Equilibrium chem

Question

The Ka of a monoprotic weak acid is 0.00244. What is the percent ionization of a 0.137 M solution of this acid?

Answer 1

Let HA be the monoprotic acid.

HA ==> H⁺ + A^-

Ka = [H⁺][A^-] / [HA]

0.00244 = (x)(x) / 0.137 - x (ignore -x to simplify the calculations and avoid using the quadratic)

0.00244 = x2 / 0.137

x² = 3.34x10^-4

x = 1.82x10^-2 M = [H⁺] = [A^-] NOTE: this is 13.3 % of the value 0.137 so ignoring it above was not valid.

Repeating the calculation we have...

0.00244 = (x)(x) / 0.137 - x

x² = 3.34x10^-4 - 2.44x10^-3x

x² + 2.44x10^-3x - 3.34x10^-4 = 0

x = 0.0171 M = [H⁺] = [A^-]

Percent ionization = 0.0171 M / 0.137 M (x100%) = 12.5% ionized