Answer
$[OH^-] = 3\times 10^{- 3}M$
$[H_3O^+] = 3.333\times 10^{- 12}M$
pH = 11.48
pOH = 2.52
Work Step by Step
1. Since $Ca(OH)_2$ is a strong base, and it has 2 $OH^-$ in each molecule:
$[OH^-] = 2 * [Ca(OH)_2] = 2 * 1.5 \times 10^{-3} = 3 \times 10^{-3}M$
2. Calculate pOH and pH:
$pOH = -log[OH^-]$
$pOH = -log( 3 \times 10^{- 3})$
$pOH = 2.523$
$pH + pOH = 14$
$pH + 2.523 = 14$
$pH = 11.477$
3. Now, find $[H_3O^+]$
$[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 3 \times 10^{- 3} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 3 \times 10^{- 3}}$
$[H_3O^+] = 3.333 \times 10^{- 12}$
