The Ka of a monoprotic weak acid is 7.30 × 10-3. What is the percent ionization of a 0.181 M solution of this acid?
HA ---------------------------> H+ + A-
0.181 0 0
0.181-x x x
Ka = [H+][A-]/[HA]
7.30 x 10^-3 = x^2 / 0.181 -x
x^2 + 7.30 x 10^-3 x - 1.32 x 10^-3 =0
x = 0.0329
percent ionization = 0.0329 x 100 / 0.181
= 18.2 %
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