Question

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

Estimate the solubility of M(OH)₂ in a solution buffered at pH=14.00.

(At pH=7.00, its solubility is 0.03M, and at pH=10.0, its solubility is 2.03 x 10^-8 M.

Answer 1

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

Ksp = (M^2+)(OH^-)^2 = 3E-16 = 3*10^-16
Kf = [M(OH)4^2-)/(OH^-)^2 = 0.05

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

at pH 7, (H^+) = 1E-7 and OH^- = 1E-7
(Mg^2+) = Ksp/(1E-7)^2 = about 0.03 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.

pH 14 (pOH = 0).[Note: don't confused with pOH = 0, that is (OH^-) = 1.0M ]

pH= - log(H+)

At pH=14

H+ = 1E-14 = 1*10^-14

OH - =1.0M

So, for pH= 14, solubility = Kf/[OH-]

=0.05/1

=0.05