A certain buffer solution sontains equal concentration of `X^(-)` and `HX`. The `K_(a)` for `HX` is `10^(-8)`. The of the buffer is

Options

A. 3
B. 8
C. 11
D. 14

Answer

To find the pH of the buffer solution containing equal concentrations of \( X^- \) and \( HX \), we can use the Henderson-Hasselbalch equation. Here’s a step-by-step solution:

### Step 1: Identify the components of the buffer
The buffer solution consists of:
- A weak acid \( HX \)
- Its conjugate base \( X^- \)

### Step 2: Write the Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation is given by:
\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{Base}]}{[\text{Acid}]}\right) \]
In our case, the base is \( X^- \) and the acid is \( HX \).

### Step 3: Determine the concentrations
According to the problem, the concentrations of \( X^- \) and \( HX \) are equal. Therefore:
\[ [\text{Base}] = [X^-] \quad \text{and} \quad [\text{Acid}] = [HX] \]
This implies:
\[ \frac{[\text{Base}]}{[\text{Acid}]} = 1 \]

### Step 4: Substitute into the equation
Substituting the values into the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pKa} + \log(1) \]
Since \( \log(1) = 0 \), the equation simplifies to:
\[ \text{pH} = \text{pKa} \]

### Step 5: Calculate pKa
To find \( \text{pKa} \), we use the relationship:
\[ \text{pKa} = -\log(K_a) \]
Given that \( K_a = 10^{-8} \):
\[ \text{pKa} = -\log(10^{-8}) = 8 \]

### Step 6: Final calculation of pH
Now substituting \( \text{pKa} \) back into the pH equation:
\[ \text{pH} = 8 + 0 = 8 \]

### Conclusion
Thus, the pH of the buffer solution is:
\[ \text{pH} = 8 \]

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