Question

What is the fraction of association (?) for the following potassium propionate solutions? Ignore activities. The Ka of propanoic acid is 1.34 × 10-5.

(a) 4.00 × 10-1 M K(C2H5CO2) ****the answer is not 4.31e-5

(b) 4.00 × 10-2 M K(C2H5CO2) ****the answer is not 1.36e-4

(c) 4.00 × 10-12 M K(C2H5CO2) ***** the answer is not 1.365e-1

Answer 1

The dissociation constant Ka represent the dissociation of acid as follows :

C2H5CO2H + H2O <-----------------> C2H5CO2^-1 + H3O^+1

This means for association of acid the value will be just inverse of this. So, for association/formation of acid the reaction can be written as:

K(C2H5CO2) + H2O <--------------> C2H5COOH + K^+1 + OH^-1

And for this the constant K can be written as:

K = 1/Ka = 1/(1.34×10^-5) = 7.463 × 10^4

And K = [C2H5CO2H][K^+1][OH^-1]/[K(C2H5CO2)]

Using ICE table

[K(C2H5CO2)] . [C2H5CO2^-1] . [OH-]. [K+]

I. 0.4. 0. 0. 0

C . -x . +x . +x . +x

E . 0.4 - x . x . x . x

Putting all the value in the expression

7.463×10^4 = x^3/(0.4 - x)

x = 0.399999

So, the concentration of associated K(C2H5CO2) = 0.4 - x

= 0.4 - 0.399999 = 0.000001

Fraction alpha = associated / total concentration

= 0.000001/0.4 = 2.5 ×10^-6

Similaly can be calculated for the other parts