`pH` of `0.1M` solution `HCl` is changed by `0.3` unit due dilution, calculate change in osmotic pressure if temperature of solution is `300K. (R = 1//12" " "litre atm"//K//"mole")`

Options

A. `2.5atm`
B. `5 atm`
C. `7.5 atm`
D. `1.25 atm`

Explanation

To solve the problem, we need to calculate the change in osmotic pressure due to the dilution of a 0.1 M HCl solution, where the pH changes by 0.3 units. We will use the formula for osmotic pressure and the information provided.

### Step-by-Step Solution:

1. **Determine the Initial Concentration of HCl:**
- The initial concentration of HCl is given as \( C_1 = 0.1 \, M \).

2. **Calculate the Initial pH:**
- Since HCl is a strong acid, it dissociates completely in solution. Therefore, the concentration of \( H^+ \) ions is equal to the concentration of HCl.
- The initial pH can be calculated as:
\[ \text{pH} = -\log[H^+] = -\log(0.1) = 1 \]

3. **Determine the New pH After Dilution:**
- The pH changes by 0.3 units, so the new pH is:
\[ \text{New pH} = 1 - 0.3 = 0.7 \]

4. **Calculate the New Concentration of HCl:**
- The concentration of \( H^+ \) ions corresponding to the new pH can be calculated as:
\[ [H^+] = 10^{-\text{pH}} = 10^{-0.7} \approx 0.1995 \, M \]
- Therefore, the new concentration of HCl after dilution is approximately \( C_2 \approx 0.1995 \, M \).

5. **Calculate the Change in Concentration:**
- The change in concentration \( \Delta C \) is:
\[ \Delta C = C_1 - C_2 = 0.1 - 0.1995 \approx -0.0995 \, M \]
- (Note: The concentration has decreased due to dilution).

6. **Calculate the Osmotic Pressure:**
- The formula for osmotic pressure \( \Pi \) is given by:
\[ \Pi = i \cdot C \cdot R \cdot T \]
- For HCl, the van't Hoff factor \( i = 2 \) (since it dissociates into \( H^+ \) and \( Cl^- \)).
- Using the initial concentration \( C_1 = 0.1 \, M \):
\[ \Pi_1 = 2 \cdot 0.1 \cdot \frac{1}{12} \cdot 300 \]
- Calculate \( \Pi_1 \):
\[ \Pi_1 = 2 \cdot 0.1 \cdot 25 = 5 \, \text{atm} \]

7. **Calculate the New Osmotic Pressure:**
- Using the new concentration \( C_2 \approx 0.1995 \, M \):
\[ \Pi_2 = 2 \cdot 0.1995 \cdot \frac{1}{12} \cdot 300 \]
- Calculate \( \Pi_2 \):
\[ \Pi_2 = 2 \cdot 0.1995 \cdot 25 \approx 9.975 \, \text{atm} \]

8. **Calculate the Change in Osmotic Pressure:**
- The change in osmotic pressure \( \Delta \Pi \) is:
\[ \Delta \Pi = \Pi_2 - \Pi_1 = 9.975 - 5 = 4.975 \, \text{atm} \]

### Final Answer:
The change in osmotic pressure is approximately \( 4.975 \, \text{atm} \).