A hydrogen electrode placed in a solution containing sodium acetate and acetic acid in the ratio of `x:y` and `y:x` has an electrode potential value `E_(1)` and `E_(2)` volts, respectively, at `25^(@)C`. The `pK_(a)` value of acetic acid is _____

Explanation

To find the \( pK_a \) value of acetic acid based on the given electrode potentials \( E_1 \) and \( E_2 \) for the hydrogen electrode in solutions of sodium acetate and acetic acid, we can follow these steps:

### Step 1: Write the Nernst Equation
The Nernst equation for the hydrogen electrode can be written as:
\[ E = E^0 - \frac{0.059}{n} \log \left( \frac{[H^+]}{1} \right) \]
Since \( E^0 \) for the hydrogen electrode is 0, we simplify it to:
\[ E = -0.059 \log [H^+] \]

### Step 2: Relate \( E_1 \) and \( E_2 \) to pH
Using the relationship between \( E \) and pH, we can express \( E_1 \) and \( E_2 \) in terms of pH:
\[ E_1 = -0.059 \times pH_1 \]
\[ E_2 = -0.059 \times pH_2 \]

### Step 3: Write the pH in terms of \( pK_a \)
For a buffer solution of acetic acid and sodium acetate, we can use the Henderson-Hasselbalch equation:
\[ pH_1 = pK_a + \log \left( \frac{[A^-]}{[HA]} \right) \]
For the ratio \( x:y \) (where \( [A^-] = [\text{sodium acetate}] \) and \( [HA] = [\text{acetic acid}] \)):
\[ pH_1 = pK_a + \log \left( \frac{x}{y} \right) \]

For the ratio \( y:x \):
\[ pH_2 = pK_a + \log \left( \frac{y}{x} \right) \]

### Step 4: Combine the equations
Now we can express \( pH_1 \) and \( pH_2 \):
\[ pH_1 = pK_a + \log \left( \frac{x}{y} \right) \]
\[ pH_2 = pK_a + \log \left( \frac{y}{x} \right) \]

### Step 5: Add the two pH equations
Adding both equations:
\[ pH_1 + pH_2 = 2pK_a + \log \left( \frac{x}{y} \right) + \log \left( \frac{y}{x} \right) \]
The logarithmic terms cancel out:
\[ \log \left( \frac{x}{y} \cdot \frac{y}{x} \right) = \log(1) = 0 \]
Thus:
\[ pH_1 + pH_2 = 2pK_a \]

### Step 6: Relate \( pH \) to \( E \)
Substituting \( pH_1 \) and \( pH_2 \) in terms of \( E_1 \) and \( E_2 \):
\[ -\frac{E_1}{0.059} - \frac{E_2}{0.059} = 2pK_a \]
This simplifies to:
\[ pK_a = -\frac{E_1 + E_2}{2 \times 0.059} \]

### Final Expression
Thus, the \( pK_a \) value of acetic acid can be calculated using:
\[ pK_a = -\frac{E_1 + E_2}{0.118} \]