A certain buffer solution contains equal concentartion of `X^(Theta)` and `HX`. The `K_(b)` for `X^(Theta)` is `10^(-10)`. The `pH` of the buffer is

Options

A. `4`
B. `7`
C. `10`
D. `14`

Explanation

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

### Step 1: Understand the buffer system
The buffer solution consists of a weak acid \( HX \) and its conjugate base \( X^- \). The \( K_b \) value for the base \( X^- \) is given as \( 10^{-10} \).

### Step 2: Relate \( K_b \) and \( K_a \)
We know that the relationship between \( K_a \), \( K_b \), and \( K_w \) (the ion product of water) is given by:
\[ K_w = K_a \times K_b \]
where \( K_w = 1 \times 10^{-14} \) at 25°C.

### Step 3: Calculate \( K_a \)
From the equation above, we can rearrange it to find \( K_a \):
\[ K_a = \frac{K_w}{K_b} \]
Substituting the values:
\[ K_a = \frac{1 \times 10^{-14}}{10^{-10}} = 1 \times 10^{-4} \]

### Step 4: Calculate \( pK_a \)
Next, we need to find \( pK_a \):
\[ pK_a = -\log(K_a) = -\log(1 \times 10^{-4}) = 4 \]

### Step 5: Use the Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation for a buffer solution is:
\[ pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right) \]
Since the concentrations of \( X^- \) (the base) and \( HX \) (the acid) are equal, we have:
\[ \frac{[X^-]}{[HX]} = 1 \]
Thus, \( \log(1) = 0 \).

### Step 6: Substitute values into the equation
Now substituting \( pK_a \) and the log term into the equation:
\[ pH = pK_a + 0 = 4 + 0 = 4 \]

### Conclusion
The pH of the buffer solution is \( 4 \).