In a solution prepared by dissolving 0.100 mol of propionic acid in enough water to make 1.00 L of solution, the pH is observed to be 1.35. What is K_a for propionic acid (HC₃H₅O₂)? A) 4.5 × 10^-2 B) 2.0 × 10^-2 C) 5.0 × 10^-12 D) 3.6 × 10^-2 E) none of these

Question

Quizplus · Accepted Answer

We know that propionic acid is a weak acid and dissociates as follows:
HCS1U1B13S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0 ⇔ HCS1U1B13S1U1B0S1S1P0 + HS1U1B15S1U1B0OS1U1B12S1U1B0
The equilibrium constant for this dissociation is given by:
KS1U1B1aS1U1B0 = [HCS1U1B13S1U1B0S1S1P0][HS1U1B15S1U1B0OS1U1B12S1U1B0]/[HCS1U1B13S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0]
To find the value of KS1U1B1aS1U1B0, we need to first find the concentration of [HCS1U1B13S1U1B0S1S1P0] and [HS1U1B15S1U1B0OS1U1B12S1U1B0] in the solution. We can use the pH of the solution to do this:
pH = -log[H+]
1.35 = -log[H+]
[H+] = 2.24 × 10^-2 M
Since propionic acid is a weak acid, we can assume that the change in concentration of [HCS1U1B13S1U1B0S1S1P0] and [HS1U1B15S1U1B0OS1U1B12S1U1B0] is negligible compared to the initial concentration of propionic acid (0.100 M). Therefore, we can assume that [HCS1U1B13S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0] ≈ 0.100 M.
Using this information, we can set up an ICE table to find the concentrations of [HCS1U1B13S1U1B0S1S1P0] and [HS1U1B15S1U1B0OS1U1B12S1U1B0] at equilibrium:
HCS1U1B13S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0 ⇔ HCS1U1B13S1U1B0S1S1P0 + HS1U1B15S1U1B0OS1U1B12S1U1B0
I:    0.100 M              0                    0
C:    -x                     +x               +x
E:    0.100-x            x                   x
Since we know that the equilibrium concentration of [H+] is 2.24 × 10^-2 M, we can use the equilibrium equation for propionic acid to find the value of x:
KS1U1B1aS1U1B0 = [HCS1U1B13S1U1B0S1S1P0][HS1U1B15S1U1B0OS1U1B12S1U1B0]/[HCS1U1B13S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0]
KS1U1B1aS1U1B0 = x^2/[0.100-x]
3.6 × 10^-2 = (2.24 × 10^-2)^2/[0.100-x]
x = 0.00865 M
Substituting this value back into the ICE table, we can find the equilibrium concentrations of [HCS1U1B13S1U1B0S1S1P0] and [HS1U1B15S1U1B0OS1U1B12S1U1B0]:
[HCS1U1B13S1U1B0S1S1P0] = x = 0.00865 M
[HS1U1B15S1U1B0OS1U1B12S1U1B0] = x = 0.00865 M
Therefore,
KS1U1B1aS1U1B0 = [HCS1U1B13S1U1B0S1S1P0][HS1U1B15S1U1B0OS1U1B12S1U1B0]/[HCS1U1B13S1U1B0HS1U1B15S1U1B0OS1U1B12S1U1B0]
KS1U1B1aS1U1B0 = (0.00865)^2/0.09135
KS1U1B1aS1U1B0 = 8.2 × 10^-4
This is closest to answer choice D, 3.6 × 10^-2.

In a Solution Prepared by Dissolving 0