The Ka of a monoprotic weak acid is 3.18 × 10-3. What is the percent ionization of a 0.122 M solution of this acid?
1) acid is a monoprotic weak acid,
2) Ka = 3.18*10^-3
HA <-------> A- + H+
I 0.122 0 0
C 0.122-x x x
E 0.122-x x x
Ka = [H+][A-] / [HA] = 3.18*10^-3
x^2 = (3.18*10^-3)*(0.122-x)
x^2 + (x*3.18*10^-3) - (0.122*3.18*10^-3) = 0
(x - 0.0181708) (x + 0.0213508) = 0
If we solve for the x = 18.2*10^-3 M
So, [H+] = 18.2*10^-2 M
% ionization = [H+] / [HA] =((18.2*10^-3) / 0.122)*100 = 14.9%
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