A standard solution of FeSCN2+ is prepared by combining 9.00 mL of 0.200 M Fe(NO3)3 w/1.00 mL of 0.0020 M KSCN. The equilibrium concentration of FeSCN2+ ([FeSCN2+]std) for this standard solution is assumed to be
Fe^+3 + SCN^- ==> FeSCN^+2
(Fe^+3) in the final solution before reaction is 0.200 x (9 mL/10 mL) = ?? M
(SCN^-) in the final solution before reaction is 0.00200 x (1.00 mL/10.00 mL) = xx M
The limiting reagent must be SCN^-; therefore, the (FeSCN^+2) = (SCN^-) in the final solution after reaction
2. 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, the equilibrium concentration of SCN– will be
b = 1 or whatever number it is. I assume its a constant value anyway.
c = 0.00060
solve for a, the absorptivity constant.
Then A =abc for the original solution.
A = 0.520, you know a from the previous calculation, solve for c.
Mark, I can't make heads or tails of question 3 but if you will use A=abc as I did in question 2, I'm sure you will be able to work the problem yourself.
About question 3.
We're trying to find the concentration of Fe in M (mols/Liter).
So for the first trial, we have 0.0020M FE(NO3)3 at 5ml, 0.0020M KSCN at 2ml, and water at 3ml.
remember to convert ml to L
if we multiply 0.0020 mols Fe(NO3)3/1 L by 0.005L, we get 0.0020 mols Fe(NO3)3. to get rid of mols of Fe(NO3)3, we just multiply by 1 mol Fe/1 mol Fe(NO3)3 because there is only one. Then we are left with 0.00001 mol Fe. To get it in molarity, we're supposed to take the mols and divide it by L of solution. so 0.005L+0.002L+0.003L=0.01L
so if we take 0.00001mols Fe and divide that by 0.01 L soln, we get 0.001 M Fe. And that is the answer. It's right cuz I just solved it and got it right.
Thanks leslie :)
whats the answer to number 1?
The answer to number 1 is .0002
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