The % degree of dissociation of 0.05M  $N{H}_3$  at  ${25}^{o}C$  in a solution of   $pH=11$ will be

C=0.05M
C=5 $\times {10}^{-2}$ M
If pH =11  $\Rightarrow pOH=3and\left[O{H}^{-}\right]={10}^{-3}$   $\therefore \left[O{H}^{-}\right]=C.\alpha$

$\alpha =\frac{\left[O{H}^{-}\right]}{C}=\frac{{10}^{-3}}{5\times {10}^{-2}}$ 

 $$\begin{gathered} \alpha =0.2\times {10}^{-1} \alpha =0.02 \\ \therefore percentageof\alpha =0.02\times 100=2\% \end{gathered}$$