20 mL of a solution of sulphuric acid neuitralise 21.2 mL of 30% soution of sodium carbonate. How much water should be added to 100 mL of this solution to bring down its down its strength to decinormal?

Explanation

Step1. Calculation of normality of sulphuric acid solution
Strength of `Na_(3)CO_(3)` solution `= 30 g//100 mL = 300 gL^(-1)`
Normality of `Na_(2)CO_(3)` solution `= ("Strength")/("Equivalent mass")=((300 gL^(-1)))/((53"g equiv"^(-1)))`
`=5.66 "equiv. L"^(-1) =5.66N`
`N_(1)xxV_(1)-=N_(2)xxV_(2)or N_(1)=(N_(2)xxV_(2))/(V_(1))`
`H_(2)SO_(4))" "(Na_(2)CO_(3))`
`N_(1)=((5.66N)xx(21.2"mL"))/((20.0mL))=6.0N`
Step II. Calculation of volume of water to be added
Normality of `H_(2)SO_(4)` solution `= 6.0 N`
Volume of `H_(2)SO_(4)` solution `= 100 mL`
Normality of diluted `H_(2)SO_(4)` solution `= 0.1 N`
Volume of diluted `H_(2)SO_(4)` solution may be calculated as :
`underset(("Diluted"H_(2)SO_(4)))(N_(1)xxV_(1))=underset(("Concentrated" H_(2)SO_(4)))(N_(2)xxV_(2))`
`V_(1)=(N_(2)xxV_(2))/(N_(1))=((6.0N)xx(100mL))/((0.1N))=6000mL`
Volume of water to be added `= (6000-100)=5900` mL.