Question

A solution is 0.020 M in each Ca²⁺ and Cd²⁺. Adjusting the pH of the solution to which of the following values would achieve the best separation by precipitation of the hydroxides? solubility product constants

Best pH =  ---Select--- 13.0 12.5 12.0 11.5 11.0 10.5 10.0 9.5 9.0 8.5 8.0

What would be the resulting concentrations of the ions if the solution were adjusted to this pH?

[Ca²⁺] =  M

[Cd²⁺] =

M

Answer 1

The Solution containing 0.020M each Ca²⁺ and Cd²⁺ .

   The resulting solution has pH = 12.0

   so concentration of H⁺ ion, [H⁺] = antilog (-pH)   [since pH = -log H⁺ ]

                                                   = antilog (12.0 )

                                                  = 1*10^-12 M

                        Therefore [OH^-] = K_w / [H⁺]         since  K_w = [H⁺]*[OH^-]

                                                 = 1*10^-14 / 1*10^-12

                                                = 1*10^-2

       We know the reactions between Ca²⁺ and Cd²⁺ with OH^-

                   Ca^{2+ _{(aq) + 2 OH}-} (aq) --------> Ca (OH)₂

                 Cd^{2+ _{(aq) + 2 OH}-} (aq) --------> Cd (OH)₂

   

     K_sp of     Ca (OH)₂ = 6.5*10^-6

               so        6.5*10^-6 = [Ca²⁺] * [ OH^-] ²

                                        = [Ca²⁺]* [1*10^{-2 _]2}

                            [Ca²⁺] =  6.5*10^-6/ [1*10^{-2 _]2}

                                                    ^{_{= 6.5*10}-2} M

                                     = 0.065 M

K_sp of     Cd(OH)₂ = 2.5*10^-14

               so        2.5*10^-14  = [Cd²⁺] * [ OH^-] ²

                                        = [Cd²⁺]* [1*10^{-2 _]2}

                            [Cd²⁺] =  2.5*10^-14 / [1*10^{-2 _]2}

                                                     ^{_{= 2.5*10}-10}M

     Therefore   [Ca²⁺] = 0.065M and   [Cd²⁺] = 2.5*10^-10 M