The pH of a solution containing 0.1 M `CH_3COONa and 0.1 M (C_2H_5COO)_2` Ba will be `K_a(CH_3COOH)=2xx10^(-5). K_a(C_2H_5COOH)=1.5xx10^(-5)`:

Options

A. 8.13
B. 9.13
C. 10.18
D. 11.18

Answer

To find the pH of a solution containing 0.1 M sodium acetate (CH₃COONa) and 0.1 M barium propanoate ((C₂H₅COO)₂Ba), we can use the Henderson-Hasselbalch equation, which is suitable for buffer solutions.

### Step-by-Step Solution:

1. **Identify the Components**:
- Sodium acetate (CH₃COONa) dissociates in solution to give acetate ions (CH₃COO⁻).
- Barium propanoate ((C₂H₅COO)₂Ba) dissociates to give propanoate ions (C₂H₅COO⁻).

2. **Determine the Concentrations**:
- The concentration of acetate ions [CH₃COO⁻] = 0.1 M.
- The concentration of propanoate ions [C₂H₅COO⁻] = 0.1 M.

3. **Use the Henderson-Hasselbalch Equation**:
The pH of a buffer solution can be calculated using the formula:
\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{Base}]}{[\text{Acid}]}\right) \]
Here, we can consider acetate as the base and propanoate as the acid.

4. **Calculate pK_a Values**:
- For acetic acid (CH₃COOH), \( K_a = 2 \times 10^{-5} \)
\[ \text{pK}_a(\text{CH}_3\text{COOH}) = -\log(2 \times 10^{-5}) \approx 4.70 \]
- For propanoic acid (C₂H₅COOH), \( K_a = 1.5 \times 10^{-5} \)
\[ \text{pK}_a(\text{C}_2\text{H}_5\text{COOH}) = -\log(1.5 \times 10^{-5}) \approx 4.83 \]

5. **Substituting Values into the Henderson-Hasselbalch Equation**:
Since we have both acetate and propanoate at equal concentrations (0.1 M), we can substitute:
\[ \text{pH} = \text{pK}_a(\text{CH}_3\text{COOH}) + \log\left(\frac{[CH_3COO^-]}{[C_2H_5COO^-]}\right) \]
Since \([CH_3COO^-] = [C_2H_5COO^-]\), the log term becomes zero:
\[ \text{pH} = 4.70 + \log(1) = 4.70 \]

6. **Final pH Calculation**:
The pH of the solution is approximately 4.70.

### Final Answer:
The pH of the solution is **4.70**.