Chapter 6

Question 1

Using equation 5.9, show that three extractions with 5-mL portions of a solvent give better recovery than a single extraction with 15mL of solvent when K = 0.5

Answer 1

Equation 5.9

F_A = (V_o / (KV_x + V_o))ⁿ

For the 3 extractions, the equation would look like:

F_A = (V_o / ((0.5 x 5) + V_o))³

For the 1 extraction, the equation would look like this:

F_A = (V_o / ((0.5 x 15) + V_o))¹

The cubed denominator (3 extractions) will be larger than the single denominator (1 extractions):

F_A = (V_o / ((0.5 x 5) + V_o))³> FA = (V_o / ((0.5 x 15) + V_o))¹

Question 2

Assume that the partition coefficient, K_f, for partitioning of compound A between diethyl ether and water is 3; that is, A preferentially partitions into ether.

a) Given 400 mL of an aqueous solution containing 12 g of compound A, how many grams os A could be removed from the solution by a single extraction with 200 mL of diethyl ether

Answer 2

Equation 5.9

F_A = (V_o / (KV_x + V_o))ⁿ

F_A = (400 / ((3 x 200) + 400))¹

F_A = 0.4 (40% remaining)

Fraction removed = 1 - 0.4 = 0.6 (60%)

Mass removed = (0.6)(12g) = 7.2 g compound A

Question 3

Assume that the partition coefficient, Kf, for partitioning of compound A between diethyl ether and water is 3; that is, A preferentially partitions into ether.

b) How many total grams of A can be removed from the aqueous solution with three successive extractions of 67 mL each?

Answer 3

Equation 5.9

F_A = (V_o / (KV_x + V_o))ⁿ

F_A = (400 / ((3 x 67) + 400))³

F_A = 0.295 (29.5%) remaining

Fraction removed = 1 - 0.295 = 0.705 (70.5%)

Mass removed = (0.705) x (12g) = 8.46 g compound A