The `pH` of a `0.1M` solution of `NH_(4)OH` (having dissociation constant `K_(b) = 1.0 xx 10^(-5))` is equal to

Options

A. `10`
B. `6`
C. `11`
D. `12`

Explanation

To find the pH of a 0.1 M solution of NH₄OH (ammonium hydroxide) with a dissociation constant \( K_b = 1.0 \times 10^{-5} \), we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of ammonium hydroxide in water can be represented as:
\[ \text{NH}_4\text{OH} \rightleftharpoons \text{NH}_4^+ + \text{OH}^- \]

### Step 2: Set up the expression for \( K_b \)
The expression for the base dissociation constant \( K_b \) is given by:
\[ K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_4\text{OH}]} \]

### Step 3: Define the initial concentrations and changes
Let the initial concentration of NH₄OH be \( C = 0.1 \, M \). At equilibrium, if \( \alpha \) is the degree of dissociation, then:
- Concentration of NH₄OH at equilibrium = \( C(1 - \alpha) \)
- Concentration of NH₄⁺ at equilibrium = \( C\alpha \)
- Concentration of OH⁻ at equilibrium = \( C\alpha \)

### Step 4: Substitute into the \( K_b \) expression
Substituting the equilibrium concentrations into the \( K_b \) expression gives:
\[ K_b = \frac{C\alpha \cdot C\alpha}{C(1 - \alpha)} = \frac{C^2 \alpha^2}{C(1 - \alpha)} = \frac{C\alpha^2}{1 - \alpha} \]

### Step 5: Assume \( \alpha \) is small
Since \( K_b \) is small, we can assume \( \alpha \) is small compared to 1, so \( 1 - \alpha \approx 1 \). Thus, we can simplify the equation to:
\[ K_b \approx C\alpha^2 \]

### Step 6: Solve for \( \alpha \)
Substituting the values into the equation:
\[ 1.0 \times 10^{-5} = 0.1 \alpha^2 \]
\[ \alpha^2 = \frac{1.0 \times 10^{-5}}{0.1} = 1.0 \times 10^{-4} \]
\[ \alpha = \sqrt{1.0 \times 10^{-4}} = 1.0 \times 10^{-2} \]

### Step 7: Calculate the concentration of OH⁻
The concentration of OH⁻ ions at equilibrium is:
\[ [\text{OH}^-] = C\alpha = 0.1 \times 1.0 \times 10^{-2} = 1.0 \times 10^{-3} \, M \]

### Step 8: Calculate pOH
Now, we can calculate the pOH:
\[ pOH = -\log[\text{OH}^-] = -\log(1.0 \times 10^{-3}) = 3 \]

### Step 9: Calculate pH
Using the relationship between pH and pOH:
\[ pH + pOH = 14 \]
\[ pH = 14 - pOH = 14 - 3 = 11 \]

### Final Answer
The pH of the 0.1 M solution of NH₄OH is \( \boxed{11} \).