Answer
$[Ca^{2+}]=0.30\,M$
$[OH^{-}]=0.60\,M$
Work Step by Step
We find:
$Ca(OH)_{2}\rightarrow Ca^{2+}+2OH^{-}$
$[Ca^{2+}]=0.30\,M\,Ca(OH)_{2}\times\frac{1\,M\,Ca^{2+}}{1\,M\,Ca(OH)_{2}}=0.30\,M\,Ca^{2+}$
$[OH^{-}]=0.30\,M\,Ca(OH)_{2}\times\frac{2\,M\,OH^{-}}{1\,M\,Ca(OH)_{2}}=0.60\,M\,OH^{-}$
