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.
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