Five grams of Compound A is dissolved in 90 mL of water. The distribution coeffi

Question

Five grams of Compound A is dissolved in 90 mL of water. The distribution

coefficient for Compound A between hexanes and water is 5 (in other words, K

Compound

A(hexanes/water)= 5.)

5a.How much of Compound A will be in the hexanes if you extract it from the water one

time with 90 mL of hexanes?

5b. How much of Compound A will be in the hexanes if you extract it from the water

with three sequential extractions using 30 mL of hexanes each time, and then combine

the hexanes extracts?

5c. Compare the amount extracted in hexanes by single extraction (90 mL) versus 3

times extraction (3x30 mL). Which is more efficient?

Explanation / Answer

a) K = 5 = xg in 90mL/(5-x) g in 90 mL

5 = x/(5-x)

25 - 5x = x

6x = 25

x = 25/6 = 4.1667 grams

b) Using the three extraction method

5 = 3x1/(5-x1)

25 - 5x1 = 3x1

8x1 = 25

x1 = 25/8 = 3.125 gms

5 = 3x2/(1.875-x2)

9.375 = 8x2

x2 = 9.375/8 = 1.171 grams

5 = 3x3/(0.704-x3)

3.52 = 8x3

x3 = 3.52/8 = 0.44

Mass Extracted = 0.44 + 1.171 + 3.125 = 4.736

c) Hence the three times extraction method is more efficient then the single extraction method

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