If a certain buffer solution contains equal concentration of `X^-` and HX. Then the pH of buffer will be `(K_b for X^- is 10^-10)`

Options

A. 10
B. 4
C. 5
D. 11

Answer

To solve the problem, we need to determine the pH of a buffer solution containing equal concentrations of \( X^- \) and \( HX \), given that the \( K_b \) for \( X^- \) is \( 10^{-10} \).

### Step-by-Step Solution:

1. **Identify the relationship between \( K_a \) and \( K_b \)**:
- We know that \( K_a \times K_b = K_w \), where \( K_w = 10^{-14} \) at 25°C.
- Given \( K_b = 10^{-10} \) for \( X^- \), we can find \( K_a \) for \( HX \) using the formula:
\[ K_a = \frac{K_w}{K_b} = \frac{10^{-14}}{10^{-10}} = 10^{-4} \]

**Hint**: Remember that \( K_w \) is the ion product of water, which is \( 10^{-14} \) at room temperature.

2. **Calculate \( pK_a \)**:
- The \( pK_a \) is calculated using the formula:
\[ pK_a = -\log(K_a) = -\log(10^{-4}) = 4 \]

**Hint**: The \( pK_a \) is simply the negative logarithm of the acid dissociation constant \( K_a \).

3. **Use the Henderson-Hasselbalch equation**:
- The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation:
\[ pH = pK_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]
- Since the concentrations of the salt \( X^- \) and the acid \( HX \) are equal, the ratio \( \frac{[\text{Salt}]}{[\text{Acid}]} = 1 \).
- Therefore, \( \log(1) = 0 \).

**Hint**: When the concentrations of the acid and its conjugate base are equal, the log term becomes zero.

4. **Final Calculation of pH**:
- Substituting into the equation:
\[ pH = pK_a + 0 = 4 + 0 = 4 \]

**Hint**: The pH of the buffer solution is simply equal to the \( pK_a \) when the concentrations of the acid and conjugate base are equal.

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