General Chemistry: Principles and Modern Applications (10th Edition)

Published by Pearson Prentice Hal
ISBN 10: 0132064529
ISBN 13: 978-0-13206-452-1

Chapter 16 - Acids and Bases - Example 16-2 - Relating [H(3)0+],[OH],pH, and pOH - Page 706: Practice Example A

Answer

$[H^+] = 1.41 \times 10^{-3}$ $[OH^-] = 7.08 \times 10^{-12}$

Work Step by Step

Equations that will be used: $[H^+] = 10^{-pH}$ $[OH^-] = 10^{-pOH}$ $pH + pOH = 14$ 1. Calculate the $[H^+]$ $[H^+] = 10^{-2.85} = 1.41 \times 10^{-3}$ 2. Find the pOH $2.85 + pOH = 14$ $pOH = 14 - 2.85 = 11.15$ 3. Calculate $[OH^-]$ $[OH^-] = 10^{-11.15} = 7.08 \times 10^{-12}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

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.