Answer
$2.5\times10^{-5}$
Work Step by Step
Using $
pH=-\log[H_3O^+]
$ or the formula for the pH of a solution, with $pH=
4.6
$, then
\begin{align*}\require{cancel}
4.6&=-\log[H_3O^+]
\\
4.6&=-\log_{10}[H_3O^+]
&(\text{use }\log b=\log_{10}b)
\\
4.6&=\log_{10}[H_3O^+]^{-1}
&(\text{use }\log_b x^y=y\log_b x)
.\end{align*}
Since $\log_b y=x$ implies $y=b^x$, then the equation above implies
\begin{align*}\require{cancel}
10^{4.6}&=[H_3O^+]^{-1}
\\
\left(10^{4.6}\right)^{-1}&=\left([H_3O^+]^{-1}\right)^{-1}
\\
10^{-4.6}&=[H_3O^+]
\\
[H_3O^+]&=10^{-4.6}
.\end{align*}
Using the laws of exponents, the equation above is equivalent to
\begin{align*}\require{cancel}
[H_3O^+]&=\dfrac{1}{10^{4.6}}
\\\\&=
\dfrac{1}{10^{0.6+4}}
\\\\&=
\dfrac{1}{10^{0.6}\times10^4}
\\\\&=
\dfrac{1}{10^{0.6}}\times10^{-4}
\\\\&=
0.25\times10^{-4}
\\\\&=
2.5\times10^{-5}
.\end{align*}
Hence, the hydrogen ion concentration, $[H_3O^+]$, of orange juice is $
2.5\times10^{-5}
$.
