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.
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.
To approach these questions, you need to understand the concepts of solubility product constant (Ksp) and complex formation constant (Kf). Let's address each question step by step:
a. To determine the molar solubility of lead ion (Pb2+) in solution, you need to use the solubility product constant (Ksp). The equilibrium equation for the dissolution of PbI2(s) in water is:
PbI2(s) ⇌ Pb2+(aq) + 2I-(aq)
The expression for the solubility product constant (Ksp) is derived from the equilibrium equation, where the concentrations of all species are raised to the power equal to their stoichiometric coefficient:
Ksp = [Pb2+][I-]^2
Given that the Ksp value is 9.8 × 10^-9, you can set up an equation using the molar solubility of PbI2, represented as x, as follows:
9.8 × 10^-9 = x * (2x)^2
Solving this equation will give you the molar solubility of PbI2, which represents the concentration of Pb2+ in solution.
b. When iodide ion (I-) is added to a solution containing the precipitate PbI2, an unexpected reaction occurs, where PbI2(s) reacts with 2I-(aq) to form the complex ion [PbI4]2-(aq). The equilibrium equation for this reaction is:
PbI2(s) + 2I-(aq) ⇌ [PbI4]2-(aq)
The expression for the formation constant (Kf) is similar to the one for Ksp, but it represents the equilibrium constant for the formation of the complex ion:
Kf = [PbI4]2- / ([Pb2+][I-]^2)
Given that the Kf value is 3.0 × 10^4, and the concentration of iodide ion (I-) is 1.25 M, you can set up an equation using the concentration of Pb2+ (as a complex), represented as x, as follows:
3.0 × 10^4 = x / (1.25^2 * 2)
Solving this equation will give you the concentration of Pb2+ in solution as a complex.
In summary, to approach these questions, you need to set up the necessary equilibrium equations and use the appropriate equilibrium constants (Ksp and Kf) to solve for the unknown concentrations.