Answer
$\text{pH }5.9$
Work Step by Step
Using $
pH=-\log[H_3O^+]
$ or the formula for the pH of a solution, with $H_3O^+=
1.3\times10^{-6}
$, then
\begin{align*}\require{cancel}
pH&=-\log(1.3\times10^{-6})
\\&=
-\left(\log1.3+\log10^{-6}\right)
&(\text{use }\log_b (xy)=\log_b x+\log_b y)
\\&=
-(\log1.3-6\log10)
&(\text{use }\log_b x^y=y\log_b x)
\\&=
-\left(\log1.3-6(1)\right)
&(\text{use }\log10=\log_{10}10=1)
\\&=
-\log1.3+6
.\end{align*}
Using a calculator, the value of $
\log1.3
$ is approximately $
0.1139
$. Hence, the equation above is equivalent to
\begin{align*}
pH&\approx-0.1139+6
\\&\approx
5.9
.\end{align*}
Hence, Tuna has $
\text{pH }5.9
$.
