Answer
$[OH^-] = 1.9\times 10^{- 4}M$
$[H_3O^+] = 5.263\times 10^{- 11}M$
pH = 10.28
pOH = 3.721
Work Step by Step
1. Since $KOH$ is a strong base: $[OH^-] = [KOH] = 1.9 \times 10^{-4}M$
2. Calculate pOH and pH:
$pOH = -log[OH^-]$
$pOH = -log( 1.9 \times 10^{- 4})$
$pOH = 3.721$
$pH + pOH = 14$
$pH + 3.721 = 14$
$pH = 10.279$
3. Now, find $[H_3O^+]$
$[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 1.9 \times 10^{- 4} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 1.9 \times 10^{- 4}}$
$[H_3O^+] = 5.263 \times 10^{- 11}$
