For complete neutralisation of 25 mL of `Na_(2)CO_(3)` solution (specific gravity `1.25g*mL^(-1)`), 32.9 mL of HCl solution containing 109.5g of the acid per litre is required. Calculate the volume of 0.84(N) `H_(2)SO_(4)` that will be neutralised by 125g of `Na_(2)CO_(3)` solution.

Explanation

`N_(HCl)=(109.5)/(36.5xx1)=3` millieqv. Of `Na_(2)CO_(3)`=millieqv. Of HCl
Now, `Nxx25=32.9xx3" "thereforeN_(Na_(2)CO_(3))=3.948`
`therefore` Volume of `Na_(2)CO_(3)` solution`=(125)/(1.25)=100mL`
`therefore`milliequivalent of `H_(2)SO_(4)`=milliequivalent of `Na_(2)CO_(3)`
or, `0.84xxV=100xx3.948," "therefore`Volume of `H_(2)SO_(4)=470mL`.