The Ka of a monoprotic weak acid is 6.58 × 10-3. What is the percent ionization of a 0.171 M solution of this?
The Ka of a monoprotic weak acid is 6.58 × 10-3. What is the percent ionization of a 0.171 M solution of this acid?
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HA <--> H+ + A-
Ka = 6.58X10^-3 = [H+][A-]/[HA]
6.58X10^-3 = x^2 / 0.171-x
1.125X10^-3 -6.58X10^-3 x = x^2
x^2 + 6.58X10^-3 x - 1.125X10^-3 = 0
x = 0.0304
% ionization = 0.0304 / 0.171 X 100 = 17.7% ionized