a.
.........PbI2 ==> Pb^2+ + 2I^-
I........solid.....0.......0
C.........-x.......x.......2x
E........solid.....x.......2x
Sustitute the E line into Ksp expression and solve for x = (Pb^2+)
b. You need Kc for the reaction and you can get that from the sum of the two reactions of Ksp and Kf.
PbI2 --> Pb^2+ + 2I^-.......Ksp
Pb^2+ + 4I^- ==> [PbI4]^2-.....Kf
---------------------------------
PbI2 + 2I^- --> [PbI4]^2-....Kc = Ksp*Kf
I assume that 1.25M for I^- is the equilibrium concentration.
Then Kc = [PbI4]^2-/(I^-)^2
Substitute and solve for the complex.
Super confusing question im not sure how to approach
a. Determine the concentration of lead ion in solution (the molar solubility) if PbI2(s) is in
equilibrium with water. Ksp = 9.8*10-9
b. From part a, it should be clear that if Pb2+(aq) and I-(aq) are added together in solution, a solid precipitate of PbI2 will form, which is yellow. What may be unexpected is what happens
when addition iodide ion is added. One would expect the equilibrium to shift right and lead to a greater amount of precipitate formed. However, the solution turns clear with no precipitate
visible. The reaction that occurs is PbI2(s) + 2I-(aq) [PbI4]2-(aq). Determine the concentration of Pb2+ in solution (as a complex) if the iodide concentration is 1.25 M. Kf ([PbI4]2-) = 3.0*104. Note; this phenomenon is somewhat rare in insoluble compounds. For example, adding Cl- to AgCl(s) does not lead to greater solubility.
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