1.325 g of anhydrous sodium carbonate are dissolved in water and the solution made up to 250 mL. On titration 25 mL of this solution neutralise 20 mL of a solution of suphuric acid. How much water should be added to 450 mL of this acid solution to make it exactly `N//12` ?

Explanation

Eq. mass of `Na_(2)CO_(3)=("Mol.mass")/(2)=(106)/(2)=53`
250 mL of the sodium carbonate solution contains = 1.325 g
1000 mL of the sodium carbonate solution contains
`=(1.325g)/(250)xx1000=5.300` g
Normality of `Na_(2)CO_(3)` solution` =("Strength "(g//L))/("Eq. mass")`
`=(5.30)/(53)=(1)/(10)N`
Applying `underset((Na_(2)CO_(3)))(N_(1)V_(1))-=underset((H_(2)SO_(4)))(N_(2)V_(2))`
`(1)/(10)xx25=N_(2)xx20`
`N_(2)=(25)/(10xx20)=(1)/(8)`
Applying `underset(("Before dilution"))(N_(B)V_(B))-=underset(("After dilution"))(N_(A)V_(A))`
`=(1)/(8)xx450=(1)/(12)xxV_(A)`
`V_(A)=(450xx12)/(8)=675 mL`
Water to be added for dilution =(675-450)=225 mL