Question #37594

2 Answers
May 27, 2017

None of these options are correct.

Explanation:

PbI_2(s) rightleftharpoonsPb^(2+) + 2I^-PbI2(s)Pb2++2I

And K_"sp"=[Pb^(2+)][I^-]^2Ksp=[Pb2+][I]2 (Note that PbI_2PbI2 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 PbI_2(s)PbI2(s) as SS, and given the stoichiometry.........

K_"sp"=(S)xx(2S)^2=4S^3Ksp=(S)×(2S)2=4S3

So S=""^3sqrt(K_"sp"/4)=""^3sqrt((8.0xx10^-9)/4)=1.26xx10^-3*mol*L^-1S=3Ksp4=38.0×1094=1.26×103molL1

See here for another example of a solubility equilibrium.

May 27, 2017

None of the options are correct

n(PbI_2)=1.3*10^-3" mol"n(PbI2)=1.3103 mol

Explanation:

The equilibrium reaction is PbI_2 " <--> " Pb^(2+)+2I^-PbI2 <--> Pb2++2I

Let x=[Pb^(2+)]x=[Pb2+]

Then due to the stoichiometry

[I^-]=2x[I]=2x

For a saturated solution

K_(sp)=8.0*10^-9=[Pb^(2+)]*[I^-]^2=x(2x)^2Ksp=8.0109=[Pb2+][I]2=x(2x)2

8.0*10^-9=4x^38.0109=4x3

x=((8.0*10^-9)/4)^(1/3)x=(8.01094)13

x=[Pb^(2+)]=1.3*10^-3" M"x=[Pb2+]=1.3103 M

n(Pb^(2+))=C*V=1*1.3*10^-3" mol"n(Pb2+)=CV=11.3103 mol

As one mole of PbI_2PbI2 releases one mole of Pb^(2+)Pb2+, the number of moles of PbI_2PbI2 dissolved is the same as n(Pb^(2+))n(Pb2+)

n(PbI_2)=1.3*10^-3" mol"n(PbI2)=1.3103 mol

This is saying that you can dissolve up to 1.3*10^-3" mol"1.3103 mol of PbI_2PbI2 in 1 L of water, which is the solubility. Any more PbI_2PbI2 added will remain a solid.