Answer
$\text{pH }11.6$
Work Step by Step
Using $
pH=-\log[H_3O^+]
$ or the formula for the pH of a solution, with $H_3O^+=
2.5\times10^{-12}
$, then
\begin{align*}\require{cancel}
pH&=-\log(2.5\times10^{-12})
\\&=
-\left(\log2.5+\log10^{-12}\right)
&(\text{use }\log_b (xy)=\log_b x+\log_b y)
\\&=
-(\log2.5-12\log10)
&(\text{use }\log_b x^y=y\log_b x)
\\&=
-\left(\log2.5-12(1)\right)
&(\text{use }\log10=\log_{10}10=1)
\\&=
-\log2.5+12
.\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+12
\\&\approx
11.6
.\end{align*}
Hence, Ammonia has $
\text{pH }11.6
$.
