Question

Iron (III) ions form a complex with thiocyanate according to the equation
Fe^3+(aq) + 6 SCN-(aq) ⇄ Fe(SCN)6 ^3-(aq) Kf = 3.20x10^3

If 0.001 moles of Fe(NO3)3 and 0.2 moles of NaSCN are dissolved in enough water to create a 1.00L of solution, what's the concentration of Fe(SCN)6 3- (in M)?

I did:
..................Fe^3+(aq) + 6 SCN-(aq) ⇄ Fe(SCN)6 ^3-(aq)
Initial.........0.001................0.2....................0
change ... +x......................+6x.................-x
equilib......0.001+x~0.001...0.2+6x~ 0.2....-x

Kf = [Fe (SCN)6 ^3-]/ [Fe^3+] [SCN-]^6

3.20x10^3 = (-x)/ ( (0.001)(0.2)^6 )

I got x = -2.1 x 10^-4 M, but it's wrong.

Answer 1

First, your answer of a negative concentration you KNOW can't be right. When have you ever had a solution that had less than nothing as a concentration. Doesn't make sense.
Looking at the equation, you can see that the limiting reagent (LR) in this case is Fe^3+. Therefore, all of the Fe^3+ will be used because the Kf is so large. I would answer that [Fe(SCN)6]^3- = 0.001 M because all of the initial Fe^3+ has formed the complex and by Le Chatelier's Principle the excess SCN^- helps drive the reaction to the right.
Also, I should point out that your ICE chart is going the wrong way. If you have zero Fe(SCN)6^3- it can't be forming the reactants.