Question

$1.13g$ of an ammonium salt was boiled with $100mL$ of $N$ $NaOH$ solution till the evolution of ammonia gas was given off. $60mL$ of $N$ $H_2{SO}_4$ were required to neutralize the excess of $NaOH$ present in remaining solution. Find out the % of ${NH}_3$ in the salt.

Answer 1

$NaOH+H_{2}S{O}_4\to \underset{\left(salt\right)}{N{a}_{2}S{O}_4}+H_{2}O$

$\therefore$ volume of $H_{2}S{O}_4$ used $=$ vol. of NaOH(excess)

$=60$ml

$\therefore$ volume of NaOH that reacted with ammonium salt

$=100-60=40$ml

$\underset{ammoniumsalt}{N{H}_4^{+}}+NaOH\to \underset{\left(ammonia\right)}{N{H}_3}+H_{2}O+N{a}^{+}$

$\therefore$ volume of ammonium salt$=$vol. of NaOH reacted with it

$=40$ml

So, amount of $N{H}_3$ in $40$ml of $N{H}_3$

$=\frac{eq.massofN{H}_3\times 40}{1000}$

$=\frac{17\times 40}{1000}=0.68$g

$\therefore$ present of $N{H}_3$ in ammonium salt$=\frac{0.68}{1.13}\times 100$

$=60.17\%$.