Question

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

Answer 1

.......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.

Answer 2

thank you!

Answer 3

^ this made no sense at all

Answer 4

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

Answer 5

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

Answer 6

For pH = 14.0 it'll be the same as your Ksp so 4e-16

Answer 7

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

Answer 8

wrong, sir. it is not the ksp...gosh how stupid.

Answer 9

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

Answer 10

Roxxie: you are a genius!!!!!

Answer 11

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

Answer 12

You're correct Roxxie!!!! Genius!