A monoprotic acid with a Ka of 2.92 × 10-5 has a partition coefficient of 4.2 (favoring octanol) when distributed between water and octanol. Find the formal concentration of the acid in each phase when 100 mL of 0.10 M aqueous acid is extracted with 31 mL of octanol at (a) pH 4.00 and (b) pH 8.00.
To solve this problem, we need to use the equation for partition coefficient (P) and the equations that relate pH and the concentration of the acid to the dissociation of the acid (Ka).
The equation for partition coefficient (P) is given by:
P = [acid]octanol / [acid]water
Where [acid]octanol and [acid]water are the concentrations of the acid in octanol and water, respectively.
Since we are given the partition coefficient (P = 4.2), we can rearrange the equation to solve for [acid]octanol:
[acid]octanol = P * [acid]water
To find the concentration of the acid in each phase, we need to use the equation that relates pH and the concentration of the acid to the dissociation of the acid. The equation is derived from the acid dissociation constant (Ka) and is given by:
[H+] = √(Ka * [acid])
With this equation, we can find the concentration of H+ ions in the aqueous phase, which will allow us to calculate the concentration of the acid in the water phase.
Now let's solve the problem for each scenario:
(a) pH 4.00:
First, we need to find the concentration of H+ ions in the water phase.
[H+] = 10^(-pH)
[H+] = 10^(-4.00) = 1.00 × 10^(-4) M
Now, using the equation [acid]octanol = P * [acid]water, we can determine the concentration of the acid in octanol:
[acid]octanol = 4.2 * [acid]water = 4.2 * 0.10 M = 0.42 M
So, the formal concentration of the acid in each phase when 100 mL of 0.10 M aqueous acid is extracted with 31 mL of octanol at pH 4.00, is 1.00 × 10^(-4) M in water and 0.42 M in octanol.
(b) pH 8.00:
Using the same approach, we can find the concentration of H+ ions in the water phase:
[H+] = 10^(-pH)
[H+] = 10^(-8.00) = 1.00 × 10^(-8) M
Now, let's calculate the concentration of the acid in octanol:
[acid]octanol = 4.2 * [acid]water = 4.2 * 0.10 M = 0.42 M
So, the formal concentration of the acid in each phase when 100 mL of 0.10 M aqueous acid is extracted with 31 mL of octanol at pH 8.00, is 1.00 × 10^(-8) M in water and 0.42 M in octanol.