An aqueous mixture at room temperature is 0.1 M with respect to ammonium chloride and 0.01 M with respect to `NH_4 OH,pK_b ` of aqueous ammonia as base is 5. The pH of the mixture is nearly

Options

A. 7.5
B. 6.8
C. 6.5
D. `8.0`

Answer

To solve the problem, we need to determine the pH of a solution that contains both ammonium chloride (NH4Cl) and ammonium hydroxide (NH4OH). This is a buffer solution made up of a weak base (NH4OH) and its conjugate acid (NH4Cl).

### Step-by-Step Solution:

1. **Identify the Components**:
- The solution contains:
- Ammonium chloride (NH4Cl) at a concentration of 0.1 M.
- Ammonium hydroxide (NH4OH) at a concentration of 0.01 M.

2. **Understand the Buffer System**:
- NH4Cl dissociates in water to give NH4⁺ (the conjugate acid).
- NH4OH is a weak base that partially dissociates to give NH4⁺ and OH⁻ ions.

3. **Use the Henderson-Hasselbalch Equation**:
- The pH of a buffer solution can be calculated using the formula:
\[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \]
- However, since we have pK_b for NH4OH, we need to convert it to pK_a:
\[ \text{pK}_a + \text{pK}_b = 14 \]
- Given that pK_b = 5, we find:
\[ \text{pK}_a = 14 - 5 = 9 \]

4. **Plug in the Concentrations**:
- The concentration of the base (NH4OH) is 0.01 M.
- The concentration of the acid (NH4Cl) is 0.1 M.
- Now, substituting these values into the Henderson-Hasselbalch equation:
\[ \text{pH} = 9 + \log \left( \frac{0.01}{0.1} \right) \]

5. **Calculate the Logarithm**:
- The ratio \(\frac{0.01}{0.1} = 0.1\).
- Therefore, \(\log(0.1) = -1\).
- Now substituting this back into the equation:
\[ \text{pH} = 9 + (-1) = 8 \]

6. **Final Result**:
- The pH of the mixture is approximately 8.

### Conclusion:
The pH of the mixture is nearly 8.