Posted by For Dr. Bob on .
The question asks, "Butanoic acid has a partition coeff of 3.0 when distributed b/t water and benzene. Find the formal concentration of butanoic acid in each phase when 100 mL of 0.10 M aqueous butanoic acid is extracted with 25 mL of benzene at pH 10.00."
I know that the first step is to find D, where D = K(Ka)/[(Ka+[H+)].
So I calculated:
D = 3.0(1.52x10^-5)/[(1.52x10^-5+10^-10)].
The answer should have been D = 1.97 x 10^-5, but I didn't get this.
You answered: If the question is worded correctly, then Kd water/benzene = 3.0
However, the distribution coefficient USUALLY is listed as organic/water.
So my question is, what do I need to do to get 1.97 x 10^-5? Where did I go wrong?
Th question is asking for the concn of butanoic acid in the water phase and in the benzene phase? Is the answer you are looking up either of these concentrations or does it say specificially that D = 1.97....?
For Dr. Bob,
Yes. That's all I am given on the question. The answer key says that D = 1.97 x 10^-5.