Answer
The pH of this solution is $11.477$.
Work Step by Step
Since $Ba(OH)_2$ is a strong base, and it has 2 $OH^-$ in each molecule:
$[OH^-] = 2 * [Ba(OH)_2] = 2 * 0.0015 = 0.0030M$
1. Calculate the pOH.
$pOH = -log[OH^-]$
$pOH = -log( 3 \times 10^{- 3})$
$pOH = 2.523$
2. Now, use the pOH value to find the pH.
$pH + pOH = 14$
$pH + 2.523 = 14$
$pH = 11.477$
