Answer to Question #195354 in Chemistry for Kyle

Question

A 0.50 M monoprotic acid has a pH of 1.0. Calculate it's percent in ionization. Express in three significant figure.

Answer 1

Solution:

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