Question

Calculate the molar solubility of Fe(OH)₂ in a buffer solution where the pH has been fixed at the indicated values. K_sp = 7.9 x 10^-16.

(a) pH 8.0

(b) pH 11.0

(c) pH 13.6

HopHelpCh17N6

Answer 1

a)

pH = 8.0

pOH = 14 -pH

pOH = 6

[OH-] = 1.0 x 10^-6 M

Fe(OH)2 -------------------> Fe+2 + 2OH-

S   1.0 x 10^-6 M

Ksp = [Fe+2][OH-]^2

7.9 x 10^-16 = (S) ( 1.0 x 10^-6 )^2

S = 7.9 x 10^-4 M

molar solubility = S = 7.9 x 10^-4 M

b)

pH = 11.0

pOH = 3

[OH-] = 1.0 x 10^-3 M

Ksp = [Fe+2][OH-]^2

7.9 x 10^-16 = (S) ( 1.0 x 10^-3 )^2

S = 7.9 x 10^-10 M

molar solubility = S = 7.9 x 10^-10 M

c)

pH = 13.6

pOH = 0.4

[OH-] = 10^-0.4 = 0.398 M

Ksp = [Fe+2][OH-]^2

7.9 x 10^-16 = (S) ( 0.398 )^2

S = 4.98 x 10^-15 M

molar solubility = S = 4.98 x 10^-15 M