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
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.
To find the concentration of butanoic acid in each phase (water and benzene), we need to consider the partition coefficient and the initial concentration of butanoic acid.
The partition coefficient (P) is defined as the ratio of the concentration of butanoic acid in benzene (Cbenzene) to the concentration in water (Cwater), so we have:
P = Cbenzene/Cwater
For part (a), when the pH is 4.00, we need to make some assumptions. Butanoic acid is a weak acid, so it will partially dissociate in water to form butanoate ions (C4H7O2-) and hydrogen ions (H+). At pH 4.00, we assume that most of the butanoic acid will be in the undissociated form.
First, calculate the initial concentration of butanoic acid in water (Cwater) using the given volume and molarity:
Cwater = (0.10 M) x (20.0 mL/1000 mL) = 0.002 M
Next, use the partition coefficient to find the concentration of butanoic acid in benzene (Cbenzene):
P = Cbenzene/Cwater
3.0 = Cbenzene/0.002
Cbenzene = 3.0 x 0.002 = 0.006 M
So, the concentration of butanoic acid in water is 0.002 M, and in benzene is 0.006 M for part (a).
For part (b), when the pH is 10, we assume that the butanoate ion will be the prevailing form in water since the pH is higher than the pKa of butanoic acid. Therefore, the concentration of butanoic acid in water (Cwater) is negligible at pH 10.
Using the initial concentration of butanoic acid in water (Cwater) = 0.002 M and the partition coefficient (P = 3.0), we can find the concentration of butanoic acid in benzene (Cbenzene):
P = Cbenzene/Cwater
3.0 = Cbenzene/0 (negligible concentration of butanoic acid in water at pH 10)
Cbenzene = 3.0 x 0 = 0 M
So, the concentration of butanoic acid in water is negligible, and in benzene is 0 M for part (b).
In summary:
(a) When the pH is 4.00, the concentration of butanoic acid is 0.002 M in water and 0.006 M in benzene.
(b) When the pH is 10, the concentration of butanoic acid is negligible in water and 0 M in benzene.