Introductory Chemistry (5th Edition)

Published by Pearson
ISBN 10: 032191029X
ISBN 13: 978-0-32191-029-5

Chapter 14 - Acids and Bases - Exercises - Problems - Page 524: 80

Answer

(a) 1.35 : Basic solution. (b) 11.51 : Acidic solution. (c) 4.27 : Basic solution (d) 1.92 : Basic solution.

Work Step by Step

pOH < 7: Basic solution pOH = 7: Neutral solution pOH > 7: Acidic solution 1. Calculate the pOH value: $pOH = -log[OH^-]$ $pOH = -log(0.045)$ pOH = 1.35 2. Calculate the pOH value: $pOH = -log[OH^-]$ $pOH = -log(3.1 \times 10^{-12})$ pOH = 11.51 3. Calculate the pOH value: $pOH = -log[OH^-]$ $pOH = -log(5.4 \times 10^{-5})$ pOH = 4.27 4. Calculate the pOH value: $pOH = -log[OH^-]$ $pOH = -log(0.012)$ pOH = 1.92
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.