Solubility Equilibria

 

40) m = 0.54 g PbI2          V = 1.00 L          T = 25°C

 

      [PbI2] = n/V = 0.54 g PbI2/ x 1 mol PbI2/461.00 g PbI2/1.00 L = 1.2 x 10-3 M

      [Pb2+] = 1.2 x 10-3 mol PbI2/L x 1 mol Pb2+/1 mol PbI2 = 1.2 x 10-3 M

      [I-] = 1.2 x 10-3 mol PbI2/L x 2 mol I-/1 mol PbI2 = 2.4 x 10-3 M

            

               PbI2(s)     Pb2+(aq)     +      2I-(aq)

      [ ]i                                   0                      0

      [ ]c                          +1.2 x 10-3      +2.4 x 10-3           

      [ ]e                            1.2 x 10-3        2.4 x 10-3

 

      Ksp = [Pb2+] x [I-]2 = 1.2 x 10-3 x (2.4 x 10-3)2 = 6.9 x 10-9