This is what i posted up a while back... i got to a using

0.520= a (.202M) and got a= 2.57

but now i don't know where to go from there with the ICE

The standard solution of FeSCN2+ (prepared by combining 9.00 mL of 0.200 M Fe(NO3)3 w/1.00 mL of 0.0020 M KSCN) has an absorbance of 0.520.

If a trial's absorbance is measured to be 0.275 and its initial concentration of SCN– was 0.00060 M, what is the equilibrium concentration of SCN–?

A = abc

A = 0.520
a = A/c (we will forget b since that same cell length was used for both standard AND unknown BUT remember this is an arbitrary a and not a real a as a constant.
a= 0.520/0.0002 = 2,600. right? check my thinking. The c you are interested in is c of the FeSCN^+2 since that is the colored complex you are measuring. Since
Fe(NO3)3 + KSCN ==> FeSCN^+2 + 3NO3^-
So 0.0020 M KSCN x 0.0010 L (1 mL) = 2 x 10^-6 mols so 2 x 10^-6 mols of the FeSCN^+2 will be formed and that is present in 0.010 L so the molarity of the complex is 2 x 10^-6/10-2 = 2 x 10^-4 M.
Now A = abc for the trial's run.
A = 0.275
A/a = c = 0.275/2600 = 1.057 x 10^-4 M
Now, I'm not sure of the problem next BECAUSE I don't know what the SCN^- is. The problem states that it is 0.00060 M for its initial concentration. If that is the concn then 6 x 10^-4 - 1.11 x 10^-4 = ?? is the amount of SCN^- remaining unreacted in the unknown sample. However, if the 0.00060 M is the concn BEFORE it was diluted (I don't know if this sample was treated the same as the standard or not). I suspect the actual meaning of the problem is to use it as I did above as 6.0 x 10^-4 and it isn't diluted from that figure. You really don't need the ICE; I thought you might understand it better if we went through that scheme.
Fe^+3 + SCN^- ==> FeSCN^+2
Initial:
(SCN^-) = 0.00060
(FeSCN^+2) = 0

change:
(SCN^-) = -x
(FeSCN^+2) = +1.11 x 10^-4 [which means x is 1.11 x 10^-4]

Equilibrium:
(FeSCN^-) = 0 + 1.11 x 10^-4 = 1.11 x 10^-4
(SCN^-) = 6.0 x 10^-4 - 1.11 x 10^-4 = ??

THANK YOU SOOOO MUCH!!

To find the equilibrium concentration of SCN–, you will need to use the concept of Beer's Law and the reaction quotient (Qc). Here's how you can proceed:

1. Start by writing the balanced chemical equation for the reaction between Fe3+ and SCN– to form FeSCN2+:
Fe3+ + SCN– → FeSCN2+

2. Based on the balanced equation, determine the stoichiometry. In this case, for every 1 mole of FeSCN2+ formed, 1 mole of SCN– is consumed.

3. Calculate the initial concentrations of Fe3+ and SCN– using the given information. The volume of the Fe(NO3)3 solution is 9.00 mL, and its concentration is 0.200 M, so:
Initial [Fe3+] = (0.200 M) * (9.00 mL / 10.00 mL) = 0.180 M

The volume of the KSCN solution is 1.00 mL, and its concentration is 0.0020 M, so:
Initial [SCN–] = (0.0020 M) * (1.00 mL / 10.00 mL) = 0.00020 M

4. Using Beer's Law, you can relate the concentration of the FeSCN2+ complex and its absorbance. Recall that Beer's Law states that absorbance is directly proportional to the concentration of the absorbing species. The equation can be written as:
A = εbc,
where A is the absorbance, ε is the molar absorptivity, b is the path length (which is usually 1 cm in this case), and c is the concentration.

From the given data, you have:
A = 0.275 (trial's absorbance)

5. Next, calculate the equilibrium concentration of FeSCN2+. Since the stoichiometry is 1:1, the equilibrium concentration of FeSCN2+ is also 0.275 M.

6. Now, use the equilibrium concentration of FeSCN2+ to calculate the change in concentration of SCN–. Since 1 mole of SCN– is consumed for every 1 mole of FeSCN2+ formed, the change in [SCN–] is equal to the equilibrium concentration of FeSCN2+. Hence, Δ[SCN–] = [FeSCN2+] = 0.275 M.

7. To find the equilibrium concentration of SCN–, subtract the change in concentration from the initial concentration:
Equilibrium [SCN–] = Initial [SCN–] - Δ[SCN–] = 0.00020 M - 0.275 M = -0.27480 M.

Note: The negative sign here indicates that the concentration of SCN– is zero at equilibrium. Consequently, the reaction has consumed all the SCN– ions, leading to the formation of the FeSCN2+ complex.

So, the equilibrium concentration of SCN– is zero (0 M).