421-821-chapter-13-14

421-821-chapter-13-14 - Application of UV/Vis Spectroscopy...

Info iconThis preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon
Application of UV/Vis Spectroscopy Common Problems : a) Mixtures : - blank samples often contain multiple absorbing species. - the absorbance is the sum of all the individual absorbances A= A 1 + A 2 +A 3 + … = ε 1 bc + ε 2 bc + ε 3 bc … - substances in both the blank and sample which absorb can be “blanked out” in both double and single beam spectrometers.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
If blank absorbance too high: - dilute the sample - use a different λ where the analyte absorbs more relative to the interference. - use a different method of separation But , If the blank absorbance is high, P o will decrease too much, the response will be slow and the results inaccurate λ scan of substance Large blank absorbance
Background image of page 2
b) Instrumental Deviations from Beer’s Law : - stray light (already discussed). - polychromatic light (more then a single λ ) since all instruments have a finite bandpass, a range of λ ’s p & are sent through the sample. ε may be different for each λ Deviation’s from Beer’s law at high concentration
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Illustration of Deviation from Beer’s law: Let us say that exactly 2 wavelengths of light were entering the sample λ = 254 nm ε 254 =10,000 λ = 255 nm ε 255 = 5,000 let P o = 1 at both λ ’s What happens to the Beer’s law plot as c increases? A = p A 254 + A 255 A = log P o /P (total) = log (P o 254 + P o 255 )/(P 254 + P 255 ) At the individual λ ’s: A 254 = ε 254 bc = log P o 254 /P 254 10 ε bc = P o 254 /P 254 P 254 = P o /10 ε bc = P o 10 - ε bc For both together: A = log (P o 254 + P o 255 )/(P o 254 10 - ε 254 bc + P o 255 10 - ε 255 bc)
Background image of page 4
Since P o 254 = P o 255 =1: A = 2.0/(P o 254 10 - ε 254 bc + P o 255 10 - ε 255 bc) A = 2.0/(10 -10,000x1.0xc + 10 -5000x1.0xc ) C A (actual) A(expected) 10 -6 M 0.0075 0.0075 10 -5 M 0.074 0.075 10 -4 M 0.068 0.75 10 -3 M 5.3 7.5 0 1 2 3 4 5 6 7 8 2 10 -4 4 10 -4 6 10 -4 8 10 -4 1 10 -3 Concentration (M) Negative Deviation A = bc ε The results are the same for more λ ’s of light. The situation is worse for greater differences in ε ’s (side of absorption peak, broad bandpass) Always need to do calibration curve! Can not assume linearity outside the range of linearity curve!
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
HIn K a H + + In - Red, λ =600nm colorless phenolphthalein: If solution is buffered, then pH is constant and [HIn] is related to absorbance. b) Chemical Deviations from Beer’s Law : - Molar absorptivity change in solutions more concentrated than 0.01M due to molecular interactions Beer’s law assumes species are independent electrolytes may also cause this problem ε is also affected by the index of refraction - association, dissociation, precipitation or reaction of analyte c in Beer’s law is the concentration of the absorbing species. commonly use the analytical concentration – concentration of all forms of the species.
Background image of page 6
C HIn [HIn] [In - ] [HIn]/[In - ] 10 -5 8.5x10 -7 9.2x10 -6 0.0924 10 -4 3.8x10 -5 6.2x10 -5 0.613 10 -3 7.3x10 -4 2.7x10 -4 2.70 But, if unbuffered solution, equilibrium will shift depending on total analyte concentration K a example: if K a = 10 -4 H + + In - HIn C HIn Expected Actual “Apparent” deviation since can be accounted for by chemical equilibrium
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Isosbestic point At the isosbestic point in spectra : A = ε b([HIn] + [In - ]) But, if unbuffered solution, equilibrium will shift depending on total analyte concentration example: if K a = 10 -4 HIn K a H + + In -
Background image of page 8
Image of page 9
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 38

421-821-chapter-13-14 - Application of UV/Vis Spectroscopy...

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online