E ) The distribution coefficient , Ko ( Cether / C water ) , for an organic substance X at room temperature is 13. What relative volume of ether to water should be used for the extraction of 94 percent of X from a water solution in single extraction ?

The extraction/distribution coefficient of an arbitrary solvent to water for a given substance is expressed as the mass concentration of the substance in the arbitrary solvent (C₁) divided by the mass concentration of the substance in water (C₂).

K = (C₁/C₂)

Let the initial mass of the organic substance X in water be 1 g (it could be any mass basically, it is just to select a right basis, since we are basically working with percentages here).

If 94% of the organic substance X is extracted by ether in a single extraction, 0.94 g ends up in ether and 0.06 g of the organic substance X that remains in water.

Let the volume of ether required be x mL.

Let the volume of water required be y mL.

Relative volume of ether to water that should be used for the extraction = (x/y)

Mass concentration of the organic substance X in ether = (0.94/x)

Mass concentration of organic substance X in water = (0.06/y)

The distribution coefficient , Ko (Cether / C water), for an organic substance X at room temperature is 13.

13 = (0.94/x) ÷ (0.06/y)

13 = (0.94/x) × (y/0.06)

13 = (15.667y/x)

(x/y) = (15.667/13) = 1.205

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