Question

1.

a) Calculate the solubility of Pb3(PO4)2 in water.

b) Calculate the solubility of Hg2Br2 in water and in 0.300 M CaBr2. Explain why the solubilities are different in water and in 0.300 M CaBr2.

Answer 1

   a. Pb3(PO4)2 ----------> 3Pb⁺² + 2PO4^3-

                                           3s       2s

    Ksp           = [Pb+2]³[PO4^3-]²

     1*10^-54    = (3s)³ *(2s)²

                    = 108s⁵

s⁵                = 1*10^-54/108

                    = 0.0926*10^-55

s                  = 4.5*10^-12 M

solubility of Pb3(PO4)2 is 4.5*10^-12 M

b.            Hg2Br2 ---------> Hg₂²⁺ + 2Br^-

                                         0           0.3

                                        s           0.3+2s

          Ksp    = [Hg₂²⁺][Br-]²

           6.4*10^-23   = s*(0.3+2s)²

              s             = 1.4*10^-17

due to commom ion effect solubilities are different in water and inCaBr2.