Answer
$pH = 12.90$
Work Step by Step
1. Since these are 2 strong bases, determine the concentration given by each one
$[OH^-]_{KOH} = [KOH] = 0.050M$
$[OH^-]_{Ba(OH)_2} = [Ba(OH)_2] * 2 = 0.015M * 2 = 0.030M$
2. Sum these numbers:
$0.050M + 0.030M = 0.080M$
3. Now, calculate the pH:
$pOH = -log[OH^-]$
$pOH = -log( 0.080)$
$pOH = 1.10$
$pH + pOH = 14$
$pH + 1.10 = 14$
$pH = 12.90$
