Ko/a = amt org layer/amt H2O layer.
K is 8
MeCl2 is 30 mL
H2O is 30 mL
Caffeine is 10g
Let X = amount extracted with org layer, then 10-x = amount left in H2O layer.
8 = (x/30)/(10-x)/30
You can cancel the 30 in the denominator. Solve for x.
You can do it three times for three extractions or you can use the following:
fn = [1+Kd*(Vo/Va)]-n
fn = fraction solute REMAINING in H2O layer
Kd is 8 in this case
Vo = volume organic
Va = volume H2O
n = number of extractions
Caffeine has a distribuation coefficient of about 8 between Me Cl2 and water. Given 30ml of water containing 10 grams of caffeine.
a. How much caffeine can be extracted with one 30mL portion of MeCl2?
b. How much caffeine can be extracted with three 10mL portions of Me Cl2?
c. Which is more efficient?
1 answer
1 answer