Question

The percentage degree of dissociation of $0.05M$ ${NH}_3$ at ${25}^{o}C$ in a solution of $pH=11$ is :

Answer 1

$\underset{C_o{C}_o(1-\alpha )}{{NH}_{4}OH}\rightleftharpoons \underset{0C_o\left(\alpha \right)}{{NH}_4^{+}}+\underset{0C_o\left(\alpha \right)}{{OH}^{-}}$
Given $pH=11$;
$\therefore$ $\left[H^{+}\right]={10}^{-11}$
or $\left[{OH}^{-}\right]={10}^{-3}$
or $C_o\alpha ={10}^{-3}$
or $0.05\alpha ={10}^{-3}$ [$C_o=0.05$]
$\therefore$ $\alpha ={10}^{-3}/0.05=2\times {10}^{-2}$ or $2$%