Answer
$[OH^-] = 8.7\times 10^{- 5}M$
$[H_3O^+] = 1.149\times 10^{- 10}M$
pH = 9.94
pOH = 4.06
Work Step by Step
1. Since KOH is a strong acid: $[OH^-] = [KOH] = 8.7 \times 10^{-5}M$
2. Calculate pOH and pH:
$pOH = -log[OH^-]$
$pOH = -log( 8.7 \times 10^{- 5})$
$pOH = 4.06$
$pH + pOH = 14$
$pH + 4.06 = 14$
$pH = 9.94$
3. Now, find $[H_3O^+]$
$[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 8.7 \times 10^{- 5} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 8.7 \times 10^{- 5}}$
$[H_3O^+] = 1.149 \times 10^{- 10}$
