3) I would look at adjusting the pH such that most of the compound is in the unionized form.
2)
Kd = (concn org phase)/(concn H2O phase)
I would choose a convenient value to start with such as 10 grams. And choose a convenient volume for phase 1 of say 100 mL and phase 2 of 50 mL.
Then 2 = (x/50)/[(10-x)/100] and solve for x which will be the grams in the organic phase(2). 10-x will be what is left in the H2O phase(1). Then calculate the percent extracted. Follow with the second, third, and fourth extractions, calculating percent extraction after each.
I'm so confused as to how to even begin the problem and where to put whats given to me into an equation.
1) If the value of Kp is 0.5 for the distribution of a compound between pentane (solvent 2) and water (solvent 1), and equal volumes of the two solvents were used, how many extractions of the aqueous layer will be required to recover at least 90% of the compound?
2) Calculate the % of a compound that can be removed from liquid phase 1 by using ONE to FOUR extractions with a liquid phase 2. Assume that Kp = 2 and the volume of phase 2 equals to 50% that of phase 1.
3) A slightly polar organic compound distributes between diethyl ether and water with a partition coefficient equal to 3 (in favor of the ether). What simple method can be used to increase the partition coefficient? Explain.
2 answers
For #1. I'm not certain how solvent 1 and solvent 2 work; I ASSUME K is org/H2O = 0.5. If that is the other way around, you simply change the nomenclature but work the problem the same way.
Choose a number like 10 g for the compound to be extracted and 100 mL for the equal volumes of material.
2 = (org)/(H2O) = (x/100)/[(10-x)/100] and solve for x for the organic phase (I get 3.33 g so the amount left in the water phase is 10-3.33 = 6.667 and the percent extracted is (3.33/10)100 = 33.3%. Therefore, you can get 66.7% if you extract twice or 99.9% if you extract three times. The problem say AT LEAST 90% so the answer is 3 extractions as a minimum. Remember to make sure K is 0.5 for the way I've worked it; otherwise, K would be 2 if "solvent 1" and "solvent 2" mean something different.
Choose a number like 10 g for the compound to be extracted and 100 mL for the equal volumes of material.
2 = (org)/(H2O) = (x/100)/[(10-x)/100] and solve for x for the organic phase (I get 3.33 g so the amount left in the water phase is 10-3.33 = 6.667 and the percent extracted is (3.33/10)100 = 33.3%. Therefore, you can get 66.7% if you extract twice or 99.9% if you extract three times. The problem say AT LEAST 90% so the answer is 3 extractions as a minimum. Remember to make sure K is 0.5 for the way I've worked it; otherwise, K would be 2 if "solvent 1" and "solvent 2" mean something different.
1 answer