Calculate the volume of 0.1 N `H_(2)SO_(4)` required to neutralise 200 ml. of 0.2 N NaOH solution.
It is an acid base neutralisation reaction.
Hence, at the neutralisation point.
Number of equivalents of acid = Number of equivalents of base.

Explanation

Formulae `N_(1)V_(1)=N_(2)V_(2)`
`0.1xxV_(1)`=0.2xx200 `N_(1)=0.1` N
`V_(1)`=?
`V_(2)`=200" ml"
`N_(2)`=0.2 N