Solubility Equilibria

52) (a) [Co(NO₃)₂] = 0.020 M          pH = 8.5          K_sp = 1.3 x 10^-15           

            Co(NO₃)₂(aq) Co²⁺(aq) + 2NO₃^-(aq)

            [Co²⁺] = 0.020 mol Co(NO₃)₂/L x 1 mol Co²⁺/1 mol Co(NO₃)₂ = 0.020 M

            pH + pOH = 14.00

            pOH = 14.00 - pH = 14.00 - 8.5 = 5.5

            [OH^-] = 10^-pOH = 10^-5.5 = 3 x 10^-6 M

            Co(OH)₂(s) Co ²⁺(aq)   +   2OH^-(aq)  

             Q = [Co ²⁺] x [OH^-]²

             Q = 0.020 x (3 x 10^-6)² = 2 x 10^-13

             Q > K_sp, therefore, Co(OH)₂ will precipitate.

       (b) [AgNO₃] = 0.010 M          [NaIO₃] = 0.015 M          K_sp = 3.1 x 10^-8

            V_AgNO3 = 100. mL             V_NaIO3 = 10. mL

             AgNO₃(s) Ag⁺(aq) + NO₃^-(aq)

             NaIO₃(s) Na ⁺(aq) + IO₃^-(aq)

             [AgNO₃] = n/V

             n_Ag+ = 0.010 mol AgNO₃/L x 1 mol Ag⁺/1 mol AgNO₃ x 100. mL x 1 L/10³ mL = 1.0 x 10^-3 mol Ag⁺

             n_IO3- = 0.015 mol NaIO₃/L x 1 mol IO₃^-/1 mol NaIO₃ x 10. mL x 1 L/10³ mL = 1.5 x 10^-4 mol IO₃^-

             [Ag⁺] = 1.0 x 10^-3 mol Ag⁺/(110. mL x 1 L/10³ mL) = 9.1 x 10^-3 M

             [IO₃^-] = 1.5 x 10^-4 mol IO₃^-/(110. mL x 1 L/10³ mL) = 1.4 x 10^-3 M

             AgIO₃(s) Ag ⁺(aq)   +   IO₃^-(aq)

             Q = [Ag ⁺] x [IO₃^-]

             Q = 9.1 x 10^-3 x 1.4 x 10^-3 = 1.3 x 10^-5

             Q > K_sp, therefore, AgIO₃ will precipitate.