Question

Solid Lead nitrate is dissolved in 1 litre of water. The solution was found to boil at \(100.15\degree C\). When 0.2 mol of NaCl is added to the resulting solution, it was observed that the solution froze at \(–0.8\degree C\). The solubility product of PbCl₂ formed is ________ \(\times 10^{–6}\) at 298 K. (Nearest integer) Given: \(Kb = 0.5 K \text{ kg mol}^{–1}\) and \(Kf = 1.8 K \text{ kg mol}^{–1}\).
Assume molality to be equal to molarity in all cases.

Hint:

For colligative property problems:
• Use boiling or freezing point changes to determine molality.
• Consider dissociation factors for ionic compounds to adjust molality for total particles.
• For solubility products, express ion concentrations in terms of molarity and solve for Ksp.

Answer 1

Correct Answer: 13

Step 1: Determine the molality from boiling point elevation:
The boiling point elevation \(\Delta T_b\) is given by:
\[\Delta T_b = K_b \times m,\]
where \(m\) is the molality of the solution. Substituting the values:
\[0.15 = 0.5 \times m \quad \implies \quad m = \frac{0.15}{0.5} = 0.3~\text{mol/kg}.\]
Step 2: Total molality after adding NaCl:
NaCl dissociates into two ions (\(\text{Na}^+\) and \(\text{Cl}^-\)). When 0.2 mol of NaCl is added to the solution, the molality increases by \(0.2 \times 2 = 0.4~\text{mol/kg}\)
Total molality after adding NaCl is:
\[m_{\text{total}} = 0.3 + 0.4 = 0.7~\text{mol/kg}.\]
Step 3: Freezing point depression:
The freezing point depression \(\Delta T_f\) is given by:
\[\Delta T_f = K_f \times m_{\text{total}}.\]
Substituting the given freezing point depression and \(K_f\):
\[0.8 = 1.8 \times m_{\text{total}} \quad \implies \quad m_{\text{total}} = \frac{0.8}{1.8} = 0.444~\text{mol/kg}.\]
Step 4: Solubility product of PbCl\(_2\):
Lead chloride (PbCl\(_2\)) dissociates as:
\[\text{PbCl}_2 \leftrightharpoons \text{Pb}^{2+} + 2\text{Cl}^-.\]
Let \(s\) be the molarity of PbCl\(_2\) in solution. The concentrations of the ions at equilibrium are:
\[[\text{Pb}^{2+}] = s, \quad [\text{Cl}^-] = 2s.\]
The solubility product \(K_{sp}\) is:
\[K_{sp} = [\text{Pb}^{2+}] \times [\text{Cl}^-]^2 = s \times (2s)^2 = 4s^3.\]
From freezing point depression, the effective molality of ions is \(0.444~\text{mol/kg}\). Using the relation for ionic dissociation:
\[m_{\text{total}} = s + 2s = 3s \quad \implies \quad s = \frac{0.444}{3} = 0.148~\text{mol/L}.\]
Step 5: Calculate \(K_{sp}\):
Substituting \(s = 0.148\) into the expression for \(K_{sp}\):
\[K_{sp} = 4 \times (0.148)^3 = 4 \times 0.00323 = 0.01292 \quad \text{or} \quad 13 \times 10^{-6}.\]
Final Answer: \(13\).