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Determine extinction coefficients in glass for Fe2+/Fe3+

  1. Aug 28, 2013 #1
    Hello everybody,

    I want to determine the extinction coefficients of Fe2+ and Fe3+ in glass.
    There are literature data (e.g. Weyl's book "coloured glass"), so I know what kind of curves I should expect. As I am studying a slightly different soda-lime-silicate system, I want to recalculate the curves.

    I have glasses with different Fe2+/Fe3+ ratio and total iron concentration [Fe].
    I have measured the absorbance of two glasses and I have solved a simple two equations system starting from Lambert-Beer law: A = Ʃ εCd
    where A is absorption, ε is the extinction coefficient, C is the concentration and d is the thickness of my glass.

    Unfortunately, I get negative values in one of the two extinction coefficient curves. Obviously, this doesn't make sense.

    Anyone sees what is wrong in my reasoning? I can't figure it out.

    Thank you very much in advance,
    Mark

    PS For a close look at the system I have made, see the attachment (.pdf)
     

    Attached Files:

  2. jcsd
  3. Aug 28, 2013 #2

    chemisttree

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    Try solving for Fe+3 in terms of Fe+2 and total Fe.
     
  4. Aug 29, 2013 #3
    Thank you chemisttree for replying.
    However, I don't get what you mean. How can I solve in terms of total Fe?
    In the system I have the concentrations of the two absorbing species:
    Fe3+ (CFe3+) and Fe2+ (CFe2+)

    Where should total Fe appear in the equations?
     
  5. Aug 29, 2013 #4

    Borek

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    You said you know total Fe for the glass, so you can write an equation for the mass balance:

    Fe2+ + Fe3+ = Ftotal
     
  6. Aug 29, 2013 #5
    You are right! I understand.
    But even if I substitute CFe3+ with (CFetotal - CFe2+), I don't see how it would solve the problem. At the end it is always the same value.
     
  7. Aug 29, 2013 #6
    the unknown variable are the ε for the two ionic species.
    so adding the mass balance equation does not add any value.
     
  8. Aug 29, 2013 #7

    chemisttree

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    Sooo, you're not even going to try it my way?
     
  9. Aug 30, 2013 #8
    I am sorry chemisttree, maybe I hadn't explain well myself. I didn't want to be disrespectful.
    I have tried to use your advice. But I don't see how.
    Substituting ferric concentration with the subtraction of ferrous from total iron concentration, the final solution remain unchanged.

    I attached the spectra of the two glasses I am using. Also, I have plot quickly (read as "I haven't add units and axis names") the absorption coefficients I obtained for Fe2+ and Fe3+. You see that εFe3+ is negative, which doesn't have any physical meaning.

    Below you find the concentration data for both GlassA and GlassB.

    GlassA:
    CFe2+ = 0.056 wt%
    CFe3+ = 0.125 wt%
    CFetotal = 0.181 wt%

    GlassB:
    CFe2+ = 0.127 wt%
    CFe3+ = 0.285 wt%
    CFetotal = 0.412 wt%

    If I substitute in GlassA CFe3+ = CFetotal - CFe2+, I get the same value: 0.181 - 0.056 = 0.125 wt%.

    I would appreciate if you could keep helping me.
    Thank you.
     

    Attached Files:

  10. Sep 4, 2013 #9
    Nobody has an idea on what's wrong?
     
  11. Sep 4, 2013 #10

    chemisttree

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    For the attached spectra, what value of extinction coefficient did you use for Fe+2 and Fe+3? Did you hold the extinction coefficient for Fe+2 constant? Or did it vary with wavelength?
     
  12. Sep 4, 2013 #11
    they both vary with wavelength. In the 2-equations system I have attached to my first thread, the unknowns are the two extinction coefficients.
     
  13. Sep 4, 2013 #12

    chemisttree

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    So you have the spectra in the form of wavelength and absorbance? Can't help you without raw data.
     
  14. Sep 5, 2013 #13
    Yes, my data are absorbance as a function of wavelength. NOw I don't have access to the hard drive where data data are. I will post the raw data later on today so you can have a look.
    thank you again!
     
  15. Sep 6, 2013 #14
    sorry for the delay. here is the file with the absorbance for both glasses.
     

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