10 " mL of " 2 M HCl and 20 " mL of " 1 M `HNO_3` and V volume of `5M H_2SO_4` are mixed together and the solution was made upto 5 L. 10 " mL of " this acid solution exactly neutralises 28.6 " mL of " `Na_2CO_3` solution containing 1 g of `Na_2CO_4.10H_2O` in 100 " mL of " water. Calculate the amount of `SO_4^(2-)` ions in grams present in solution.

Explanation

Mw of `Na_2CO_3.10H_2O=2xx23+12+3xx16+10xx18=286g`
Ew of `Na_2CO_3.10H_2O("valency factor"=2)=(286)/(2)=143g`
Strength of `Na_2CO_3.10H_2O=(1xx1000)/(100)=10gL^(-1)`
`N_(Na_(2)CO_(3))=("Strength")/(Ew)=(10)/(143)`
total mEq of mixture of acid in 5 L of solution`=10xx2xx1+20xx1xx1xxVxx5xx2` (n-factor)
`=20+20+10V=40+10V`
Now m" Eq of "acid in 10 mL solution
`=m" Eq of "Na_2CO_3` used for it
`=N_(Na_(2)CO_(3))xxV(mL)=(10)/(143)xx28.6`
`=2mEq(i n 10mL)`
`=(2xx5000)/(10)mEq(i n 5L)`
`=1000mEq`
m" Eq of "acid in 5 L solution`=1000mEq`
`40+10V=1000`
`10V=960`
`10V-=960-=m" Eq of "H_2SO_4-=m" Eq of "SO_4^^(2-)`
`m" Eq of "SO_4^(2-)=(weight)/(Ew)xx10^(3)=960`
`(weightxx10^(3))/((96)/(2))=960`
Weight of `SO_4^(2-)=(960xx48)/(1000)=4.6g`