Question #37594

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Question #37594

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Answer 1 · 2017-05-27T05:20:03

None of these options are correct.

Explanation:

$P b I_{2} (s) ⇌ P b^{2 +} + 2 I^{-}$ $PbI_2(s) rightleftharpoonsPb^(2+) + 2I^-$

And $K_{sp} = [P b^{2 +}] {[I^{-}]}^{2}$ $K_"sp"=[Pb^(2+)][I^-]^2$ (Note that $P b I_{2}$ $PbI_2$ does not appear in the equation in that as a solid it does not participate in the equilibrium, as it CANNOT express a concentration.)

And given the prior equation, we can represent the solubility of $P b I_{2} (s)$ $PbI_2(s)$ as $S$ $S$ , and given the stoichiometry.........

$K_{sp} = (S) \times {(2 S)}^{2} = 4 S^{3}$ $K_"sp"=(S)xx(2S)^2=4S^3$

So $S =^{3} \sqrt{\frac{K_{sp}}{4}} =^{3} \sqrt{\frac{8.0 \times 10^{- 9}}{4}} = 1.26 \times 10^{- 3} \cdot m o l \cdot L^{- 1}$ $S=""^3sqrt(K_"sp"/4)=""^3sqrt((8.0xx10^-9)/4)=1.26xx10^-3*mol*L^-1$

See here for another example of a solubility equilibrium.

Answer 2 · 2017-05-27T05:53:02

None of the options are correct

$n (P b I_{2}) = 1.3 \cdot 10^{- 3} mol$ $n(PbI_2)=1.3*10^-3" mol"$

Explanation:

The equilibrium reaction is $P b I_{2} <--> P b^{2 +} + 2 I^{-}$ $PbI_2 " <--> " Pb^(2+)+2I^-$

Let $x = [P b^{2 +}]$ $x=[Pb^(2+)]$

Then due to the stoichiometry

$[I^{-}] = 2 x$ $[I^-]=2x$

For a saturated solution

$K_{s p} = 8.0 \cdot 10^{- 9} = [P b^{2 +}] \cdot {[I^{-}]}^{2} = x {(2 x)}^{2}$ $K_(sp)=8.0*10^-9=[Pb^(2+)]*[I^-]^2=x(2x)^2$

$8.0 \cdot 10^{- 9} = 4 x^{3}$ $8.0*10^-9=4x^3$

$x = {(\frac{8.0 \cdot 10^{- 9}}{4})}^{\frac{1}{3}}$ $x=((8.0*10^-9)/4)^(1/3)$

$x = [P b^{2 +}] = 1.3 \cdot 10^{- 3} M$ $x=[Pb^(2+)]=1.3*10^-3" M"$

$n (P b^{2 +}) = C \cdot V = 1 \cdot 1.3 \cdot 10^{- 3} mol$ $n(Pb^(2+))=C*V=1*1.3*10^-3" mol"$

As one mole of $P b I_{2}$ $PbI_2$ releases one mole of $P b^{2 +}$ $Pb^(2+)$ , the number of moles of $P b I_{2}$ $PbI_2$ dissolved is the same as $n (P b^{2 +})$ $n(Pb^(2+))$

$n (P b I_{2}) = 1.3 \cdot 10^{- 3} mol$ $n(PbI_2)=1.3*10^-3" mol"$

This is saying that you can dissolve up to $1.3 \cdot 10^{- 3} mol$ $1.3*10^-3" mol"$ of $P b I_{2}$ $PbI_2$ in 1 L of water, which is the solubility. Any more $P b I_{2}$ $PbI_2$ added will remain a solid.

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Question #37594

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