Answer
$[H_3O^+] = 0.25M$
$[OH^-] = 4 \times 10^{-14}M$
$pH = 0.6021$
Work Step by Step
1. Since HCl is a strong acid:
$[H_3O^+] = [HCl] = 0.25M$
2. Calculate the pH value
$pH = -log[H_3O^+]$
$pH = -log( 0.25)$
$pH = 0.6021$
3. Use $K_w$ to calculate $[OH^-]$:
$[H_3O^+] * [OH^-] = Kw = 10^{-14}$
$ 0.25 * [OH^-] = 10^{-14}$
$[OH^-] = \frac{10^{-14}}{0.25}$
$[OH^-] = 4 \times 10^{- 14}$
