# Chemical Forums

## Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Soraya on March 27, 2010, 04:00:06 AM

Title: Beer Lambert's law but confusing....
Post by: Soraya on March 27, 2010, 04:00:06 AM
Okay I just wanted to see if all I need to do with this problem is apply Beer-Lambert's law. I am having particular problem with the units of the extinction coefficient. I'm used to them being in M. Here is the problem:

20mg of BSA in freeze dried form is dissolved in 2ml of water. 0.04ml of this is added to 0.96ml water. The A280 of the resulting solution relative to the water blank was 0.237. The extinction coefficient for BSA is 0.66 (mg/ml)-1,  cm-1.

??? I'm very lost..

a. calculate the concentration of the original solution of BSA
b. comment on the answer - why is there a discrepancy? If so what could be the reason?
Title: Re: Beer Lambert's law but confusing....
Post by: Francopper on March 28, 2010, 01:14:19 PM
Hi. As the absorbance is dimensionless, and the law states that A=EbC, the extinction coefficient needs to have the appropriate units so that they can be simplified when solving the equation. When the concentrations are given in molar, M, the extinction coefficient units have to be 1/(M.cm). Since M is mol/L, 1/(M.cm)=L/(mol.cm).
In the case of this problem instead of being mol/L it is mg/mL, but that should not confuse you since there's no difference at all when calculating. See:
a. 0.237 = 0.66(mg/mL)-1.cm-1. b. C = 0.66 mL/(mg.cm) . b . C, assuming b=1 cm (that should be part of the data), then it is
0.237 = 0.66 mL/mg . C, ergo C = 0.359 mg/mL.
The original solution was diluted 24 times (0.96 mL/0.04 mL = 24), so the original concentration would be C=0.359 mg/mL * 24 = 8.616 mg/mL.
b. To see if there's a discrepancy with the real concentration, calculate the concentration using the mass of BSA and the volume of water used. C= 20mg/2mL = 10 mg/mL. There's a discrepancy because, as you may know, Beer-Lambert's law does not work under any concentrations, that's to say the lineal dependency of absorbance vs. concentration works in a certain range of concentrations; when too diluted or too concentrated the law is useless. This has to be such case due to the discrepancy found. At least that's what comes to my mind.
Cheers