Concentration of solution \( =0.1\)

Degree of ionisation \( = 2\% = \frac{2}{{100}} = 0.02\)

Ionic product of water \( = 1 \times {10^{ - 14}}\)

Concentration of \([{H^ + }]\)= Concentration of solution \(\times\) degree of ionisation \( = 0.1 \times 0.02 = 2 \times {10^{ - 3}}M\)

Concentration of \([O{H^ - }] = \frac{{{\rm{Ionic product of water}}}}{{[{H^ + }]}}\)

\( = \frac{{1 \times {{10}^{ - 14}}}}{{2 \times {{10}^{ - 3}}}} = 0.5 \times {10^{ - 11}} = 5 \times {10^{ - 12}}M\).