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Mansoura National University
            Pharm D-Clinical Pharmacy Program                 Level 1             Pharm. Anal. Chem. 1 (PC 101)


                                                           Quiz

                   1.  Calculate the molar solubility of Ag2SO4 in one liter of water.  Ksp = 1.4
                            -5
                       x 10
                                                                                                      -5
                   2.  Calculate the Ksp value of BaSO4 which has a solubility of 3.9 x 10
                       mole/L at 25°c



                                         Solubility Product Constant (Ksp)

             ➢  It is the equilibrium constant for the equilibrium between an ionic solid and its
                 saturated solution.
               A B (s)                 a A (aq) + b B(aq)
                 a  b
                           a    b
               Ksp = [A]  [B]


               Quiz


          •  Estimate the solubility of Ag2CrO4 in pure water if the solubility
                                                                              -12
              product constant for silver chromate is 1.1 x 10 .
                                                                                   2-
                                                                  +
                                  Ag2CrO4(s)             2 Ag (aq) + CrO4 (aq)
                                                             + 2
                                                                         2-
                                                 Ksp = [Ag ] [CrO4 ]
                   •  Substitute  the  equilibrium  amounts  and  the  K  into  the  equilibrium
                                                                                 sp
                       expression and solve for x.
                                                            -12       2
                                                   1.1 x 10  = [2x] [x]
                                                                   -5
                                                    x = 6.50 x 10  M
                                                                                           -3
                                                                                -5
          •   The solubility of barium sulphate at 298 K is  1.05 x 10  mol dm . Calculate
              the solubility product.
               The equilibrium is:

                                                                                +2
                                                                                            -2
                                                               BaSo4                             Ba    +    So4
               Notice that each mole of barium sulphate dissolves to give 1 mole of barium
               ions and 1 mole of sulphate ions in solution.


               That means that:



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