Answer
$0.520$
Work Step by Step
First, we need to calculate the concentration of $HCl$ in the solution. If the solution is $1.09\%\ HCl$ by mass then there are $\frac{0.0109g\ HCl}{1g\ Solution}$ we can multiply this by the given density of the solution to get the grams $HCl$ per milliliter then convert this to grams per liter and divide by the the molar mass of HCl to get moles per liter which is the concentration we want.
$\frac{0.0109g\ HCl}{1g\ solution}\times\frac{1.01g\ solution}{1mL\ solution}\times\frac{1000\ mL}{1\ L}\times\frac{1mol\ HCl}{36.46g\ HCl}= \frac{0.302mol\ HCl}{1L\ solution}= 0.302M HCl$
Now we will need to know that: $pH= -log[H_{3}O^{+}]$ and that the $[H_{3}O^{+}]$in a strong acid solution is equal to the concentration of that strong acid because strong acids completely dissociate in solution. Therefore, we can plug the concentration of strong acid in place of that of $H_{3}O^{+} $in the pH equation to get our answer.
Since $HCl$ is a strong acid:
$pH= -log(0.302)$
$pH=0.520$
