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Arrhenius Acid

The dissoluti...

Question

The dissolution of ammonia gas in water does not obey Henry's law. On dissolving, a major portion o fammonia, molecules unite with $H_{2} O$ to form ${N H}_{4} O H$ molecules. ${N H}_{4} O H$ again dissociates into ${N H}_{4}^{+}$ and ${O H}^{-}$ ions. In solution therefore, we have ${N H}_{3}$ molecules, ${N H}_{4} O H$ molecules and ${N H}_{4}^{+}$ ions and the following equilibrium exist:
${N H}_{3} (g)$ (pressure $P$ and concentration $c$ ) Initially $⇌ {N H}_{3} (l) + H_{2} O ⇌ {N H}_{4} O H ⇌ {N H}_{4}^{+} + {O H}^{-}$
Let $c_{1}$ $m o l / L$ of ${N H}_{3}$ pass in liquid state which on dissolution in water froms $c_{2}$ $m o l / L$ of ${N H}_{4} O H$ . The solution contains $c_{3}$ $m o l / L$ of ${N H}_{4}^{+}$ ions. The dissociation constant of ${N H}_{4} O H$ can be given as:

A

K_{b} = \frac{{(c_{3})}^{2}}{(c_{2} - c_{3})}

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B

K_{b} = \frac{{(c_{3})}^{2}}{c_{2}}

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C

K_{b} = \frac{c_{3}}{(c_{2} - c_{3})}

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D

K_{b} = \frac{c_{3}}{(c_{1} - c_{2})}

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Solution

The correct option is A $K_{b} = \frac{{(c_{3})}^{2}}{(c_{2} - c_{3})}$
As given,
${N H}_{4} O H c_{2} (c_{2} - c_{3}) ⇌ {N H}_{4}^{+} 0 c_{3} + {O H}^{-} 0 c_{3}$
$K_{b} = [N H_{4}^{+}] [O H^{-}] / [N H_{4} O H]$
$K_{b} = \frac{{c_{3}}^{2}}{(c_{2} - c_{3})}$

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Similar questions

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