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

Vol. of `H_(2)SO_(4)V_(1)=?`
Volume of NaOH `V_(2)=200` ml.
Normality of `H_(2)SO_(4) N_(1)=0.1`
Normality of NaOH `N_(2)` = 0.2N
`V_(1)N_(1)=V_(2)N_(2)`
`V_(1)=(V_(2)N_(2))/N_(1)=(200xx0.2)/0.1=400`ml.