[Juwanaxedqwe]

The concentration of Mg2+ in seawater is 0.052 M. At what pH will 78% of the Mg2+ be precipitated as the hydroxide salt? Ksp for Mg(OH)_2 is 8.9x10^-12.

0.78*0.052 = 0.04056M

Ksp = [Mg][2OH]^2
8.9x10^-12 = [0.04056]*[4x^2]
x = 7.40655
1x10^-14 / 2x = 1.48131x10^-10
-log(1.48131x10^-10) = 9.17 = pH

Am I doing something wrong? I'd appreciate the help, thanks.

[Borek]

Ksp = [Mg][2OH]^2

This is not correct. Close, but no banana.

[Juwanaxedqwe]

Going by the ICE method, doesn't 2 mols of hydroxide form per mol of Mg(OH)_2 that ionizes in a solution?
So, 2x of hydroxide forms giving us a Ksp formula that should like:
Ksp = x*(2x)^2

but since I already know 0.04056 mols of Mg^2+ is used, it would be:
Ksp = .04056*(2x)^2
x being the amount of hydroxide used to compute the Ksp, no?

I made a typo in my first post,
"x = 7.40655" should equal 7.40655x10^-6

[enahs]

Write down the complete reaction and what the Ksp of that reaction means, and then reread your question.

[Juwanaxedqwe]

MgOH_2(s)   Mg²⁺_(aq) + 2OH^-_(aq) at equilibrium

Accordingly, Ksp = [Mg²⁺][OH^-]²

The only other possibility I can think of is:
8.9*10^-12 = .04056*x²
whereby x = 1.481310257*10^-5
[H₃O⁺][OH^-] = Kw = 10^-14
and then, 10^-14/2x since 2mols of hydroxide is ionized which = 3.37539*10^-10
and then, -log(3.37539*10^-10) = 9.47

[enahs]

Does the question ask you about going from Mg(OH)₂ (s) to Mg²⁺ (aq) + 2 OH^- (aq) or to go from Mg²⁺ (aq) + 2 OH^- (aq) to Mg(OH)₂(s)?

[Juwanaxedqwe]

I considered the latter but I ended up getting a ridiculous value for OH^-.
If it was inverted, would 1/Ksp =.04056*4x²?

[Borek]

Ksp = [Mg²⁺][OH^-]²

And that's a correct equation to use.

But I see you tried to stick some 2 into your calculations later. No need for that. The only place where 2 from Mg(OH)₂ matters is in [Mg²⁺][OH^-]². You are not interested in stoichiometry, only in concentrations.

[enahs]

I was hoping if you looked at it the other way you might figure it out, sometimes it helps looking at the same thing from a different way. If you work it backwards, you should still get the exact same mathematical answer because remember it is products over reactants, so you should have your new really large Ksp value = 1/0.04056 * x².

[Juwanaxedqwe]

You are not interested in stoichiometry, only in concentrations.

I learned something new, thank you!