Standardization of Acid Solution: Calculate the normality of a solution of `H_(2)SO_(4)` if 40.0 mL of the solution reacts completely with 0.364 gram of `Na_(2)CO_(3)`. Strategy: Refer to the balanced equation. We are given the mass of `Na_(2)CO_(3)`, so we convert grams of `Na_(2)SO_(3)` to equivalents of `Na_(2)CO_(3)`, then to equivalents of `H_(2)SO_(4)`, which let us calculate the normality of the `H_(2)SO_(4)` solution.

Explanation

Since `H_(2)SO_(4)` is a dibasic acid, 1 mol `H_(2)SO_(4)` contains `2 eq. H_(2)SO_(4)`. By law of equivalence, there must be equal number of equivalents of all reactants and products in a balanced chemical equation.
The balanced equation for the reaction of `H_(2)SO_(4)` with `Na_(2)CO_(3)` interpreted in terms of equivalents, is
`underset(1 mol 2 eq.)(H_(2)SO_(4))+underset(1 mol 2 eq.)(Na_(2)CO_(3))rarr underset(1 mol 2 eq.)(Na_(2)SO_(4))+underset(1 mol 2 eq.)(CO_(2))+underset(1 mol 2 eq.)(H_(2)O)`
So, `1 eq. Na_(2)CO_(3)= 1/2 mol Na_(2)CO_(3)=53 g`
First we calculate the number of equivalents of `Na_(2)CO_(3)` in the sample
no. eq. `Na_(2)CO_(3)=(mass of Na_(2)CO_(3))/("gram equivalent mass of " Na_(2)CO_(3))`
`(0.364 g Na_(2)CO_(3))/(53.0 g Na_(2)CO_(3)//eq. Na_(2)CO_(3))`
`=6.86xx10^(-2) eq. Na_(2)CO_(3)`
Because no. eq. `H_(2)SO_(4)= no . eq. Na_(2)CO_(3)`, we can write
`L_(H_(2)SO_(4))xxN_(H_(2)SO_(4))=6.86xx10^(-3) eq. H_(2)SO_(4)`
`N_(H_(2)SO_(4))=(6.86xx10^(-3) eq. H_(2)SO_(4))/(L_(H_(2)SO_(4))`
`=(6.86xx10^(-3) eq. H_(2)SO_(4))/(0.040 L)`
`=0.172 N H_(2)SO_(4)`