Calculating the pH given number of moles of HI and 500. ml of water
Simple questions are often interesting. 

(1) Most importantly, volumes are rarely additive. So the notion of adding the volume of the HI and the volume of water just doesn't work. 

(2) In an answer to a question about hydroiodic acid,  Curt F. showed a chart of concentration and density. For the $\pu{20 ^\circ C}$ values the liters of water in a liter of the acid solution can easily be calculated. So for the OP to get an error of 1% or less, the molarity would have to be about 0.2 molar or less. 

\begin{array}{|c|c|c|c|}
\hline
\text{Concentration, %}(w/w) &amp; \text{Density, }\pu{kg/L}(\text{@ }\pu{20 ^\circ C}) &amp;  \text{Molarity} &amp; \pu{L(water)/L(acid)}  \\ \hline
5.2 &amp; 1.0342 &amp; 0.420 &amp; 0.980 \\ \hline
10.8 &amp; 1.0812 &amp; 0.922 &amp; 0.964\\ \hline
16.4 &amp; 1.1226 &amp; 1.44 &amp; 0.938 \\ \hline
22.4 &amp; 1.1765 &amp; 2.06 &amp; 0.913 \\ \hline
27.2 &amp; 1.2333 &amp; 2.62 &amp; 0.898 \\ \hline
33.1 &amp; 1.2918 &amp; 3.34 &amp; 0.864 \\ \hline
38.7 &amp; 1.3605 &amp; 4.12  &amp; 0.834\\ \hline
42.9 &amp; 1.4208 &amp; 4.77 &amp; 0.811 \\ \hline
48.7 &amp; 1.5072 &amp; 5.74 &amp; 0.773 \\ \hline
53.0 &amp; 1.5913 &amp; 6.59 &amp; 0.748 \\ \hline
57.0 &amp; 1.6933 &amp; 7.55 &amp; 0.728 \\ \hline
\hline
\end{array}

You can see from the red line that the data is not linear.