Question

Calculate the equilibrium constant Kc for the net reaction show below

AgI(s) + 2NH3(aq) <> Ag(NH3)2+(aq) + I-(aq)

For AgI, Ksp=8.3*10^-17

For Ag(NH3)2+ , Kf=1.5*10^7

Answer 1

AgI which is insoluble in water dissolves in aqueous ammonia solution due to the formation of the soluble complex [Ag(NH3)₂] I

The various equilibria involved in this process are

(i)   AgI(s) = Ag⁺(aq) + I^- (aq)

The equilibrium constant for this reaction is the solubility product designated by the symbol K_sp

                    K_sp = [Ag⁺][I^-]

(ii) Ag⁺ + 2NH₃ = [Ag(NH₃)₂]⁺

                 [Ag(NH₃)₂]⁺ is a complex ion

The equilibrium constant for the complex formation is denoted by the symbol K_f and is given by

        K_f = [[Ag(NH₃)₂]⁺] / [Ag⁺][NH₃]²

Adding reaction (i) and (ii) we get

AgI(s) + 2NH₃   = [Ag(NH₃₎₂]⁺ + I^- (aq)

The equilibrium constant for the overall reaction K_c is given by

K_c = [[Ag(NH₃)₂]⁺] [I^-] / [NH₃]²

    Multiplying both numerator and denominator by [Ag+] and rearranging

K_c = [Ag⁺] [[Ag(NH₃)₂]⁺][I^-] / [Ag⁺] [NH₃]²

       =[Ag⁺] [I^-]    x [[Ag(NH₃)₂]⁺] /[Ag⁺] [NH₃]²

K^c = K_sp .K_f

K_sp = 8.3x10^-17

K_f = 1.5x10⁷

K_c = 8.3 x10^-17 x 1.5 x10⁷

  K_c= 1.25 x 10^-9