Question

Composition of Solution for Determing k

Test Tube	0.0025 M Fe(NO3)3	0.0025 M KSCN	0.10 M HNO3	Results of Spectro	Absorbance
6	1.0 ml	1.0 ml	5.0 ml	--->	.109
7	1.0 ml	1.5 ml	4.5 ml		.183
8	1.0 ml	2.0 ml	4.0 ml		.278
9	1.0 ml	2.5 ml	3.5 ml		.295
10	1.0 ml	3.0 ml	3.0 ml		.337

Test Tube	Starting Fe3+	Starting SCN-	Equlibrium Fe(SCN)2+	Eq. Fe3+	Eq. SCN-
6
7
8
9
10

Complete the above chart with all concentrations in M. * Coulmns 1 and 2 dilution occured for both Fe(NO3)3 "Fe3+" and KSCN "SCN-", Column 3 determine FeSCN2+ using A=kc, Columns 4 and 5 create and ICE chart for each test tube, Column 6 determine the value of K for each test tube of Equilibrium for Fe(SCN)2+, Fe3+, SCN-

Answer 1

In order to do this, I need the value of E (or k as you put here). With that value, I can calculate concentration of FeSCN2+ and then, the other values.

In another problem, I solved earlier, I check that the E = 3550. I'm gonna use this, to do calculation example here, so you can have an idea of how it's done this. If your E value is different, just change that value in the calculations to get the real results.

Assuming E = 3550 (M cm)^-1 and b = 1 cm (cell length), let's calculate concentration:
C = A / Eb
C7 = 0.183 / 3550 = 5.15x10^-5 M

Now, let's calculate the innitial concentrations of the reactants involved:
[Fe³⁺] = 0.0025 * (1/7) = 3.57x10^-4 M
[SCN^-] = 0.0025 * (1.5/7) = 5.35x10^-4 M

The ICE for this reaction:
r: Fe³⁺ + SCN^- <--------> FeSCN²⁺
i. 3.57x10^-4  5.35x10^-4   0
e. (3.57x10^-4-5.15x10^-5) (5.35x10^-4-5.15x10^-5) (5.15x10^-5)

Kc = 5.15x10^-5 / (3.57x10^-4-5.15x10^-5) * (5.35x10^-4-5.15x10^-5)
Kc = 348.66

Fe³⁺]eq = 3.57x10^-4 - 5.15x10^-5 = 3.06x10^-4 M
[SCN]eq = 5.35x10^-4 - 5.15x10^-5 = 4.84x10^-4 M

That'w how you do this. Do the same with solution 6, 8, 9 and 10.

Hope this helps