Question

21. (a) Determine the molar solubility of Fe(OH)3 in pure water. Ksp = 2.79 × 10-39 for Fe(OH)3.  (b) Determine the molar solubility of Fe(OH)3 if the pH of the solution is 8.0. (c) Determine the molar solubility of Fe(OH)3 if the pH of the solution is 2.0.

Can someone go over with me the work for this problem? The answer is given I just dont know how they got those answers.

Answer: (a) S or x = 1.01 × 10-10 M; (b) S or x = 2.79 × 10-21 M; (c) S or x = 0.00279 M

Answer 1

(a)

Pure water so pH = 7 and is constant

so [OH^-] = 10^-7

[Fe³⁺] = 2.79 10^-39 / [OH^-]³

= 2.79 10^-39 / 10^-21

        = 2.79 10^-18 M

so solubility is 2.79 10^-18 mol/L

(b)

Now pH = 8

so

[OH^-] = 10^-6

[Fe³⁺] = 2.79 10^-39 / [OH^-]³

= 2.79 10^-39 / 10^-18

      = 2.79 10^-21 M

(c)

Now pH = 2

so

[OH^-] = 10^-12

[Fe³⁺] = 2.79 10^-39 / [OH^-]³

= 2.79 10^-39 / 10^-36

      = 2.79 10^-3 M

= 0.00279 M