Answer
pH = 10.079
Hydronium concentration: $8.333 \times 10^{-11}M$
Work Step by Step
- KOH is a strong base, therefore:
$[OH^-] = [KOH] = 1.2 \times 10^{-4}M$
1. Calculate the pOH, then find the pH.
$pOH = -log[OH^-]$
$pOH = -log( 1.2 \times 10^{- 4})$
$pOH = 3.921$
$pH + pOH = 14$
$pH + 3.921 = 14$
$pH = 10.079$
2. Calculate the $[H_3O^+]$
$[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 1.2 \times 10^{- 4} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 1.2 \times 10^{- 4}}$
$[H_3O^+] = 8.333 \times 10^{- 11}$
