Answer
(a) $1.32 \times 10^{-2}M$
(b) $6.60 \times 10^{-3}M$
Work Step by Step
1. Find the concentration of $[OH^-]$:
$[OH^-] = 10^{-pOH}$
$[OH^-] = 10^{-1.88} = 1.32 \times 10^{-2}M$
(a) Knowing that $[OH^-] = 1.32 \times 10^{-2}$ and KOH is a strong base:
$[OH^-] = [KOH] = 1.32 \times 10^{-2}M$
(b) The $Ba(OH)_2$ is a strong base too, but, it has 2 $OH^-$ for each base molecule, therefore, the concentration of $OH^-$ is the double of the base concentration:
$[OH^-] = [Ba(OH)_2] * 2$
$1.32 \times 10^{-2} \div 2 = [Ba(OH)_2]$
$[Ba(OH)_2] = 6.60 \times 10^{-3}M$
