Question

A weak acid (HA) has a pKa of 4.666. If a solution of this acid has a pH of 4.294, what percentage of the acid is not ionized? (Assume all H in solution came from the ionization of HA.)

Please help me out...I am seriously not sure where to begin. I converted pka to Ka but then I'm not sure where to go from there or if I am on the right track. Any detailed help of how to solve the problem would be appreciated. I don't want just the answer. Thank you!

Answer 1

               HA(aq) <==> H+(aq) + A-(aq)

pH of weak acid = 1/2(pKa - log C)

   (4.294) = (1/2)(4.666 - log C)

Initial concentration of Acid   = 1.19*10^(-4) M

Ka of acid = 10^(-pka)   = 10^(-4.666) = 2.158*10^(-5)

Ka = [A-][H+]/[HA] = 2.158*10^(-5)

[H+] = [A-] = x

[HA] = 1.19*10^(-4) - x

x^2/((1.19*10^(-4))-x) = (2.158*10^(-5))

x = 4.1*10^(-5) M = [H+] = [A-]

concentration of acid dissociated = (4.1*10^(-5) M

percentage of acid dissociated =concentration of acid dissociated / Initial concentration of Acid *100

percentage of acid dissociated   = (4.1*10^(-5) /(1.19*10^(-4)))*100

                                          = 34.45%