Answer
The equilibrium constant for this reaction is equal to:
$K_c = 1.80 \times 10^{-11}$
Work Step by Step
The dissolution of solid magnesium hydroxide in water is described by this reaction:
$Mg(OH)_2(s) \lt -- \gt Mg^{2+}(aq) + OH^-(aq)$
Balancing the reaction:
$Mg(OH)_2(s) \lt -- \gt Mg^{2+}(aq) +2 OH^-(aq)$
Now write the $K_c$, you just need to put the concentrations of the products divided by the concentrations of the reactants.
** Remember: For $K_c$, pure solids (s) and liquids(l) are omited in the expression. So, in this case, $Mg(OH)_2(s)$ will not appear.
** The balanced coefficients are the exponents of the concentrations of their respective compounds:
$K_c = \frac{[Products]}{[Reactants]} = \frac{[Mg^{2+}][OH^-]^2}{1} = (1.65 \times 10^{-4})(3.30 \times 10^{-4})^2 = 1.80 \times 10^{-11}$
