Question

$2.26g$ of an ammonium salt were treated with $100mL$ of normal $NaOH$ solution and boiled till no more of ammonia gas was given off. The excess of $NaOH$ solution left over required $60mL$ normal sulphuric acid. Calculate the percentage of ammonia in the salt.

Answer 1

$60mL$ normal $H_{2}S{O}_4\equiv 60mL$ normal $NaOH$
This, $(100-60)mL$ normal $NaOH$ were consumed by ammonium salt.
So, $40mL$ normal $NaOH\equiv 40mL$ normal $N{H}_3$
Amount of $N{H}_3$ in $40mL$ normal $N{H}_3$
$=\frac{Eq.massofN{H}_3\times 40}{1000}$
$=\frac{17\times 40}{1000}=0.68$
So, % of ammonia in the ammonium salt $=\frac{0.68}{2.26}\times 100$
$=30.09$.