Answer
$4.0\times10^{-6}$
Work Step by Step
Using $
pH=-\log[H_3O^+]
$ or the formula for the pH of a solution, with $pH=
5.4
$, then
\begin{align*}\require{cancel}
5.4&=-\log_{10}[H_3O^+]
&(\text{use }\log b=\log_{10}b)
\\
5.4&=\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^{5.4}&=[H_3O^+]^{-1}
\\\\
\left(10^{5.4}\right)^{-1}&=\left([H_3O^+]^{-1}\right)^{-1}
\\\\
\dfrac{1}{10^{5.4}}&=[H_3O^+]
&(\text{use }a^{-1}=\dfrac{1}{a})
\\\\
[H_3O^+]&=\dfrac{1}{10^{5.4}}
.\end{align*}
Using the laws of exponents, the equation above is equivalent to
\begin{align*}\require{cancel}
[H_3O^+]&=\dfrac{1}{10^{0.4+5}}
\\\\&=
\dfrac{1}{10^{0.4}\cdot10^5}
&(\text{use }a^m\cdot a^n=a^{m+n})
\\\\&=
\dfrac{1}{10^{0.4}}\cdot\dfrac{1}{10^5}
\\\\&\approx
0.3981\cdot\dfrac{1}{10^5}
\\\\&\approx
0.3981\times10^{-5}
\\&\approx
3.981\times10^{-6}
\\&\approx
4.0\times10^{-6}
.\end{align*}
Hence, the hydonium ion concentration, $[H_3O^+]$, of $\text{
spinach
}$ is approximately $
4.0\times10^{-6}
$.
