A hydrogen electrode placed in a buffer solution of `CH_(3)COONa` and `CH_(3)COOH` in the ratios of `x:y` and `y:x` has electrode potential values `E_(1)` volts and `E_(2)` volts, respectively at `25^(@)C` The `pK_(a)` values of acetic acid is `(E_(1)` and `E_(2)` are oxidation potential)

Options

A. `(E_(1)+E_(2))/(0.118)`
B. `(E_(2)-E_(1))/(0.118)`
C. `-(E_(1)+E_(2))/(0.118)`
D. `(E_(1)-E_(1))/(0.118)`

Explanation

To find the pKa value of acetic acid in terms of the electrode potentials \( E_1 \) and \( E_2 \) for the hydrogen electrode placed in buffer solutions of acetic acid and sodium acetate, we can follow these steps:

### Step 1: Understand the Electrode Potential Equation
The Nernst equation for the hydrogen electrode can be expressed as:
\[ E = E^0 - \frac{0.059}{n} \log \left( \frac{1}{[H^+]} \right) \]
where \( n \) is the number of electrons transferred (which is 1 for the hydrogen electrode).

### Step 2: Write the Equations for \( E_1 \) and \( E_2 \)
For the first buffer solution with the ratio \( x:y \):
\[ E_1 = E^0 - 0.059 \log \left( \frac{1}{[H^+]_1} \right) \]
For the second buffer solution with the ratio \( y:x \):
\[ E_2 = E^0 - 0.059 \log \left( \frac{1}{[H^+]_2} \right) \]

### Step 3: Simplify the Equations
Since \( E^0 \) is the same for both cases (as it is a standard hydrogen electrode), we can rearrange the equations:
\[ E_1 = E^0 - 0.059 \log \left( [H^+]_1 \right) \]
\[ E_2 = E^0 - 0.059 \log \left( [H^+]_2 \right) \]

### Step 4: Relate \( [H^+] \) to pH and pKa
In a buffer solution, the pH can be expressed using the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right) \]
For the first solution (ratio \( x:y \)):
\[ \text{pH}_1 = \text{pKa} + \log \left( \frac{x}{y} \right) \]
For the second solution (ratio \( y:x \)):
\[ \text{pH}_2 = \text{pKa} + \log \left( \frac{y}{x} \right) \]

### Step 5: Substitute pH in Terms of \( E_1 \) and \( E_2 \)
Using the relation \( \text{pH} = -\log [H^+] \):
\[ -\log [H^+]_1 = \text{pKa} + \log \left( \frac{x}{y} \right) \]
\[ -\log [H^+]_2 = \text{pKa} + \log \left( \frac{y}{x} \right) \]

### Step 6: Add the Two Equations
Adding the two equations gives:
\[ -\log [H^+]_1 - \log [H^+]_2 = 2 \text{pKa} + \log \left( \frac{x}{y} \cdot \frac{y}{x} \right) \]
Since \( \log \left( \frac{x}{y} \cdot \frac{y}{x} \right) = 0 \):
\[ -\log [H^+]_1 - \log [H^+]_2 = 2 \text{pKa} \]

### Step 7: Relate to Electrode Potentials
From the electrode potential equations:
\[ E_1 + E_2 = 0.059 \cdot 2 \text{pKa} \]
Thus, we can express pKa as:
\[ \text{pKa} = \frac{E_1 + E_2}{0.059 \cdot 2} = \frac{E_1 + E_2}{0.118} \]

### Final Answer
The pKa value of acetic acid in terms of \( E_1 \) and \( E_2 \) is:
\[ \text{pKa} = \frac{E_1 + E_2}{0.118} \]