Answer
(a) $[LiOH] = 6.02 \times 10^{- 5}M$
(b) $[Ba(OH)_2] = 3.01 \times 10^{-5}M$
Work Step by Step
1. Find $[OH^-]$
pH + pOH = 14
9.78 + pOH = 14
pOH = 4.22
$[OH^-] = 10^{- 4.22}$
$[OH^-] = 6.02 \times 10^{- 5}M$
------
(a) Since LiOH is a strong base: $[LiOH] = [OH^-] = 6.02 \times 10^{- 5}M$
(b) Since $Ba(OH)_2$ is a strong base with 2 $OH^-$:
$[OH^-] = [Ba(OH)_2] * 2$
$ 6.02 \times 10^{- 5} = [Ba(OH)_2] * 2$
$\frac{ 6.02 \times 10^{- 5}}{2} = [Ba(OH)_2]$
$[Ba(OH)_2] = 3.01 \times 10^{-5}M$
