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Offline Halogen876

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Beer's Law Question
« on: December 09, 2016, 03:51:56 PM »
Hello,

I am having trouble with a Beer's Law problem. Here is the question:

A spectrophotometer is set to a given wavelength but actually passes two separate wavelengths, A and B, each with the same intensity. The sample absorbs both wavelengths, but to different extents. The molar absorptivity of A is 6.00E3 and the molar absorptivity of B is 1.00E4. The path length is 1.00cm.

(a) If c=1.00E-4M, what is the actual absorbance? (Answer = 0.755)
(b) What is the ideal absorbance if the molar absorptivity is taken as the average of the two? (Answer = 0.8 )
(c) What concentration of absorber will produce an actual absorbance of 0.500? (Answer = 6.49E-5 M)
(d) What concentration would be needed in the ideal case? (Answer = 6.25E-5 M)

I have solved part (b) no problem using Beer's Law. However, I am stuck on the other 3 parts. I think there may be ratios involved but I am unsure how to set it up. I would really appreciate any assistance. Thanks!

Offline Borek

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Re: Beer's Law Question
« Reply #1 on: December 09, 2016, 04:51:42 PM »
I would assume amount of light that got through the solution (and was measured) is a sum of both individual signals.

Not that I am convinced it is the right approach,
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Offline Halogen876

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Re: Beer's Law Question
« Reply #2 on: December 09, 2016, 05:10:59 PM »
Thank you for your reply. I had thoughtof approaching it that way. When I do that, this is what I get:

A = 6.00E3 x 1.00E-4 + 1.00E4 x 1.00E-4 = 1.6

...The answer is supposed to be 0.755 though so I must be missing something... Any more ideas would be much appreciated! Thanks!

Offline mjc123

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Re: Beer's Law Question
« Reply #3 on: December 12, 2016, 04:58:34 AM »
What fraction of the A light is transmitted? What fraction of the B light? What overall fraction of the total light is transmitted?

Offline Halogen876

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Re: Beer's Law Question
« Reply #4 on: December 14, 2016, 09:23:34 AM »
Thank you for your response. I have solved part (d) of the problem using my answer to part (b) so I am left with parts (a) and (c). I tried doing a ration of the 2 molar absorptivities (6E3 / 1E4) . I got 0.6 as the ratio. By using the ratio with the concentartion, I got an answer close to the given answer but not exact. Not sure if I am missing something. Here is what I did:

6E3x(.6x1E-4) + 1E4(.4X1E-4) = 0.76

The given answer is 0.755 though so I'm not sure why I'm not exactly on that.

Is this what you mean by the fraction of the light? I think I may be on the right track but maybe missing something. Thanks!


Offline mjc123

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Re: Beer's Law Question
« Reply #5 on: December 14, 2016, 12:52:52 PM »
No, that's not right at all. The concentration is 1.00e-4. The absorbances are not additive because of the log scale. Transmitted intensity is additive.
Suppose the spectrometer had two slits, one of which passed light of wavelength A, the other B, and the transmitted light from both, after passing through the sample, was combined at the detector. Suppose the incident intensity at each slit is I0.
What is the total intensity at the detector with no sample in place?
With the sample in place, what is the intensity of the transmitted A light?
And of the transmitted B light?
What is the total intensity of transmitted light reaching the detector?
What is the apparent absorbance?

Offline Halogen876

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Re: Beer's Law Question
« Reply #6 on: December 14, 2016, 06:20:40 PM »
Thanks. I think I've figured out part (a) now. So I found each individual absorbance (0.6 and 1.00). I found the two transmittance values for those (0.25119 and 0.1). Added together those give 0.35119. Divided by 2 (due to the equal intensity) I get 0.176. The absorbance that comes from that is 0.755 (the correct answer). I hope that is all right!

I am still stuck on part (c). I tried setting it up as a ratio:

0.5/0.755 = x/1E-4 but this did not work. Not sure why that isn't appropriate to do here since absorbance and concentration are directly proportional. Thanks!

Offline mjc123

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Re: Beer's Law Question
« Reply #7 on: December 15, 2016, 04:54:21 AM »
I would have tried that, but it's not quite right because the two absorbances are not additive - you have to go  via the transmittance.

Offline Halogen876

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Re: Beer's Law Question
« Reply #8 on: December 15, 2016, 01:27:13 PM »
This is what I have tried now but I am getting stuck at a certian point:

Working backwards from how I solved part (a):

A-0.5 (given) so T=10-0.5=0.316

0.316x2=0.632 (to account for the two wavelengths of equal intensity)

0.632=TA+TB

-logTA=6E3xC
-logTB=1E4XC

-logTA/6E3=-logTB/1E4

log(0.632-TB)/6E3=log(TB)/1E4

I now have an equation with only one unknown but I don't know how to solve for it, given the complex log expression I have. I don't know if my work is correct or not up to this point but, if it is, I don't know where to go from here.

Offline mjc123

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Re: Beer's Law Question
« Reply #9 on: December 16, 2016, 08:22:57 AM »
You can't solve it analytically. You have to do it numerically. Set up a formula in Excel for calculating A from TB. Insert various values of TB till you get the right answer (or use the Solver tool).

Offline Halogen876

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Re: Beer's Law Question
« Reply #10 on: December 16, 2016, 09:20:09 AM »
Thanks for clarifying that. I think I will be ok from here. Thank you so much for all your help with this and everything else. You have helped me a lot over the past couple of years and I appreciate it all very much.

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