Answer
The $K_{sp}$ for $Mg(OH)_2$ at that temperature is equal to $1.8 \times 10^{-11}$
Work Step by Step
1. Calculate the molar mass:
24.31* 1 + 2 * ( 16* 1 + 1.01* 1 ) = 58.33g/mol
2. Calculate the number of moles
9.6 mg = 0.0096 g
$n(moles) = \frac{mass(g)}{mm(g/mol)}$
$n(moles) = \frac{ 9.6\times 10^{- 3}}{ 58.33}$
$n(moles) = 1.646\times 10^{- 4}$
3. Find the concentration in mol/L:
$1.646 \times 10^{-4}$ mol in 1L: $1.646 \times 10^{-4} M$
4. Write the $K_{sp}$ expression:
$ Mg(OH)_2(s) \lt -- \gt 1Mg^{2+}(aq) + 2OH^{-}(aq)$
$ K_{sp} = [Mg^{2+}]^ 1[OH^{-}]^ 2$
5. Determine the ion concentrations:
$[Mg^{2+}] = [Mg(OH)_2] * 1 = [1.646 \times 10^{-4}] * 1 = 1.646 \times 10^{-4}$
$[OH^{-}] = [Mg(OH)_2] * 2 = 3.292 \times 10^{-4}$
6. Calculate the $K_{sp}$:
$ K_{sp} = (1.646 \times 10^{-4})^ 1 \times (3.292 \times 10^{-4})^ 2$
$ K_{sp} = (1.646 \times 10^{-4}) \times (1.083 \times 10^{-7})$
$ K_{sp} = (1.782 \times 10^{-11})$
