The pH of 0 . 01   N lime water is

Lime water, $\mathrm{Ca}{\left(\mathrm{OH}\right)}_2$, is strong base solution So, according to law of equivalence.

$\left[{\mathrm{OH}}^{-}\right]=\mathrm{N}={10}^{-2} \mathrm{N}$
$\mathrm{pOH}=-\log \left[{\mathrm{OH}}^{-}\right]=-\log {10}^{-2}=2$
$\mathrm{pH}=14-\mathrm{pOH}=14-2$
$\mathrm{pH}=12$