Determine the molar solubility if PbI2(s) is in equilibrium with water. Ksp = 9.

Question

Determine 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 molar solutbility of PbI2(s) if the iodide concentration is increased to 6.25 M. Kf ([PbI4]2-) = 3.0*104for the reaction Pb2+(aq) + 4I-(aq) [PbI4]2-(aq) Note; this phenomenon is somewhat rare in insoluble compounds. For example, adding Cl- to AgCl(s) does not lead to greater solubility.

Explanation / Answer

Setting up the solubility equilbrium,

Pbl2 (s) ----------> Pb2+ + 2 I-   Ksp = [Pb2+][I-]2

ICE table:

  Ksp = [Pb2+][I-]2  = s(2s)2 =4s3

4s3= 9.8*10-9

s = 0.00135 M

Molar solubility of PbI2 = s = 0.00135 M

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