The equilibrium constant for the ionization of `RNH_(2)` (g) in water as
`RNH_(2)(g)+H_(2)O(l)hArrRNH_(3)^(+)(aq)+OH^(-)(aq)`
is `8xx10^(-6) at 25^(@)C.` find the pH of a solution at equilibrium when pressure of `RNH_(2)`(g) is `0.5` bar :

Options

A. `~~12.3`
B. `~~11.3`
C. `~~11.45`
D. None

Answer

To find the pH of a solution at equilibrium when the pressure of `RNH2` (g) is `0.5` bar, we will follow these steps:

### Step 1: Write the equilibrium expression
The ionization reaction is given as:
\[ RNH_2(g) + H_2O(l) \rightleftharpoons RNH_3^+(aq) + OH^-(aq) \]

The equilibrium constant \( K_c \) for this reaction is given by:
\[ K_c = \frac{[RNH_3^+][OH^-]}{[RNH_2]} \]

### Step 2: Set up the initial conditions
Initially, we have:
- Pressure of `RNH2` = 0.5 bar
- Concentration of `RNH3^+` = 0
- Concentration of `OH^-` = 0

### Step 3: Define changes at equilibrium
Let \( x \) be the concentration of `RNH3^+` and `OH^-` produced at equilibrium. Therefore, at equilibrium:
- Pressure of `RNH2` = 0.5 bar (remains the same)
- Concentration of `RNH3^+` = \( x \)
- Concentration of `OH^-` = \( x \)

### Step 4: Convert pressure to concentration
Using the ideal gas law, we can convert the pressure of `RNH2` to concentration:
\[ [RNH_2] = \frac{P}{RT} \]
Assuming \( R = 0.0831 \, \text{L bar K}^{-1} \text{mol}^{-1} \) and \( T = 298 \, \text{K} \):
\[ [RNH_2] = \frac{0.5}{0.0831 \times 298} \approx 0.0202 \, \text{mol/L} \]

### Step 5: Substitute into the equilibrium expression
Now substituting into the equilibrium expression:
\[ K_c = \frac{x \cdot x}{0.0202} = \frac{x^2}{0.0202} \]
Given \( K_c = 8 \times 10^{-6} \):
\[ 8 \times 10^{-6} = \frac{x^2}{0.0202} \]

### Step 6: Solve for \( x \)
Rearranging gives:
\[ x^2 = 8 \times 10^{-6} \times 0.0202 \]
\[ x^2 = 1.616 \times 10^{-7} \]
\[ x = \sqrt{1.616 \times 10^{-7}} \approx 1.27 \times 10^{-4} \, \text{mol/L} \]

### Step 7: Calculate pOH
Since \( x \) represents the concentration of `OH^-`:
\[ [OH^-] = 1.27 \times 10^{-4} \]
Now, calculate pOH:
\[ pOH = -\log(1.27 \times 10^{-4}) \]
Using logarithm properties:
\[ pOH \approx 4 - \log(1.27) \approx 4 - 0.10 \approx 3.90 \]

### Step 8: Calculate pH
Using the relationship \( pH + pOH = 14 \):
\[ pH = 14 - pOH = 14 - 3.90 = 10.10 \]

### Final Answer
The pH of the solution at equilibrium is approximately **10.10**.

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