Calculate the solubility (in mol/L) of Fe(OH)3 (Ksp = 4.0 x 10^-38) in each of the following situations: (A) Pure water (assume that the pH is 7.0 and remains constant).
(B) A solution buffered at pH = 5.0.
(C) A solution buffered at pH = 11.0.

(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

a boiling water reactor does not boil the water. a pressurized water reactor boils the water faster. a pressurized water reactor is heated directly by the core. a pressurized water reactor does not use fission.