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 `NH_(4),OH` molecules. A portion of the latter again dissociates into `NH_(4)^(+)`, and `OH^(-)` ions. In solution thercfore, we have `NH_(3)`, molecules, `NH_(4),OH` molecules and `NH_(4)^(+)`, ions and the following equilibrium exist:
`NH_(3(g))` (pressure P and concentration c) `rarr NH_(3(l)) + H_(2),Orarr NH_(4),OH rarr NH_(4)^(+) rarr,+OH^(-)`
Let `c_(1)`,mol/L of `NH_(3)`, pass in liquid state which on dissolution in water forms `c_(2)` mol/ L of `NH_(4),OH.` The solution contains `c_(3)`mol/L of `NH_(4)^(+)`, ions.
If P is the partial pressure of ammonia at equilibrium, then which of the following is constant?

Options

A. `P/c_(1)`
B. `P/c_(0)`
C. `P/c_(2)`
D. `P/c_(3)`

Answer

To solve the problem, we need to analyze the dissolution of ammonia in water and the equilibria involved. Here’s a step-by-step breakdown of the solution:

### Step 1: Understanding the Dissolution Process
When ammonia gas (NH₃) dissolves in water, it does not simply obey Henry's law due to the formation of ammonium hydroxide (NH₄OH) and its subsequent dissociation into ammonium ions (NH₄⁺) and hydroxide ions (OH⁻).

### Step 2: Writing the Relevant Equilibria
The dissolution can be represented by the following equilibria:
1. NH₃(g) ⇌ NH₃(l) + H₂O
2. NH₃(l) + H₂O ⇌ NH₄OH
3. NH₄OH ⇌ NH₄⁺ + OH⁻

### Step 3: Defining Concentrations
Let:
- \( c_1 \) = concentration of NH₃ in liquid state (mol/L)
- \( c_2 \) = concentration of NH₄OH formed (mol/L)
- \( c_3 \) = concentration of NH₄⁺ ions (mol/L)

### Step 4: Applying Henry's Law
According to Henry's law, the concentration of a gas in a liquid is proportional to its partial pressure. The relationship can be expressed as:
\[ C = K_H \cdot P \]
Where:
- \( C \) is the concentration of the dissolved gas (NH₃).
- \( K_H \) is the Henry's law constant.
- \( P \) is the partial pressure of the gas.

### Step 5: Identifying the Constant
From the above relationship, we can rearrange it to express the Henry's law constant:
\[ K_H = \frac{C}{P} \]
In our case, since \( C \) is the concentration of NH₃, we can substitute \( C \) with \( c_1 \) (the concentration of NH₃ in the liquid state):
\[ K_H = \frac{c_1}{P} \]

### Conclusion
Since \( K_H \) is a constant for a given gas at a specific temperature, it implies that the ratio of the concentration of NH₃ to its partial pressure remains constant. Therefore, the correct answer to the question is that the Henry's law constant \( K_H \) is constant.