Above what Fe2 concentration will Fe(OH)2 precipitate from a buffer solution that has a pH of 8.49?

Question

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Above what Fe2 concentration will Fe(OH)2 precipitate from a buffer solution that has a pH of 8.49? The Ksp of Fe(OH)2 is 4.87×10-17.

Answer 1

Fe(OH)₂ ==> Fe²⁺ + 2OH^-

K_sp = 4.87x10^-17 = [Fe²⁺][OH^-]²

From pH = 8.49, the pOH = 14 - pH = 14 - 8.49

pOH = 5.51, thus [OH^-] = 1x10^-5.51 = 3.09x10^-6 M

From the value of K_sp and knowing the [OH^-], one can now calculate the [Fe2⁺]

4.87x10^-17 = [Fe²⁺][3.09x10^-6]²

[Fe²⁺] = 4.87x10^-17/(3.09x10^-6)²

[Fe²⁺] = 5.10x10^-6 M

 

 

 

 

 

Above this concentration, a precipitate will form.