Question

What is the pH of a 0.5 m solution of sodium propionate, NaC₃H₅O₂ at 25 degrees C?

(For propionic acid, HC₃H₅O₂, K_a = 1.3 x 10^-5 at 25 degrees C)

Answer 1

The Ka of acid will be

CH3CH2COOH --> CH3COO- + H+

Ka = [CH3COO-] [ H+] / [CH3COOH]

The sodium acetate will hydrolyze as

CH3CH2COO-Na+ + H2O --> CH3CHCOOH + Na+ + OH-

Kh = Kwater / Ka

Kh = 10^-14 / 1.3 x 10-5

Kh = 0.769 X 10^-9

Kh = [CH3CH2COOH] [ OH-] / [CH3CH2COO-]

Kh = x^2 / 0.5-x

0.769 X 10^-9 = x^2 / 0.5-x

We may ignore x in denominator as Kh is very low

0.769 X 10^-9 = x^2 / 0.5

x^2 = 0.385 X 10^-9

x= 1.96 X 10^-5

[OH-]= 1.96 X 10^-5

pOH = -log[OH-] = 4.71

pH = 14-4.71 = 9.29