I understand that the 2.7 of problem #1 is the distribution coefficient(all of these are equilibrium type problems) but I don't know the system you are using for "solvent 1" and "solvent 2". Am I correct that Korganicphase/aqueousphase = 2.7 or is it the reciprocal of that.
Here is the formula to use for #1.
fn = [1+K*(Vo/Va)]-n
fn = fraction remaining in the aq phase after n extractions.
K = distribution constant organic/aqeuous
Vo = volume of the organic phase
Va = volume of the aqueous phase
n = # extractions.
For the other problems, use
Ko/a = (concn org layer)/(concn aq layer)
A) A compound distributes between water (solvent 1) and benzene (solvent 2) with Kp = 2.7. If 1.0g of the compound
were dissolved in 100mL of water, how much compound could be extracted by three 10-mL portions of benzene?
B) 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?
C) 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.
D) 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.
I have no idea how to approach these problems. Any help is much appreciated!
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I am Dr. Henary and I am very mad after seeing this. I will use a thicker accent for future lectures because of you. Have a good day.
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