Chemistry just kicked in, yo.

This has me pissed. I need to use the solubility product for Fe(OH)3 (2.6 x 10^-39) to calculate the pH at which it will precipitate from a .2 M Fe solution.

Thank you chemistry lab for not giving me a sample problem, and thank you chemistry class for not being up to speed with the lab.

Here's what I have so far... the only thing is, where do I plug the molarity in? Should I just be using the Ksp value and the .2 M as the concentration of Fe with that times three and cubed for [OH]?

Ksp Fe(OH)3 = 2.6 x 10^-39
 Fe(OH)3 (s) ↔ Mg3+(aq) + 3OH-(aq)
Ksp = [Fe3+][OH-]3 = 2.6 x 10-39
Ksp = (s)(3s)3 = 27s4 = 2.6 x 10-39
s =  9.9061 x 10-11 M
[OH-] = 3s = 2.97183 x 10^-10
pOH = -log(2.97183 x 10^-10) = 9.53
pH = 4.47

JEWS
JEWS
Fe(OH)3 (s) ↔ Mg3+(aq) + 3OH-(aq)
wat
JEWS
JEWS
>Ksp = (s)(3s)3 = 27s4 = 2.6 x 10-39
wrong, use .2M= [Fe3+] here, (it'll actually be (.2M +s), but .2 will be much much greater than s, so we can approximate)
JEWS
JEWS
Proceed as you did before, with that change
JEWS
JEWS

Whoops, yeah that Mg should read Fe.

so...

(.2)(3s)3
5.4s^(3)
s =  7.8378 x 10^-14
[OH-] = 3s = 2.35134 x 10^-13 M
pOH = -log(2.35134 x 10^-13) = 12.63
pH = 1.37

Alright, and by my prior post's reasoning the pH of free Fe when reduced to 10^-7 should go like this:

Ksp = (10-7)(3s)3 =  2.7 x 10-6 s3 =  2.6 x 10-39
s = 9.875 x 10-12
[OH-] = 3s = 2.96268 x 10-11   
pOH = -log(2.96268 x 10-11) = 10.53
pH = 3.47

amirite?

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Cool ASCII, bro.

Is #4 correct?