Question

Above what Fe2 concentration will Fe(OH)2 precipitate from a buffer solution that has a pH of 9.96? The Ksp of Fe(OH)2 is 4.87×10-17.

Answer 1

we have below equation to be used:

pH = -log [H+]

9.96 = -log [H+]

log [H+] = -9.96

[H+] = 10^(-9.96)

[H+] = 1.096*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)/(1.096*10^-10)

[OH-] = 9.12*10^-5 M

At equilibrium:

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

   s 9.12*10^-5 + 2s

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

4.87*10^-17=(s)*(9.12*10^-5+ 2s)^2

Since Ksp is small, s can be ignored as compared to 9.12*10^-5

Above expression thus becomes:

4.87*10^-17=(s)*(9.12*10^-5)^2

4.87*10^-17= (s) * 8.317*10^-9

s = 5.855*10^-9 M

Answer: 5.86*10^-9 M