Question

The Ksp of Cadmium II phosphate in distilled water at 25 degrees celsius is 2.5x10^-33. The complex ion equilibrium formation constant for the tetraaminocadmiumate complex ion Cd(NH3)4^2+ is 1.3x10^7. Determine the equilibrium constant for the dissolution of cadmium II phosphate in a solution of 6.00M Ammonia.

Answer 1

Solubility equation of Cd₃(PO4)₂ is

Cd₃(PO4)₂ = 3 Cd^2+​ + 2 (PO4)^3-

Ksp =(3S)³ (2S)² = 108 S^5=2.5 x 10^-33

S =5√(2.5x 10^-33)/108=1.23 x 10^-7

Formation equation of Cd complex is

Cd^2+​ + 4 NH3  = Cd(NH3)4^2+

The complex ion equilibrium formation constant  

= [Cd(NH3)4^2+] / [Cd^2+]  [NH3] ^4=  1.3x10^7

Dissolution equation is

Cd₃(PO4)₂ + 12 NH3  = 3 Cd(NH3)4^2+  + 2 (PO4)^3-

for 6.00 M divide equation by 2

0.5 Cd₃(PO4)₂ + 6 NH3  = 1.5 Cd(NH3)4^2+  +(PO4)^3-

So dissolution equilibrium is  

[Cd(NH3)4^2+]^1.5 [ PO4^3-] /[NH3] ^6 [Cd^2+​ ]

So dissolution equilibrium is 1.5 xcomplex ion equilibrium formation constant x( [ PO4^3-]= S)

= 1.5 x 1.3 x 10^7 x1.23 x 10^-7 = 2.3985