Separation • 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− q ) m / V2 one phase to a second based on K = = the relative solubility of the []S 1 qm/ V1 analyte in two immiscible liquids. Fraction remaining in V ’ q = 1 phase 1 after 1 extraction V+ KV [S]2 (1− q ) m / V2 1 2 K = = n []S 1 qm/ V1 Fraction remaining in V ’ n 1 phase 1 after n extractions q= q&)')( ⋅ q ⋅... q = K: the partition coefficient for distribution of S between n V1 + KV2 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. in phase 2 C 100 mL extractions with toluene? D = = 2 Total conc. in phase 1 C 100 1 q = =0.062 ≈ 6% • A basic amine whose neutral form, B, has partition 100+ (3)(500) coefficient, K, between phases 1 and 2. The conjugated 5 acid BH+ is soluble only in phase 1 and the acid 100 ’ q = =0.00098 ≈ 0.1% 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 Total conc. in phase 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 Total conc. in phase 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 [][][]B BH + Ka H + B KKa 3.0× 1.0× 10 1 + 1 + pH =10.00: D = = = 2.73 Ka+[] H + 1.0× 10−9 + 1.0 × 10−10 V 50 + 1 []HA 2 KH[] q = = = 0.15 15% D = = = Kα HA − + V1 + DV2 50+ 2.73 × 100 [][]HA1 + A 1 Ka+[] H −9 − + KKa 3.0× 1.0 × 10 [][]AH1 []HA 2 pH =8.00 : D = = = 0.273 (Ka = ; K = ) Ka+[] H + 1.0× 10−9 + 1.0 × 10−8 []HA 1 []HA 1 V 50 1 : fraction of the species (P.191) q = = = 0.65 65% V1 + DV2 50+ 0.273 × 100 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% V1 + DV2 100 +D ×100 D = 19 KH[]+ 30.0× [H + ] D = = = 19 Ka+ [] H + 1.0× 10−5 + [H + ] [H + ]= 1.73 × 10−5 M pH = 4.76 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) 3.