use:
pH = -log [H+]
8.83 = -log [H+]
[H+] = 1.479*10^-9 M
use:
[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)/(1.479*10^-9)
[OH-] = 6.761*10^-6 M
At equilibrium:
Fe(OH)2 <---->
Fe2+
+ 2
OH-
s
6.761*10^-6 + 2s
Ksp = [Fe2+][OH-]^2
4.87*10^-17=(s)*(6.761*10^-6+ 2s)^2
Since Ksp is small, s can be ignored as compared to
6.761*10^-6
Above expression thus becomes:
4.87*10^-17=(s)*(6.761*10^-6)^2
4.87*10^-17= (s) * 4.571*10^-11
s = 1.065*10^-6 M
So,
[Fe2+] = s = 1.065*10^-6 M
Answer: 1.06*10^-6 M
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