Question

In a simple solvent extraction experiment, how much (i.e., the %E) of an unknown analyte is extracted into the organic layer, if the K for the analyte is 8.518? Assume the organic layer has a density less than water and the aqueous and organic layer are 263.4mL each.

Answer 1

Let the total weight of the substance (Analyte) be = A gms. and x gms. be the weight of the substance extracted in the organic layer , then the amount of the substance left in aqueous layer = (A - x ) gms. Therefore,

concentration in organic layer = x / 263.4

concentration in aqueous layer = ( A -x ) / 263.4

Now C_organic  /  C_aqueous   = K

So , { x /263.4 } / ( A - x ) = 8.518

therefore , x = 0.8949 A

Thus, 89.49 % of the analyte is extracted.

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