## Chemistry 12th Edition

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