ANALYTICAL CHEMISTRY CHEM 2001
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This 35 page Class Notes was uploaded by Javonte Swift on Tuesday October 13, 2015. The Class Notes belongs to CHEM 2001 at Louisiana State University taught by S. Soper in Fall. Since its upload, it has received 9 views. For similar materials see /class/223106/chem-2001-louisiana-state-university in Chemistry at Louisiana State University.
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Date Created: 10/13/15
Lecture 12 Analytical Separations Common techniques 0 Precipitation Solubility 0 Metal Chelates General Solvent Extraction Common Separation Techniques Mechanical Phase Separation Precipitation Difference in solubility of the compounds formed li luidsolid Distillation Difference in volatility of compounds gasliquid Extraction Difference in solubility in 2 immiscible liquids Kli Auidli Auid Ion Exchange Difference in stability of reactants with ionexchange resin liquid solid C hrom atovral hquot Difference in rate of movement of a solute through a stationary phase Electrophoresis Difference in migration rate in an electrical eld gradient Precipitation Sel arations based ul on the solubility differences of various components The analyte is converted into an insoluble substance a precipitate which can then be isolated Examples of how to separate Vary pH Use metal chelators PbN032 K2004 PbCrO4 K2Cr04 aq PbN032 aq lt3 PbCr04 s KNo3 aq Examples of Precipitation Separations Based on pH Reagent Substances Separated Substances Not Precipitated strong Concentrated HN03 or SiOz Nb205 stnos M t th t I d Oxidizing Acids HCO4 Sb205 Ta205 wos Pbo2 OS 0 ermea mg a among 39 39 CH CO HCH CO N weak39y o c39d39c 3 2 3 2 a AIOH3 CrOH3 FeOH3 Most divalent metal cations Solutlon buffer Weakly Basic 380 ARCH SCOH3 Alkali and alkaline earth metals Mn2 NH NH Cl B ff Solution 3 4 u 939 CrltOHgt3 FeOHgt3 002 Ni2 Cu2 Zn2 Ag Cd2 39 M td39 t t 39 Strongly Basic Concentrated NaOH os Iva en mea Ions 3 2 Alkali metals AI Zn ox anions Solutlon rare WNW c3 y Precipitation using Metal Chelators One way to separate metal ions from each other is to complex one using an organic ligand and then extract it using an organic solvent Common Ligands 039 NH4 r O wa N C SH N OH Cupferron 6 Dithizone 8hydroxyquinone Metal Chelators The ligands Each ligand loses a proton when it binds to a metal on Example of 8hydroxyquinone A13 3 Al HO Separations can be pH dependent Distillation Distillation separates a mixture of two or more substances into its components by applying heat The separation is based on differences in the boiling points of the components evaporation 11 1u1d heat val orlzatlon condensatlon Example of a Distillation Apparatus 1 DistilllatiDn F Condenser ww mnm f1 1 w H7 g39 V39 1quot IL 39 39 39 IZZ quot 37 g I Fractmnatmg column Hquot 39q Receiving ask Distillation ask k in I SJ General Solvent Extraction Concepts Extraction is the transfer of a solute between two immiscible liquids This technique is known as solvent extraction General Solvent Extraction Concepts Extraction is the transfer of a solute bewteen two immiscible liquids This technique is known as solvent extraction For example Pour V Analyte2 Solve t i V Basic Solvent Extraction The anald te 1 artitions between two solvents Analyte in solvent 2 Analyte in solvent 1 K Analyte2 Analyte1 Basic Solvent Extraction The anald te 1 artitions between two solvents Analyte in solvent 2 V Analyte in solvent 1 K Analyte2 Analyte1 If K is large then the analyte moves fr0m phase 1 to phase 2 lf K is small then the analyte remains in phase 1 V v A i gt i Harris Quantitative cr ml Analysis 3quot1 ed A The yellow uranyl nitrate U02NO32 N layer B The U02NO32is distributed between both layers after shaking C The U02NO32 is in the diethyl ether layer Solvent Extraction 0 Common Organic Solvents Less dense than water benzene diethd l ether hexane etc More dense than water 0 Chlorofouu dichlorornethane quotWon tetrachloride Extraction can be enhanced with Multiple extractions pH Metal Chelators SOLVENT EXTRACTION EQUILIBRIA Gibbs Phase