# Chemistry DrBob

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–?

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1. 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 = ??

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2. THANK YOU SOOOO MUCH!!

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posted by Amy

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