What is the fraction of association?

Question

What is the fraction of association for each potassium propionate solution? The Ka of propanoic acid is 1.34 x 10^-5

2.00 x 10^-1 M K(C2H5COO2) = ?

2.00 x 10^-2 M K(C2H5COO2) = ?

2.00 x 10^-12 M K(C2H5COO2) = ?

Answer 1

The following ignores activities.

K(C2H5COO)will ionize completely to K⁺ and C2H5COO^- The C2H5COO^- will act as a base when in H2O as follows:

C2H5COO^- + H2O ==> C2H5COOH + OH^-

Let B^- be the base and HB the conjugate acid . The fractional association for the base is given by

[HB] / [B^-]

We are given [B^-] as it will be the concentration of the potassium salt. So now we need to find [HB]. This is done as follows:

Kb = [HB][OH^-] / [B^-] = (x)(x)/2x10^-1 - x

Kb = 1x10^-14/1.34x10^-5= 7.46x10^-10

7.46x10^-10 = x²/2x10^-1

x = [HB] = 6.11x10^-5

Fractional association = 6.11x10^-5/2x10^-1 = 3.05 x 10^-4

Repeat for the other 2 concentrations. Note however that in the Kb expression we ignored x in the denominator as it will be small relative to 2x10^-1. You will not be able to do that for the 3rd concentration and will have to use the quadratic equation.