20 mL of a solution of `H_2SO_4` neutralises 21.2 mL of 30% solution (w/v) of `Na_2CO_3`. How much water should be added to each 100 mL of the solution to bring down its strength to decinormal ?

Answer

500 mL

Explanation

The corresponding equation is:
`underset(105.99 g)(Na_(2)CO_(3)) + underset(98.076 g)(H_(2)SO_(4)) to Na_(2)SO_(4) + H_(2)O + CO_(2)`
The amount of `Na_2CO_3` present in 21.2 mL solution.
`=3/100 xx 21.2 = 0.63 g`
`therefore 105.99 g` of `Na_(2)CO_(3)` react with `H_(2)SO_(4) = 98.076 g`
`therefore 0.636 g` of `Na_(2)CO_(3)` will react with `H_(2)SO_(4)`
`=98.076/105.99 xx 0.636 = 0.588 g`
This much `H_(2)SO_(4)` is present in 20 mL.
`therefore w=(NEV)/1000`
`therefore 0.588 = (N xx 49.038 xx 20)/1000`
(Eq. wt. of `H_(2)SO_(4) = (98.076)/2 = 49.038`)
`therefore N=0.599 = 0.6`
Therefore, the normality of the given `H_2SO_4` solution is 0.6 N. Suppose, v mL of water are required to be added to 100 mL of it to make it decinormal `N/10`
`therefore 0.6 xx 100 = 1/10 xx (100 + v)` which gives v = 500 mL
Hence, 500 mL of water should be added to each 100 mL of the solution of `H_2SO_4` to bring down its strength to decinormal.