Answer
$[OH^-] = 8.77\times 10^{- 3}M$
$[H_3O^+] = 1.14\times 10^{- 12}M$
pH = 11.94
pOH = 2.057
Work Step by Step
1. Since LiOH is a strong base: $[OH^-] = [LiOH] = 8.77 \times 10^{-3}M$
2. Calculate pOH and pH:
$pOH = -log[OH^-]$
$pOH = -log( 8.77 \times 10^{- 3})$
$pOH = 2.057$
$pH + pOH = 14$
$pH + 2.057 = 14$
$pH = 11.943$
3. Now, find $[H_3O^+]$
$[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 8.77 \times 10^{- 3} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 8.77 \times 10^{- 3}}$
$[H_3O^+] = 1.14 \times 10^{- 12}$
