A mmonia is highly soluble gas in water and gives a alkaline solution of `NH_(4) OH`. What volume of `NH_(3)` gas at `STP` will be required to the passed in `100 mL` of `0.5 M H_(2) SO_(4)` to bring down its strength to `0.25 M`?
(For titrations with aqueous `NH_(3)`, it is assumed that `NH_(4) OH` dissociates to 100% extent)
a. 2.24 L b. 1.68 L c. 1.12 L d. 0.56 L`

Explanation

c. Inittially mEq of `H_(2) SO_(4) = 100 xx 0.5 xx 2 = 100`
mEq of `H_(2) SO_(4)` required `= (0.25 xx 2) xx 100 = 50`
mEq of `H_(2) SO_(4)` used ` = 100 - 50`
= mEq of `NH_(4) OH`
= mmol of `NH_(4) OH`
(`n` factor = 1)
mmol of `NH_(3)` = mmol of `NH_(4) OH = 50`
`V_(NH_(3)` at `STP = 50 xx 10^(-3) xx 22.42 = 1.12 L`