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

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