posted by UCI student on .
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 0.510. If a trial's absorbance is measured to be 0.250 and its initial concentration of SCN− was 0.00050 M, the equilibrium concentration of SCN− will be...
ive tried for a while, but i keep on not getting the correct answer.
you don't need to just give it to me, but could someone show me the steps and reasoning needed to get the proper answer?
i think you have omitted part of the procedure but the following may get you started. For the standard:
Fe^+3 + SCN^- ==> FeSCN^+2
We had 9.00 mL 0.200 M Fe^+3.
1.00 mL 0.002 M SCN^-
Clearly, SCN^- is the limiting reagent and the iron(III) is in excess.
moles FeSCN^- formed = M x L = 0.002 M x 0.001 L = 2 x 10^-6 moles.
What is the concn? M = moles/L. You have the 2 x 10^-6 moles in 10 mL (it was diluted 1:10) so the M = 2 x 10^-6/0.010 = 2 x 10^-4 M.
The next step is to evaluare a in the equation A = abc where A = absorbanace, a is the absorptivity constant, b is the cell length but we will ignore that since you are using the same cell (or at least the same length cell) in both standard and sample alike, and c is the concentration. [Technically, the constant a is epsilon and is the molar absorptivity constant when c is measured in moles/L but we'll just continue to call it a).
So A = a*c
0.510 = a*2 x 10^-4
a = 2.55 x 10^3 but check confirm this yourself.
Now we get to the point that I don't know how the trial run was treated. hope this is enough to get you started.
yea, that's about as far as i got. then i tried an ice table, but im not getting the right answers still
You missed my first pick-up line. I don't know how the "trial runs" were treated. I assume you mean by "trial's absorbance" the absorbance of the unknown sample. How was the sample treated before the absorbance was measured?
turns out I was just formatting my answer wrong each time, I get it. nvm