The molar solubility of lead(II) iodide, PbI2(s), is 1.45 x 10-3 M at 20 oC and 6.85 x 10-3 M at 80 oC. A. (2 pts.) What is the Ksp of PbI2(s) at 20 oC? B. (2 pts.) Determine the Ksp at 80 oC. C. (3 pts.) Use the information from parts A and B to determine the ?Ho. D. (3 pts.) What is ?So for this reaction at 20 oC? E. (2 pts.) Calculate the value of ?Go from your obtained enthalpy and entropy values at 20 oC. F. (2 pts.) At what temperature range is this dissociation spontaneous? Problem #1 continued on next page... G. (3 pts.) Determine the value of Gibbs Free Energy at 20 oC when [Pb+2] = 4.1 x 104 M and [I-] = 5.2 x 103 M. Do these conditions make this reaction more or less spontaneous at this temperature? Briefly explain. H. (2 pts.) In what ways does the Debye-Hückel limiting law provide a more accurate determination of the solubility product constant for PbI2(s)? Briefly explain. I. (3 pts.) A separate experiment is conducted in which 4.82 x 10-5 M CaI2(aq) is added to a solid solution of PbI2 at 20 oC. Qualify how the molar solubility changes. Does this addition favor the dissociation of the solid? Briefly explain. J. (4 pts.) Finally, in another experiment, a student mixes 40.0 mL of 3.00 M Pb(NO3)2(aq) with 20.0 mL of 2.00 x 10-3 M NaI(aq) at 20 oC. Does PbI2(s) precipitate from the solution? Briefly explain why or why not.