The number of moles of  $Ca{\left(OH\right)}_2$ required to prepare  $250ml$ of solution with  $(P^H=14)$ assuming complete ionization) is

$P^H=14 \left[H^{+}\right]={10}^{-14}$  1 mole of  $Ca{\left(OH\right)}_2\to$ 2 mole of  $O{H}^{-}$ ions

$\therefore Ca{\left(OH\right)}_2$ required  $=0.5mole/litre$  $1000ml\to 0.5mole$   $250ml\to x$

$x=0.5\times \frac{250}{1000} x=\frac{0.5}{4}=0.125mole$