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.
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?
5 answers
The Ans is 1.6 E -2 but using that approach doesn't work
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.
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.
They used 1.8E-11
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.
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.