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 the solution at equilibrium when the pressure of `RNH2(g)` is `0.5 bar`, we can follow these steps:

### Step 1: Calculate the concentration of `RNH2(g)`

Using the ideal gas law, we can find the concentration of `RNH2(g)`:

\[ C = \frac{P}{RT} \]

Where:
- \( P = 0.5 \, \text{bar} \)
- \( R = 0.0821 \, \text{L bar/(mol K)} \)
- \( T = 298 \, \text{K} \)

Substituting the values:

\[ C = \frac{0.5 \, \text{bar}}{0.0821 \, \text{L bar/(mol K)} \times 298 \, \text{K}} = \frac{0.5}{24.4758} \approx 0.0204 \, \text{mol/L} \]

### Step 2: Set up the equilibrium expression

The equilibrium expression for the ionization of `RNH2` is given by:

\[ K_c = \frac{[RNH_3^+] [OH^-]}{[RNH_2]} \]

Given that \( K_c = 8 \times 10^{-6} \).

### Step 3: Define the changes at equilibrium

Let \( \alpha \) be the amount that ionizes. At equilibrium, we have:

- Concentration of `RNH2` = \( 0.0204 - \alpha \)
- Concentration of `RNH3^+` = \( \alpha \)
- Concentration of `OH^-` = \( \alpha \)

### Step 4: Substitute into the equilibrium expression

Substituting these values into the equilibrium expression:

\[ 8 \times 10^{-6} = \frac{\alpha \cdot \alpha}{0.0204 - \alpha} \]

Assuming \( \alpha \) is small compared to \( 0.0204 \), we can simplify this to:

\[ 8 \times 10^{-6} = \frac{\alpha^2}{0.0204} \]

### Step 5: Solve for \( \alpha \)

Rearranging gives:

\[ \alpha^2 = 8 \times 10^{-6} \times 0.0204 \]

Calculating the right side:

\[ \alpha^2 = 1.632 \times 10^{-7} \]

Taking the square root:

\[ \alpha = \sqrt{1.632 \times 10^{-7}} \approx 4.04 \times 10^{-4} \, \text{mol/L} \]

### Step 6: Calculate \( [OH^-] \)

Since \( [OH^-] = \alpha \):

\[ [OH^-] = 4.04 \times 10^{-4} \, \text{mol/L} \]

### Step 7: Calculate \( pOH \)

Using the formula for \( pOH \):

\[ pOH = -\log[OH^-] = -\log(4.04 \times 10^{-4}) \approx 3.39 \]

### Step 8: Calculate \( pH \)

Using the relation \( pH + pOH = 14 \):

\[ pH = 14 - pOH = 14 - 3.39 \approx 10.61 \]

### Final Answer

The pH of the solution at equilibrium is approximately **10.61**.

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