What is the percent ionization of a monoprotic weak acid solution that is 0.131 M? The acid-dissociation (or ionization) constant, Ka, of this acid is 4.92 x 10-10.

Answer 1

Let the weak monoprotic acid be represented by HA.

HA ==> H⁺ + A^-

The percent ionization will be the [H⁺] or [A^-] divided by the [HA] (x100%). So we need to find the [H⁺] which will be the same as the [A^-]. We can predict that the % ionization will be small since the Ka is so small.

Ka = [H⁺][A^-] / [HA]

4.92x10^-10 = (x)(x) / 0.131 - x (assume x is small relative to 0.131 and ignore it as an approximation)

4.92x10^-10 = x² / 0.131

x² = 6.45x10^-11

x = 8.03x10^-6 (note this is very small relative to 0.131 so above assumption was valid)

[H⁺] = [A^-] = 8.03x10^-6 M

% ionization = 8.03x10^-6 M / 0.131 M (x100%) = 0.00613% (very low, indeed)