The van't Hoff factor i for an infinitely dilute solution of ${\mathrm{NaHSO}}_4$ is: 

If the solution is infinitely dilute, ${\mathrm{NaHSO}}_4$ would dissociate completely. 

$\therefore {\mathrm{NaHSO}}_4\to {\mathrm{Na}}^{+}+{\mathrm{HSO}}_4^{-}$

${\mathrm{HSO}}_4^{-}$ would further dissociate as: 

${\mathrm{HSO}}_4^{-}\to {\mathrm{H}}^{+}+{{\mathrm{SO}}_4^{-}}^2$

$\mathrm{i}=3$