$[O{H^ - }] = C\,.\,\alpha $ ; $C = \frac{1}{{10}}M$, $\alpha = 0.2$
$[O{H^ - }] = \frac{1}{{10}} \times 0.2 = 2 \times {10^{ - 2}}\,M$
$pOH = - \log \,\,[O{H^ - }]$$ = \log \,\,[2 \times {10^{ - 2}}]$; $pOH = 1.7$
$pH = 14 - pOH$ $ = 14 - 1.7 = 12.30$.