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_{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 $y=b^x$ implies $\log_b y=x,$ then the equation above is equivalent to
\begin{align*}\require{cancel}
10^{4.6}&=[H_3O^+]^{-1}
\\\\
\left(10^{4.6}\right)^{-1}&=\left([H_3O^+]^{-1}\right)^{-1}
\\\\
\dfrac{1}{10^{4.6}}&=[H_3O^+]
&(\text{use }a^{-1}=\dfrac{1}{a})
\\\\
[H_3O^+]&=\dfrac{1}{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^{0.6+4}}
\\\\&=
\dfrac{1}{10^{0.6}\cdot10^4}
&(\text{use }a^m\cdot a^n=a^{m+n})
\\\\&=
\dfrac{1}{10^{0.6}}\cdot\dfrac{1}{10^4}
\\\\&\approx
0.2512\cdot\dfrac{1}{10^4}
\\\\&\approx
0.2512\times10^{-4}
\\&\approx
2.512\times10^{-5}
\\&\approx
2.5\times10^{-5}
.\end{align*}
Hence, the hydonium ion concentration, $[H_3O^+]$, of $\text{
bananas
}$ is approximately $
2.5\times10^{-5}
$.
