To calculate the pH of the solution using the hydrogen electrode potential, we can use the Nernst equation for the hydrogen electrode, which relates the electrode potential to the pH of the solution:
\[ E = E^{\circ} - \frac{RT}{nF} \ln(Q) \]
For a hydrogen electrode, we simplify this to:
\[ E = E^{\circ} - \frac{0.0591}{n} \log{[\text{H}^+]} \]
where:
- \( E \) is the measured electrode potential.
- \( E^{\circ} \) is the standard electrode potential for the hydrogen electrode (which is 0 V).
- \( R \) is the universal gas constant (8.314 J/(mol K)).
- \( T \) is the temperature in Kelvin (assumed to be 298 K unless otherwise stated).
- \( F \) is Faraday’s constant (96485 C/mol).
- \( n \) is the number of electrons transferred in the half-reaction (which is 2 for the hydrogen electrode).
- \([\text{H}^+]\) is the concentration of protons in mol/L.
At \(25^\circ C (298 K)\), the equation simplifies to:
\[ E = 0 - 0.0591 \log{[\text{H}^+]} \]
Rearranging the equation gives:
\[ \log{[\text{H}^+]} = -\frac{E}{0.0591} \]
Substituting the measured potential \(E = -0.2364 , \text{V}\):
\[ \log{[\text{H}^+]} = -\frac{-0.2364}{0.0591} \approx 4.005 \]
Now, we convert from logarithmic form to concentration:
\[ [\text{H}^+] = 10^{4.005} \approx 10^{4} \approx 10^{-4.005} , \text{mol/L} \]
Now we can find the pH:
\[ \text{pH} = -\log{[\text{H}^+]} \approx 4.005 \]
Thus, the pH of the solution is approximately 4.01.