Answer
$[OH^-] = 1\times 10^{- 3}M$
$[H_3O^+] = 1 \times 10^{- 11}M$
pH = 11
pOH = 3
Work Step by Step
1. Since $Ca(OH)_2$ is a strong base with 2 $OH^-$ in each molecule:
$[OH^-] = 2 * [Ca(OH)_2]$
$[OH^-] = 2 * 5 \times 10^{-4} = 10^{-3}$
2. Calculate pOH and pH:
$pOH = -log[OH^-]$
$pOH = -log( 1 \times 10^{- 3})$
$pOH = 3$
$pH + pOH = 14$
$pH + 3 = 14$
$pH = 11$
3. Now, find $[H_3O^+]$
$[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 1 \times 10^{- 3} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 1 \times 10^{- 3}}$
$[H_3O^+] = 10^{- 11}$
