Answer
$[H_3O^+] = 0.0548M$
$[OH^-] = 1.825 \times 10^{-13}M $
$pH = 1.261$
$pOH = 12.739$
Work Step by Step
1. HCl is a strong acid, so : $[HCl] = [H_3O^+] = 0.0548M:$
2. Now, find the pH and pOH:
$pH = -log[H_3O^+]$
$pH = -log( 0.0548)$
$pH = 1.261$
$pOH + pH = 14$
$pOH = 14 - pH = 14 - 1.261$
$pOH = 12.739$
3. And the $[OH^-]$:
$[H_3O^+] * [OH^-] = Kw = 10^{-14}$
$ 5.48 \times 10^{- 2} * [OH^-] = 10^{-14}$
$[OH^-] = \frac{10^{-14}}{ 5.48 \times 10^{- 2}}$
$[OH^-] = 1.825 \times 10^{- 13}M$
