Calculate the percent ionization of ha in a 0.10 m solution.

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 that we can neglect it and 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 = (1.8x10^-4)/(0.10) x 100% = 0.18%

explanation:

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