Calculate the pH of a buffer when the reduction potential of hydrogen electrode placed in the buffer solution is found to be `-0.413V`.

Options

A. 10
B. 4
C. 7
D. 12

Answer

To calculate the pH of a buffer solution given the reduction potential of the hydrogen electrode, we can follow these steps:

### Step 1: Understand the relationship between the reduction potential and pH
The Nernst equation for the hydrogen electrode can be expressed as:
\[ E = E^0 - \frac{0.0591}{n} \log \left( \frac{1}{[H^+]}\right) \]
Where:
- \( E \) is the reduction potential of the hydrogen electrode.
- \( E^0 \) is the standard reduction potential (for the hydrogen electrode, \( E^0 = 0 \) V).
- \( n \) is the number of electrons transferred (for the hydrogen ion reduction, \( n = 1 \)).
- \( [H^+] \) is the concentration of hydrogen ions.

### Step 2: Substitute the known values into the equation
Given:
- \( E = -0.413 \, \text{V} \)
- \( E^0 = 0 \, \text{V} \)
- \( n = 1 \)

Substituting these values into the Nernst equation:
\[ -0.413 = 0 - \frac{0.0591}{1} \log \left( \frac{1}{[H^+]}\right) \]

### Step 3: Rearranging the equation
Rearranging the equation gives:
\[ -0.413 = -0.0591 \log \left( \frac{1}{[H^+]}\right) \]

### Step 4: Simplifying the equation
This can be simplified to:
\[ 0.413 = 0.0591 \log \left( [H^+] \right) \]

### Step 5: Isolate the logarithm
Dividing both sides by \( 0.0591 \):
\[ \log \left( [H^+] \right) = \frac{0.413}{0.0591} \]

### Step 6: Calculate the value
Calculating the right side:
\[ \log \left( [H^+] \right) \approx 6.98 \]

### Step 7: Convert logarithm to concentration
Taking the antilog:
\[ [H^+] = 10^{-6.98} \]

### Step 8: Calculate the pH
Since pH is defined as:
\[ \text{pH} = -\log [H^+] \]
We can substitute:
\[ \text{pH} = 6.98 \]

### Conclusion
Thus, the pH of the buffer solution is approximately 7.