A 0.25mol solution of CH3COOH is found to be 13% ionized.calculate the value of the acid equilibrum for the acid
I assume you meant 0.25M (not mol) solution....
0.25 x 0.13 = estimated 0.0325.
.......HAc ==> H^+ + Ac^-
I......0.25.....0.....0
C.......-x......x.....x
E.....0.25-x....x.....x
Ka = (H^+)(Ac^-)/(HAc)
x = 0.0325M
Substitute into Ka expression and solve for Ka.