Question

Calculate the number of moles of ammonia needed to dissolve 0.0010 mole of silver chloride in 4000 mL of solution at 25 degrees Celsius. K_sp for AgCl is 1.8 x 10^-10 and the dissociation constant for [Ag(NH₃)₂]⁺ is 6.3 x 10^-8

Answer 1

Given kSP for AgCl is 1.8x 10^-10 and Kd for complex= 6.3 x 10^-8

Thus kf for the complex = 1x 10⁸/6.3

The equilibria are

    AgCl(s) -------> Ag⁺ + Cl^-                             Ksp

Ag⁺ 2NH₃ ---------> [Ag(NH3)]⁺                        Kf

overall equilibrium is

AgCl(s) + 2NH3 ---------> [Ag(NH3)]⁺ + Cl-      Keq = Ksp x Kf

Thus Keq= Ksp x Kf = [Ag(NH3)]⁺ [Cl-] /[NH3}²

   Since all AgCl is dissolved [Cl-] = [Ag(NH3)]⁺ = 0.001M

      AgCl(s) + 2NH3 --------->    [Ag(NH3)]⁺ +       Cl-

                                        x                                0.001/4              0.001/4    equilibrium concentrations since in 4L solution

   Hence Keq = o.001 x0.001 /x² = 1.8 x 10 ^-10 x 10⁸/ 6.3

Thus x = 0.00467M

Since 0.002M is already used for complexation the required moles of ammonia are

     = 0.00467 + 0.002 =0.00667M