When 0.1 M `Pb(NO_(3))_(2)(aq)` solution is titrated with 0.1 M Kl(aq) solution at 300 K temperature then what will be the osmotic pressure of final solution (in atm) at equivalence point?
[Use : R=0.08 L-atm `mol^(-1)``K^(-1)`]

Answer

To solve the problem of finding the osmotic pressure of the final solution when 0.1 M \( Pb(NO_3)_2(aq) \) is titrated with 0.1 M \( KI(aq) \) at 300 K, we will follow these steps:

### Step 1: Write the Reaction
The reaction between lead(II) nitrate and potassium iodide can be written as:
\[ Pb(NO_3)_2 + 2KI \rightarrow PbI_2(s) + 2KNO_3 \]
At the equivalence point, lead iodide \( PbI_2 \) precipitates out, leaving \( KNO_3 \) in solution.

### Step 2: Determine Moles of Reactants
Assuming we use \( V \) liters of \( Pb(NO_3)_2 \):
- Moles of \( Pb(NO_3)_2 \) = \( 0.1 \, \text{mol/L} \times V = 0.1V \)
- Moles of \( KI \) required = \( 2 \times 0.1V = 0.2V \)

### Step 3: Total Volume at Equivalence Point
The total volume of the solution at the equivalence point will be the sum of the volumes of both solutions:
\[ \text{Total Volume} = V + 2V = 3V \]

### Step 4: Calculate Moles of \( KNO_3 \) Produced
From the stoichiometry of the reaction:
- Moles of \( KNO_3 \) produced = \( 2 \times \text{moles of } Pb(NO_3)_2 = 2 \times 0.1V = 0.2V \)

### Step 5: Calculate Concentration of \( KNO_3 \)
The concentration of \( KNO_3 \) in the final solution is given by:
\[ C = \frac{\text{Moles of } KNO_3}{\text{Total Volume}} = \frac{0.2V}{3V} = \frac{0.2}{3} \, \text{mol/L} \]

### Step 6: Determine the Van 't Hoff Factor (i)
For \( KNO_3 \), it dissociates into:
\[ KNO_3 \rightarrow K^+ + NO_3^- \]
Thus, \( i = 2 \) (since it produces 2 ions).

### Step 7: Calculate Osmotic Pressure
The formula for osmotic pressure \( \Pi \) is:
\[ \Pi = i \times C \times R \times T \]
Substituting the values:
- \( i = 2 \)
- \( C = \frac{0.2}{3} \, \text{mol/L} \)
- \( R = 0.08 \, \text{L-atm/mol-K} \)
- \( T = 300 \, \text{K} \)

Thus,
\[ \Pi = 2 \times \left(\frac{0.2}{3}\right) \times 0.08 \times 300 \]

### Step 8: Simplify the Calculation
Calculating step-by-step:
1. \( \frac{0.2}{3} = 0.0667 \, \text{mol/L} \)
2. \( 0.0667 \times 2 = 0.1333 \, \text{mol/L} \)
3. \( 0.1333 \times 0.08 = 0.01067 \, \text{L-atm/mol} \)
4. \( 0.01067 \times 300 = 3.2 \, \text{atm} \)

### Final Answer
The osmotic pressure of the final solution at the equivalence point is:
\[ \Pi = 3.2 \, \text{atm} \]