When an aqueous solution of an organic compound is shaken with an immiscible org

Question

When an aqueous solution of an organic compound is shaken with an immiscible organic solvent, such
as diethyl ether, the solute distributes itself between the two phases. When the two phases separate
into two distinct layers, an equilibrium will have been established such that the ratio of the
concentrations of the solute in each solvent defines a constant, K, called the distribution coefficient (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 diethyl ether/water system is 3.0.
If you were given a solution containing 8.0 g of X in 500 mL of water, and wanted to extract
compound X into diethyl ether, show that it would be more effective to extract X using three 50 mL
aliquots of diethyl ether rather than a single 150 mL aliquot. (HINT: Determine how much of X
would remain in the aqueous solution in each case.)

Explanation / Answer

Distribution coefficient K = amount of solute in organic phase/amount of solute in aqueous phase

K = diethyl ether/water = 3

We have 8 g of solute in 500 mL water

Extraction method 1.

Extraction with 3 50 mL portions of diethylether

let x g of solute remains in aqueous phase after first extraction, then

8-x = g of solute in organic phase

So, Feeding the value in K we get,

K = 3 = [(8-x)/50]/(x/150]

3x/150 = (8-x)/50

150x = 1200 - 150x

x = 4 g

So, the amount of solute remaining in aqueous phase = 4 g

amount of solute in diethylether phase = 8-4 = 4 g

Now, let x g of solute remains in aqueous phase after second extraction then,

4-x = g of solute in organic phase.

Repeating the same procedure,

3 = [(4-x)/50]/(x/150)

3x/150 = 4-x/50

150x = 600 - 150x

x = 2 g of solute remains in aqueous (H2O) phase

amount of solute extracted in organic diethylether phase = 4-2 = 2 g

Now, let x of solute remains after third extraction in aqueous phase then,

amount of solute extracted in organic phase = 2-x g

Feedin values,

3 = [(2-x)/50]/(x/150)

3x/150 = 2-x/50

150x = 300 - 150x

x = 1 g of solute remaining in water phase after thid extraction

amount of solute extracted in organic phase = 2-1 = 1 g

Total amount of solute extracted in diethylether phase = 4+2+1 = 7 g

Extraction Method 2.

Extraction with 1 150 mL portion of diethylether

Now lets say we do a single extraction with 150 mL of diethylether

then, say x g of solute reamains in water after extraction then,

8-x = g of solute extracted in organic phase

Feed values,

3 = [(8-x)/150]/(x/150)

3x/150 = 8-x/150

3x = 8-x

x = 2 g of solute remains in water phase

amount of solute extracted in diethylether phase = 8-2 = 6 g

So we can clearly see that multiple extraction method 1 is more efficient (total amount of solute extracted is 7 g) as opposed to single extraction (total amount of solute extracted is 6 g) method 2.

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