Answer
a bog
Work Step by Step
Using $
pH=-\log[H_3O^+]
$ or the formula for the pH of a solution, with $H_3O^+=
2.5\times10^{-2}
$, then
\begin{align*}\require{cancel}
pH&=-\log(2.5\times10^{-2})
\\&=
-\left(\log2.5+\log10^{-2}\right)
&(\text{use }\log_b (xy)=\log_b x+\log_b y)
\\&=
-(\log2.5-2\log10)
&(\text{use }\log_b x^y=y\log_b x)
\\&=
-\left(\log2.5-2(1)\right)
&(\text{use }\log10=\log_{10}10=1)
\\&=
-\log2.5+2
.\end{align*}
Using a calculator, the value of $
\log2.5
$ is approximately $
0.3979
$. Hence, the equation above is equivalent to
\begin{align*}
pH&\approx-0.3979+2
\\&\approx
1.6021
.\end{align*}
Note that a pH value of $6.0$ to $7.5$ is a rich fen, a pH value of $3.0$ to $6.0$ is a poor fen, and a pH value of less than $3.0$ is a bog. Since the pH solution of water from the given wetland area is approximately $
1.6021
$, then the wetland area is considered a bog.
