Question

Kc is 1.67x10^20 at 25*C for the formation of iron (III) oxalate complex ion: Fe3+(aq)+3 C2O4^2->[Fe(C2O4)3]^3-. If 0.0200 M Fe3+ is initially mixed with 1.00 M oxalate ion, what is the concentration of Fe3+ ion at equilibrium?

Answer 1

as per reaction 3 C2O4^2- combines with 1 Fe3+

Hence for 0.02 M Fe3+ the C2O4^2- combined or reacted = 3 x 0.02 = 0.06

Hence C2O4^2- left = 1-0.06 = 0.94

[Fe(C2O4)3]^3- fomed = 0.02 M

now we have reverse equilibrium

                                  [Fe(C2O4)3]3-       <----->   Fe3+ (aq)   +   3C2O4^3- (aq)

Initial                              0.02                                0                       0.94

equilibrium                         0.02-X                             X                 0.094 + 3X

here we make approximations , since 1/Kc is very small, X is small , hence 0.02-X = 0.02 and 0.94+ 3X is nearlt 0.94

now 1/Kc = [Fe3+] [C2O4^2-]^3 / [Fe(C2O4)3]3-

1/ ( 1.67 x 10^20 ) = ( X) ( 0.94^3) / ( 0.02)

X = [Fe3+] = 1.44 x 10^-22 M