Consider an amphoteric hydroxide, M(OH)2(s), where M is a generic metal.

M(OH)2 (s) <--> M^2+ (aq) + 2OH- (aq) Ksp=4e-16
M(OH)2 (s) + 2OH- (aq) <--> M(OH)4 ^2- (aq) Kf=0.06

Estimate the solubility of M(OH)2 in a solution buffered at pH= 7.0, 10.0, and 14.0

I'm not sure how to do this problem, please help

For the pH of 14 the answer will be the Kf which is 0.06 for this problem

thank you!

I got the first two right but I cant figure out ph=14

^^ yeah how do you do it when the pH=14?

its not though. i put that and it was wrong...

Roxxie: you are a genius!!!!!

Roxxie: I know it's been 6 years and I've never met you, but I love you

You're correct Roxxie!!!! Genius!

^ this made no sense at all

.......M(OH)2(s) ==> M^2+ + 2OH^-

.......M(OH)2(s) + 2OH^- ==> M(OH)4^2-

Ksp = (M^2+)(OH^-)^2 = 4E-16
Kf = [M(OH)4^2-)/(OH^-)^2 = 0.06

Solubility = (M2+) + [M(OH)4^2-]

at pH 7, (H^+) = 1E-7 and OH^- = 1E-7
(Mg^2+) = Ksp/(1E-7)^2 = about 0.04 M
[Mg(OH)2^2-] = 0.06*(1E-7)^2 = 6E-16; therefore, clearly the complex ion is not the predominant factor at pH = 7. You can follow through at pH = 10 (pOH = 4) and pH 14 (pOH = 0).[Note: don't confused with pOH = 0, that is (OH^-) = 1.0M
These are ESTIMATES of the solubility.