What is the molar solubility of Mg(OH)2 in a basic solution with a pH of 12.00? Ksp for Mg(OH)2 is 5.6 × 10-12.Group of answer choices5.6 × 10-10 M5.6 × 10-8 M1.1 × 10-4 M2.4 × 10-6 M

To solve this problem, we need to understand that the solubility product constant, Ksp, is given by the product of the concentrations of the ions in the solution. For Mg(OH)2, this is [Mg2+][OH-]^2.

First, we need to find the concentration of OH- ions in the solution. We can do this using the pH value given. The pOH is given by 14 - pH, so the pOH is 14 - 12 = 2. The concentration of OH- ions is then 10^-pOH = 10^-2 = 0.01 M.

Next, we substitute these values into the Ksp expression. We let the molar solubility of Mg(OH)2 be x. This means that the concentration of Mg2+ ions is also x, and the concentration of OH- ions is 0.01 + 2x (because each Mg(OH)2 gives two OH- ions). So we have:

Ksp = [Mg2+][OH-]^2 5.6 × 10^-12 = x(0.01 + 2x)^2

Assuming that 2x is much smaller than 0.01 (which is reasonable if the solubility is small), we can approximate this as:

5.6 × 10^-12 = x(0.01)^2 5.6 × 10^-12 = x(0.0001) x = 5.6 × 10^-12 / 0.0001 x = 5.6 × 10^-8 M

So the molar solubility of Mg(OH)2 in this solution is approximately 5.6 × 10^-8 M.