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
___ M
(b) pH 10.7
___ M
(c) pH 13.6
___ M

HopHelpCh17N6

Answer 1

1. pH = 8

Ksp = [Fe⁺²][OH^-]²

Let Fe⁺² = X

pH + pOH = 14

8 + pOH = 14

so pOH = 6

pOH = -

So = - 6

So [OH^-] = 1 * 10^-6

So Ksp = [X] * [10^-6]^2

7.9 * 10^-16 = X * [10^-6]^2

So, X = 7.9 * 10^-4 M = molar solubility.

2. pH = 10.7

Ksp = [Fe⁺²][OH^-]²

Let Fe⁺² = X

pH + pOH = 14

10.7 + pOH = 14

so pOH = 3.3

pOH = -

So = - 3.3

So [OH^-] = 5.011 * 10^-4

So Ksp = [X] * [5.011 * 10^-4]^2

7.9 * 10^-16 = [X] * [5.011 * 10^-4]^2

So, X = 3.146 * 10^-9 M = molar solubility.

3. pH = 13.6

Ksp = [Fe⁺²][OH^-]²

Let Fe⁺² = X

pH + pOH = 14

13.6 + POH = 14

so POH = 0.4

POH = -

So = - 0.4

So [OH^-] = 0.398

So Ksp = [X] * [0.398]^2

7.9 * 10^-16 = [X] * [0.398]^2

So, X = 4.987 * 10^-15 M = molar solubility.