Question #37594

2 Answers
May 27, 2017

Answer:

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

Answer:

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.