Precipitation of Ions

• For PbCl2, (0.100)[Cl-]2 = 1.6 x 10-5

  [Cl-] = 0.0126 M to begin precipitating PbCl2.
  

• ForAgCl, [Ag+](0.0126) = 1.70 x 10-10

  [Ag+] = 1.35 x 10-8 M left when PbCl2 begins to precipitate.
  

• % Ag+ left = 100 x 1.35 x 10-8/0.100

  = 1.35 x 10-5%
  

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