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 Extraction Efficiency • The transfer of an analyte from one phase to a second • A solute S has a partition coefficient of 3 between based on the relative solubility of the analyte in two toluene and water. If you have 100 mL of a 0.010 M immiscible liquids. solution of S in water.(1)What fraction of the solute remains in H O after a 500 mL extraction with toluene? []S 2 (1− q ) m / V2 2 K = = (2) What fraction of the solute remains in H2O after a 5- []S 1 qm/ V1 100 mL extractions with toluene? n ⎛ V ⎞ 100 q = ⎜ 1 ⎟ q = =0.062 ≈ 6% ⎜ ⎟ 100+ (3)(500) ⎝V1+ KV 2 ⎠ 5 At equilibrium: K: the partition coefficient for distribution of ⎛ 100 ⎞ q = ⎜ ⎟ =0.00098 ≈ 0.1% S between the two phases; q: the fraction of S remaining in ⎝100+ (3)(100) ⎠ phase 1; n: the # of extractions. It is more efficient to do several small extractions If q = 1/4, then 1/4 remains in phase 1 after one extraction than one big extraction. pH Effects pH Effects • The charge changes of an acid or base is dependent on pH. • K for an amine B is 3.0 and the Ka for BH+ is 1.08×10-9. If • Distribute coefficient (D): an alternate form of the 50.00 mL of 0.010 M aqueous amine is extracted with 100 partition coefficient. mL of solvent, calculate the % remaining the in aqueous phase in M at (1) pH 10.00; (2) pH 8.00. Total conc. in phase 2 C D = = 2 Total conc. in phase 1 C 1 KKa 3.0× 1.0 × 10−9 pH =10.00 : D = + = −9 −10 = 2.73 []B 2 KKa Ka+[] H 1.0× 10 + 1.0 × 10 D = + = + = Kα B [][][]B1 + BH 1 Ka+ H V1 50 + q = = =0.15 ⇒ 15% [][]BH1 []B 2 (Ka = + ; K = ) V1+ KV 2 50+ 2.73 × 100 []BH 1 []B 1 KKa 3.0× 1.0 × 10−9 + []HA KH[] pH =8.00 : D = + = −9 −8 = 0.273 D = 2 = = Kα Ka+[] H 1.0× 10 + 1.0 × 10 [][]HA+ A− Ka+[] H + HA 1 1 V 50 [][]AH− + []HA q = 1 = =0.65 ⇒ 65% (Ka = 1 ; K = 2 ) α: fraction of the species (P.191) V1+ KV 2 50+ 0.273 × 100 []HA 1 []HA 1 1 Extraction with a Metal Chelator Extraction with a Metal Chelator •Usually neutral complexes can be extracted into organic •Commonly used: dithizone, 8-hydroquinoline, and solvents. Charged complexes (e.g. MEDTA2-) are not very cupferron. soluble in organic solvents. •Crown ethers can extract alkali metal ions and can bring •Commonly used: dithizone, 8-hydroquinoline, and them into non-polar solvents. cupferron. Chromatography Extraction with a • A separation process based on the various partitioning Metal Chelator coefficients of different solutes between the two phases. •Each ligand can be presented • Involving the interaction of solute(s) and two phases as a weak acid, HL. • Mobile phase: A gas or liquid that moves through the column. •Mn+ is in the aqueous phase • Stationary phase: A solid or liquid that remains in place. 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. 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 (1) Adsorption chromatography - Solute is adsorbed on the surface of -GC the stationary phase (solid). - The stronger a solute adsorbs, the - the partitioning of solutes between a mobile phase (gas) longer it takes to travel through the chromatography column and bonded liquid stationary phase 2 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 (3) Ion-exchange chromatography (4) Molecular Size exclusion chromatography - ionic interactions to separate ions. - a stationary phase of anions will -size exclusion, gel filtration, or gel separate cations and vice versa. permeation chromatography -separate molecules by size -large molecules pass through faster (they do not get caught up in pores) Type of Chromatography The Chromatogram • Based on the mechanism of interaction of the solute •A plot of detector response with time. with the stationary phase •Volume flow rate (flow rate): vol. of solvent pass through the column (5) Affinity chromatography -Specific interactions of one kind of solute molecule to a second •Linear flow rate: the length of the column passed though by the molecular that is covalently attached to the stationary phase solvent -Most selective (e.g. use antibodies to select out one