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5 mL of 8 N HNO(3), 4.8 mL of 5N HCl and...

`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.

Text Solution

Verified by Experts

The correct Answer is:
`(6.5376 g)`
Molecular weight of `Na_(2)CO_(3).10H_(2)O = 286`
`rArr` Molarity of carbonate solution `= (1)/(286) xx (1000)/(100) = 0.035`
`rArr` Normality of carbonate solution `= 2xx 0.035 = 0.07 N`
In acid solution : Normality of `HNO_(3) = (8 xx 5)/(2000) = 0.02`
Normality of `HCl = (5 xx 4.8)/(2000) =0.012`
Let normality of `H_(2)SO_(4)` in final solution be N.
`rArr (N + 0.02 +0.012)xx 30 = 0.07 xx 42.9`
`rArr N = 0.0681`
`rArr` Gram equivalent of `SO_(4)^(2-)` in `2L` solution `= 2 xx 0.0681`
`= 0.1362`
`rArr` Mass of `SO_(4)^(2-)` in solution `= 0.1362 xx (96)/(2) = 6.5376 g`
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