`5 mL` of `8 N HNO_(3), 4.8 mL` of `5 N HCl`, and a certain volume of `17 m H_(2) SO_(4)` are mixed together and made upto `2 L`. `30 mL` of the acid mixture exactly neutralises `42.9 mL` of `Na_(2 CO_(3)` solution containing `0.1 g` of `Na_(2) CO_(3). 10 H_(2) O` in `10 mL` of water. Calculate:
The amount (in g) of the sulphate ions in the solution.

Explanation

For `H_(2)SO_(4)`, n factor`=2`
Let there be `n m" Eq of "H_(2)SO_(4)`.
m" Eq of "all acids in `2L=(5xx8)+(48.5xx5)+n`
`=64+n`
Normality`=((6n+n))/(2000)`
Normality of `Na_(2)CO_(3).10H_(2)O` is
`(1)/((286)/(2))xx(1)/(100)xx1000=0.07`
`thereforeN_(1)V_(1)=N_(2)V_(2)`
or `((64+n))/(200)xx30=0.07xx42.9`
or `n=136.2 m" Eq of "H_(2)SO_(4)`
Moles of `H_(2)SO_(4)=(136.2)/(2)xx10^(-3)`
Mass of `SO_(4)^(2-)=(136.2)/(2)xx10^(-3)xx96`
`=6.5376g`