Answer
$ K_{sp} (MOH)= (2.3 \times 10^{-9})$
Work Step by Step
1. Calculate the hydroxide ion concentration:
pH + pOH = 14
9.68 + pOH = 14
pOH = 4.32
$[OH^-] = 10^{-pOH}$
$[OH^-] = 10^{- 4.32}$
$[OH^-] = 4.8 \times 10^{- 5}$
2. Write the $K_{sp}$ expression:
$ MOH(s) \lt -- \gt 1M^{+}(aq) + 1OH^{-}(aq)$
$ K_{sp} = [M^{+}]^ 1[OH^{-}]^ 1$
3. Determine the ion concentrations (considering a pure solution):
$[M^{+}] = [OH^-] = 4.8 \times 10^{-5}M$
4. Calculate the $K_{sp}$:
$ K_{sp} = (4.8 \times 10^{-5})^ 1 \times (4.8 \times 10^{-5})^ 1$
$ K_{sp} = (4.8 \times 10^{-5}) \times (4.8 \times 10^{-5})$
$ K_{sp} = (2.3 \times 10^{-9})$