Rule Multiple Solvent Extractions Simultaneous Multiple Extractions GIBBS PHASE RULE P F C 2 where P of phases F of degrees of freedom variance of components For Solvent Extraction we have 1 Two essentially mumsquot W mm nd we solute distributed between them therefore 2 At constant T temp and p pres F unity RECALL SOLVENT EXTRATION SYSTEM Distribution Law The ratio of concentration is invariant 1e independent or Otal concentration Consider the distribution equilibrium ofa nonionic solute A aq ltgt A org KD AL Aa not true when both liquid phases are saturated With solute LECTURE 13 DISTRIBUTION LAW AND MULTIPLE EXTRACTIONS Distribution Ratio D D is constant independent of the volume ratio total conc39 n in organic phase total conc39 n in aqueous phase Percent Extracted E depends on volume ratio mmoles of solute in organic phase fraction extracted total mmoles of solute E 84 V x 100 SOVO SaVa E amp D VaVO 100D IfVa V0 then E D1 If extraction is not 100 then multiple extractions can oe useo to quantitatively remove sciute Multiple Extractions Assume D25 and Va10 mL a Extract with 10 mL organic or b with 2 x 5 mL organic E EM962 DVaV0 25lO10 b E M 926 therefore left with 74 25105 Second extraction 0926 x 74 685 Total E 926 685 994 Conclusion For a given volume multiple cxuacuuus are ucuer LECTURE 14 CRAIG COUNTERCURRENT EXTRACTIONS SIMULTANEOUS MULTIPLE EXTRACTIONS Simultaneous Multiple Solvent Extractions RESERVOIR D 1 2 3 p fraction of solute in mobile upper phase p DCDC 1 when VI VOVa10 and D solute mobile phase C solute stationary phase q traCtIon OT SOIUte In Stationary lower phase q 1DC1 note that p q 10 Let n number of transfers n 0 no transfer All solute starts in tube 0 All p tubes have stationary phase After equilibration Tube 1 is filled with a volume of solvent from the reservoir and q the solute distributes between the upper and lower phase at 0 equilibrium n 1 one tranSfer First transfer n1 The p fraction from upper mobile phase of tube 0 is transferred into tube 1 qp p2 q2 Total solute in tube 0 pq q2qpqq CID Total solute in tube 1 p2 pqppqp n 2 second transfer In2 2 2 3 qp p Afterequlibration q3 2qu qr2 0 1 2 Total solute in tube 0 pq2 q3 q2p q 02 Total solute in tube 1 2qp22pq22pqpq2pq JUIulU III tuuu L 13 39 JnIZ V2p q p2 LECTURE 1 5 CRALU Cu U N 1 ERCURRENT EXTRACTIONS CONTINUED PRELUDE TO CHROMATOGRAPHY At this point you should get the idea Without the need for more transfer The amount of solute in each tube leads to the binomial expansion q pn Where each term in the expansion represents the fraction of solu rrticular tub e1 fo two transfers we have q p2 q2 2pq p2 When 11 gets large enough it leads to a Gaussian Distribution 1 e np r2 2npq 2 npq Where f is the fraction of solute in tube number r This equation relates to the normal Gaussian distribution that we talked about in statistics square root of npq 5standard deviation and np true value rmaX This equation relates to the normal Gaussian distribution that we talked about in statistics square root of npq 5standard uevratron and np true varue rm After n transfers solute distribution may look like r tube number This equation relates to the normal Gaussian distribution that we talked about in statistics square root of npq 5standard devranon and np true va1ue rmaX After n transfers solute distribution may look like r tube number A large sigma means a broader peak ie solute spreads out more Remember that i 5 on the most probable value includes 68 of solute molecules 1 2 5 includes 955 ofsolule molecules and 1 3 5 includes 999 rmaXA rmaXB 124039A 4GB Resolution 2 R The term 40A peak width for solute A Le i 20 95 of solute A molecules Same is true for B and therefore 124oA 4GB is the average peak width Example Let s see how resolution works Assume two solutes A and B DCA 38 and DCB 31 The larger DC forA means A migrates faster than B 1 PA 0792 PB 23 20756 48 41 qul OOO792O208 qB 20244 If n 1000a very large number of tubes l then rmaXA 1000 X 0792 792 and rmaxB 1000 X 0756 756 y In the present case r tube number 5 A 1000 x 0792 x 0208 2 128 5 B 1000 x 0756 x 0244 2 136