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Beer's law

  1. Apr 5, 2014 #1
    1. The problem statement, all variables and given/known data

    photo.jpg



    3. The attempt at a solution

    So I'm at a loss here. I'm not given any units.... I know the concentration is C in Beer's law (awesome name btw). But what is my εb? Is it % transmittance?

    So then for A

    A= (.055)(.0001)
    What does this even mean?


    Ops this was supposed to go to the chemistry forums.
     
  2. jcsd
  3. Apr 5, 2014 #2
    What is the question they are asking? Incidentally, Beer is the guy's name.

    Let me guess the question. They give you the transmittance of an unknown solution, and they want you to find its concentration.

    Chet
     
  4. Apr 5, 2014 #3

    Borek

    User Avatar

    Staff: Mentor

    εb doesn't matter here.

    This is about constructing calibration curve, C vs A. Form the Beer's law (AKA Lambert-Beer's law) you know the dependence is linear (A=kC), so it is enough to determine the value of k - and it doesn't matter what are its components.
     
  5. Apr 5, 2014 #4
    Have your graphics program plot the concentration (y axis) as a function of the transmissivity (x axis). Have your graphics program use a semi-log scale for transmissivity). The graph will come out to be a straight line. Have your graphics program fit an equation to this straight line. This will express the concentration as a linear function of log of transmissivity. This will constitute your calibration.

    Chet
     
  6. Apr 5, 2014 #5
    That's where I kind of get lost.
    So my Cs are obviously given. Now I found a formula for A which was
    A=log(1/%T) so I found my absorbance numbers.

    Thing is, they increase with a lower T as shown by the above equation. I should get a line that has a y-intercept of approximately 0, my line goes the wrong way, it has negative slope :/.
     
  7. Apr 5, 2014 #6
    What does your plot of log T as a function of C look like? It should be a straight line.

    Chet
     
  8. Apr 6, 2014 #7
    It is a straight line but the slope is negative and it has a big y-intercept....

    I get absorbances to be the following
    A 1.3
    B .646
    C .335
    D .165
    E .082
     
  9. Apr 7, 2014 #8
    Is it possible to show us your graph?

    Chet
     
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