The neutralization of \(NaOH\) by \(H _{2} SO _{4}\) takes place as follows \(H _{2} SO _{4}+2 NaOH \longrightarrow Na _{2} SO _{4}+ H _{2} O\)
For complete neutralization Equivalents of acid = equivalents of base Equivalents of \(NaOH =\) moles \(\times\) acidity \(=1 \times 1=1\) Equivalents of \(H _{2} SO _{4}=\frac{x}{98} \times 2=\frac{x}{49}\) (Mol. mass of \(H _{2} SO _{4}=98\) ) Putting the values \(1 \times 1 =\frac{x}{49}\) \(\Rightarrow x =49\, g\) but \(H _{2} SO _{4}\) is \(70 \%\) let \(y g 70 \% H _{2} SO _{4}\) is required \(\frac{70}{100} \times y =49\) \(\Rightarrow y =70\, g\)