UV spec

# UV spec - CH 13-14: UV-Visible Spectroscopy CH 13: 13-1,...

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CH 13-14: UV-Visible Spectroscopy CH 13: 13-1, 13-2, 13-5, 13-7, 13-26 CH 14: 14-1, 14-6, 14-10, 14-15 Special Project #3 Due Friday October 29

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A closer look at the Beer-Lambert Law Consider an infinitesimally thin slice (1 molecule thin) of a solution containing an absorbing species. Let the thin slice be denoted by S, and the number of absorbing species in the slice dn. S
A closer look at the Beer-Lambert Law S Assume the absorbing species is a disc and the total area of all of the absorbers in S has an area dS. The ratio of absorbing area to total area is dS/S. If we assume that every time a photon hits an absorber it is absorbed, then dS/S is the fraction of photons is absorbed in the slice.

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A closer look at the Beer-Lambert Law S Now let P be the Power entering the slice and dP the power absorbed. -dP/P is also the fraction of photons absorbed, thus -dP/P = dS/S
A closer look at the Beer-Lambert Law S To relate this expression to the number of absorbers in area S (dn), dS can be replaced with adn, where a is called the capture cross section. -dP/P = dS/S -dP/P = adn/S

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A closer look at the Beer-Lambert Law S -dP/P = adn/S P 0 P 0 n -ln P – ln P 0 = an/S -ln P/P 0 = an/S log P 0 /P = an/2.303S
A closer look at the Beer-Lambert Law S n is the total number of absorbers in the entire block. We can remove the 2-dimensional area of the slice, S by noting that the 3-dimensional volume of the block is: log P 0 /P = an/2.303S S = V cm 3 /b cm

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A closer look at the Beer-Lambert Law S log P 0 /P = anb/2.303V n/V is absorbers/cm 3 , which can be converted to mols/L using N A and 1000 cm 3 /L . log P 0 /P = 6.02 x 10 23 abc/(2.303 x 1000) Collecting the constants yields the Beer-Lambert Law: logP 0 /P = ε bc = A
What if you have two species in solution? Key Concept: Beer’s Law is additive. Problem: The Sb 3+ and Mn 2+ absorbances overlap. How can we determine the concentration of the Mn 2+ if the Sb 3+ is absorbing at the same wavelengths as the Mn 2+ ?

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What if you have two species in solution? Key Concept: Beer’s Law is additive. A total, λ 1 = A analyte 1, λ 1 + A analyte 2, λ 1 + ………. . A total, λ 1 = ε analyte 1, λ 1 bC analyte 1, λ 1 + ε analyte 2, λ 1 bC analyte 2, λ 1 + …
Concept Te\$t Consider the following data acquired on a sample containing Mn 2+ and Sb 3+ . A total, 475 nm ε Mn2+, 475 nm ε Sb3+, 475 nm ε Mn2+, 600 nm ε Sb3+, 600 nm A total, 600 nm 0.20 0.55 1350 M -1 cm -1 2500 M -1 cm -1 4100 M -1 cm -1 1100 M -1 cm -1 The concentrations of Sb 3+ and Mn 2+ are (b = 1 cm): (B) ~0.5 mM, ~0.13 mM (C) ~1 mM, ~2 mM (D) ~10 mM, ~130 mM (A) ~10 μ M, ~130 μ M

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There are two unknowns so we need two equations. A total, 475 nm = ε Sb3+, 475 nm bC Sb3+ + ε Mn2+, 475 nm bC Mn2+ A total, 600 nm = ε Sb3+, 600 nm bC Sb3+ + ε Mn2+, 600 nm bC Mn2+ 0.20 = 2500 x C Sb3+ + 1350 x C Mn2+ 0.55 = 1100 x C Sb3+ + 4100 x C Mn2+ 457 nm 600 nm C Mn2+ = (0.55 - 1100C Sb3+ )/4100 Rearranging This is inserted into the A 475 nm equation and solved for C Sb3+ .
Key Concept: The Isosbestic Point A specific wavelength at which two chemical species have the same extinction coefficient. (From Greek isos meaning same and sbestos meaning extinguish.) As one species in converted to the other, the absorbance at the isosbestic point is constant.

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## This note was uploaded on 02/23/2011 for the course CHEM 4181 at Colorado.

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UV spec - CH 13-14: UV-Visible Spectroscopy CH 13: 13-1,...

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