Question

Butyric acid is responsible for the foul smell of rancid butter. Calculate the pH for a ____________________ (12)

Butyric acid solution that is 0.25 M. Ka of butyric acid (CH3CH2CH2COOH) at 25oC is 1.48 X 10-5.

Answer 1

Butyric acid ionizes partially, so you need the pka or ka value of butyric acid, this was given in the question.

So you required to calculate the concentration of H+ (OR H3O+)

Calculation

CH3CH2CH2COOH(aq) + H2O(l) -----> CH3CH2CH2COO-(aq) + H3O+(aq)

Ka IS DISSOCIATION CONSTANT WHICH IS CONCENTRATION OF PRODUCT/CONCENTRATION OF REACTANT.

Ka= [CH3CH2CH2COO-] * [H3O+] / [CH3CH2CH2COOH]

[H2O] is not included because its change in concentration is negligible.
After partial dissociation, [CH3CH2CH2COOH] is still approximately 0.25 M.
[CH3CH2CH2COO-] = [H3O+]
Ka of CH3CH2CH2COOH= 1.48 X 10-5

1.48 X 10-5 = [H30+]^2 / 0.25
[H30]² = 1.48 X 10-5 * 0.25
= 3.7 X 10-4
[H30+]= 1.92*10^-2

pH= -log[H30+]
pH= -log(1.92*10^-2)
pH= 1.7166

pH of 0.25M butyric acid is 1.71