Solvent Extraction Analytical Separations PDF

1  Analytical Separations  Most chemical analyses of real samples require that the analyte be partially or completely separated from interfering components of the sample.  A very large part of all analytical chemistry is devoted to achieving these separations.  Most org...

1  Analytical Separations  Most chemical analyses of real samples require that the analyte be partially or completely separated from interfering components of the sample.  A very large part of all analytical chemistry is devoted to achieving these separations.  Most organic analytical chemistry is done using some kind of chromatography.  Chromatographic methods are instrumental separation methods (gas chromatography, liquid chromatography).  Solvent extraction is a simple, manual separation technique that is still widely used in the modern analytical laboratory. 2  Solvent Extraction  Solvent extracti0n is used most often to transfer the desired analytes from one matrix to another.  For example, the sample might be a sample of soil. (We then say that the matrix is soil.) Soil is not suitable for almost all instrumental methods of analysis.  A solvent extraction is done on the soil to transfer the analytes from the soil to a solvent which is suitable for instrumental measurement.  In many cases, water is also not suitable for direct introduction into the chemical instrument.  Extraction refers to the transfer of a solute from one liquid phase to another or from a solid phase to a liquid phase. 3  Solvent Extraction  The most common case is the extraction of an aqueous solution with an organic solvent.  Diethyl ether, benzene, and other hydrocarbons are common solvents that are less dense than water and form a phase that sits on top of the aqueous phase.  Chloroform, dichloromethane (a.k.a. methylene chloride) and carbon tetrachloride are common solvents that are immiscible with and denser than water. (Whenever a choice exists between CHCl3 and CCl4, the less toxic CHCl3 should be used.)  In a two-phase mixture, some of each solvent is found in both phases, but one phase is predominantly water and the other phase is predominantly organic. 4  Solvent Extraction  The volumes of each phase after mixing are not exactly equal to the volumes that were mixed.  For simplicity, however, we will assume that the volumes of each phase are not changed by mixing.  Suppose that solute S is partitioned between phases 1 and 2. 5  Solvent Extraction  The partition coefficient, K, is the equilibrium constant for the reaction S(in phase 1) S(in phase 2)  In dilute solution, the ratio of activities can be replaced by the ratio of concentrations. 6  Solvent Extraction  Suppose that a volume V1 of solvent 1 containing m moles of solute S is extracted with volume V2 of solvent 2.  Let q be the fraction of S remaining in phase 1 when equilibrium is achieved.  The molarity in phase 1 will therefore be qm/V1.  The fraction of total solute transferred to phase 2 will be (1 – q), and the molarity in phase 2 is (1 – q)m/V2.  Putting these values of molarity into the expression for K, Solve for q: (1) 7  Solvent Extraction  The last equation says that the fraction of solute remaining in phase 1 depends on the partition coefficient and the two volumes.  If the phases are separated and a new quantity of fresh solvent 2 is mixed with phase 1, the fraction of solute remaining in phase 1 at equilibrium will be  After n extractions with volume V2, the fraction remaining in phase 1 is (2) 22 8  Extraction (Solid – Liquid)  A weighed solid is placed in a closable container and some solvent is added.  The solid and liquid are mixed well, and the liquid is separated from the solid.  The liquid dissolves some part of the solid.  The material that dissolves interacts more strongly with the solvent than with the remaining solid, and so it transfers into the solution. 9  Extraction  We describe this preference over another phase with a partition coefficient, KD, based on the mass/volume of the ith analyte in the 2 phases:  The weaker the interaction of the analyte with the solid matrix, the easier the separation into the solvent.  This easier separation means a larger KD value. 10  Extraction  Consider an analyte with KD = 2.0.  Twice the concentration of material is found in the solvent as is found in the solid.  This is not a large ratio (with equal volumes of solvent and solid, about 33% of the analyte is left in the solid).  We might want to correct for the 66% recovery as a determinate error.  Recall that the amount left in the solid depends on the volume of extractant. 11  Extraction  If we double the amount of solvent, half as much mass will stay in the solid.  You might say “Use 10 x as much solvent, shake it, then concentrate it.”  This would result in 20 x as much analyte in the solvent as in the solid.  