protein from a •t : unretained mobile phase travels through the column in the mixture of hundreds) m minimum possible time •tr: retention time, the time for each component needed after injection of the mixture onto the column until that component reaches the detector •tr’: adjusted retention time, tr’=tr-tm •Vr: retention volume, volume of mobile phase required to elute a solute to a maximum from a column. Vr= tr*flow rate The Chromatogram Retention Parameters • Adjusted retention time ' −= ttt mrr - Time spent in the stationary phase (or ts) • Relative retention - ratio of adjusted retention times for any two components - the greater the relative retention, the greater the separation ' tr 2 α = ' tr1 • Capacity factor (or retention factor): - the longer a component is retained by the column, the greater is the capacity factor. − tt k ' = mr tm 3 • Example: Calculate the adjusted retention time and Retention Time and Partition Coefficient capacity factor for benzene and toluene in the GC experiment? Methane (as a solvent) peak is at 42 s; • The capacity factor is equivalent to the time the solute spends benzene at 251 s and toluene at 333 s. in the stationary phase over the mobile phase and can be related to the partition coefficient: CV V t− t t ' VV− k ' =s s =K s ⇔k ' = r m =r = r m CVm m Vm tm tm Vm For Benzene For Toluene • Relative retention can be related to retention time, capacity ' ' tr= t r − t m =251 − 42 = 209 s tr= t r − t m =333 − 42 = 291 s factor, and/or partition coefficient ' ' tr 2 k2 K 2 α =' =' = tr1 k1 K1 t− t 251− 42 ' t− t 333− 42 k ' = r m = = 5.0 k = r m = = 6.9 • Physical basis of chromatography: the greater the ratio of t 42 tm 42 m partition coefficients between mobile and stationary phases, the greater the separation between two components of a mixture • Example: If use the open tubular chromatography column, Scaling Up where methane (as a solvent) peak is at 42 s and benzene peak at 251 s. Calculate the partition coefficient (K) for •Analytical and preparative benzene between stationary and mobile phases and the •Keep column length constant fraction of the time benzene spends in the mobile phase. •Cross-sectional area of column ~ mass of Cross -sectional area of column analyte ~ volume flow rate (if maintain 2 2 4 2 constant linear flow rate) ~ sample volume Vm =π r1 = π(124) = 4.83 × 10 μm applied to column Cross - sectional area of coating 2 2 •If change the column length, then the mass of Vs =2π r2 × thickness = 2π (124.5)× 1 = 7.8 × 10 μm sample can be increased in proportion to the ' t− t 251− 42 V k = r m = =5.0 =K s ⇒K = 310 increase in length tm 42 Vm t− t t k ' = r m =s ⇒t = k' t s m Scaling equation: tm tm Fraction of time in mobile phase : Mass 2 Radius 2 2 t t 1 1 ⎛ ⎞ m m = ⎜ ⎟ = ' = ' = = 0.17 ts+ t m k tm+ t m k +1 5.0+ 1 Mass 1 ⎝ Radius 1 ⎠ Diffusion •One main cause of band broadening is diffusion. •Diffusion coefficient (D): measures the rate at which a substance moves randomly from a region of high concentration to a region of lower concentration. •Std deviation of diffusive band spreading: σ = 2Dt 4 Efficiency of Separation Efficiency of Separation Solutes moving through a column can spread into a Gaussian distribution with a standard deviation, σ (K is a constant). Resolution A resolution of 1.5 gives an essentially complete separation of A and B. of two peaks: Δt ΔV 0.589Δt Δt ΔV 0.589Δt Resolution = r = r = r Resolution = r = r = r wav wav w 2/1 av wav wav w 2/1 av Column Efficiency-Theoretical Plates (N) • Example: A solute with a retention time of 407 2 s has a base width of 13 s on a 12.2 m column. Plate Height σ H = Find the plate height and number of plates. L Number of Theoretical Plates 2 2 L 16tr (16)(407) 4 N = N =2 = 2 =1.57 × 10 L H w 13 L=x • Martin & Synge (1941): L Lx16 L2 N = = = L 12.2 m Treated a chromatographic H σ 2 w2 H = = = 0.78 mm column as if it were similar N 1.57× 104 to a distillation column made σ =w/4 up of numerous discrete but 16t 2 t2 5.55 t 2 contiguous narrow layers N =r = r = r (theoretical plates).