$K_f$, or the *formation constant* for a complex, is one of the common “named equilibria” you will encounter (along with $K_a, K_b, K_w,$ and $K_{sp}$. A formation reaction will always have the form: $$ \text{ion} + \text{ligand} \rightleftharpoons \text{complex}$$.

As an example of dissolution by complex ion formation, let us consider what happens when we add aqueous ammonia to a mixture of silver chloride and water. Silver chloride dissolves slightly in water, giving a small concentration of Ag^{+} ([Ag^{+}] = $1.3×10^{-5}\;M$):

However, if NH_{3} is present in the water, the complex ion, $Ag(NH_3)_2^+$, can form according to the equation:

Dissociation of a Complex Ion

Calculate the concentration of the silver ion in a solution that initially is 0.10 *M* with respect to $Ag(NH_3)_2^+$

Solution

Applying the standard ICE approach to this reaction yields the following:

Substituting these equilibrium concentration terms into the *K*_{f} expression gives

The very large equilibrium constant means the amount of the complex ion that will dissociate, *x*, will be very small. Assuming *x* << 0.1 permits simplifying the above equation:

Because only 1.1% of the $Ag(NH_3)_2^+$ dissociates into Ag^{+} and NH_{3}, the assumption that *x* is small is justified.

Using this value of *x* and the relations in the above ICE table allows calculation of all species’ equilibrium concentrations:

The concentration of free silver ion in the solution is 0.0011 *M*.

Calculate the silver ion concentration, [Ag

^{+}], of a solution prepared by dissolving 1.00 g of AgNO

_{3}and 10.0 g of KCN in sufficient water to make 1.00 L of solution. (Hint: Because

*K*

_{f}is very large, assume the reaction goes to completion then calculate the [Ag

^{+}] produced by dissociation of the complex.)

## Answer:

$2.9×10^{-22}\;M$