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
a. 0.12m
b. 0.2m
To solve this problem, we need to understand the concept of partition coefficient and how it relates to the acid dissociation constant (Ka). Let's break down the problem into steps:
Step 1: Understand the problem and the given information.
- We are given that the butanoic acid has a partition coefficient of 3.0, favoring benzene.
- The acid dissociation constant (Ka) for butanoic acid is given as 1.52 x 10^(-5).
- We have 20.0 mL of 0.10 M aqueous butanoic acid.
- We are extracting the butanoic acid with 25 mL of benzene.
Step 2: Calculate the pH-dependent concentration of H+ ions in the aqueous phase.
- In part (a), we are given that the pH is 4.00.
- Recall that pH is a measure of the concentration of H+ ions in a solution.
- We can convert the pH to H+ concentration using the formula: [H+] = 10^(-pH).
- Substituting the given pH value into the equation, we find [H+] = 10^(-4.00).
Step 3: Calculate the concentration of butanoic acid in the aqueous phase.
- We have 20.0 mL of 0.10 M aqueous butanoic acid.
- To find the concentration, we divide the number of moles by the volume: Concentration = moles/volume.
- Moles = concentration x volume. So, moles = (0.10 mol/L) x (20.0 mL / 1000 mL/L) = 0.002 mol.
Step 4: Calculate the concentration of butanoic acid in the benzene phase.
- The partition coefficient (P) is defined as the ratio of the concentration of the compound in the organic phase (benzene) to the concentration in the aqueous phase.
- In this case, P = [butanoic acid]benzene / [butanoic acid]aqueous.
- Given that P = 3.0, we can rearrange the equation to find [butanoic acid]benzene = P x [butanoic acid]aqueous = 3.0 x (0.002 mol) = 0.006 mol.
In part (a) of the problem, the concentration of butanoic acid in each phase is:
- Aqueous phase: 0.002 mol
- Benzene phase: 0.006 mol
Step 5: Repeat the process for part (b) where pH = 10.
- Find the H+ concentration using [H+] = 10^(-pH) = 10^(-10).
- Calculate the concentration of butanoic acid in the aqueous phase as before: 0.002 mol.
- Calculate the concentration of butanoic acid in the benzene phase using P x [butanoic acid]aqueous = 3.0 x (0.002 mol) = 0.006 mol.
In part (b) of the problem, the concentration of butanoic acid in each phase is:
- Aqueous phase: 0.002 mol
- Benzene phase: 0.006 mol