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 0.550. If a trial's absorbance is measured to be 0.350 and its initial concentration of SCN– was 0.0010 M, the equilibrium concentration of SCN– will be what?
Chemistry - DrBob222, Saturday, June 28, 2008 at 1:01am
Fe(NO3)3 + SCN^- ==> FeSCN^+2
mols Fe(NO3)3 = M x L
mols SCN^- = M x L.
Obviously, SCN^- is the limiting reagent. (SCN^-) = mols/L total solution OR, if my arithmetic is ok, (FeSCN^+2) = 0.0002. Check me out on that.
Then A = abc.
You know A = 0.550 for the standard. You know c = 0.0002 M. You can substitute any number for b (the cell length) as long as you use the same cell length for all calculations (I assume you used the same cell length for all measurements). That leaves a as the only unknown, which in this case is the molar absorptivity constant (which is usually written as epsilon but I can't write an epsilon with on this board). Then for the trial's value,
A = abc. You know A. You know a (from the previous calculation), you know b and you can calculate c, the equilibrium concn of the FeSCN^+2. Then set up an ICE chart to determine how much of the SCN remains. It will be 0.001 - amount FeSCN^+2 formed. Check my thinking.
Chemistry - Jennifer, Saturday, June 28, 2008 at 8:29pm
Hey I know this isn't an answer, but i have the same problem and i'm having trouble with it. If you figured it out, could you help me with the answer?
Chemistry - DrBob222, Saturday, June 28, 2008 at 11:29pm
Jenifer--I told Tom how to work the problem. If yours is like it, just follow the same instructions. However, please repost it. Just above this are a couple others by UN. Look at those, also.
Chemistry - Jennifer, Sunday, June 29, 2008 at 4:02pm
got it right!
Chemistry - Christie, Tuesday, October 7, 2008 at 4:51pm
For some reason, this isn't working out for me. Can you further help me? I am very confused.