Chem
posted by Mark on .
There are three questions in particular that I am having problems with. Here they are
1. 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
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
3.The concentration and volume of reagents combined in each trial are listed in the table below:
Trial 00020 M 00020 M H2O(mL)
Fe(NO3)3 (mL) KSCN (mL)
1 5 2 3
2 5 3 2
3 5 4 1
4 5 5 0
Initially (right after combining and before any reaction occurs) the concentration of Fe3+ in trial 1 is

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
A =abc
A=0.275
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