The volume of $0.05\text{N}$ ${\text{Na}}_2{\text{CO}}_3$  solution required to neutralize 200 mL of $0.02\text{M}$  ${\text{H}}_2{\text{SO}}_4$  solution is                               

Volume of ${\text{Na}}_2{\text{CO}}_3$ $={\text{V}}_{\text{1}}$

Molarity of ${\text{H}}_2{\text{SO}}_4$  ${\text{M}}_2=0.02$

Normality of ${\text{H}}_2{\text{SO}}_4$ ${\text{N}}_2={\text{M}}_2\times n$

Since $N=M\times n$

Then, normality is

  $N=0.02\times 2=0.04N$

Volume of${\text{H}}_2{\text{SO}}_4$  solution ,${\text{V}}_2=200\text{ mL}$

According to neutralization method,

              ${\text{N}}_1 {\text{V}}_1={\text{N}}_2{\text{V}}_2$

         $\Rightarrow {\text{V}}_1=\frac{{\text{N}}_2{\text{V}}_2}{{\text{N}}_1}$

          ${\text{V}}_{\text{1}}=\frac{0.04\times 200}{0.05}=160\text{mL}$