The Ka of a monoprotic weak acid is 3.29 × 10-3. What is the percent ionization of a 0.176 M solution of this acid?

Answer 1

............HA --> H^+ + A^-
I........0.176.....0......0
C.........-x.......x.......x
E.......0.176-x....x.......x

Ka = (H^+)(A^-)/(HA)
Substitute from the ICE chart into the Ka expression and solve for H^+.
%ionization = [(H^+)/0.176]*100 = ?