Question

What [OH-] should be maintained in a solution if, after precipitation of Mg2+ as solid magnesium hydroxide, the remaining [Mg2+] is to be at a level of 1ug-L-1?

Answer 1

Mg(OH)2(s) ==> Mg^2+ + 2OH^-
Ksp = (Mg^2+)(OH^-)^2
You want [Mg^2+] = 1 ug/L. Convert to M = moles/L
moles Mg^2+ = 1E-6 g/L x 1 mol/24.3 g = ? which you substitute in the equation above. You know Ksp, you plug in this amount for Mg in mols/L and solve for OH^-
Post your work if you get stuck.

Answer 2

The Ans is 1.6 E -2 but using that approach doesn't work

Answer 3

You just assumed that approach was wrong. You COULD HAVE posted your work and let me look at it.
What did you use for Ksp? Using a value of 5.61E-12 I found on the web my answer is 1.12E-2 but I found other values for Ksp also. Did you use 1.05E-11 for Ksp or something close to that? Let me know if you want to wrap this up tonight.

Answer 4

They used 1.8E-11

Answer 5

The correct answer, using the numbers in the problem, is as follows:
Ksp = 1.8E-11 = [Mg^2+][OH^-]^2
If [Mg^2+] = 1 ug/L that is 1E-6/24.3 = 4.12E-8 mols Mg^2+/L and
[OH^-] = sqrt (Ksp/[Mg^2+]) = sqrt (1.8E-11/4.12E-8) = 0.021 M.
I suspect there is a typo in the problem. Let me know how things turn out please.