4. A 35.00 ml sample of water run-off from the BSU coal-pile was collected and t

Question

4. A 35.00 ml sample of water run-off from the BSU coal-pile was collected and treated with an excess of KSCN to complex the Fe(I) and form a red-complex at 450 nm. After dilution with DI water to 250.00 ml, it was analyzed on a double beam spectrophotometer, and it ga a reading of 63.5%T at 450 nm in a 1.50 cm cell. A standard was also prepared by taking 10.00 ml of a 8.2 x 104 M FeCI) standard solution, adding excess KSCN, diluting to 100.0 ml, and analysis in the same 1.50 cm cell, giving a 43.2% T What is the concentration in ppm of the Fe" in the original run-off solution. (12 points) 5. There are 4 regions of Beer's Law where it fails: A, B, C, D. In your own words, pleas describe and show using the Beer's Law equation the causes of these deviations: (12 points) Cone

Explanation / Answer

4) The absorbance (A) of a solution is related to the concentration (c) of the solution by the equation

A = *c*l where = molar absorptivity of the solution and l = path length of the solution.

Let be the molar absorptivity of the complex formed between Fe3+ and KSCN and c be the concentration of Fe3+ in the diluted sample. The percent trasmittance is give, i.e, %T = 63.5 and absorbance A = 2 – log (%T) = 2 – log (63.5) = 0.1972. Therefore,

0.1972 = *c*(1.50 cm) …..(1)

Again, 10.00 mL of 8.2*10-4 M standard Fe3+ was diluted with excess KSCN giving a total volume of 100.00 mL; therefore, concentration of Fe3+ in the diluted solution is (10.00 mL)*(8.2*10-4 M)/(100.00 mL) = 8.2*10-5 M. The percent transmittance is given as %T = 43.2; therefore, A = 2 – log (%T) = 2 – log (43.2) = 0.3645. Thus,

0.3645 = *(8.2*10-5 M)*(1.50 cm) ……(2)

Dividing (2) by (1) gives

0.3645/0.1972 = (8.2*10-5 M)/c

=====> 1.8484 = (8.2*10-5 M)/c

=====> c = (8.2*10-5 M)/(1.8484) = 4.4363*10-5 M

The concentration of Fe3+ in the diluted sample is 4.4363*10-5 M; 35.00 mL of run off was diluted to 250.00 mL, hence, the dilution factor = (250.00 mL)/(35.00 mL) = 7.1429. The concentration of Fe3+ in the run-off is (4.4363*10-5 M)*(dilution factor) = (4.4363*10-5 M)*(7.1429) = 3.1688*10-4 M.

Ppm of Fe3+ in the original run-off = (3.1688*10-4 M)*(molar mass of Fe3+)*(1000 mg/1 g) = (3.1688*10-4 M)*(1 mol.L-1/1 M)*(55.845 g/1 mol)*(1000 mg/1 g) = 17.6962 mg/L 17.70 mg/L (ans).

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