The `pK_(a)` value of `NH_(3)` is 5. Calculate the pH of the buffer solution, 1 L of which contains `0.01 M NH_(4)Cl` and `0.10M NH_(4)OH `:

Options

A. 4
B. 6
C. 8
D. 10

Explanation

To calculate the pH of the buffer solution containing \(0.01 \, M \, NH_4Cl\) and \(0.10 \, M \, NH_4OH\), we will use the Henderson-Hasselbalch equation.

### Step-by-Step Solution:

1. **Identify the components of the buffer solution**:
- The buffer consists of \(NH_4Cl\) (which provides the conjugate acid \(NH_4^+\)) and \(NH_4OH\) (which acts as the base \(NH_3\)).
- Given concentrations:
- \([NH_4^+] = 0.01 \, M\) (from \(NH_4Cl\))
- \([NH_3] = 0.10 \, M\) (from \(NH_4OH\))

2. **Determine the \(pK_a\) and \(pK_b\)**:
- Given \(pK_a\) of \(NH_3\) is 5.
- Use the relation \(pK_a + pK_b = 14\) to find \(pK_b\):
\[ pK_b = 14 - pK_a = 14 - 5 = 9 \]

3. **Use the Henderson-Hasselbalch equation**:
- The equation is given by:
\[ pH = pK_a + \log\left(\frac{[Base]}{[Acid]}\right) \]
- Here, the base is \(NH_3\) and the acid is \(NH_4^+\).

4. **Substitute the values into the equation**:
- Substitute \(pK_a = 5\), \([Base] = 0.10 \, M\), and \([Acid] = 0.01 \, M\):
\[ pH = 5 + \log\left(\frac{0.10}{0.01}\right) \]

5. **Calculate the logarithm**:
- \(\frac{0.10}{0.01} = 10\)
- Therefore, \(\log(10) = 1\).

6. **Final calculation of pH**:
- Substitute back into the equation:
\[ pH = 5 + 1 = 6 \]

### Final Answer:
The pH of the buffer solution is **6**.