A 0.50 M monoprotic acid has a pH of 1.0. Calculate it's percent in ionization. Express in three significant figure.
The percent ionization for an acid is: [H3O+] / [HA] × 100%
The chemical equation for the dissociation of the monoprotic acid (HA) is:
HA(aq) + H2O(l) ⇌ A−(aq) + H3O+(aq)
Since [H3O+] = 10-pH, we find that [H3O+] = 10-1.0 = 0.1 M,
so that percent ionization is:
(0.1 / 0.5) × 100% = 20.0 %
Answer: 20.0% ionized