100 ml of 0.2 M ${\mathrm{H}}_2{\mathrm{SO}}_4$ is added to 100 ml of 0.2 M NaOH. The resulting solution will be

Meq of 0.2 M ${\mathrm{H}}_2{\mathrm{SO}}_4 = \frac{2 \times 0.2 \mathrm{M}}{1000} \times 100 = 0.04 \mathrm{meq}/\mathrm{L}$

Meq of .2 M $\mathrm{NaOH} = \frac{ 0.2 }{1000} \times 100 = 0.02 \mathrm{meq}./\mathrm{L}$

                  $\left[{\text{H}}^{+}\right] \text{left unreacted} = 0.04 -0.02=0.02$

Total volume = $200 ; \left[{\mathrm{H}}^{+}\right] = \frac{0.02}{200} = .0001={10}^{-4}\text{M}$

pH =$-\log \left[{\mathrm{H}}^{+}\right]$  = 4.

Another method:

                           Strength of ${\mathrm{H}}_2{\mathrm{SO}}_4 ={\mathrm{V}}_{\mathrm{A}}{\mathrm{N}}_{\mathrm{A}}=100\times 0.4 =40 (\mathrm{Since} {\mathrm{N}}_{\mathrm{A}}=2\times \mathrm{M})$

                           Strength of $\mathrm{NaOH} ={\mathrm{V}}_{\mathrm{B}}{\mathrm{N}}_{\mathrm{B}} =100\times 0.2=20$

      $\therefore {\mathrm{V}}_{\mathrm{A}}{\mathrm{N}}_{\mathrm{A}} > {\mathrm{V}}_{\mathrm{B}}{\mathrm{N}}_{\mathrm{B}}$  and the resultant solution is acidic.