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Lecture

16spectrophotometry3.pdf

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Department
Chemistry
Course Code
ENCH 213
Professor
Diane Beauchemin

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Applications of spectrophotometry Analysis of a mixture Spectrophotometric titration Measuring an equilibrium constant The method of continuous variation Flow injection analysis Immunoassays andAptamers Sensors based on luminescence quenching Analysis of a mixture without separation of its components • two cases ◦ spectra of pure components have substantial overlap ◦ regions exist where each component contributes the most • A= ε b[X] + ε b[Y] +… X Y • For n unknowns, need at least n wavelengths • to used n equations to solve for unknowns -1 -1 At 510 nm, the molar absorptivities for two complexes (A and B) are 36,400 and 5250 M cm , respectively. The absorbance in a 1.00 mm cuvet of a solution with [A] = 1.00 x 10 M and [B] = 2.00 x 10 M would be a) 0.469. b) 4.69. c) 0.780. Analysis of a mixture when individual spectra overlap • measure the absorbance of standard solutions at >n λ’s ◦ at least n+1 λ’s to increase accuracy • find the molar absorptivity at each λ ◦ A = εX [X] + ε b[Y] calc b guess Y guess 2 ◦ least-square analysis to minimize (A -A )calc m ▪ can be done using SOLVER tool in Excel (not in exam) Analysis of a mixture when individual spectra are resolved • determine ε of each species ◦ use enough standards to encompass the sample conc. and verify that Beer’s Law is obeyed over this range • measureA at each λ ◦ A' = ε' bXX] + ε' bYY] at λ' ◦ A'' = ε'' X[X] + ε'' Y[Y] at λ'' • solve → choose n λ’s where the spectrum of individual species is different (using standard solutions) Determination of Pd(II) and Au(III) with methromeprazine • Pd complex has absorption maximum at 480 nm, with ε = 3.55x10 M cm 3 -1 -1 4 -1 -1 • Au complex has absorption maximum at 635 nm, with ε = 1.45x10 M cm • ε of Pd at 635 nm= 5.64x10 M cm 2 -1 -1 • ε ofAu at 480 nm= 2.9x10 M cm 3 -1 -1 Assume all been • For sample,A 480nm 0.533;A 635nm= 0.590 blank subtracted • 1.00-cm cell used for all measurements How to solve At 480 nm: 3 3 • 0.533=(3.55x10 )(1.00)c +(2.9Pd10 )(1.00)c Au • cPd 0.533-[(2.96x10 )c ]/(3.Aux10 ) 3 At 635 nm: 2 4 • 0.590=(5.64x10 )(1.00)c +(1.4Pd10 )(1.00)c Au • 0.590=(5.64x10 )(0.533-2.96x10 c )/(3.55x10 ) +(1.45x10 )c 4 Au Au • cAu 0.0847- 4.70x10 c +(1.45Au0 )c 4 Au 4 -5 • cAu (0.590 - 0.0847)/(1.403x10 )= 3.60x10 M • c = 0.533-[(2.96x10 ) (3.60x10 )]/(3.55x10 )= 1.20x10 M -4 Pd When measured in a 1.00 cm cuvet, a 8.50 x 10 M solution of A exhibited absorbances of 0.129 and -5 0.764 at 475 nm and 700 nm, respectively. A 4.65 x 10 M solution of B gave absorbances of 0.567 and 0.083 at 475 nm and 700 nm, respectively. Both species were dissolved in the same solvent, and the solvent’s absorbance was 0.005 and 0.000 at 475 nm and 700 nm, respectively, in either a 1 or 1.25 cm cuvet. Calculate the concentrations of A and B in a solution that yielded the following absorbance data in a 1.25 cm cuvet: 0.502 at 475 nm and 0.912 at 700 nm. -5 -5 a) cB= 7.65 x 10 M and c = 2.40 A 10 M b) c B 2.37 x 10 M and c = 7.65 xA10 M -5 -5 -5 c) cB= 4.65 x 10 M and c = 8.50 A 10 M 1. find molar absorptivity for each analyte at each wavelength E A475= (0.129-0.005)/(8.50x10 M x 1.00cm) = 145 M cm 8.8 -1 -1 -5 -1 -1 E A700= (0.764)/(8.50x10 M x 1.00cm) = 898 M cm 8 -5 -1 -1 E B 475 (0.567-0.005)/(4.65x10 M x 1.00cm) = 120 M cm 86 E B 700 (0.083)/(4.65x10 M x 1.00cm) = 178 4.9 8M cm -1 2. make equations i) at 475 nm: 0.502-0.005 =A +A = EBb c + B b c A A A8x1.25B B B12086x1.25c A B ii) at 700 nm: 0.912 = … etc. Isosbestic points • if a chemical reaction occurs where X is converted to Y, the spectra recorded during the reaction will cross at the same points where those of the pure X and Y cross • molar absorptivity is the same • isosbestic point: ◦ ε =Xε Y ◦ c +Xc = Yonstant ◦ good evidence that only 2 principal species are present The presence of an isosbestic point indicates a) that there are at least two different compounds present in the solution being evaluated. b) that one absorbing species is being converted to another absorbing species during a chemical reaction. c) that three or more absorbing species have an intersecting absorbance at a specific wavelength. Spectrophotometric and photometric titrations • if either one or more reactants (i.e.,analyte and/or titrant) or the products of the titration reaction absorb radiation, or if an absorbing indicator is present ◦ spectrophotometric determination of the end point of the titration can be carried out • photometric titration curve = plot ofA corrvs volume of titrant ◦ A corrA measx (V+v)/V ▪ V = initial volume of solution (before titration) ▪ v = volume of titrant added Photometric titration curves • depending on the molar absorptivities of the substance titrated (ε s, the titrant (t ) and the product (εp), various titration curves can be obtained • the absorbing system(s) must obey Beer’s law in order for linear portions to be extrapolated Advantages of photometric titrations • more accurate results ◦ data from several measurements are pooled to determine the end point • experimental data are taken away from equivalence point region ◦ can also be used for reactions with no sharp end point ▪ • can be used to great advantage in titrations with EDTA and other complexing agents Example: successive EDTA titrations • Titration of Bi(III), Cu(II) by 0.10 M EDTA at 745 nm • εBiY=0, εBi=0, εEDTA=0, εCu=0, εCuY>0 Only things that absorbs • K BiY>>K CuY • complex must follow Beer’s law over the range of concentrations covered • keep ionic strength, pH,… constant • if either M and/or L also absorb, correctAso that it is only due to M m l 3+ 2+ Fe and Cu form complexes with EDTA. The ferric complex is colorless, as is Fe at the concentration involved here. The copper complex is a deep blue color (deeper than the color of Cu itself), while EDTA is colorless. The following titration curve was obtained when a solution containing Fe and Cu was titrated with EDTA using a spectrophotometer set at a wavelength of 745 nm. a) v1is the iron end point. b) v2is the iron end point. c) The difference between v and2v repr1sents the quantity of iron. d) The quantity of iron could not be calculated from a graph like this unless the quantity of co
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