Answer
$ K_{sp} (Pb(OH)_2) = (1.4 \times 10^{-15})$
Work Step by Step
1. Calculate the hydroxide ion concentration:
pH + pOH = 14
9.15 + pOH = 14
pOH = 4.85
$[OH^-] = 10^{-pOH}$
$[OH^-] = 10^{- 4.85}$
$[OH^-] = 1.41 \times 10^{- 5}$
2. Write the $K_{sp}$ expression:
$ Pb(OH)_2(s) \lt -- \gt 1Pb^{2+}(aq) + 2OH^{-}(aq)$
$ K_{sp} = [Pb^{2+}]^ 1[OH^{-}]^ 2$
3. Determine the ion concentrations:
$[OH^{-}] = [Pb(OH)_2] * 2 = 1.41 \times 10^{-5}:$
$[Pb(OH)_2] = \frac{1.41 \times 10^{-5}}{2} = 7.05 \times 10^{-6}$
$[Pb^{2+}] = [Pb(OH)_2] * 1 = [7.05 \times 10^{-6}] * 1 = 7.05 \times 10^{-6}$
4. Calculate the $K_{sp}$:
$ K_{sp} = (7.05 \times 10^{-6})^ 1 \times (1.41 \times 10^{-5})^ 2$
$ K_{sp} = (7.05 \times 10^{-6}) \times (1.99 \times 10^{-10})$
$ K_{sp} = (1.4 \times 10^{-15})$
