What is the pH of a solution that is .060M in potassium propionic and .085M in propionic acid?

Ka=1.3E-5 for the acid.

The answer is +4.73 but I don't know how to go about the problem. I thought that the potassium propionic was reacting with the acid to form their conjugate acid and base but I wa told that the acid would be reacting with water and dissociate forming the ions of the acid. I'm not sure how this works.

Use the Henderson-Hasselbalch equation. That's what you use for buffers.

pH = pKa + log (base)/(acid)
The base is the K salt and the acid is propionic acid.

To determine the pH of a solution that contains both potassium propionic (C3H5KO2) and propionic acid (C3H6O2), you need to consider the dissociation of the acid in water. The acid, propionic acid, can be represented by the formula CH3CH2COOH.

First, let's write the dissociation equation for propionic acid in water:

CH3CH2COOH + H2O ⇌ CH3CH2COO- + H3O+

The equilibrium constant for this reaction is given by Ka = [CH3CH2COO-][H3O+]/[CH3CH2COOH].

Given that the concentration of propionic acid is 0.085 M, we can assume that x is the concentration of CH3CH2COO- and H3O+ formed and (0.085 - x) is the concentration of CH3CH2COOH remaining at equilibrium.

Using the Ka expression, we can write the equation:

1.3E-5 = (x)(x)/(0.085 - x)

Since the value of x is small compared to 0.085, we can make the approximation 0.085 - x ≈ 0.085. Thus, the equation simplifies to:

1.3E-5 = x^2/0.085

Rearranging the equation:

x^2 = (1.3E-5)(0.085)

x^2 ≈ 1.1E-6

Taking the square root of both sides:

x ≈ 1.0E-3

Now, we need to find the concentration of H3O+ which is equal to x, to use it to calculate the pH of the solution.

pH = -log[H3O+]

pH = -log(1.0E-3)

pH ≈ 3.0

Therefore, the pH of the solution is approximately 3.0.