Asked by anon

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

Answers

Answered by DrBob222
.......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.
Answered by anon
thank you!
Answered by ConfusedStudent
^ this made no sense at all
Answered by Paula
I got the first two right but I cant figure out ph=14
Answered by what
^^ yeah how do you do it when the pH=14?
Answered by Person
For pH = 14.0 it'll be the same as your Ksp so 4e-16
Answered by Bobina
its not though. i put that and it was wrong...
Answered by yall are loserzz
wrong, sir. it is not the ksp...gosh how stupid.
Answered by Roxxie
For the pH of 14 the answer will be the Kf which is 0.06 for this problem
Answered by Marrtissey
Roxxie: you are a genius!!!!!
Answered by Kaitie
Roxxie: I know it's been 6 years and I've never met you, but I love you
Answered by Tito1347
You're correct Roxxie!!!! Genius!
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