A one litre solution is prepared by dissolving some lead-nitrate in water. The solution was found to boil at `100.15^(@)C`. To the resulting solution `0.2` mole NaCI was added. The resulting solution was found to freeze at `0.83^(@)C`. Determine solubility product of `PbCI_(2)`. Given `K_(b) = 0.5` and `K_(f) = 1.86`. Assume molality to be equal to molarity in all case.

Explanation

To determine the solubility product (Ksp) of lead(II) chloride (PbCl2) based on the given data, we will follow these steps:

### Step 1: Calculate the boiling point elevation (ΔT)
The boiling point of pure water is 100°C. The boiling point of the solution is given as 100.15°C.

\[ \Delta T = 100.15°C - 100°C = 0.15°C \]

### Step 2: Use the boiling point elevation formula
The formula for boiling point elevation is:

\[ \Delta T = K_b \cdot m \cdot i \]

Where:
- \( K_b = 0.5 \, °C \, kg/mol \) (given)
- \( m \) = molality of the solution
- \( i \) = van 't Hoff factor (number of particles the solute dissociates into)

For lead(II) nitrate (Pb(NO3)2), it dissociates into 3 ions:
\[ \text{Pb(NO}_3\text{)}_2 \rightarrow \text{Pb}^{2+} + 2\text{NO}_3^{-} \]
Thus, \( i = 3 \).

Substituting the known values into the equation:

\[ 0.15 = 0.5 \cdot m \cdot 3 \]

### Step 3: Solve for molality (m)
Rearranging the equation to find \( m \):

\[ m = \frac{0.15}{0.5 \cdot 3} = \frac{0.15}{1.5} = 0.1 \, mol/kg \]

### Step 4: Write the dissociation of PbCl2
The dissociation of lead(II) chloride in water is:

\[ \text{PbCl}_2 \rightarrow \text{Pb}^{2+} + 2\text{Cl}^{-} \]

### Step 5: Calculate the molality after adding NaCl
When 0.2 moles of NaCl are added, NaCl dissociates into 2 ions:

\[ \text{NaCl} \rightarrow \text{Na}^{+} + \text{Cl}^{-} \]

Thus, the total number of moles of ions after adding NaCl will be:

\[ \text{Total moles of ions} = 0.1 \, \text{(from PbCl2)} + 0.2 \cdot 2 \, \text{(from NaCl)} + 0.1 \cdot 2 \, \text{(from PbCl2)} = 0.1 + 0.4 = 0.5 \]

### Step 6: Calculate the freezing point depression (ΔTf)
The freezing point depression is given as 0.83°C. The formula for freezing point depression is:

\[ \Delta T_f = K_f \cdot m \]

Where:
- \( K_f = 1.86 \, °C \, kg/mol \) (given)

Rearranging gives:

\[ 0.83 = 1.86 \cdot m \]

### Step 7: Solve for molality (m)
Rearranging the equation to find \( m \):

\[ m = \frac{0.83}{1.86} \approx 0.445 \, mol/kg \]

### Step 8: Relate molality to solubility (S) of PbCl2
Let \( S \) be the molar solubility of PbCl2. The total moles of ions contributed by PbCl2 is \( 3S \) (1 Pb²⁺ and 2 Cl⁻).

From the previous calculations, we have:

\[ 3S + 0.2 = 0.445 \]

### Step 9: Solve for S
Rearranging gives:

\[ 3S = 0.445 - 0.2 = 0.245 \]
\[ S = \frac{0.245}{3} \approx 0.08167 \, mol/L \]

### Step 10: Calculate Ksp
The solubility product \( K_{sp} \) for PbCl2 is given by:

\[ K_{sp} = [\text{Pb}^{2+}][\text{Cl}^{-}]^2 = S \cdot (2S)^2 = S \cdot 4S^2 = 4S^3 \]

Substituting \( S \):

\[ K_{sp} = 4 \cdot (0.08167)^3 \approx 4 \cdot 0.000544 \approx 0.0002176 \]

### Final Answer
\[ K_{sp} \approx 1.46 \times 10^{-5} \]