Following equilibrium is upon dissociation of propanoic acid,
HC3H5O2 ⇄ H+ + C3H5O2-
Thus, dissociation constant of acid can be mathematically expressed as,
Ka = [tex] \frac{[H+]{C3H5O2-}}{[HC3H5O2]} [/tex]
Given: pH of HC3H5O2 = 4.2 and Ka = 1.34x10^-5
We know that, pH = -log[H+] = 4.2
∴ [H+] = 6.31 x 10^-5
From the reaction it can be seen that [H+] = [C3H5O2-]
∴ [HC3H5O2] = [H+] [C3H5O2-] /Ka = (6.31 x 10^-5)^2 / 1.34x10^-5 = 2.97 x 10^-4 M
Thus, equilibrum conc. of HC3H5O2 is 2.97 x 10^-4 M.
