Question

Consider the following reaction: Fe3+(aq)+SCN−(aq)⇌FeSCN2+(aq) A solution is made containing an initial [Fe3+] of 1.0×10^−3 M and an initial [SCN−] of 7.8×10^−4 M . At equilibrium, [FeSCN2+]= 1.7×10^−4 M .

Part A Calculate the value of the equilibrium constant (Kc).

Express your answer using two significant figures. Kc =

Answer 1

Fe3+(aq)   +. SCN−(aq) <--------------------> FeSCN2+(aq)

1.0 x 10^-3 7.8 x 10^-4 0             ----------------> initial

1.0 x 10^-4 - x     7.8 x 10^-4 -x x             ---------------->equilibrium

but x = 1.7 x 10^-4 is given

equilibrium concentrations :

[Fe+2] = 1.0 x 10^-3 - x = 1.0 x 10^-3 - 1.7 x 10^-4

   = 0.83 x 10^-3 M

[SCN-] = 7.8 x 10^-4 - x

=7.8 x 10^-4 - 1.7 x 10^-4 = 6.1 x 10^-4 M

[FeSCN2+]= 1.7×10^-4 M

equilibrium constant (Kc) = [FeSCN2+] / [Fe+2] [SCN-]

Kc = 1.7×10^-4 / (0.83 x 10^-3 ) (6.1 x 10^-4)

                                   Kc = 335.76