Question

The dissolution of ammonia gas in water does not obey Henry's law. On dissolving, a major portion of ammonia, molecules unite with $H_{2}O$ to form $N{H}_{4}OH$ molecules. $N{H}_{4}OH$ again dissociate into $N{H}_4^{+}$and $O{H}^{-}$ions. In solution therefore, we have $N{H}_3$ molecules, $N{H}_{4}OH$ molecules and $N{H}_4^{+}$ions and the following equilibrium exist:
$N{H}_3(g)$ (pressure $p$ and concentration $C$ ) initially $\rightleftharpoons N{H}_3(l)+H_{2}O\rightleftharpoons N{H}_{4}OH\rightleftharpoons N{H}_4^{+}+O{H}^{-}$
Let $C_{1}mol/L$ of $N{H}_3$ pass in solution state a part of which on dissolution in water forms $C_{2}mol/L$ of $N{H}_{4}OH$. The solution contains $C_{3}mol/L$ of $N{H}_4^{+}$ions
Total concentration of ammonia, which can be determined by volumetric analysis is equal to:

• A. $C_1+C_2$
• B. $C_1+C_2+C_3$
• C. $C_1+C_3$
• D. $C_2+C_3$

Answer 1

Answer: (B)

The intermediate solution of acid will react with all the $N{H}_3$ present in solution