`pH` of a mixture containing `0.10 M X^(-)` and `0.20 M HX` is: `[pK_(b)(X^(-))=4]`

Options

A. `4+log 2`
B. `4-log 2`
C. `10+log 2`
D. `10-log 2`

Explanation

To find the pH of the mixture containing \(0.10 \, M \, X^-\) and \(0.20 \, M \, HX\), we can use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the concentrations of the acid and its conjugate base.

### Step-by-Step Solution:

1. **Identify the Components:**
- We have a weak acid \(HX\) with a concentration of \(0.20 \, M\).
- We have its conjugate base \(X^-\) with a concentration of \(0.10 \, M\).
- The given \(pK_b\) of \(X^-\) is \(4\).

2. **Calculate \(pK_a\):**
- We can find \(pK_a\) using the relationship:
\[ pK_a + pK_b = 14 \]
- Therefore,
\[ pK_a = 14 - pK_b = 14 - 4 = 10 \]

3. **Apply the Henderson-Hasselbalch Equation:**
- The Henderson-Hasselbalch equation for a weak acid and its conjugate base is given by:
\[ pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right) \]
- Here, \([A^-] = [X^-] = 0.10 \, M\) and \([HA] = [HX] = 0.20 \, M\).

4. **Substitute Values into the Equation:**
- Plugging in the values:
\[ pH = 10 + \log\left(\frac{0.10}{0.20}\right) \]
- Simplifying the ratio:
\[ \frac{0.10}{0.20} = 0.5 \]
- Thus, we have:
\[ pH = 10 + \log(0.5) \]

5. **Calculate \(\log(0.5)\):**
- We know that \(\log(0.5) = -0.301\) (approximately).
- Therefore:
\[ pH = 10 - 0.301 = 9.699 \]

6. **Final Result:**
- Rounding to two decimal places, the pH of the mixture is approximately:
\[ pH \approx 9.70 \]