Answer
$pH = 6.215$
$[OH^-] = 1.639 \times 10^{- 8}M$
This solution is acidic.
Work Step by Step
$[H_3O^+] * [OH^-] = Kw = 10^{-14}$
$ 6.1 \times 10^{- 7} * [OH^-] = 10^{-14}$
$[OH^-] = \frac{10^{-14}}{ 6.1 \times 10^{- 7}}$
$[OH^-] = 1.639 \times 10^{- 8}M$
$pH = -log[H_3O^+]$
$pH = -log( 6.1 \times 10^{- 7})$
$pH = 6.215$
$[H_3O^+] > [OH^-]$: Acidic solution
Assuming $25 ^{\circ}C$: pH < 7 : Acidic solution.
