A solution containing `H^(+)` and `D^(+)` ions is equilibrium with a mixture of `H_(2)` and `D_(2)` gases at `25^(@)C`. If the particle pressure of both the gases is 1.0 atm, find the value of `log""([D^(+)])/([H^(+)])`.
(Given : `E_(D^(+)//D_(2))^(0) = -0.003V, (2.303 RT)/(F) = 0.0591`)

Explanation

To solve the problem, we need to find the value of \( \log\left(\frac{[D^+]}{[H^+]}\right) \) for a solution in equilibrium with a mixture of \( H_2 \) and \( D_2 \) gases at a pressure of 1.0 atm each.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The equilibrium reaction can be written as:
\[ D_2 + 2H^+ \rightleftharpoons 2D^+ + H_2 \]
This indicates that \( D^+ \) and \( H^+ \) ions are produced from the gases \( D_2 \) and \( H_2 \).

2. **Use the Nernst Equation**:
At equilibrium, the cell potential \( E \) is 0. The Nernst equation is given by:
\[ E = E^0 - \frac{2.303RT}{nF} \log\left(\frac{[D^+]^2}{[H^+][D_2]}\right) \]
Here, \( n = 1 \) (one electron transfer), \( R = 8.314 \, \text{J/mol·K} \), \( T = 298 \, \text{K} \) (25°C), and \( F = 96485 \, \text{C/mol} \).

3. **Substituting Known Values**:
Given \( E^0_{D^+/D_2} = -0.003 \, \text{V} \) and \( 2.303 \frac{RT}{F} = 0.0591 \):
\[ 0 = -0.003 - \frac{0.0591}{1} \log\left(\frac{[D^+]^2}{[H^+][D_2]}\right) \]

4. **Rearranging the Equation**:
Rearranging gives:
\[ 0.003 = 0.0591 \log\left(\frac{[D^+]^2}{[H^+][D_2]}\right) \]

5. **Solving for the Logarithm**:
Divide both sides by \( 0.0591 \):
\[ \log\left(\frac{[D^+]^2}{[H^+][D_2]}\right) = \frac{0.003}{0.0591} \]
Calculate the right side:
\[ \frac{0.003}{0.0591} \approx 0.0507 \]

6. **Finding the Ratio**:
Since we need \( \log\left(\frac{[D^+]}{[H^+]}\right) \), we can express:
\[ \log\left(\frac{[D^+]}{[H^+]}\right) = \frac{1}{2} \log\left(\frac{[D^+]^2}{[H^+][D_2]}\right) - \log\left(\frac{[D_2]}{[H^+]}\right) \]
Since \( P_{H_2} = P_{D_2} = 1.0 \, \text{atm} \), we can assume \( [D_2] = [H_2] \) in the solution, hence:
\[ \log\left(\frac{[D_2]}{[H^+]}\right) = 0 \]
Therefore:
\[ \log\left(\frac{[D^+]}{[H^+]}\right) = \frac{1}{2} \times 0.0507 \approx 0.02535 \]

### Final Answer:
Thus, the value of \( \log\left(\frac{[D^+]}{[H^+]}\right) \) is approximately \( 0.02535 \).