Question

Question #37594

Answer 1

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.

Answer 2

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.