Lecture_overheads_-_Ch22

Lecture_overheads_-_Ch22 - CHEM 320 Quantitative Analysis...

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CHEM 320 – Quantitative Analysis Ch. 22 – Introduction to Analytical Separations 22-1 Solvent Extraction Partitioning between 2 Phases (K) Distribution Coefficient (D) Extraction with Metal Chelator Dithiazone Chelation/Extraction/Spectrophotometric Method for Determination of Pb in Water 22-2 What is Chromatography? Adsorption Chromatography Partition Chromatography Ion Exchange Chromatography Size Exclusion Chromatography Affinity Chromatography 22-3 A Plumber’s View of Chromatography (t r , α , k’) 22-4 Efficiency of Separation (R, H, N) 22-5 Why Bands Spread (H = A + B/u + Cu) Multiple Flow Path term (A) Longitudinal Diffusion Term (B/U) Finite Equilibration Time Term (Cu)
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22-1 Solvent Extraction Partitioning of a solute ( S ) between two immiscible phases [ ] [ ] 1 2 S S S S A A K 1 2 = Partition coefficient ( K ) Defines equilibrium for the partitioning (separation) of a substance between two phases (liquid/liquid, gas/liquid, gas/solid, etc.) A S 2 = activity of solute (S) in phase 2 [ S ] 2 = conc of solute (S) in phase 2
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[ ] [ ] - = = 1 2 1 2 V m q V m q) (1 S S K 1 V m q - 2 ) 1 ( V m q 1 V m Single Extraction 0 q = fraction of moles of solute S remaining in phase 1 at equilibrium Final Initial [S] 1 [S] 2 + = 2 1 1 KV V V q Rearrange to solve for q Value of q (0 to 1) depends on V 1 , V 2 , and K } Example – Calculate q for extraction of an analyte from a 100 mL aqueous solution into 500 mL of toluene given K toluene//water = 3.00 q = 100/ [100+3(500)] = 0.0625 94% of analyte is in toluene and 6% is still in aqueous phase
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Multiple Extractions fraction in phase 1 at equilibrium (q value depends on solute and solvents) 6.25% 16 1 4 1 4 1 q q = = = 0.0977% 10 x 9.77 4 1 KV V V q 4 5 5 2 1 1 5 = = = + = - n 2 1 1 n KV V V q + = after n extractions } after 5 extractions } after 2 extractions } 0.25 1/4 3(100)) 100/(100 q = = + = Example Calculate fraction of solute after multiple 100 mL extractions
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Example Solvent Extraction You perform a liquid/liquid extraction on a 50.00 mL aqueous wastewater sample that contains an unknown concentration of nitrophenol (NP, C 6 H 5 NO 2 ). The organic phase volume is 20.00 mL. The partition coefficient ( K ) for the extraction of NP in water to NP in the organic layer is 10.00. 1. Calculate q for this extraction 2. If [NP] = 5.00 ppb in the organic phase at equilibrium, calculate the total mass of NP in the system 3. Calculate the original NP concentration in the wastewater sample Useful Equations [ ] [ ] 10.00 NP NP K 1 2 = = + = 2 1 1 KV V V q Partition coefficient Fraction remaining in Phase 1 at equil
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[ NP ] 2 [ NP ] 1 Phase 1 (aqueous wastewater sample) Phase 2 (organic)
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This note was uploaded on 11/28/2011 for the course CHEM 321 taught by Professor P.palmer during the Spring '11 term at S.F. State.

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Lecture_overheads_-_Ch22 - CHEM 320 Quantitative Analysis...

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