Calculate the pH  of  ${\text{10}}^{\text{-2}}{\text{\hspace{0.17em}\hspace{0.17em}N\hspace{0.17em}\hspace{0.17em}H}}_{\text{2}}{\text{SO}}_{\text{4}}$

 ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\to {\text{SO}}_{\text{4}}^{\text{2-}}{\text{+2H}}^{\text{+}}$
 ${\text{Before \hspace{0.17em}ionisation 10}}^{\text{-2 }}\text{N 0 0}$
 ${\text{After ionisation 0 10}}^{-2}{\text{ 10}}^{-2}$
 $\because {\text{1\hspace{0.17em}equivalent\hspace{0.17em}of\hspace{0.17em}H}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{\hspace{0.17em}gives\hspace{0.17em}1\hspace{0.17em}equivalent\hspace{0.17em}of\hspace{0.17em}H}}^{\text{+}}$
 $\therefore \left[{\text{H}}^{\text{+}}\right]{\text{=10}}^{\text{-2}}$
 $\text{Now, pH=}-{\text{logH}}^{+}\text{=}-{\text{log10}}^{-2}\text{=2}$