In a buffer solution containing equal concentration of `B^(-)` and `HB`, the `K_(b)` for `B^(-)` is `10^(-10)`. The `pH` of buffer solution is

Options

A. 10
B. 7
C. 6
D. 4

Explanation

To find the pH of the buffer solution containing equal concentrations of \( B^- \) and \( HB \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information:**
- The concentration of \( B^- \) (conjugate base) is equal to the concentration of \( HB \) (weak acid).
- The \( K_b \) for \( B^- \) is \( 10^{-10} \).

2. **Use the Henderson-Hasselbalch Equation:**
The pH of a buffer solution can be calculated using the formula:
\[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \]
Since the concentrations of \( B^- \) and \( HB \) are equal, we can express this as:
\[ \text{pH} = \text{pK}_a + \log(1) \]
Since \( \log(1) = 0 \), we have:
\[ \text{pH} = \text{pK}_a \]

3. **Calculate \( pK_a \):**
To find \( pK_a \), we can use the relationship between \( pK_a \) and \( pK_b \):
\[ pK_a + pK_b = 14 \]
First, we need to calculate \( pK_b \):
\[ pK_b = -\log(K_b) = -\log(10^{-10}) = 10 \]
Now, substituting \( pK_b \) into the equation:
\[ pK_a = 14 - pK_b = 14 - 10 = 4 \]

4. **Final Calculation of pH:**
Since we established that \( \text{pH} = \text{pK}_a \):
\[ \text{pH} = 4 \]

### Conclusion:
The pH of the buffer solution is **4**.