Find the pH of the following acid solution
`Pt"|"H_(2)(1" bar")"|"H^(+) ("aq. acid")"||"H^(+) (1M) "|"H_(2)(g, 1 "bar")"|"Pt`
The measured e.m.f. of the cell = 0.182 V.

Explanation

To find the pH of the given acid solution using the provided information, we can follow these steps:

### Step 1: Understand the Cell Setup
The cell is represented as:
\[ \text{Pt} | \text{H}_2 (1 \text{ bar}) | \text{H}^+ (\text{aq. acid}) || \text{H}^+ (1 \text{ M}) | \text{H}_2 (g, 1 \text{ bar}) | \text{Pt} \]

In this cell:
- The left side (anode) involves hydrogen gas and hydrogen ions from the acid solution.
- The right side (cathode) has a standard hydrogen ion concentration of 1 M.

### Step 2: Use the Nernst Equation
The Nernst equation is given by:
\[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \]

For the hydrogen electrode, the standard cell potential \( E^{\circ}_{\text{cell}} \) is 0 V. The reaction at the anode is:
\[ \text{H}_2 \rightarrow 2\text{H}^+ + 2e^- \]

### Step 3: Determine the Number of Electrons (n)
In this reaction, 1 mole of \( \text{H}_2 \) produces 2 moles of \( \text{H}^+ \), so:
- \( n = 2 \) (since 2 electrons are involved in the oxidation of hydrogen).

### Step 4: Set Up the Nernst Equation
Substituting the known values into the Nernst equation:
\[ E_{\text{cell}} = 0 - \frac{0.0591}{2} \log \left( \frac{[\text{H}^+]^2}{1} \right) \]

Given that \( E_{\text{cell}} = 0.182 \, \text{V} \), we can rearrange the equation:
\[ 0.182 = - \frac{0.0591}{2} \log \left( [\text{H}^+]^2 \right) \]

### Step 5: Simplify the Equation
This simplifies to:
\[ 0.182 = -0.02955 \log \left( [\text{H}^+]^2 \right) \]

### Step 6: Solve for \([\text{H}^+]\)
Taking the logarithm:
\[ \log \left( [\text{H}^+]^2 \right) = -\frac{0.182}{0.02955} \]

Calculating the right side:
\[ \log \left( [\text{H}^+]^2 \right) \approx -6.15 \]

Using properties of logarithms:
\[ 2 \log \left( [\text{H}^+] \right) = -6.15 \]
\[ \log \left( [\text{H}^+] \right) = -3.075 \]

### Step 7: Calculate \([\text{H}^+]\)
Now, we can find \([\text{H}^+]\):
\[ [\text{H}^+] = 10^{-3.075} \approx 0.00084 \, \text{M} \]

### Step 8: Calculate pH
Finally, the pH is calculated as:
\[ \text{pH} = -\log [\text{H}^+] \]
\[ \text{pH} \approx 3.079 \]

Thus, the pH of the acid solution is approximately **3.079**.

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