Question #37594

2 Answers
May 27, 2017

None of these options are correct.

Explanation:

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

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

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

So 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.

May 27, 2017

None of the options are correct

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

Explanation:

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

Let x=[Pb^(2+)]

Then due to the stoichiometry

[I^-]=2x

For a saturated solution

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

8.0*10^-9=4x^3

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

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

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

As one mole of PbI_2 releases one mole of Pb^(2+), the number of moles of PbI_2 dissolved is the same as n(Pb^(2+))

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

This is saying that you can dissolve up to 1.3*10^-3" mol" of PbI_2 in 1 L of water, which is the solubility. Any more PbI_2 added will remain a solid.