Question

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

Answer 1

Following equilibrium is upon dissociation of propanoic acid,

HC3H5O2 ⇄ H+ + C3H5O2-

Thus, dissociation constant of acid can be mathematically expressed as,
Ka =
`\frac{[H+]{C3H5O2-}}{[HC3H5O2]}`

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.