Answer
$pOH = 1.40$
$pH = 12.60$
Work Step by Step
1. $Ba(OH)_2$ is a strong base with 2 $OH^-$ in each molecule:
$[OH^-] = 2 * [Ba(OH)_2] = 2* 0.020M = 0.040M$
2. Calculate pOH and pH:
$pOH = -log[OH^-]$
$pOH = -log( 0.040)$
$pOH = 1.40$
$pH + pOH = 14$
$pH + 1.40 = 14$
$pH = 12.60$
