## Chemistry (4th Edition)

This solution has $1.98 \times 10^{-3}$ moles, and its pOH is 0.443.
$n(moles) = C(M) * V(L)$ - 5.50ml = 0.0055L (ml to L - Divide by 1000) $n(moles) = 0.360M * 0.0055L$ $n(moles) = 1.98 \times 10^{-3}$ -------- - Since KOH is a strong base: $[KOH] = [OH^-] = 0.360M$ $pOH = -log[OH^-]$ $pOH = -log(0.360)$ $pOH = 0.443$