Question

What happens to the equilibrium constant in a complexation reaction?

Answer 1

Hint: The equilibrium constant of a chemical reaction is the value of its reaction quotient at chemical equilibrium, a state approached by a dynamic chemical system after sufficient time has elapsed at which its composition has no measurable tendency towards further change.

Complete Step By Step Answer:
A complexation reaction has a unique equilibrium constant called a formation constant, $ {K_f} $ .Generally, complexation reactions are favorable, and as such, $ {K_f} $ , overall is usually large.
Complexation reactions may be multistep.
Take the following reaction for example:
 $ {[Cu{({H_2}O)_4}]^{2 + }}(aq) + 4N{H_3}(aq) \rightleftharpoons {[Cu{(N{H_3})_4}]^{2 + }}(aq) + 4{H_2}O(l) $
This, although it seemingly could occur in one step, actually, it happens in multiple steps, and there is one formation constant for the displacement of each aqua ligand on the tetraaquacopper(II) complex.
There exists a formation constant as an equilibrium constant for each individual step.
We would separate the steps to incorporate the ammine ligands by displacing one aqua ligand at a time.
 $ {[Cu{({H_2}O)_4}]^{2 + }}(aq) + 4N{H_3}(aq) \rightleftharpoons {[Cu{(N{H_3})_4}]^{2 + }}(aq) + 4{H_2}O(l) $
 $ \Rightarrow {K_f} = \dfrac{{\left[ {{{[Cu{{({H_2}O)}_3}N{H_3}]}^{2 + }}} \right]}}{{\left[ {{{[Cu{{({H_2}O)}_4}]}^{2 + }}} \right][N{H_3}]}} $

Next step, displace another:
 $ \Rightarrow {K_{f_2}} = \dfrac{{\left[ {{{[Cu{{({H_2}O)}_2}{{(N{H_3})}_2}]}^{2 + }}} \right]}}{{\left[ {{{[Cu{{({H_2}O)}_3}N{H_3}]}^{2 + }}} \right][N{H_3}]}} $

Next step, displace another:
 $ \Rightarrow {K_{f_3}} = \dfrac{{\left[ {{{[Cu({H_2}O){{(N{H_3})}_3}]}^{2 + }}} \right]}}{{\left[ {{{[Cu{{({H_2}O)}_2}{{(N{H_3})}_2}]}^{2 + }}} \right][N{H_3}]}} $

And finally, the fourth step!
 $ \Rightarrow {K_{f_4}} = \dfrac{{\left[ {{{[Cu{{(N{H_3})}_4}]}^{2 + }}} \right]}}{{\left[ {{{[Cu({H_2}O){{(N{H_3})}_3}]}^{2 + }}} \right][N{H_3}]}} $
 $ {K_{f_4}} < {K_{f_3}} < {K_{f_2}} < {K_{f_1}} $ . As each ammine ligand displaces one aqua ligand, there is one less aqua ligand on the complex. The less aqua ligands after each step, it tends to be entropically more difficult to displace each consecutive aqua ligand, provided no other variables change.

The overall formation constant can be acquired from multiplying those of each individual, consecutive step.
 $ \Rightarrow {K_{foverall}} = {\beta _4} = \dfrac{{\left[ {{{[Cu{{(N{H_3})}_4}]}^{2 + }}} \right]}}{{\left[ {{{[Cu{{({H_2}O)}_4}]}^{2 + }}} \right]{{[N{H_3}]}^4}}} $

Therefore, we have:
 $ \Rightarrow {\beta _4} = {K_{f_1}}{K_{f_2}}{K_{f_3}}{K_{f_4}} $ .

Note:
The equilibrium constant is independent of the initial analytical concentrations of the reactant and product species in the mixture. Thus, given the initial composition of a system, known equilibrium constant values can be used to determine the composition of the system at equilibrium. However, reaction parameters like temperature, solvent, and ionic strength may all influence the value of the equilibrium constant.