If the ph of hc3h5o2 is 4.2 and the ka 1.34x10^-5, what is the equilibrium concentration

Following equilibrium is upon dissociation of propanoic acid,

HC3H5O2    ⇄     H+    +    C3H5O2-

Thus, dissociation constant of acid can be mathematically expressed as,
Ka = 

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.