F06. C334
Acid-Base Extractions. Partition (or Distribution) Coefficient
Refer also to pp. 58-60 in your lab text
Partition (or distribution) coefficient kp:
kp = solubility of compound A in solvent 2 / solubility of compound A in solvent 1
or kp = (g/ml) organic layer / (g/ml) aqu. layer
Example:
Solubility of compound A in ether: 20.0 g/100 ml
Solubility of compound A in water: 10.0 g/100 ml
kp ?
kp = 20.0 g/100 ml ether / 10.0 g/100 ml water = 2.00
Problem 1: How much A total is extracted, if 100 ml of an aqueous solution of
2.5 grams of A are extracted twice with 100 ml of ether, using a fresh portion
of ether each time?
a) First extraction:
kp : (x g / 100ml) organic layer / [ (2.5 g - x)] / 100ml) aqu. layer = 2.00
Solving for x: x = 2.00 (2.5 –x) = 5.0 - 2.00 x ; 3.0 x = 5.0; x = 1.7
1.7 g of A are extracted in the ether layer after the first extraction.
Therefore, (2.5 – 1.7) g or 0.8 g are left in the aqueous layer.
b) Second extraction:
kp : (x g / 100ml) organic layer / [ (0.8 g - x)] / 100ml) aqu. layer = 2.00
Solving for x: x = 2.00 (0.8 –x) = 1.6 -2.00 x ; 3.0 x = 1.6; x = 0.53
0.53 g of A are extracted in the ether layer after the second extraction.
Total of A extracted: (1.7 + 0.53) g = 2.2 g
Problem 2: How much A would be extracted if instead of two consecutive
extractions as in Problem 1 only one extraction, but with a volume of 200
ml of ether were performed?
kp : (x g / 200ml) organic layer / [ (2.5 g - x)] / 100ml) aqu. layer = 2.00
Solving for x: x = 2.00 [2 (2.5 –x)] = 2.00(5.0 -2x) ; 5.0 x = 10.0; x = 2.0
In Problem 2 only 2.0 g of A is extracted which is less than in problem 1!