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HA ==> H+ + A-
So %-ionization = [H+]/[HA] x 100%
Ka = [H+][A-]/[HA] = 3.2 x 10^-7
Let [H+] = x. Then [A-] = x and [HA] = 0.10 – x.
But x is so small .
So that we can neglect and we can say that [HA] = 0.10.
Ka = (x)(x)/(0.10) = 3.2 x 10^-7
X^2 = 3.2 x 10^-8, so x = 1.8 x 10^-4 = [H+]
%-Ionization = (1.8×10^-4)/(0.10) x 100%
= 0.18%
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