Chemistry - Ionic Equilibrium Question with Solution | TestHub

ChemistryIonic EquilibriumSolubility ProductMedium2 minPYQ_2018

ChemistryMediumstatement

The minimum volume of water required to dissolve 0.1 g lead (II) chloride to get a saturated solution ( $K_{s p}$ of ${P b C l}_{2} = 3.2 \times 10^{- 8}$ ;atomic mass of $P b = 207 u$ ) is:

Options:

Answer:

Solution:

${{(K}_{s p})}_{P b {C l}_{2}} = 32 \times 10^{- 9}$

$\begin{matrix} {PbCl}_{2} & = & {Pb}^{2 +} \\ s \end{matrix} \begin{matrix} + & 2 {Cl}^{-} \\ 2 s \end{matrix}$

$K_{s p} = [P b^{2 +}] {[2 C l^{-}]}^{2}$

$K_{s p} = 4 s^{3} = 32 \times 10^{- 9}$

$s^{3} = 8 \times 10^{- 9}$

$s = 2 \times 10^{- 3} M$

$\frac{w}{M . w} \times \frac{1}{V_{L}} = 2 \times 10^{- 3}$

$\frac{0.1}{278} \times \frac{1}{V_{L}} = 2 \times 10^{- 3}$

$V_{L} = \frac{0.1 \times 1000}{278 \times 2} = 0.18 L$

Stream:JEESubject:ChemistryTopic:Ionic EquilibriumSubtopic:Solubility Product

⏱ 2mℹ️ Source: PYQ_2018

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