Chapter 4 - Inverse, Exponential, and Logarithmic Functions - 4.4 Evaluation Logarithms and the Change-of-Base Theorem - 4.4 Exercises - Page 457: 36

$3.16 \times 10^{-7}$

We are given that $pH=-\log [H_3O^{+}]$ and $pH=6.5$ or, $-6.5=\log [H_3O^{+}]$ The above expression can be written as: $[H_3O^{+}]=10^{-6.5}$ Now, we will evaluate the result by using calculator . Therefore, our answer is: $10^{-6.5}=3.16 \times 10^{-7}$