Answer
$\text{pH }4.3$
Work Step by Step
Using $
pH=-\log[H_3O^+]
$ or the formula for the pH of a solution, with $H_3O^+=
5.0\times10^{-5}
$, then
\begin{align*}\require{cancel}
pH&=-\log(5.0\times10^{-5})
\\&=
-\left(\log5.0+\log10^{-5}\right)
&(\text{use }\log_b (xy)=\log_b x+\log_b y)
\\&=
-(\log5.0-5\log10)
&(\text{use }\log_b x^y=y\log_b x)
\\&=
-\left(\log5.0-5(1)\right)
&(\text{use }\log10=\log_{10}10=1)
\\&=
-\log5.0+5
.\end{align*}
Using a calculator, the value of $
\log5.0
$ is approximately $
0.6990
$. Hence, the equation above is equivalent to
\begin{align*}
pH&\approx-0.6990+5
\\&\approx
4.3
.\end{align*}
Hence, Sodium bicarbonate has $
\text{pH }4.3
$.
