alpha+\mathrm{C} \alpha  +\mathrm{C} \alpha$ Concentration at  equili. in molarity :

${lcc}\mathrm{C}-\mathrm{C} \alpha $

$  (0+\mathrm{C} \alpha) \ \  (0+\mathrm{C} \alpha)  $

$ =\mathrm{C}(1-\alpha) \ \  =\mathrm{C} \alpha $

$  =\mathrm{C} \alpha$

$ \text { where, } \mathrm{C}=\text { Initial concentration of base in molarity. }$

$\alpha=$ The ionization degree of base

$\therefore$ Amount of ionization $\mathrm{MOH}=\mathrm{C} \alpha$

$\therefore$ The concentration decrease in base when equilibrium is reached $=\mathrm{C} \alpha \mathrm{M}$

$\therefore$ At equilibrium concentration of $\mathrm{MOH}=(\mathrm{C}-\mathrm{C} \alpha)=\mathrm{C}(1-\alpha)$

$\quad$ Equilibrium $\left[\mathrm{M}^{+}\right]=\left[\mathrm{OH}^{-}\right]=\mathrm{C} \alpha \mathrm{M}$

On equilibrium reaction in solution of $\mathrm{MOH}$ base.

$\mathrm{K}=\frac{\left[\mathrm{M}^{+}\right]\left[\mathrm{OH}^{-}\right]}{[\mathrm{MOH}]}=\left[\begin{array}{c}\text { Ionization constant } \\ \mathrm{K}_{b} \text { of weak base }\end{array}\right]$

$\therefore \mathrm{K}_{b}=\frac{\left[\mathrm{M}^{+}\right]\left[\mathrm{OH}^{-}\right]}{[\mathrm{MOH}]}=\frac{\left[\mathrm{OH}^{-}\right]^{2}}{[\mathrm{MOH}]} \ldots(\mathrm{Eq} \cdot-\mathrm{i})$

and $\left[\mathrm{OH}^{-}\right]=\sqrt{\mathrm{K}_{b} \cdot[\mathrm{MOH}]} \quad \ldots($ Eq.-$ii$)

So, $K_{b}=\frac{(C \alpha)}{C(1-\alpha)}=\frac{C \alpha^{2}}{1-\alpha} \quad$...(Eq.-$iii$)

This equation-($i$, $ii$) and $(iii)$ are ionization constant equations of ionic equilibrium of weak base.