Question

determine the molar solubility of BaF2 in a solution containing 0.0870 m LiF. Ksp (BaF2)=1.6x10^-5

Answer 1

Let the molar solubility of BaF2 be x

                                                      BaF2 ---------> Ba²⁺      +     2 F^-

                                Initial               -                    0                     0.0870

                               Change            -                    + x                  + 2x

                               Equilibrium       -                     x                  ( 0.0870 + 2x)

                                  [Ba²⁺] x [ F-]² = K_sp

                                 (x) ( 0.0870 + 2x ) = 1.6 x 10^-5

                                 2x² + 0.0870 x = 0.000016

                                 2x² + 0.0870x - 0.000016 = 0

                                   x = 0.000032   ( using quadratic formula)

                                  x = 3.2 x 10^-5

The required molar solubility is 3.2 x 10^-5 M.