Calculate pH of 0.002 N   ${\mathrm{NH}}_4\mathrm{OH}$ having  2%  dissociation (log 4 = 0.6)

${\mathrm{NH}}_4\mathrm{OH}$ is a weak base and partially dissociated

                                  ${\mathrm{NH}}_4\mathrm{OH} \rightleftharpoons {\mathrm{NH}}_4^{+} + {\mathrm{OH}}^{-}$

Concentration              1                    0            0

before dissociation

Concentration            $1-\alpha$                $\alpha$          $\alpha$

after dissociation

$\mathrm{C}=2\times {10}^{-3} ; \mathrm{\alpha }=\frac{2}{100}$  $\therefore \left[O{H}^{-}\right] = C\alpha = 2 \times {10}^{-3} \times \frac{2}{100} = 4 \times {10}^{-5} M$

$\mathrm{pOH} = -\log \left[{\mathrm{OH}}^{-}\right]$

        $$\begin{gathered} = -\log 4 \times {10}^{-5} = 4.4 \\ \mathrm{pH}=14-4.4 = 9.6 \end{gathered}$$