This may not be the easiest way but it's what I would do.
First, calculate (H^+), (butanoate), and (butanoic acid) in the 0.1 M solution at pH of 4.00. Then since I like to work in grams with these things I would convert the butanoic acid part into gram and for any number let's just call that y.
Then Ko/a = 3.0 = (x/25)/(y-x)/20.
Solve that equation for x and y-x
Then you can change grams in each phase to any other concentration unit you wish. Do the same thing for pH 10.
I think the idea here is that the pH of 10 leaves essentially all of the butanoic acid as the acid while the pH 4 leaves a little less as the acid and a little more as the ion. The ion will not be extracted by the benzene since it is ionic.
Butanoic acid (Ka = 1.52 x 10-5) has a partition coefficient of 3.0 (favouring benzene) when
distributed between water and benzene. Find the concentration of butanoic acid in each phase when
20.0 mL of 0.10 M aqueous butanoic acid is extracted with 25 mL of benzene at
(a) pH 4.00
(b) pH 10
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