At equilibrium, the value of [H+] in a 0.260M solution of an unknown acid is 0.00404M . Determine the degree of ionization and the Ka of this acid.

Question

At equilibrium, the value of [H⁺] in a 0.260M solution of an unknown acid is 0.00404M . Determine the degree of ionization and the K_a of this acid.



Answer 1

Let HA represent the weak acid.

HA <==> H⁺ + A^-

If [H⁺] = 0.00404 M, then [A^-] also = 0.00404 M. From these data, we can calculate the Ka...

Ka = [H⁺][A^-] / [HA] = (0.00404)² / 0.260

Ka = 6.28x10^-5

Degree of ionization = 0.00404 / 0.260 = 0.0155 or 1.55% ionized