A 0.490 M solution of an unknown monoprotic weak acid, HA, is 2.80% ionized. What is the value of Ka for HA?

Answer 1

HA ==> H^+ + A^-

Ka = (H^+)(A^-)/(HA)
You know HA is 0.490. If it is ionized 2.80%, then H^+ and A^- must be 0.490 x 0.0280 = ??. Substitute into Ka expression and solve.