What is the molar solubility of Fe(OH)2 when buffered at pH of 9.00? [Fe(OH)2 : Ksp...

Question

Question

What is the molar solubility of Fe(OH)2 when buffered at pH of 9.00? [Fe(OH)2 : Ksp = 7.9 x 10^-16]

science chemistry

Answer 1

Answer #1

Fe(OH)₂ ---------> Fe²⁺ + 2OH^-

pH = 9

pOH = 14- pH = 14 - 9

pOH = 5

[OH] = 10^-pOH = 10^-5

-------------------------------------

K_s_p = [Fe²⁺].[OH^-]²

[Fe²⁺] = (K_s_p) / ([OH^-]²)

Therefore, Molar solubility is given by:

[Fe²⁺] = (7.9 x 10^-16) / (10^-5)²

[Fe²⁺] = 7.9 x 10^-6 M -------------------(ANSWER)

Answer 2

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