Answer
$$pH = 11.52 $$
Work Step by Step
1. Calculate the molar mass:
$ Mg(OH)_2 $ : ( 24.31 $\times$ 1 )+ ( 1.008 $\times$ 2 )+ ( 16.00 $\times$ 2 )= 58.33 g/mol
2. Use the information as conversion factors to find the molarity of this solution:
$$\frac{ 9.63 mg \space Mg(OH)_2 }{100 \space mL \space solution} \times \frac{1 \space g}{1000 \space mg} \times \frac{1 \space mol \space Mg(OH)_2 }{ 58.33 \space g \space Mg(OH)_2 } \times \frac{1000 \space mL}{1 \space L} = 1.65 \times 10^{-3} \space M$$
3. Since each $Mg(OH)_2$ has 2 $OH^-$
$[OH^-] = 2[Mg(OH)_2] = 2(1.65 \times 10^{-3} \space M) = 3.30 \times 10^{-3} \space M$
4. Calculate the pH:
$$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 3.30 \times 10^{-3} } = 3.0 \times 10^{-12} \space M$$
$$pH = -log[H_3O^+] = -log( 3.0 \times 10^{-12} ) = 11.52 $$
