Answer
$1.0\times10^{-2}$
Work Step by Step
Using $
pH=-\log[H_3O^+]
$ or the formula for the pH of a solution, with $pH=
2.0
$, then
\begin{align*}\require{cancel}
2.0&=-\log_{10}[H_3O^+]
&(\text{use }\log b=\log_{10}b)
\\
2.0&=\log_{10}[H_3O^+]^{-1}
&(\text{use }\log_b x^y=y\log_b x)
.\end{align*}
Since $y=b^x$ implies $\log_b y=x,$ then the equation above is equivalent to
\begin{align*}\require{cancel}
10^{2.0}&=[H_3O^+]^{-1}
\\\\
\left(10^{2.0}\right)^{-1}&=\left([H_3O^+]^{-1}\right)^{-1}
\\\\
\dfrac{1}{10^{2.0}}&=[H_3O^+]
&(\text{use }a^{-1}=\dfrac{1}{a})
\\\\
[H_3O^+]&=\dfrac{1}{10^{2.0}}
.\end{align*}
Using the laws of exponents, the equation above is equivalent to
\begin{align*}\require{cancel}
[H_3O^+]&=1\times10^{-2}
.\end{align*}
Hence, the hydonium ion concentration, $[H_3O^+]$, of the $\text{
human gastric contents
}$ is $
1.0\times10^{-2}
$.
