Question

2.20 g of an ammonium salt was boiled with 75 mL of NaOH till the emission of ammonia gas ceased. The excess of unused NaOH solution required 70 mL of N/2 sulphuric acid for neutralisation. Calculate the percentage of ammonia in the salt.

Answer 1

$(N{H}_4{)}_{x}A+NaOH\to (Na{)}_{x}A+N{H}_3+H_{2}O$

Volume of NaOH consumed by $H_{2}S{O}_4$ = y

$\therefore N\times Y=\frac{N}{10}\times 70$

Y= 35mL

$\therefore$ Out of 75mL , 35mL consumed by $H_{2}S{O}_4$

Hence, 40 mL consumed by $N{H}_3$.

Amount of $N{H}_3$ in N NaOH = $\frac{\text{Eq. mass of}N{H}_3\times 40}{1000}$

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

Percentage of ammonia in salt = $\frac{0.68}{20}\times 100$ = 30.9 %