1. Why didn\'t you need to know the molar absorptivity of the equilibrium mixtur

Question

1. Why didn't you need to know the molar absorptivity of the equilibrium mixture or the path length for the calculation you did in this experiment?

2. For each determination of the absorbance approximately 5mL of the solution was used. What would have been the effect on your determination of the absorbance if samples of approximately 6mL volumes had been used?

3. Frank and Oswalt report a molar absorptivity of 4700 L mol-1 cm-1 for the thiocyanatoiron (III) ion. What absorbance would you expect for a solution that is 1.0x10^-4M in thiocyanatoiron(III) ion, if the path lenth is 1.00cm?

4. At a given temperature, the equilibrium constant for the system studied in this experiment is 1.40x10^2. Suppose 100.0mL of 2.00x10^-3M KSCN was mixed with 100.0mL of 2.00x10^-3M Fe(NO3)3. At equalibrium, what a molar concentration of the thipcyanatoiron(III) ion would you expect?

Explanation / Answer

1. If I am assuming it right (since no description of the experiment performed is given), then absorbance of the thiocyanatoiron(III) is measured at different concentrations, which can serve as several data points for the Beer's law plot for equation A = kcl [ k =molar absorptivity coefficient, c = concentration, l = path length]

Path length in all cases is used standardly as 1 cm. Hence we do not need to use it in the calculation.

And, molar absorptivity coefficient can be calculated from the slope of the linear plot of absorbance (A) vs. concentration (c). Hence, we do not need that value given separately either.

2. As we can see in the Beer-Lambert's law, the absorbance depends on the concentration of the solution, not the amount (moles) of substance you are dealing with. Hence, irrespective of the use of 5 mL or 6 mL of the same solution, the concentration remains the same for all cases, with no as such effect on absorbance measurements.

3. molar absorptivity (k) = 4700 L mol-1 cm-1

Concentration (c) = 1.0x10^-4M

cell length (l) = 1cm

Hence, absorbance (A) = kcl = 4700 L mol-1 cm-1 * 1.0x10^-4M * 1cm = 0.470

4.                             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

[concentration of both Fe3+ and SCN- gets to half of its original concentration since from 100 mL, each of them gets diluted to 200 mL of total volume]

Hence, equilibrium constant K = [Fe(SCN)2+]/[Fe3+][SCN-] = 1.40x10^2

                                  i.e. x/(1.00 * 10^-3 - x)2 = 1.40x10^2

                                   i.e. x = 1.40x10^2 x2 - 0.28 x + 1.4 * 10^-4

                                  i.e. 1.40x10^2 x2 - 1.28 x + 1.4 * 10^-4 = 0

                                   Hence, x = 1.11 * 10^-4

So, molar concentration of thiocyanatoiron(III) ion i.e. [Fe(SCN)2+] at equilibrium is 1.11 * 10^-4 M

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