Composition of Solution for Determing k Complete the above chart with all concen

Question

Composition of Solution for Determing k

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-

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

Explanation / Answer

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

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