Question

Above what Fe2 concentration will Fe(OH)2 precipitate from a buffer solution that has a pH of 9.44? The Ksp of Fe(OH)2 is 4.87x10-17

Answer 1

we have below equation to be used:

pH = -log [H+]

9.44 = -log [H+]

log [H+] = -9.44

[H+] = 10^(-9.44)

[H+] = 3.631*10^-10 M

we have below equation to be used:

[OH-] = Kw/[H+]

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

[OH-] = (1.0*10^-14)/[H+]

[OH-] = (1.0*10^-14)/(3.631*10^-10)

[OH-] = 2.754*10^-5 M

At equilibrium:

Fe(OH)2 <----> Fe2+ + 2 OH-

   s 2.754*10^-5 + 2s

Ksp = [Fe2+][OH-]^2

4.87*10^-17=(s)*(2.754*10^-5+ 2s)^2

Since Ksp is small, s can be ignored as compared to 2.754*10^-5

Above expression thus becomes:

4.87*10^-17=(s)*(2.754*10^-5)^2

4.87*10^-17= (s) * 7.585*10^-10

s = 6.42*10^-8 M

Answer: 6.42*10^-8 M