Question

a) You are given 200 mL of an aqueous solution contianing 10 grams of compound A, from which are asked to separate A. How many grams of A could be removed in a single extraction with 100 mL of ethyl ether? Asummer that the ratio of the concentration of A in ethyl ether to its concentration in water (i.e., the distribution coefficient for A in ethyl ether/water) is equal to 3.0.

b) How many total grams of A can be removed with three successive extractions of 33 mL of ethyl ether for each rinse? [Suggestion: For each extraction let x equal the weight extracted into the ether layer.]

Note: in case (a) the concentration in the ether later is x/100, and the water later is (10-x)/200: the ratio of these quantities is equal to the distribution coefficient.

Answer 1

1. For the 1^st extraction with 100ml of diethyl ether

Let x= mass of A extracted into the ether layer, 10-x= mass of A remaining in the aqueous layer.

Hence distribution coefficient= (mass of A/ volume of ethyl ether)/ (mass of A/volume of water)= 3

Hence (x/100)/(10-x)/200=3

2x/(10-x)= 3

2x= 30-3x, 5x=30, x=6gm

6 gm ofA is extracted into the ethyl ether phase.

1. When 33ml of ethyl ether is used. Let x= mass of A extracted into the ether phase, 10-x= mass of ether in water phase

Hence x/33/(10-x)/200 =3

200x/33*(10-x)=3

6.1x= 30-3x

9.1x= 30, x= 3.3 gm, 10-3.3= 6.7 gm of A will be there in the aqueous phase which will be taken for subsequent extraction. When 33ml of ethyl ether is used for second extraction

x/33(6.7-x)/200=3,x = mass of A extracted

6.1x= 3*(6.7-x)

9.1x= 3*6.7, x= 2.21 gm, mass of A remaining the water phase= 6.7-2.1= 4.6 gm

For the third extraction

(x/33)/(4.6-x)/200 = 3

6.1x= 3*(4.6-x)

9.1x= 3*4.6, x= 1.52 gm

Total of A extracted= 3.3+2.21+1.52=7.03 gm