Find the pH of a 2 litre solution which is 0.1 M each with respect to `CH_(3)COOH` and `(CH_(3)OO)_(2)Ba, (Ka =1.8 xx 10^(-5))`.

Explanation

To find the pH of a 2-liter solution that is 0.1 M in both acetic acid (CH₃COOH) and barium acetate [(CH₃COO)₂Ba], we can follow these steps:

### Step 1: Identify the components
We have:
- Acetic acid (weak acid) with a concentration of 0.1 M.
- Barium acetate (which dissociates to give acetate ions, CH₃COO⁻) with a concentration of 0.1 M.

### Step 2: Determine the concentrations of the acid and its conjugate base
Since we have a 2-liter solution:
- The concentration of acetic acid (CH₃COOH) = 0.1 M
- The concentration of acetate ions (from barium acetate) = 0.1 M

### Step 3: Use the Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation for a buffer solution is given by:
\[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]
Where:
- \([\text{A}^-]\) is the concentration of the conjugate base (acetate ion)
- \([\text{HA}]\) is the concentration of the weak acid (acetic acid)

### Step 4: Calculate pKₐ
Given \( K_a = 1.8 \times 10^{-5} \), we can calculate pKₐ:
\[ \text{pK}_a = -\log(K_a) = -\log(1.8 \times 10^{-5}) \]
Calculating this gives:
\[ \text{pK}_a \approx 4.74 \]

### Step 5: Substitute values into the Henderson-Hasselbalch equation
Now, substituting the values into the equation:
- \([\text{A}^-] = 0.1\) M (from barium acetate)
- \([\text{HA}] = 0.1\) M (from acetic acid)

So, we have:
\[ \text{pH} = 4.74 + \log \left( \frac{0.1}{0.1} \right) \]

### Step 6: Simplify the equation
Since \(\log(1) = 0\):
\[ \text{pH} = 4.74 + 0 = 4.74 \]

### Final Answer
Thus, the pH of the solution is approximately **4.74**.

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