`100mL` of `HCl` gas at `25^(@)C` and `740mm` pressure is dissolved in `1L` of `H_(2)O`. Calculate the `pH` of solution. Given vapour presure of `H_(2)O` at `25^(@)C` is `23.7mm`.

Explanation

To solve the problem of calculating the pH of a solution formed by dissolving 100 mL of HCl gas at 25°C and 740 mm pressure in 1 L of water, we can follow these steps:

### Step 1: Calculate the Effective Pressure of HCl
First, we need to account for the vapor pressure of water at 25°C, which is given as 23.7 mm. The effective pressure of HCl gas can be calculated as:

\[ P_{\text{HCl}} = P_{\text{total}} - P_{\text{H2O}} = 740 \, \text{mm} - 23.7 \, \text{mm} = 716.3 \, \text{mm} \]

### Step 2: Convert Pressure to Atmospheres
Next, we convert the effective pressure of HCl from mm to atmospheres since the ideal gas law uses pressure in atmospheres. We know that 1 atm = 760 mmHg.

\[ P_{\text{HCl (atm)}} = \frac{716.3 \, \text{mm}}{760 \, \text{mm/atm}} = 0.943 \, \text{atm} \]

### Step 3: Use the Ideal Gas Law to Find Moles of HCl
Using the ideal gas law \( PV = nRT \), we can solve for the number of moles (n) of HCl.

- \( R = 0.0821 \, \text{L atm/(mol K)} \)
- \( T = 25°C = 298 \, \text{K} \)
- \( V = 100 \, \text{mL} = 0.1 \, \text{L} \)

Substituting the values into the equation:

\[ n = \frac{PV}{RT} = \frac{(0.943 \, \text{atm})(0.1 \, \text{L})}{(0.0821 \, \text{L atm/(mol K)})(298 \, \text{K})} \]

Calculating this gives:

\[ n \approx \frac{0.0943}{24.4758} \approx 0.00385 \, \text{mol} \]

### Step 4: Calculate the Molarity of HCl in the Solution
Now, we need to find the molarity (M) of HCl in the 1 L solution. Molarity is defined as moles of solute per liter of solution.

\[ M = \frac{n}{V} = \frac{0.00385 \, \text{mol}}{1 \, \text{L}} = 0.00385 \, \text{M} \]

### Step 5: Calculate the pH of the Solution
Since HCl is a strong acid, it dissociates completely in water. Therefore, the concentration of \( H^+ \) ions will be equal to the molarity of HCl.

\[ [H^+] = 0.00385 \, \text{M} \]

Now, we can calculate the pH using the formula:

\[ \text{pH} = -\log[H^+] = -\log(0.00385) \]

Calculating this gives:

\[ \text{pH} \approx 2.414 \]

### Final Answer
The pH of the solution is approximately **2.414**.

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