Question

Fe^3+(aq) + L (aq) ⇌ FeL^3+(aq)
Kf= 4.68x10^25

A 1.00 L solution of 1.00g/L of Fe^3+(aq) has a 1.00 mol of L added. What is the equilibrium concentration of iron (III), (i.e., [Fe^3+]eq) in the final solution?

Answer 1

1.00 g/L Fe = 1/atomic mass Fe = ? M = approx 0.002 but that's just an estimate. You should get a more accurate answer.
1,00 M of L initially.

.............Fe^3+ + L ==> FeL^3+
I...........0.002.....1.00........0
Since Kf is such a huge number, ALL of the Fe^3+ will be converted to the conplex ion so you have this at equilibrium (for the first step).
C........-0.002....-0.002......0.002
E............0........0.00998....0.002 and this becomes a new problem which i will state as, what is the Fe^3+(aq) concentration present at equilibrium when 0.002M FeL^3+ dissociates.
..............FeL^3+ ==> Fe^3+ + L
I.............0.002............0...........0
C...........-x..................x............x
E..........0.002-x..........x.............x

Kf = (FeL^3+)/(Fe^3+)(L) = 4.68E25
Substitute the last E line into the Kf expression and solve for x = (Fe^3+).
Post your work if you get stuck. Note that Kf expression is written from the first equilibrium above and not the second one. The second one is a dissociation and not a formation.

Answer 2

The answer I got using quadratic formula to find x= 1.95x10^-14. is this a reasonable answer. I'm not 100% since i don't have the answer key for this.

Answer 3

Do i use the formation ICE table?
For the answer x= 1.95x10^-14 i used the dissociation ICE table

Answer 4

Shouldn't I just use the formation ICE table...then it'll be [Fe^3+]=[FeL3+]/Kf* [L]?

Answer 5

I noted that so you wouldn't get confused and you got confused anyway. You use Kf.
Kf = (FeL^3+)/(Fe^3+)(L)
Kf is 4.68E25
(Fe^3+) = x
(L) = 1 - (1/55.85)
(FeL^3+) = (1/55.85) - x
I get something like 1E-28 but that is only a close answer