Question

Determine the degree of dissociation (in %) of $0.05M$ $N{H}_3$ at $298K$ in a solution of $pH=11$.

Answer 1

$N{H}_{4}OH\rightleftharpoons N{H}_4^{\oplus }+\stackrel{⊝}{O}H$
$(1-\alpha )$ $\alpha$ $\alpha$

Given, pH = 11
$\therefore \left[H^{+}\right]={10}^{-11}M$

$\therefore \left[O{H}^{-}\right]={10}^{-3}=C\alpha$

Since, $C=0.05M$

$\therefore \alpha =\frac{{10}^{-3}}{C}=\frac{{10}^{-3}}{0.05}=2\times {10}^{-2}or2\%$