A 10.0 ml of `(NH_(4))_(2) SO_(4)` solution was treated with excess of `NaOH`. The ammonia evolved was absorbed in 50 ml of 0.1 N HCl. The excess HCl required 20 ml of 0.1 N. `NaOH`. Calculate the strength of `(NH_(4))_(2) SO_(4)` in the solution.

Explanation

`(NH_(4))_(2)SO_(4)+2NaOHrarr Na_(2)SO_(4)+2Nh_(4)OH`
`NH_(4)OH rarr NH_(3)+H_(2)O`
`"Excess "HCl+NaOH rarr NaCl +H_(2)O`
`NaOH` used by excess `HCl="20 ml of 0.1 N"`
`"Excess HCl "="20 ml of 0.1 N"`
`HCl" used by "NH_(3)="30 ml of 0.1 N"`
`"Ammonia produced "= "30 ml of 0.1 N"`
`=" 30 ml of 0.1 M"`
`(NH_(4))_(2)SO_(4)" used "="30 ml of 0.05 M"`
`"Mass of "(NH_(4))_(2)SO_(4)=(30xx0.05)/(1000)`
`=0.198g`
`"Thus "(NH_(4))_(2)SO_(4)="0.198 g/litre"`