Lead iodide is a relatively insoluble compound. It can dissolve from the solid phase into the aqueous phase with a Keq of 0.00027. If the concentration of lead iodide in the aqueous phase 0.125 M, then:

Question

a. we need more information to determine the progress of the reaction.

b. the system is at equilibrium.

c. the reaction is currently dissolving lead iodide into solution.

d. the reaction is currently condensing lead iodide out of solution.

Answer 1

.........PbI2 ==> Pb^2+ + 2I^-
I........solid.....0.......0
C........solid.....x.......2x
E........solid.....x.......2x

Ksp = (Pb^2+)(I^-)^2
0.00027 = (x)(2x)& = 4x^3
so x = solubility PbI2 = about 0.04M
WHICH IS THE MAXIMUM AMOUNT OF PbI2 in solution. Frankly, I don't know how the concentration of PbI2 became 0.125 M but the answer should be obvious from that.