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Question #ed5f3

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Answer 1 · 2017-05-19T04:35:25

$P b C l_{2} (s) ⇌ P b^{2 +} + 2 C l^{-}$ $PbCl_2(s)rightleftharpoonsPb^(2+) + 2Cl^(-)$

Explanation:

The solubility expression for lead chloride can be written using the $solubility product$ $"solubility product"$ , another equilibrium constant, that must be measured........

At $298 \cdot K$ $298*K$ ,

$K_{sp} = [P b^{2 +}] {[C l^{-}]}^{2} = 1.7 \times 10^{- 4}$ $K_"sp"=[Pb^(2+)][Cl^-]^2=1.7xx10^-4$ .

In a saturated solution: $K_{sp} = (S) {(2 S)}^{2} = 4 S^{3}$ $K_"sp"=(S)(2S)^2=4S^3$

$S =^{3} \sqrt{\frac{1.7 \times 10^{- 4}}{4}} = 3.49 \times 10^{- 2} \cdot m o l \cdot L^{- 1}$ $S=""^(3)sqrt((1.7xx10^-4)/4)=3.49xx10^-2*mol*L^-1$ ; which gives a gram solubility of under $10 \cdot g \cdot L^{- 1}$ $10*g*L^-1$ .

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