The solubility of PbCl 2 in water is 0.01 M at 2 5 ∘ C . The maximum concentration of Pb 2 + in 0.1 M NaCl will be

K s p of PbCl₂ = 4 s 3 = 4 × ( 0.01 ) 3 = 4 × 1 0 − 6 . Concentration of Pb²⁺ in NaCl solution = [ P b 2 + ] = [ C l − ] 2 K s p = 1 0 − 2 4 × 1 0 − 6 = 4 × 1 0 − 4 M.

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Chemistry - Ionic Equilibrium Question with Solution | TestHub

ChemistryIonic EquilibriumSolubility ProductMedium2 minQB

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The solubility of $PbCl_{2}$ in water is 0.01 M at $2 5^{\circ} C$ . The maximum concentration of $Pb^{2 +}$ in 0.1 M NaCl will be

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Answer:

Solution:

$K_{s p}$ of PbCl₂ = $4 s^{3} = 4 \times (0.01)^{3} = 4 \times 1 0^{- 6}$ .

Concentration of Pb²⁺ in NaCl solution = $[P b^{2 +}] = \frac{K _{s p}}{[ C l ^{-} ] ^{2}} = \frac{4 \times 1 0 ^{- 6}}{1 0 ^{- 2}} = 4 \times 1 0^{- 4}$ M.

Stream:JEESubject:ChemistryTopic:Ionic EquilibriumSubtopic:Solubility Product

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