Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning
ISBN 10: 1285199049
ISBN 13: 978-1-28519-904-7

Chapter 14 - Acids and Bases - Questions for Review and Thought - Topical Questions - Page 652b: 33d

Answer

$pH = 9.362$ $[H_3O^+] = 4.348 \times 10^{- 10}M$ This solution is basic.

Work Step by Step

$[OH^-] * [H_3O^+] = Kw = 10^{-14}$ $ 2.3 \times 10^{- 5} * [H_3O^+] = 10^{-14}$ $[H_3O^+] = \frac{10^{-14}}{ 2.3 \times 10^{- 5}}$ $[H_3O^+] = 4.348 \times 10^{- 10}M$ $pH = -log[H_3O^+]$ $pH = -log( 4.348 \times 10^{- 10})$ $pH = 9.362$ $[OH^-] > [H_3O^+]$: Basic solution: Assuming $25 ^{\circ}C$: pH > 7: Basic solution.
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