But you run the risk of loss of analyte during the solvent removal step.  There is a better way, and that is to use the same amount or less solvent but do a number of sequential extractions. 12  Extraction  If the partition coefficient is very large (>100), a single extraction will probably remove a solute from one phase to the other.  However, it can be shown that for a given amount of extracting solvent, it is more effective to use it in several small portions rather than in a single large batch. 13  Extraction (Liquid-Liquid)  A standard method for separation  Simple and inexpensive  Two liquids are immiscible – they do not dissolve in each other.  After they are intimately mixed – such as by shaking – separate into 2 layers.  The 2 layers are then segregated by draining one of them off until the interface is reached. 14  Extraction (Liquid-Liquid)  Extraction is a very common laboratory procedure used when isolating or purifying a product. Organic chemistry employs solid-liquid, liquid-liquid, and acid-base extractions. It is very common for organic products synthesized in a reaction to be purified by liquid-liquid extraction.  A separatory funnel is used for this process. In this procedure, the organic product is isolated from inorganic substances. The organic product will be soluble in an organic solvent (organic layer) while the inorganic substances will be soluble in water (aqueous layer). 15  Extraction (Liquid-Liquid)  The organic solvent used for extraction must meet a few criteria: 1. Should readily dissolve substance to be extracted. 2. Should not react with the substance to be extracted. 3. Should not react with or be miscible with water (the usual second solvent). 4. Should have a low boiling point so it can be easily removed from the product.  Common extraction solvents are diethyl ether and methylene chloride. 16  Solvent Extraction  For example, assume that 4 g of butyric acid is to be extracted from 500 mL of water with 500 mL of ether.  The partition coefficient for this system is 3 at 25°C.  If the ether is used in a single batch: where X is the weight of butyric acid remaining in the water layer. Thus X=1 and 3 g are extracted into the ether layer. 17 Solvent Extraction 18  Solvent Extraction  However, if the ether is used in 2 successive 250 mL portions, for the first one:  In this case, X = 1.6 and 2.4 g are found in the ether layer, which is then removed.  In the second extraction with the remaining 250 mL portion of ether: 19  Solvent Extraction  An additional 0.96 g is extracted by the ether and 0.64 g remain in the water.  Thus, a total of 3.36 g have been extracted, a significant improvement. 20  Solvent Extraction  Similar calculations would show that if the ether had been divided into five 100 mL portions, only 0.23 g of the original 4 g of butyric acid would remain in the aqueous phase after the fifth extraction, and the combined ether extracts would contain 3.77 g. 21  pH Effects  Suppose that the solute being partitioned between phases 1 and 2 is an amine with a base constant Kb.  Let’s also assume that BH+ is soluble only in the aqueous phase (1).  Suppose that the neutral form, B, has partition coefficient, K, between the phases.  The distribution coefficient, D, is defined as which becomes 22  pH Effects  Combining the relationships K = [B2]/[B1] and Ka = [H+][B]/[BH+] = Kw/Kb with the last equation leads to  The distribution of solute between the two phases will be pH-dependent. 23 23  pH Effects  This figure shows the effect of pH on the distribution ratio in example 23.  When you want to extract a base into water, it is clearly desirable to use a low enough pH to convert it to BH+.  By the same reasoning, to extract an acid (HA) into water, you should use a high enough pH to convert the acid to A24-.  pH Effects  We have seen that the distribution coefficient for a base is given by  It can also be shown that for an acid, the distribution coefficient is 25  Chelating Ligands  Metal ions are Lewis acids, in that they can share electron pairs donated by ligands, which are therefore Lewis bases.  Cyanide is called a monodentate ligand because it binds to a metal ion through only one atom (the carbon atom).  Most transition metal ions have room to bind 6 ligand atoms.  A ligand that can bind to a metal ion through more than one ligand atom is said to be multidentate.  It is also called a chelating ligand. 26  Extraction with a Metal Chelator  One scheme for separating metal ions from each other is to selectively complex one ion using an organic ligand and extract it into an organic solvent.  Three ligands commonly used for this purpose are 27  Metal Chelators  Each ligand can be represented as a weak acid, HL, which loses one proton when it binds to a metal ion through the atoms shown in bold type.  Each of these ligands can react with many different metal ions, but some selectivity is achieved by controlling the pH.  