Answer:
(A) 1.962x10^-10 M solubility in pure water
(B) 4.0 x 10^-33 M solubility
(C) 4.0 x 10^-27 M solubility
Explanation:
(A) Fe(OH)3 would give (Fe3+) and (3OH-)
Ksp = [Fe^3+][OH-]^3 = 4.0 x 10^-38
Let y = [Fe^3+]
Let 3y = [OH-]
4x10^-38 = (y)(3y)^3
4x10^-38 = 27y^4
y^4 = 4x10^-38 ÷ 27
y^4 = 1.481 x 10^-39
y = 1.962x10^-10 M solubility in pure water
(B) pH = 5.0
5.0 = - log [OH-]
-5.0 = log [OH-]
[OH-] = 10^-5.0 = 1.0 x 10^-5 M
So, Ksp = [Fe^3+][OH-]^3 = 4.0 x 10^-38
[Fe^3+][1.0 x 10^-5] = 4.0 x 10^-38
[Fe^3+] = 4.0 x 10^-38 ÷ 1.0 x 10^-5
= 4.0 x 10^-33 M solubility
(C) pH = 11.0
11.0 = - log [OH-]
-11.0 = log [OH-]
[OH-] = 10^-11.0 = 1.0 x 10^-11 M
So, Ksp = [Fe^3+][OH-]^3 = 4.0 x 10^-38
[Fe^3+][1.0 x 10^-11] = 4.0 x 10^-38
[Fe^3+] = 4.0 x 10^-38 ÷ 1.0 x 10^-11
= 4.0 x 10^-27 M solubility
