Answer
- This solution has $1.98 \times 10^{-3}$ moles of $KOH$.
- Its pOH is 0.44.
Work Step by Step
-
** $5.5 ml = 5.5 \times 10^{-3} L$
$n (moles) = Concentration (M) * Volume (L)$
$n (moles) = 0.360 * 5.50 \times 10^{-3} = 1.98 \times 10^{-3}$
-
$pOH = -log[OH^-]$
Since $KOH$ is a strong base:
$[OH^-] = [KOH] = 0.360M$
$pOH = -log(0.360) = 0.44$
