Solid Lead nitrate is dissolved in 1 litre of water. The solution was found to boil at 100.15°C. When 0.2 mol of NaCl is added to the resulting solution, it was observed that the solution froze at –0.8°C. The solubility product of formed is ______ at 29

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Solution

Q.
Solid Lead nitrate is dissolved in 1 litre of water. The solution was found to boil at 100.15°C. When 0.2 mol of NaCl is added to the resulting solution, it was observed that the solution froze at –0.8°C. The solubility product of ${PbCl}_{2}$ formed is ______ $\times 10^{- 6}$ at 298 K. (Nearest integer)

Given: $K_{b} = 0.5$ K kg ${mol}^{- 1}$ and $K_{f} = {1.8 K kg mol}^{- 1}$ . Assume molality to be equal to molarity in all cases. [2023]

Ans.
(13)

$Pb {({NO}_{3})}_{2} \to {Pb}^{2 +} + 2 {NO}_{3}^{-}$

$Δ T_{b} = i k_{b} m$

$(100.15 - 100) = 3 \times 0.5 \times m$

$0.15 = 1.5 \times m$

$m = \frac{0.15}{1.5} = 0.1$

$No. of moles of {Pb}^{+ 2} = 0.1$

$NaCl$ $\to$ ${Na}^{+}$ $+$ ${Cl}^{-}$

$(0.2) (0.2) (0.2)$

$\begin{matrix} {Pb}^{2 +} & + & 2 {Cl}^{-} & \to & {PbCl}_{2} ↓ \\ (0.1) & (0.2) \\ (0.1 - α) & (0.2 - 2 α) & 2 α \end{matrix}$

$Δ T_{f} = i \times k_{f} \times m$

$0.8 = 1.8 [\frac{0.1 - α + 0.2 - 2 α + 0.2 + 0.2}{1}]$

$0.8 = 1.8 [0.7 - 3 α]$

$α = \frac{1.8 \times 0.7 - 0.8}{1.8 \times 3} = 0.085$

$[{Pb}^{+ 2}] = 0.1 - 0.085 = 0.015$

$[{Cl}^{-}] = 0.2 - 0.085 \times 2 = 0.03$

${PbCl}_{2} ⇌ {Pb}^{2 +} + 2 {Cl}^{-}$

$K_{s p} = [{Pb}^{+ 2}] {[{Cl}^{-}]}^{2}$

$K_{s p} = 13.5 \times 10^{- 6}$

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