Equilibrium help please!!!

Question

At a certain temperature K = 1.1 103 L/mol for the reaction given below. Calculate the concentrations of Fe3 , SCN-, and FeSCN2 at equilibrium if 0.027 mol of Fe(NO3)3 is added to 1.0 L of 0.11 M KSCN. (Neglect any volume change.)

Fe3 (aq) SCN -(aq) FeSCN2 (aq)

i found FeSCN2 to be 0.027M but i have no idea how to find the other values please help!

Answer 1

Make an ICE table.

Fe^3+(aq) + SCN-(aq) <--> FeSCN^2+(aq)

___________________________________

I..0.027......... 0.11...................0

C - x.............. -x ....................+x

E 0.027-x .......0.11-x ...............x

K = [FeSCN^2+] / [Fe^3+][SCN-] = 1.1E3

1.1E3 = x / (0.022-x)(0.11-x)

Solving for x, we get x = 0.02177562mol = 0.02177562M

Plug x back into the "equilibrium line" on the ICE table to find each concentration at equilibrium.

Next time, please indicates CHARGES and SIGNS appropriately.

[Answer: see above]