The dissolution of ammonia gas in water does not obey Henry's law. On dissolving, a major portion of ammonia, molecules unite with \( \mathrm{H}_{2} \mathrm{O} \) to form \( \mathrm{NH}_{4} \mathrm{OH} \) molecules. \( \mathrm{NH}_{4} \mathrm{OH} \) again dissociates into \( \mathrm{NH}_{4}^{+} \)and \( \mathrm{OH}^{-} \)ions. In solution therefore, we have \( \mathrm{NH}_{3} \mathrm{molecules,}^{2} \), \( \mathrm{NH}_{4} \mathrm{OH} \) molecules and \( \mathrm{NH}_{4}^{+} \)ions and the following equilibrium exist :
\( \mathrm{NH}_{3}(g) \) (pressure \( P \) and concentration c) Initially \( \rightleftharpoons \mathrm{NH}_{3}(l)+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{NH}_{4} \mathrm{OH} \rightleftharpoons \mathrm{NH}_{4}^{+}+\mathrm{OH}^{-} \)
Let \( c_{1} \mathrm{~mol} / \mathrm{L} \) of \( \mathrm{NH}_{3} \) pass in liquid state which on dissolution in water forms \( c_{2} \mathrm{~mol} / \mathrm{L} \) of \( \mathrm{NH}_{4} \mathrm{OH} \). The solution contains \( c_{3} \mathrm{~mol} / \mathrm{L} \) of \( \mathrm{NH}_{4}^{+} \)ions.
Degree of dissociation of ammonium hydroxide is :
(A) \( c_{1} \)
(B) \( c_{3} / c_{1} \)
(C) \( c_{3} / c \)
(D) \( c_{3} / c_{2} \)
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