Question

Calculate the solubility (in moles per liter) of Fe(OH)3 ( Ksp = 4 x 10-38) in each of the following. a. water Solubility = mol/L b. a solution buffered at pH = 6.0 Solubility = mol/L c. a solution buffered at pH = 11.0 Solubility = mol/L

Answer 1

solution for 1st problem

used equations for this are Ksp = [Fe3+][OH-]^3 and [H+]*[OH-] = 10^-14 and pH = -log[H+]

(a)

Pure water (assume that the pH is 7.0 and remains constant).
[OH-] = 10^-7
[Fe(3+)] = 4.0x10^-38/[OH-]^3 = 4.0x10^-17 (M)
That means the solubility is 4.0x10^-17 mol/L.

(b)

A solution buffered at pH = 6.0.
[OH-] = 10^-8
[Fe(3+)] = 4.0x10^-38/[OH-]^3 = 4.0x10^-14 (M)
That means the solubility is 4.0x10^-14 mol/L.

(c)

A solution buffered at pH = 11.0.
[OH-] = 10^-3
[Fe(3+)] = 4.0x10^-38/[OH-]^3 = 4.0x10^-29 (M)
That means the solubility is 4.0x10^-29 mol/L.

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