The e.m.f. of cell: `H_(2)(g)` |Buffer| Normal calomal electrode is `0.6885V` at `40^(@)C` when the barometric pressure is `725mm` of Hg. What is the `pH` of the solution `E_("calomal")^(@) = 0.28`.

Explanation

To solve the problem step by step, we will use the Nernst equation and the information provided in the question.

### Step 1: Write the Nernst Equation
The Nernst equation for the cell can be written as:

\[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{RT}{nF} \ln Q \]

Where:
- \( E_{\text{cell}} \) = EMF of the cell
- \( E^{\circ}_{\text{cell}} \) = Standard electrode potential
- \( R \) = Universal gas constant (8.314 J/(mol·K))
- \( T \) = Temperature in Kelvin
- \( n \) = Number of electrons transferred in the reaction
- \( F \) = Faraday's constant (96500 C/mol)
- \( Q \) = Reaction quotient

### Step 2: Convert Temperature to Kelvin
The temperature given is 40°C. To convert this to Kelvin:

\[ T = 40 + 273.15 = 313.15 \, \text{K} \approx 313 \, \text{K} \]

### Step 3: Substitute Known Values into the Nernst Equation
Given:
- \( E_{\text{cell}} = 0.6885 \, \text{V} \)
- \( E^{\circ}_{\text{cell}} = 0.28 \, \text{V} \)
- \( n = 2 \) (since 2 electrons are involved in the reaction)
- \( R = 8.314 \, \text{J/(mol·K)} \)
- \( F = 96500 \, \text{C/mol} \)

Substituting these values into the Nernst equation:

\[ 0.6885 = 0.28 - \frac{(8.314)(313)}{(2)(96500)} \ln Q \]

### Step 4: Calculate the Value of \(\frac{RT}{nF}\)
Calculating \(\frac{RT}{nF}\):

\[ \frac{(8.314)(313)}{(2)(96500)} = \frac{2600.862}{193000} \approx 0.0135 \]

### Step 5: Rearrange the Nernst Equation to Solve for \(Q\)
Rearranging the equation gives:

\[ 0.6885 = 0.28 - 0.0135 \ln Q \]

\[ 0.0135 \ln Q = 0.28 - 0.6885 \]

\[ 0.0135 \ln Q = -0.4085 \]

\[ \ln Q = \frac{-0.4085}{0.0135} \approx -30.24 \]

### Step 6: Calculate \(Q\)
Taking the exponent to find \(Q\):

\[ Q = e^{-30.24} \approx 8.5 \times 10^{-14} \]

### Step 7: Relate \(Q\) to \(H^+\) Concentration
For the reaction \(H_2(g) \rightleftharpoons 2H^+(aq) + 2e^-\), we can express \(Q\) as:

\[ Q = \frac{[H^+]^2}{P_{H_2}} \]

Given that the barometric pressure is 725 mm Hg, we convert this to atm:

\[ P_{H_2} = \frac{725}{760} \approx 0.9539 \, \text{atm} \]

Thus,

\[ Q = \frac{[H^+]^2}{0.9539} \]

### Step 8: Solve for \([H^+]\)
Substituting \(Q\) into the equation:

\[ 8.5 \times 10^{-14} = \frac{[H^+]^2}{0.9539} \]

\[ [H^+]^2 = 8.5 \times 10^{-14} \times 0.9539 \]

\[ [H^+]^2 \approx 8.1 \times 10^{-14} \]

\[ [H^+] \approx \sqrt{8.1 \times 10^{-14}} \approx 2.85 \times 10^{-7} \, \text{M} \]

### Step 9: Calculate pH
Using the formula for pH:

\[ \text{pH} = -\log[H^+] \]

Calculating pH:

\[ \text{pH} = -\log(2.85 \times 10^{-7}) \approx 6.6 \]

### Final Answer
The pH of the solution is approximately **6.6**.

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