Maximum efficiency multiple extraction or 2.71828... revisited - Journal

Maximum efficiency multiple extraction or 2.71828... revisited. Peter Lykos. J. Chem. .... Industry initiatives can help women's careers, survey shows...
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Maximum Efficiency Multiple Extraction or 2.71828

...Revisited

Peter Lykos

Illinois Institute of Technology, 3255 Dearborn Street, Chicago, IL 60616-3793 When using a separatory funnel to extract product from an aqueous reaction mixture using an organic solvent, students are advised to divide the extracting solvent sample into two smaller samples and to perform two successive extractions rather than perform a single extraction with the original sample of extracting solvent. We shall here examine that suggestion in some detail and arrive at a result that for some students is counter intuitive. Suppose we have VmL of solution containing W, g of solute and suppose fnrther we are given V mL of extracting solvent. Let K be the equilibrium constant (where concentrations are expressed in weight per unit volume) for the distribution of the given solute between the given pair of essentially immiscible solvents. Let us consider three cases: r I , I" mL of extracting solvcnr is used fur a singlc eutrachnn; r2 I 1V/2jml.of cnraetma sulvent is used for each of two suc-

where Wl and Wz are the weights of solute remaining after the first and second extractions, respectively. WI can he eliminated between the two equations to give

The ratio of Wl from Case 1to Wz from Case 2 is

which is certainly greater than one thus showing that the process in Case 1is less efficient than the process in Case 2. Case 3. The result of Case 2 may be generalized here where n successive extractions with (Vln) mL of extracting solvent will leave behind W, g of solute, i.e.,

cessive extractions;&d

(3) (V'/n) mL of extracting solvent is used for each of n sueces-

sive extractions.

Case 1. After one extraction the concentration left in the extracted phase will be WIIV and that now in the extracting phase, (W, - Wl)IV'. The ratio of those two concentrations is the distribution coefficient, K, where

Case 2. The form of the result of Case 1may be used here where W, is exactly the same as in Case 1but where W1 is not. The first extraction is now done with V'l2 mL of extracting solvent so the new WI is

and for the second extraction with (V'I2) mL of solvent

448

Journal of Chemical Education

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If we let x = nKVIV that equation for W, can be cast into the form(l+ (l/x)Y(. If we take the limit as x + we, of course, get the base of the natural logarithms, e = 2.71828...Accordingly we get the result W. = W.exp(V/KV) where W, is the smallest amount of solute that remains in the reaction mixture as n (orx) goes to infinity. The counter intuitive result is that one can never extract all of the solute from the original solution. Note Added in Proof The February 1993 issue of this Journal carries the articles "Micmscale Multiple Extraction..." [Anderson, S. W.; Jones, P. T.; Boyd, C. D. J. Chem. Educ. 1993,70,A33].The reader should note that the authors of that paper made an experimental comparison of the single extraction versus two successive extractions (only~. As one of the solutes they used, benzoic add, is known to form dimers, an additional equilibrium (beyond distribution of a solute betwen two immiscible solvents) is operating here.