The Ka of a monoprotic weak acid is 4.47 × 10-3. What is the percent ionization of a 0.184 M solution of this acid?

Answer 1

.............HA ==> H^+ + A^-
I.........0.184......0.....0
C...........-x.......x......x
E.........0.184-x....x......x

Ka = (H^+)(A^-)/(HA)
Substitute the equilibrium amounts into the Ka expression and solve for H^+).
Then % ion = (H^+)*100/0.184

Answer 2

So i got Ka = X^2/(0.184-x)

not sure what to do next.