Answer: first you have to calculate the amount ionized. We will say it is x mol / L
then % ionization = (amount ionized / initial concentration) * 100%
x can be calculated using an ice chart
HC3H5O2 Â Â Â Â Â -----> H+ Â Â + C3H5O2-
Initial HC3H5O2 = 0.250 Â
change        = -x
equilibrium     = 0.250 - x
initial H+ Â Â Â Â Â Â = 0
change        = +x
equilibrium     = x
C3H5O2- initial  = 0
change        = +x
equilibrium     = x
Ka     = [H=][C3H5O2-] / HC3H5O2]
 1.3 * 10 ^ -5   = [x][x] / (0.250 - x) Â
So 1.3 * 10 ^ -5 * (0.250 - x) Â = x ^ 2 Â
   3.25 * 10^ -6 - (1.3 * 10^-5)x = x^2  now this is a quadratic equation and you have to rearrange it and solve for x
x^2 + 1.3 * 10^-5)x - 3.25 * 10^ -6 = 0
use the equation x = {-b (+ or -)[b^2 - 4.a.c] ^ 1/2} / 2a
you should get x = 1.80 * 10 ^ -3 or x = -1.80* 10^-3
but x can not be negative..
so x = 1.80 * 10 ^ -3
so percent ionization = (1.80 * 10 ^ -3 / 0.250) * 100%
                =0.72 %
the other way which is more easier is Â
assuming that x is very small and therefore 0.250 - x is approximately equals to 0.250 Â
then 1.3 * 10^-5 = x^2 / 0.250
so x^2 Â Â = 1.3 * 10^-5 * 0.250
    x  = 1.80 * 10 ^-3
then percent ionization is = (1.80 * 10 ^ -3 / 0.250) * 100%
                =0.72 %
if the percent ionization is > 5 % you can not do that approximation. in such a case you have to solve the quadratic equation. that is why I showed both methods.
now you can do the parts b and c
b answer : percent ionization = 1.27 % Â
c answer : 2.54%
good luck
