# Question 0813e

$\text{pH} = 5$
This is a useful formula I use to work out the $\text{pH}$ of a weak acid given $\text{K"_"a}$ and $\left[\text{HA}\right]$
"pH"=1/2"pK"_"a"-1/2log["HA"]#
$= \frac{1}{2} \times 8 - \frac{1}{2} \log 0.01$
$= 5$