Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 4 - Inverse, Exponential, and Logarithmic Functions - 4.4 Evaluation Logarithms and the Change-of-Base Theorem - 4.4 Exercises - Page 457: 34

Answer

$3.98 \times 10^{-4}$

Work Step by Step

We are given that $pH=-\log [H_3O^{+}]$ and $pH=3.4$ or, $-3.4=\log [H_3O^{+}]$ The above expression can be written as: $[H_3O^{+}]=10^{-3.4}$ Now, we will evaluate the result by using calculator . Therefore, our answer is: $10^{-3.4}=3.98 \times 10^{-4}$
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