Answer to Question #195354 in Chemistry for Kyle

Question #195354

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


1
Expert's answer
2021-05-20T06:10:09-0400

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

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