# Chapter 15 - Acids and Bases - Questions and Problems - Page 659: 15.76

Please see the work below.

#### Work Step by Step

We know that $log[H_3O^+]=-PH=-3.85$ $log[H_3O^{+}]=-3.85$ $[H_3O^+]=anti log(-3.85)$ $[H_3O^+]=10^{-3.85}=1.41\times 10^{-4}M$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.