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 523: 72

Answer

(a) $[H_3O^+] = 1.7 \times 10^{- 2}M$ (b) $[H_3O^+] = 1.3 \times 10^{- 4}M$ (c) $[H_3O^+] = 3.7 \times 10^{- 9}M$ (d) $[H_3O^+] = 4.8 \times 10^{- 13}M$

Work Step by Step

We find: (a) $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 1.76}$ $[H_3O^+] = 1.7 \times 10^{- 2}M$ (b) $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 3.88}$ $[H_3O^+] = 1.3 \times 10^{- 4}M$ (c) $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 8.43}$ $[H_3O^+] = 3.7 \times 10^{- 9}M$ (d) $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 12.32}$ $[H_3O^+] = 4.8 \times 10^{- 13}M$
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.