The pH of a 0.1M `NH_(3)` solution `(K_(b)=1.8 xx 10^(-5))` is

Options

A. 11.13
B. 1
C. 13
D. none of these

Answer

To find the pH of a 0.1 M ammonia solution (NH₃) with a given Kb value of 1.8 × 10⁻⁵, we can follow these steps:

### Step-by-Step Solution:

1. **Write the Ionization Reaction**:
Ammonia (NH₃) acts as a weak base in water:
\[ \text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^- \]

2. **Set Up the Expression for Kb**:
The base dissociation constant (Kb) is given by:
\[ K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} \]
Where:
- \([\text{NH}_4^+]\) = concentration of ammonium ions
- \([\text{OH}^-]\) = concentration of hydroxide ions
- \([\text{NH}_3]\) = concentration of ammonia

3. **Assume Initial Concentrations**:
Initially, before any dissociation, we have:
- \([\text{NH}_3] = 0.1 \, M\)
- \([\text{NH}_4^+] = 0\)
- \([\text{OH}^-] = 0\)

After dissociation, let \(x\) be the concentration of \([\text{NH}_4^+]\) and \([\text{OH}^-]\) produced:
- \([\text{NH}_3] = 0.1 - x\)
- \([\text{NH}_4^+] = x\)
- \([\text{OH}^-] = x\)

4. **Substitute into the Kb Expression**:
Substitute the concentrations into the Kb expression:
\[ 1.8 \times 10^{-5} = \frac{x \cdot x}{0.1 - x} = \frac{x^2}{0.1 - x} \]

5. **Assume x is Small**:
Since Kb is small, we can assume \(x\) is much smaller than 0.1, so \(0.1 - x \approx 0.1\):
\[ 1.8 \times 10^{-5} = \frac{x^2}{0.1} \]

6. **Solve for x**:
Rearranging gives:
\[ x^2 = 1.8 \times 10^{-5} \times 0.1 = 1.8 \times 10^{-6} \]
Taking the square root:
\[ x = \sqrt{1.8 \times 10^{-6}} \approx 1.34 \times 10^{-3} \, M \]

7. **Calculate [OH⁻]**:
Thus, \([\text{OH}^-] = x = 1.34 \times 10^{-3} \, M\).

8. **Find [H⁺] Using Kw**:
We know that:
\[ K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14} \]
Therefore,
\[ [\text{H}^+] = \frac{K_w}{[\text{OH}^-]} = \frac{1.0 \times 10^{-14}}{1.34 \times 10^{-3}} \approx 7.46 \times 10^{-12} \, M \]

9. **Calculate pH**:
Finally, we calculate pH:
\[ \text{pH} = -\log[\text{H}^+] = -\log(7.46 \times 10^{-12}) \approx 11.13 \]

### Final Answer:
The pH of the 0.1 M NH₃ solution is approximately **11.13**.