Question

`5 mL` of `8 N HNO_(3), 4.8 mL` of `5N HCl` and a certain volume of `17 M H_(2)SO_(4)` are mixed together and made upto `2 "litre"`. `30 mL` of this acid mixture exactly neutralizes `42.9 mL` of `Na_(2)CO_(3)` solution containing `1 g Na_(2)CO_(3)`. `10 H_(2)O "in" 100 mL` of water. Calculate the amount of sulphate ions in `g` present in solution.

Answer 1

Meq.of `HNO_(3) = 5 xx 8 = 40`
Meq. of `HCl = 4.8 xx 5 = 24`
Meq. of `H_(2)SO_(4) = V xx 17 xx 2 = 34V`
(Let `V mL` of `H_(2)SO_(4)`)
`:.` Total Meq. of acid in `2 litre` solution `= 40+24+34V`
`= 64 + 34V`
Now Meq. of acid in `30 mL` solution = Meq. of `Na_(2)CO_(3)` used for it
`= 42.9 xx (1 xx 1000)/(286 // 2 xx 1000) = 3`
`(because N_(Na_(2)CO_(3)) = (1)/(286//2) xx (1000)/(100))`
`:.` Meq. of acid in `2 litre` solution `= (3 xx 2000)/(30) = 200`
`:. 64 + 34V = 200`
`:. 34V = 200 - 64 = 136`
Now Meq. of `H_(2)SO_(4) =` Meq. of `SO_(4)^(2-) = 34V = 136`
`:.` Meq. of `SO_(4)^(2-) = 136`
`:. (w)/(96//2) xx 1000 = 136`
`:.` Weight of `SO_(4)^(2-) = 6.528 g`