Why didn\'t you need to know the molar absorptivity of the equilibrium mixture o

Question

Why didn't you need to know the molar absorptivity of the equilibrium mixture or the path length for the calculations you did in this experiment? For each determination of the %T, approximately 5-mL of the solution was used. What would have the effect on your determination of the %T if samples of approximately 6-mL volumes had been Briefly explain. Frank and Oswalt report a molar absorptivity of 4700 L mol^-1 cm^-1 for the thiocyanatoiron(III) ion. What %T would you expect for a solution that is 1.0 Times 10 4M in thiocyanatoiron(lll) ion, if the path length is 1.00 cm? At a given temperature, the equilibrium constant for the system studied in this experiment is 1.40 Times 10^2. Suppose 100.0 mL of 2.00 Times 10^-3M KSCN was mixed with 100.0 ml of 2.00 Times 10^-3M Fe(NO_3)_3. At equilibrium, what molar concentration of the thiocyanatoiron(lll) ion would you expect?

Explanation / Answer

1. Nothing is given about the experiment.

2. There would be no such effect. Since A as well as %T depends only on concentration, not the number of moles so, it would not affect by the change of volume from 5 mL to 6 mL.

3. Absorbance A = abc
A = 4700*1*1 * 10^-4 = 0.47

A = 2- log %T = 0.47, hence T = 295.12

4. 100 mL of KSCN and 100 mL of ferric nitrate, after mixing forms 200 mL solution and the strength of both of them gets half. So, from 2.00 * 10^-3 M solution, it becomes 1.00 * 10^-3 each.

Fe3+ + SCN- <---> Fe(SCN)2+

I 1.00 * 10^-3 1.00 * 10^-3 0   

C -x -x +x

E ( 1.00 * 10^-3-x) ( 1.00 * 10^-3-x) +x

Hence, K = x/( 1.00 * 10^-3 -x)2 = 1.40 * 10^2

i.e. x = 1.40 * 10^2 (x2 - 2.00 * 10^-3)

i.e. 1.40 * 10^2 x2 - x -0.28 = 0

So, x = 0.0484

Hence, the molar concentration of the thiocyanatoiron(III) ion would be 0.0484 M

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