Answer
$1.58 \times 10^{-5}$
Work Step by Step
We are given that $pH=-\log [H_3O^{+}]$ and $pH=4.8$
or, $-4.8=\log [H_3O^{+}]$
The above expression can be written as:
$[H_3O^{+}]=10^{-4.8}$
Now, we will evaluate the result by using calculator .
Therefore, our answer is: $10^{-4.8}=1.58 \times 10^{-5}$