Answer
$[OH^-] = 0.0224M$
$[H_3O^+] = 4.464\times 10^{- 13}M$
pH = 12.35
pOH = 1.65
Work Step by Step
1. Since $Ba(OH)_2$ is a strong base, $[OH^-] = 2 * [Ba(OH)_2]$
$[OH^-] = 2 * 0.0112$
$[OH^-] = 0.0224M$
2. Calculate pOH and pH:
$pOH = -log[OH^-]$
$pOH = -log( 0.0224)$
$pOH = 1.65$
$pH + pOH = 14$
$pH + 1.65 = 14$
$pH = 12.35$
3. Now, find $[H_3O^+]$
$[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 2.24 \times 10^{- 2} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 2.24 \times 10^{- 2}}$
$[H_3O^+] = 4.464 \times 10^{- 13}M$
