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:
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.
.........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.
To determine the progress of the reaction, we can compare the current concentration of lead iodide in the aqueous phase (0.125 M) with the equilibrium constant (Keq).
In this case, the Keq is given as 0.00027. Keq is the ratio of the product concentrations to the reactant concentrations at equilibrium. If the concentration of lead iodide in the aqueous phase is lower than what is predicted by the equilibrium constant, it means the reaction has not reached equilibrium and is still proceeding in the forward direction.
Comparing the given concentration of 0.125 M with Keq = 0.00027, we see that the concentration is higher than that predicted by Keq. Therefore, the reaction is currently dissolving lead iodide into the aqueous phase.
So, the correct answer is c. The reaction is currently dissolving lead iodide into solution.