Answer
$[OH^-] = 9.6\times 10^{- 4}M$
$[H_3O^+] = 1.042\times 10^{- 11}M$
pH = 10.98
pOH = 3.02
Work Step by Step
1. Since $Sr(OH)_2$ is a strong base with 2 $OH$ in each molecule:
$[OH^-] = 2 * [Sr(OH)_2] = 2 * 4.8 \times 10^{-4} = 9.6 \times 10^{-4} M $
2. Calculate pOH and pH:
$pOH = -log[OH^-]$
$pOH = -log( 9.6 \times 10^{- 4})$
$pOH = 3.018$
$pH + pOH = 14$
$pH + 3.018 = 14$
$pH = 10.982$
3. Now, find $[H_3O^+]$
$[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 9.6 \times 10^{- 4} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 9.6 \times 10^{- 4}}$
$[H_3O^+] = 1.042 \times 10^{- 11}M$
