An unknown amount of nitric acid $(HN{O}_3)$ was added to deionized $($pure$)$ water and the reaction below occurs. If the equilibrium $pH$ was $5\mathrm{.}0,$ how much acid $(mo\frac{l}{L}$ was added?

$HN{O}_{3}harr{H}^{+}+N{O}_3^{-}$