What is `[H^(+)]` in `mol//L` of a solution that is `0.20 M` in `CH_(3)COONa` and `0.1 M` in `CH_(3)COOH`? `K_(a)` for `CH_(3)COOH` is `1.8xx10^(-5)`?

Options

A. `3.5xx10^(-4)`
B. `1.1xx10^(-5)`
C. `1.8xx10^(-5)`
D. `9.0xx10^(-6)`

Explanation

To find the concentration of hydrogen ions \([H^+]\) in a solution that is \(0.20 \, M\) in sodium acetate \((CH_3COONa)\) and \(0.10 \, M\) in acetic acid \((CH_3COOH)\), we can use the Henderson-Hasselbalch equation, which is applicable for buffer solutions.

### Step-by-Step Solution:

1. **Identify the components of the buffer solution**:
- The weak acid is acetic acid \((CH_3COOH)\).
- The conjugate base is acetate ion from sodium acetate \((CH_3COONa)\).

2. **Write the Henderson-Hasselbalch equation**:
\[ pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right) \]
where:
- \([A^-]\) is the concentration of the conjugate base (acetate).
- \([HA]\) is the concentration of the weak acid (acetic acid).

3. **Calculate \(pK_a\)**:
Given \(K_a\) for acetic acid is \(1.8 \times 10^{-5}\):
\[ pK_a = -\log(K_a) = -\log(1.8 \times 10^{-5}) \approx 4.74 \]

4. **Substitute the concentrations into the equation**:
\[ pH = pK_a + \log\left(\frac{[CH_3COO^-]}{[CH_3COOH]}\right) \]
Here, \([CH_3COO^-] = 0.20 \, M\) and \([CH_3COOH] = 0.10 \, M\):
\[ pH = 4.74 + \log\left(\frac{0.20}{0.10}\right) \]

5. **Calculate the logarithm**:
\[ \log\left(\frac{0.20}{0.10}\right) = \log(2) \approx 0.301 \]
Therefore:
\[ pH = 4.74 + 0.301 = 5.041 \]

6. **Convert pH to \([H^+]\)**:
Using the formula:
\[ [H^+] = 10^{-pH} \]
Substitute \(pH = 5.041\):
\[ [H^+] = 10^{-5.041} \approx 9.12 \times 10^{-6} \, M \]

### Final Answer:
\[ [H^+] \approx 9.12 \times 10^{-6} \, \text{mol/L} \]