Full set of Notes with Fill-Ins - San Jose State University

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Chem 155 Unit 11 208 of 316Multi-component Analyses:Consider a cell containing two analytes with overlapping absorbancespectra:Consider two absorbing species A and B:6B i( 0.25A) i+( 0.75B) i4( 0.5A) i+( 0.5B) i( 0.75A) i+( 0.25B) iA i20200 250 300 350 400w iλ MAX (A) = 233 nm, FWHM = 60 nmε A (233) = 5.3 ε A (263) = 3.2 ε A (293) = 0.72λ MAX (B) = 293 nm, FWHM = 60 nmε B (233) = 0.72 ε B (263) = 3.2 ε B (293) = 5.3Page 208 of 316
Chem 155 Unit 11 209 of 316Given:ε A (233) = 5.3 ε A (293) = 0.72ε B (233) = 0.72 ε B (293) = 5.3A(233) = ε A (233)bC A + ε B (233)bC BA(293) = ε A (293)bC A + ε B (293)bC BA(233) = 2.55A(293) = 3.47What are the concentrations of A and B?Answer: C A = 0.4, C B = 0.6ε A1 ⋅C Aε A2 ⋅C AUnknown mixture of A and B42Wavelength / nm+ ε B1 ⋅C A B 1+ ε B2 ⋅C A B 2Absorbance0200 220 240 260 280 300 320 340 360 380 400III5.3⋅C A0.72⋅C A+ 0.72⋅C B 2.55+ 5.3⋅C B 3.47III II⋅0.725.3III0.72⋅0.72 ⋅C A + 5.3⋅0.72 ⋅C B 3.47⋅0.725.3 5.35.3III0.098⋅C A+ 0.72⋅C B 0.471Substitute CA into I and II (to check):I − III5.3⋅C A+ 0.72⋅C B 2.55−( )0.098⋅C A + 0.72⋅C B 0.4715.202⋅ C A 2.079C A2.0795.2020.40III5.3⋅0.42.55 − 5.3⋅0.4C B0.720.72⋅0.4+ 0.72⋅C B 2.55+ 5.3⋅C B 3.473.47 − 0.72⋅0.4C B5.30.600.60Page 209 of 316
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Full set of Notes with Fill-Ins - San Jose State University

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