Answer
$[OH^-] = 3 \times 10^{-3}M$
$pH \approx 11.477$
Work Step by Step
1. Because $Sr(OH)_2$ is a strong base, and 1 $Sr(OH)_2$ have 2 $OH$ :
$[OH^-] = [Sr(OH)_2] \times 2 $
$[OH^-] = 1.5 \times 10^{-3} \times 2 $
$[OH^-] = 3 \times 10^{-3}$M
2. Find the pOH:
$pOH = -log[OH^-]$
$pOH = -log(3\times 10^{-3})$
$pOH \approx 2.523$
3. Find the pH:
$pH = 14 - pOH$
$pH \approx 14 - 2.523$
$pH \approx 11.477$
