Absorbance and Concentration Question

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Discussion Overview

The discussion revolves around a homework problem involving the calculation of concentration and absorbance for the protein oxyhemoglobin at different wavelengths. Participants explore the application of the Beer-Lambert Law and address uncertainties regarding path length and concentration assumptions.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant expresses confusion about how to start the problem without knowing the path length or concentration values, questioning if concentrations remain the same at different wavelengths.
  • Another participant suggests assuming a path length of 1 cm and indicates that the problem is straightforward, emphasizing the need to solve for concentration using the given absorbance and extinction coefficient.
  • A different participant confirms the use of the equation A = εc for calculating concentration and states that each wavelength has its own absorbance equation for oxyhemoglobin.
  • One participant seeks clarification on whether the path length is always assumed to be 1 cm if not stated, noting previous problems with different path lengths.
  • Another participant responds that the path length is not always 1 cm and that assumptions may vary if not specified, suggesting that it should be part of the answer.
  • Participants calculate the concentration at 415 nm as 1.87 x 10-6 M and express concern about whether this concentration is too low, with a suggestion to express it in ppm.
  • For the absorbance at 276 nm, one participant calculates it to be 0.064 using the concentration found earlier.

Areas of Agreement / Disagreement

Participants generally agree on the application of the Beer-Lambert Law and the calculations involved, but there is uncertainty regarding the assumption of path length and whether concentrations remain constant across different wavelengths. The discussion remains unresolved on these points.

Contextual Notes

Participants note the absence of specific path length information in the problem, leading to different assumptions. There is also a discussion about the implications of low concentration values and their expression in different units.

fuzzy
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Homework Statement


The protein oxyhemoglobin has the following molar absorbance coefficients:
\epsilon276 nm = 3.44 x10^{4}M^{-1}cm^{-1}
\epsilon415 nm = 1.25 x 10^{5}M^{-1}cm^{-1}
If a solutioin of oxyhemoglobin has an absorbance at 415nm of 0.234,
a) What is the concentration at 415nm?
b) What is the absorbance at 276 nm?

Homework Equations


A=\epsilonbc, where A is absorbance, \epsilon is the molar absorbance coefficient in M^{-1}cm^{-1}, b is the length of light path through the cell in cm, and c is concentration in mol/L.

(I don't know if this one is needed...)
\frac{As}{Au}=\frac{Cs}{Cu}
where As and Au are absorbance of the standard and unknown and Cs and Cu are concentration of the standard and unknown.


The Attempt at a Solution


Can someone give me a hint at where to start here? I can't see how to do the problem without at least 1 of the concentration values or a light path length. :confused: Do I assume that the concentrations remain the same at 276 and 415?
 
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Assume that the path length, b, is the same for both samples and is 1 cm. You are given the absorbance, the extinction coefficient and you are asked to solve for 'c'. Seems pretty straightforward.

The second part asks you to substitute the concentration and molar extinction coefficient at a different wavelength and determine the absorbance at that wavelength... again, fairly straightforward.

Do I assume that the concentrations remain the same at 276 and 415?

Will the concentration change when you shine a different color light on it?
 
Check your textbook about how to handle quantitative absorbance measurements. You probably need A=e*C, where A is absorbance, e is your epsilon, C is concentration; but I could be mistaken. Each wavelength has it own A=e*C for the oxyhemoglobin.

... Yes, you DO assume the concentration remains the same at both wavelengths.

For part (a), subsituting the values given for the 415 nm,
0.234 = 1.25*10^5 * C,
Find C.

As I say, this stuff is a bit murky now, so recheck your textbook.
 
Thank you for the replies.

Assume that the path length, b, is the same for both samples and is 1 cm.
So this is always the case if it's not otherwise stated? Because in other problems it was given as 0.5cm, so I was confused a bit.

Other than that I think I get it now :D
So I get
a)
c=\frac{A}{\epsilon}

c=\frac{0.234}{1.25x10^{5}}

c=1.87x10^{-6}M
This isn't too low for a concentration?

b)
A=\epsilonc
A=(3.44x10^{4})(1.87x10^{-6})
A=0.064
 
fuzzy said:
...So this is always the case (path length = 1 cm)if it's not otherwise stated? Because in other problems it was given as 0.5cm, so I was confused a bit.

No, it is not always the case. If the path length information is absent (an oversight in my opinion), you can assume anything. Assuming 1 cm. (a standard photometric cell dimension) must be part of your answer.

Other than that I think I get it now :D
So I get
a)
c=\frac{A}{\epsilon}

c=\frac{0.234}{1.25x10^{5}}

c=1.87x10^{-6}M
This isn't too low for a concentration?

It is fairly low but you can express it as 1.87 ppm as well. ppm levels are determined routinely.

b)
A=\epsilonc
A=(3.44x10^{4})(1.87x10^{-6})
A=0.064

Looks good.
 

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