Answer
(a) $[KOH] = 1.09 \times 10^{- 3}M$
(b) $[Ba(OH)_2] = 5.45 \times 10^{-4}M$
Work Step by Step
1. Find $[OH^-]$
pH + pOH = 14
11.04 + pOH = 14
pOH = 2.96
$[OH^-] = 10^{- 2.96}$
$[OH^-] = 1.09 \times 10^{- 3}M$
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(a) Since KOH is a strong base: $[KOH] = [OH^-] = 1.09 \times 10^{- 3}M$
(b) Since $Ba(OH)_2$ is a strong base with 2 $OH^-$:
$[OH^-] = [Ba(OH)_2] * 2$
$ 1.09 \times 10^{- 3} = [Ba(OH)_2] * 2$
$\frac{ 1.09 \times 10^{- 3}}{2} = [Ba(OH)_2]$
$[Ba(OH)_2] = 5.45 \times 10^{-4}M$
