• Samples are usually complex mixtures. In order to identify and quantify the components of a Ch. 23 Fundamentals of Analytical mixture, we have to separate the components in Separations the mixture. • Separation methods – Extraction – Chromatography – Electrophoresis
Solvent Extraction Solvent Extraction • The transfer of an analyte from [S]2 − /)1( Vmq 2 one phase to a second based on K == the relative solubility of the S 1 /Vqm 1 analyte in two immiscible liquids. Fraction remaining in ≈ V ’ q = ∆ 1 ÷ phase 1 after 1 extraction « + KVV ◊ [S]2 − /)1( Vmq 2 1 2 K == n S 1 /Vqm 1 Fraction remaining in ≈ V ’ n ∆ 1 ÷ phase 1 after n extractions )()'& ...qqqq =⋅⋅= ∆ ÷ K: the partition coefficient for distribution of S between n « 1 + KVV 2 ◊ the two phases at equilibrium; If q = 1/4, then 1/4 remains in phase 1 after one extraction m: the moles of S in the system K: the partition coefficient; q: the fraction of S remaining in q: the fraction of S remaining in phase 1; phase 1; n: the # of extractions.
Extraction Efficiency pH Effects • A solute S has a partition coefficient of 3 between • The charge changes of an acid or base as the pH is toluene and water. If you have 100 mL of a 0.010 M changed. solution of S in water.(1)What fraction of the solute • Distribute coefficient (D): an alternate form of the remains in H2O after a 500 mL extraction with toluene? partition coefficient. (2) What fraction of the solute remains in H O after a 5- 2 Total conc. phasein 2 C 100 mL extractions with toluene? D = = 2 1 phasein conc. Total conc. phasein 1 C 100 1 q = ≈= %6062.0 • A basic amine whose neutral form, B, has partition + )500)(3(100 coefficient, K, between phases 1 and 2. The conjugated 5 acid BH+ is soluble only in phase 1 and the acid ≈ 100 ’ q = ∆ ÷ ≈= %1.000098.0 dissociation constant is Ka. « + )100)(3(100 ◊ [B] [B] [H + ] It is more efficient to do several small extractions 2 1 K = ; Ka = + than one big extraction. B][ 1 BH ][ 1
1 pH Effects pH Effects + -9 2 phasein conc. Total conc. phasein 2 C [B] [H + ] [B] • K for an amine B is 3.0 and the Ka for BH is 1.0×10 . If D = = 2 Ka = 1 K = 2 50.00 mL of 0.010 M aqueous amine is extracted with 100 1 phasein conc. Total conc. phasein 1 C BH + ][ B][ 1 1 1 mL of solvent, calculate the % remaining the in aqueous phase in M at (1) pH 10.00; (2) pH 8.00. [B]2 KKa D = = = Kα −9 BHB + HKa + ][ B KKa × ×100.10.3 1 + 1 + pH = :00.10 D = = = 73.2 + HKa + ][ −9 ×+× 100.1100.1 −10 V 50 + 1 Ω HA][ 2 HK ][ q = = = %1515.0 D = = = Kα HA − + 1 + DVV 2 ×+ 10073.250 1 + AHA ][ 1 + HKa ][ −9 +− KKa ×× 100.10.3 1 HA ][ HA][ 2 pH :00.8 D == = = 273.0 (Ka = ; K = ) + HKa + ][ −9 ×+× 100.1100.1 −8 HA][ 1 HA][ 1 V 50 1 Ω : fraction of the species (P.191) q = = = %6565.0 1 + DVV 2 ×+ 100273.050
pH Effects Extraction with a Metal Chelator • The weak acid HA with Ka = 1.0×10-5 is to be quantitatively extracted from an aqueous sample (100 mL sample volume) using 100 mL of a suitable organic solvent, by doing a single •Usually neutral complexes can be extracted into organic extraction. The partition coefficient is 30.0. What is the solvents. Charged complexes (e.g. MEDTA2-) are not very maximum pH allowed in order that no more than 5% of the acid soluble in organic solvents. remains in the aqueous solution? •Commonly used: dithizone, 8-hydroquinoline, and V 100 cupferron. q = 1 = = %5
1 + DVV 2 100 D ×+ 100 Ω D = 19 HK + ][ × H + ][0.30 D = = = 19 + HKa + ][ −5 +× H + ][100.1 Ω H + ×= 1073.1][ −5 M Ω pH = 76.4
Extraction with a Metal Chelator Extraction with a Metal Chelator •Commonly used: dithizone, 8-hydroquinoline, and cupferron. •Each ligand can be presented •Crown ethers can extract alkali metal ions and can bring as a weak acid, HL. them into non-polar solvents. •Mn+ is in the aqueous phase and MLn is in the organic phase •The distribution coefficient (D) for metal ion extraction depends on pH and [ligand]. •By select a pH, you can bring the metal into either phase.
2 Chromatography Type of Chromatography • A separation process based on the various partitioning coefficients of different solutes between the two phases. • Based on the mechanism of interaction of the solute • Involving the interaction of solute(s) and two phases; operates on with the stationary phase the same principle as extraction, but one phase is held in place while the other moves past it. • Mobile phase: A gas or liquid that moves through the column. (1) Adsorption chromatography - Solute is adsorbed on the surface of • Stationary phase: A solid or liquid that remains in place. the stationary phase (solid). - The stronger a solute adsorbs, the longer it takes to travel through the chromatography column
Type of Chromatography Type of Chromatography • Based on the mechanism of interaction of the solute • Based on the mechanism of interaction of the solute with the stationary phase with the stationary phase
(2) Partition chromatography (3) Ion-exchange chromatography - GC -Retention is based on the - the partitioning of solutes attraction between solute ions and between a mobile phase (gas) charged sites bound to the and bonded liquid stationary stationary phase (ionic phase interactions to separate ions). - A stationary phase of cations will separate anions and vice versa.
Type of Chromatography Type of Chromatography • Based on the mechanism of interaction of the solute • Based on the mechanism of interaction of the solute with the stationary phase with the stationary phase (5) Affinity chromatography -Specific interactions of one kind of solute molecule to a second (4) Molecular Size exclusion molecular that is covalently attached to the stationary phase chromatography -Most selective (e.g. use antibodies to select out one protein from a -size exclusion, gel filtration, or gel mixture of hundreds) permeation chromatography -separate molecules by size -large molecules pass through faster (they do not get caught up in pores)