What would happen to the formation constant?

Question

How would the formation constant for [Cd(NH3)4 2+] be affected (increase, decrease, or no change) if not all of the CdC2O4 precipitate reacted with NH3? Justify your answer.

Answer 1

Hello Chowdhury

 

The complex ion formation reaction can be written as:

 

CdC₂O₄ (s) ↔ Cd²⁺ (aq) + C₂O₄^2- (aq)

 

Cd²⁺ (aq) + 4NH₃ (aq) ↔ Cd(NH₃)₄²⁺ (aq)

 

Cadmium oxalate (CdC₂O₄ (s)) has a very low solubility in water, but some of it will dissociate to form aqueous Cd²⁺ and C₂O₄^2- ions.

 

In the 2nd equilibrium reaction, as NH₃ is added, it forms the soluble Cd(NH₃)₄²⁺ complex ion.  This causes the concentration of Cd²⁺ in the first equilibrium reaction to decrease causing more insoluble CdC₂O₄ to dissociate as per LeChaterlier's principle.  As more and more NH₃ is added, this should eventually cause CdC₂O₄ to completely dissolve.

 

However, if not all CdC₂O₄ reacts with NH₃, then this will result in less Cd(NH₃)₄²⁺ forming, thereby decreasing its formation constant.