Question

The dissolution of ammonia gas in water does not obey Henry's law. On dissolving, a major portion of ammonia molecules unites ${{H}_{2}}O$ to form $N{{H}_{4}}OH$ molecules.$N{{H}_{4}}OH$ again dissociates 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 $N{{H}_{3}}$ passing in liquid state which on dissolution in water forms ${{c}_{2}}$mol/L of $N{{H}_{4}}^{+}$ ions.
If P is the partial pressure of ammonia at equilibrium, then which of the following is constant?
A.$\dfrac{P}{{{c}_{1}}}$
B.$\dfrac{P}{{{c}_{2}}}$
C.$\dfrac{P}{{{c}_{3}}}$
D.$\dfrac{P}{{{c}_{4}}}$

Answer 1

Hint: Before solving this question, we should know that Ammonia gas reacts with water rather than getting dissolved in it. Since it gets dissolved which hampers Henry’s law. Ammonia has a very strong hydrogen bonding with water rather than simple intermolecular forces which means it can easily get soluble in water and henry law cannot handle it.

Complete answer:
William Henry in 1803 gave this gas law which states “The amount of a given gas at constant temperature dissolves in a given type and the volume of liquid is directly proportional to the partial pressure of the gas in equilibrium with that liquid.” In short solubility of gas proportionate to the partial pressure of the gas above the liquid.
If we see Henry’s law: It is a $\propto \,{{P}_{N{{H}_{3}}}}$
Here is the amount of gas that is dissolved per unit volume of solvent.
Therefore, $\,{{P}_{N{{H}_{3}}}}\propto \,\,{{c}_{N{{H}_{3}}}}$
                   ${{P}_{N{{H}_{3}}}}\,=\,{{K}_{H}}\,{{c}_{N{{H}_{3}}}}$
Here, P is the partial pressure of ammonia gas in the atmosphere above liquid
c is the concentration of ammonia gas
K is the henry law constant
 ${{K}_{H}}=\dfrac{{{P}_{N{{H}_{3}}}}}{{{c}_{N{{H}_{3}}}}}$
          = $\dfrac{P}{{{c}_{1}}}$

So, Option (A) $\dfrac{P}{{{c}_{1}}}$ is correct.

Note:
The few limitations of Henry’s law are: If the molecules are not in the state of equilibrium then this law does not hold. If gases are under extremely high pressure then this law fails. If the gas and the solution react with each other take part in a chemical reaction then also Henry’s law does not hold.