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Chp2_BeersExp[1]

# Chp2_BeersExp[1] - (a note all values now pertain only to...

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Beer’s Law Problem See problem 2.40, page 60 of text for a statement of the problem. In this problem, we have an example of the solute AND the solvent both absorbing the 520nm light. In this case, Beer’s Law becomes: solvent solvent solute so A C A A A + = + = ε ln Thus, for solution #1: we have: 0.449 = A solute + 0.012 or that A solute = 0.437 . Likewise for solution #2, we have A solute = 0.356 . In the lecture overheads, I called these “corrected” values of absorbance. We now use A solute = ε lC to answer the questions posed by the problem. (a) Calculate the molar absorbtivity assuming a 1cm path length. (b) Determine C for the second solution
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Unformatted text preview: (a) note: all values now pertain only to solute and using “exponential” notation. 3 57 . 1 4 78 . 2 00 . 1 437 . 1 1 1 1 + =-× = = = e e C A C A (b) M e e A C C A 4 27 . 2 3 57 . 1 00 . 1 356 . 2 2 2 2-= + × = = = Another approach: here, we calculate C2 without first calculating the molar absorbtivity. 2 1 2 1 2 1 C C C C A A = = NOTE: This is NOT C 1 V 1 = C 2 V 2 Thus: 4 26 . 2 437 . 356 . 4 78 . 2 1 2 1 2-= ×-= × = e e A A C C as before....
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