Most complexes that can be extracted into organic solvents must be neutral.  Charged complexes, such as Fe(EDTA)- or Fe(1,10-phenanthroline)32+, are not very soluble in organic solvents. 28  Metal Chelators  Let’s derive an equation for the distribution coefficient of a metal between the two phases.  We will assume that essentially all of the metal in the aqueous phase is in the form Mn+ and that all of the metal in the organic phase is in the form MLn. 29  Metal Chelators  We’ll define the partition coefficients for the ligand and complex as follows: where org refers to the organic phase.  The distribution coefficient we want is 30  Metal Chelators  It can be shown that  This equation says that the distribution coefficient for metal ion extraction depends on the pH and the ligand concentration.  Because distribution of the metal between the two phases is pH-dependent, you can select a pH to bring the metal into either phase.  Since the various equilibrium constants are different for each metal, it is often possible to pick a pH where D is large for one metal and small for another. 31  Metal Chelators  For example, this figure shows that Cu2+ could be separated from Pb2+ and Zn2+ by extraction with dithizone at pH 5. 32 Distillation - Mixture of 2 liquids  Benzene (C6H6) and Toluene (C6H5-CH3) are organic solvents that have similar molecular structures.  Benzene (b.p. 80.1°C) is more volatile than toluene (b.p. 110.6°C): at 20 °C, the vapour pressure of pure benzene, PB°, is 75 mm Hg, and the vapour pressure of pure toluene, PT°, is 22 mm Hg.  A mixture of benzene and toluene will have different compositions in the liquid and vapour phases.  Because the liquid and vapour phases have different compositions, it is possible to separate a binary liquid solution into its 2 pure components. 33  Separation of a mixture  This can be done by a process called fractional distillation.  If we vapourize a small amount of the mixture, the vapour will be richer in benzene than the original liquid was.  If we now draw off and condense that vapour, we will have a new liquid with the same composition as the vapour, i.e., richer in benzene than the original liquid.  A second vapourization of this new liquid will produce a vapour still richer in benzene.  If we arrange a very large number of successive vapourizations and condensations, we can eventually produce a vapour that is pure benzene. 34  Distillation Column  The apparatus for carrying out this succession of many vapourizations and condensations is called a distillation column.  The flask at the bottom holds the liquid mixture to be separated.  The packing material in the column, which may be steel wool or glass coils provides cooler surfaces to condense the vapour as it rises in the column. 35  Fractional Distillation  As the vapour rises and reaches the cooler packing material, it condenses.  Some of the liquid drops down into the pot again, but some stays in the column, and as warm vapour reaches it again, revapourizes and rises a little higher in the column before it recondenses.  The progress of the vapour upward through the column takes a reasonably long period of time. 36  Fractional Distillation  The vapour that comes out of the top of the column (called the distillate) is pure benzene, and is condensed and led away from the material in the pot, into a receiving vessel.  The liquid (called the residue) that remains in the pot at the end of the distillation, is toluene.  The distillate is the more volatile of the 2 liquids and the residue is the less volatile. 37  Liquid-Vapour Equilibrium  This graph is useful in explaining fractional distillation. 38  Fractional Distillation Notice that the graph starts at a high temperature, 110.6 °C, the boiling point of toluene, and ends at a lower temperature, 80.0 °C, the boiling point of benzene. The vapour curve lies above the liquid curve in this diagram. 39  Fractional Distillation A benzene-toluene solution with xbenz = 0.30 boils at 98.6 °C and is in equilibrium with a vapour in which xbenz = 0.51. Imagine extracting some of this vapour and cooling it to the point where it condenses to a liquid. The new liquid will have xbenz = 0.51 and is the conclusion of stage 1. Now imagine repeating the process, vaporizing a solution with xbenz = 0.51 and condensing the vapour. 40 Fractional Distillation The new liquid at the end of stage 2 has xbenz = 0.71. By repeating the cycle, the vapour becomes progressively richer in benzene. 41  Distillation of Petroleum  Petroleum is a mixture of hundreds or thousands of alkanes up to 40 or more carbons.  It also contains sulfur-, oxygen-, and nitrogen-containing compounds.  It is refined by distillation.  Gasoline range organics (GRO): C6 – C10, BP 60 - 220°C  Diesel range organics (DRO): C10 – C25, BP 170 - 400 °C 42