When an aqueous solution of an organic compound is shaken withan immiscible orga

Question

When an aqueous solution of an organic compound is shaken withan immiscible organic solvent, such as diethyl ether, the solutedistributes itself between the two phases. When the two phasesseparate into two distinct layers, an equilibrium will have beenestablished such that the ratio of the concentrations of the solutein each solvent defines a constant, K, called the distributioncoefficient (or partition coefficient). K = concentration of solute in solvent A, e.g., diethyl ether(g . L-1) concentration of solute in solvent B, e.g., water (g .L-1) The distribution coefficient for compound X in the diethylether/water system is 3.0. If you were given a solution containing8.0 g of X in 500 mL of water, and wanted to extract compound Xinto diethyl ether, show that it would be more effective to extractX using three 50 mL aliquots of diethyl ether rather than a single150 mL aliquot. (HINT: Determine how much of X would remain in theaqueous solution in each case.) Please help! When an aqueous solution of an organic compound is shaken withan immiscible organic solvent, such as diethyl ether, the solutedistributes itself between the two phases. When the two phasesseparate into two distinct layers, an equilibrium will have beenestablished such that the ratio of the concentrations of the solutein each solvent defines a constant, K, called the distributioncoefficient (or partition coefficient). K = concentration of solute in solvent A, e.g., diethyl ether(g . L-1) concentration of solute in solvent B, e.g., water (g .L-1) The distribution coefficient for compound X in the diethylether/water system is 3.0. If you were given a solution containing8.0 g of X in 500 mL of water, and wanted to extract compound Xinto diethyl ether, show that it would be more effective to extractX using three 50 mL aliquots of diethyl ether rather than a single150 mL aliquot. (HINT: Determine how much of X would remain in theaqueous solution in each case.) Please help!

Explanation / Answer

For a single 150ml aliquot: Let x be the amount of X left in water. K = ((8.0-x)/150)/(x/500) = 3 x = 80/19 = 4.2 g For three 50 ml aliquot: First try: 3 = ((8 - x)/50)/(x/500) x = 80/13 = 6.2 g Second try: 3 = ((6.2 - x)/50)/(x/500) x = 62/13 = 4.8 g Third try: 3 = ((4.8 - x )/50)/(x/500) x = 48/13 = 3.7 g Hence with three aliquots of 50ml each we can extract more than onealiquot of 150 ml.

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