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 solve the problem step by step, we will follow the outlined procedure based on the information given in the question.

### Step 1: Calculate the change in boiling point (ΔTb)
The normal boiling point of water is 100°C. The boiling point of the solution is given as 100.15°C.

\[ \Delta T_b = T_b(\text{solution}) - T_b(\text{pure water}) = 100.15°C - 100°C = 0.15°C \]

### Step 2: Use the boiling point elevation formula to find molality (m)
The formula for boiling point elevation is:

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

Where:
- \(i\) is the van 't Hoff factor (number of particles the solute breaks into),
- \(K_b\) is the ebullioscopic constant (given as 0.5),
- \(m\) is the molality.

For lead nitrate \((Pb(NO_3)_2)\), it dissociates into 1 \(Pb^{2+}\) and 2 \(NO_3^-\), giving a total of 3 particles. Thus, \(i = 3\).

Plugging in the values:

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

Solving for \(m\):

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

### Step 3: Calculate the moles of \(Pb(NO_3)_2\) in the solution
Since the volume of the solution is 1 liter, the molarity is equal to the molality. Therefore, the moles of \(Pb(NO_3)_2\) in the solution is:

\[ \text{Moles of } Pb(NO_3)_2 = 0.1 \text{ moles} \]

### Step 4: Write the reaction of \(Pb(NO_3)_2\) with \(NaCl\)
The reaction between lead nitrate and sodium chloride is:

\[ Pb(NO_3)_2 + 2NaCl \rightarrow PbCl_2 + 2NaNO_3 \]

### Step 5: Determine the moles of \(NaCl\) and the resulting moles of products
We have 0.2 moles of \(NaCl\) added. The stoichiometry of the reaction shows that 1 mole of \(Pb(NO_3)_2\) reacts with 2 moles of \(NaCl\). Therefore, the limiting reagent is \(Pb(NO_3)_2\).

After the reaction:
- Moles of \(Pb(NO_3)_2\) remaining = 0.1 - 0.1 = 0 (all reacted)
- Moles of \(NaCl\) remaining = 0.2 - 0.2 = 0 (all reacted)
- Moles of \(PbCl_2\) formed = 0.1

### Step 6: Calculate the freezing point depression (ΔTf)
The freezing point depression is given as:

\[ \Delta T_f = T_f(\text{pure water}) - T_f(\text{solution}) = 0 - 0.83 = -0.83°C \]

### Step 7: Use the freezing point depression formula to find molality
The formula for freezing point depression is:

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

Where \(K_f\) is given as 1.86. Here, \(i\) for \(NaNO_3\) is 2 (since it dissociates into 2 ions).

Plugging in the values:

\[ 0.83 = 2 \cdot 1.86 \cdot (2 \cdot 0.2 + 3S) \]

Where \(S\) is the solubility of \(PbCl_2\). Rearranging gives:

\[ 0.83 = 3.72 \cdot (0.4 + 3S) \]

Solving for \(S\):

\[ 0.83 = 1.488 + 11.16S \]
\[ 11.16S = 0.83 - 1.488 \]
\[ 11.16S = -0.658 \]
\[ S = \frac{-0.658}{11.16} = 0.059 \text{ mol/kg} \]

### Step 8: Calculate the solubility product \(K_{sp}\)
The solubility product \(K_{sp}\) for \(PbCl_2\) is given by:

\[ K_{sp} = [Pb^{2+}][Cl^-]^2 \]

Where:
- \([Pb^{2+}] = S = 0.059\)
- \([Cl^-] = 2S = 2 \times 0.059 = 0.118\)

Thus,

\[ K_{sp} = (0.059)(0.118)^2 = 0.059 \cdot 0.013924 = 0.000820 \]

### Final Answer
The solubility product \(K_{sp}\) of \(PbCl_2\) is approximately:

\[ K_{sp} \approx 8.20 \times 10^{-4} \]