Answer
$[H_3O^+] = 0.015M$
$[OH^-] = 6.667 \times 10^{-13}M$
$pH = 1.824$
Work Step by Step
1. Since $HNO_3$ is a strong acid:
$[H_3O^+] = [HNO_3] = 0.015M$
2. Calculate the pH Value
$pH = -log[H_3O^+]$
$pH = -log( 0.015)$
$pH = 1.824$
3. Use $K_w$ to find the hydroxide concentration:
$[H_3O^+] * [OH^-] = Kw = 10^{-14}$
$ 1.5 \times 10^{- 2} * [OH^-] = 10^{-14}$
$[OH^-] = \frac{10^{-14}}{ 1.5 \times 10^{- 2}}$
$[OH^-] = 6.667 \times 10^{- 13}$
