\(\therefore\left[ H ^{+}\right]=10^{-12}\,M\)

\(\therefore\left[ OH ^{-}\right]=10^{-2}\,M\)

\(\therefore\left[ Ca ( OH )_2\right]=5 \times 10^{-3}\,M\)

\(5 \times 10^{-3}=\frac{\text { milli moles of } Ca ( OH )_2}{100\,mL }\)

\(\text { milli moles of } Ca ( OH )_2=5 \times 10^{-1}\)

\(\text { Ans. }=5\)