"The Ksp of Mg(OH)2 in water is 1.2x10^-11. If the Mg2+ concentration is 1.2x10^-5, what is the pH at which Mg(OH)2 just begins to precipitate?"

So, I get that I need to find out the [OH-]. So, since the Mg2+ concentration is already given and we know that Mg(OH)2 dissociates into 1 Mg and 2OH's, I multiplied the [Mg2+] by 2 since there are two OH's for every one Mg2+. Essentially:

Mg(OH)2 --> Mg2+ + 2OH-

x --> x + 2x

But according to Kaplan's explanation, the correct way to do it is: [OH-] = square root of Ksp/[Mg2+]. I see how this is derived from

Ksp = [Mg2+][OH-]^2, but don't you still have to account for the 2:1 ratio of OH to Mg?

I solved the problem similarly to if the question were asking for the concentrations of the ions of a saturated solution of Mg(OH)2:

You'd figure out the Mg2+ concentration by solving for 'x' in Ksp= 1.2x10^-11 = 4x^3. Once you get x, you plug it back in to find [OH-] knowing that

x ---> x + 2x.

I feel like I must be missing something really basic =( Help!!!

***I understand that after you find the [OH], you have to find the pOH and then from there find the pH. But my problem lies in the first step of calculating the OH concentration.

And the correct answer is pH 11. I guessed that since I figured OH is being dissolved, the solution will probably be basic, but I still don't really understand hoow to do it.

Thanks!!!!!!!!!