Chapter 4 - Inverse, Exponential, and Logarithmic Functions - 4.4 Evaluation Logarithms and the Change-of-Base Theorem - 4.4 Exercises - Page 457: 33

$7.94 \times 10^{-8}$

We are given that $pH=-\log [H_3O^{+}]$ and $pH=7.1$ or, $-7.1=\log [H_3O^{+}]$ The above expression can be written as: $[H_3O^{+}]=10^{-7.1}$ Now, we will evaluate the result by using calculator . Therefore, our answer is: $10^{-7.1}=7.94 \times 10^{-8}$