Question

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

Answer 1

$pH=11$

$\therefore pOH=14-pH=14-11=3$

$\therefore \left[{OH}^{-}\right]={10}^{-pOH}={10}^{-3}$

$\therefore$ Degree of dissociation $(\propto )=\frac{\left[{OH}^{-}\right]}{\left[{NH}_3\right]}\times 100$%

$=\frac{{10}^{-3}}{0.05}\times 100$%$=2$%