Answer
$pH = 9.362$
$[H_3O^+] = 4.348 \times 10^{- 10}M$
This solution is basic.
Work Step by Step
$[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 2.3 \times 10^{- 5} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 2.3 \times 10^{- 5}}$
$[H_3O^+] = 4.348 \times 10^{- 10}M$
$pH = -log[H_3O^+]$
$pH = -log( 4.348 \times 10^{- 10})$
$pH = 9.362$
$[OH^-] > [H_3O^+]$: Basic solution:
Assuming $25 ^{\circ}C$: pH > 7: Basic solution.
