`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.

Explanation

Let the volume of 17 M ( i.e ., 34 N) `H_(2)SO_(4)` solution be v mL .
` :. ` total m.e of the acid mixture = `8 xx 5 + 5 xx 4.8 + 34 v " " ..(Eqn.1)`
= `(64 + 34v)`
` :.` normal of the mixture = `(m.e)/( " total volume (mL)") " "…(Eqn .1)`
` = (64 + 34 v)/(2000) `
m.e of 30 mL of this acid mixture ` = (64+ 34v)/2000 xx 30`
Now , normality of `Na_(2)CO_(3) . 10 H_(2)O ` solution = `(g//litre)/("eq.wt")`
` = 10/43 `
` {{:( "grams/litre of "Na_(2)CO_(3).10H_(2)O=10),(" and eq. wt"=(mol.wt)/2 = 286 /2 ):}}`
` :. ` m.e of 42.9 mL of `Na_(2)CO_(3).10H_(2)O ` solution = `10/143 xx 42.9`
Thus , m.e of 30 mL of acid mixture
= m.e of `42.9 " mL of " Na_(2)CO_(3) . 10 H_(2)O ` solution .
` :. ` m.e of N . (i.e ., 17 M) `H_(2)SO_(4) = 34 xx 68/17 " " ...(Eqn.1)`
` = 136 `
` :. ` equivalent of `H_(2)SO_(4) = 136/1000 = 0.136 " " ...(Eqn.3)`
` :. " equivalent of "SO_(4)^(2-) = 0.136 " "....(Eqn . 7ii)`
weight of `SO_(4)^(2-) = eq xx eq .wt of SO_(4)^(2-) " " ....(Eqn.4i)`
` = 0.136 xx 48 `
` = 6.528 g `
`( " eq. wt of " SO_(4)^(2-)= (" ionix wt")/("valency")= 96/2 = 48 )`