Chemistry: Molecular Approach (4th Edition)

Published by Pearson
ISBN 10: 0134112830
ISBN 13: 978-0-13411-283-1

Chapter 16 - Section 16.5 - Autoionization of Water and pH - For Practice - Page 735: 16.3

Answer

a. pH = 8.02: Basic solution. b. pH = 11.85: Basic solution.

Work Step by Step

a. $$pH = -log[H_3O^+] = -log( 9.5 \times 10^{-9} ) = 8.02 $$ Since pH $\gt$ 7, the solution is basic. b. $$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 7.1 \times 10^{-3} } = 1.4 \times 10^{-12} \space M$$ $$pH = -log[H_3O^+] = -log( 1.4 \times 10^{-12} ) = 11.85 $$ Since pH $\gt$ 7, the solution is basic.
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