Question

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?

Answer #1

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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