Complex Ion Formation Effects on Solubility — Overview

Let'a talk about the how the formation of complex ions affect solubility. But first, we will quickly debrief on what a complex ion is.

A complex ion is a Lewis acid-base complex that consists of a central metal cation, usually a transition metal, and one or more surrounding ligands. Transition metal ions have a tendency to form complex ions because they have more than a single oxidation state.

The positively charged metal ion acts as the Lewis acid and the ligand acts as a Lewis base, and this Lewis acid-base relationship allows for the formation of a complex ion. The multiple oxidation state property of the metal ions is responsible for their ability to effectively act as Lewis acids in many different reactions with electron donor molecules.

The formation of a complex ion has an affect on solubility. In salts, introducing a ligand that is able to form a stable complex will increase its solubility. This is due to the metal cations that are removed from the solubility equilibria in order to form the complex ion.

Let's take a look at an example to better understand what this means:

Suppose we have a beaker of a saturated solution of copper (II) sulfide, $CuS$ , which has a $K_{sp}$ of 6.3x $10^{-36}$ .

Remember that that $K_{sp}$ , known as the solubility product constant, is the equilibrium constant for a slightly soluble ionic compound dissolving in an aqueous solution.

$CuS_{s} \Leftrightarrow Cu^{2+}_{(aq)} + S^{2-}_{aq}$

with

$K_{sp} = 6.3 x 10^{-36} @ 25^{o}C$

indicating that the equilibrium of this solution lies very far to the left.

At the bottom of the beaker, some undissolved $CuS_{s}$ remains. Aqueous ammonia, a monodentate ligand, is then added to the solution. The trace amount of copper cations, $Cu^{+}_{(aq)}$ , present in the solution begin to form a complex ion, $Cu(NH_{3})_{4}^{2+}$ , with the new ligand.

$Cu^{2+}_{aq} + 4NH_{3(aq)} \Leftrightarrow Cu(NH_{3})_{4(aq)}^{2+}$

with

$K_{f}= 1.1 x 10^{13}$

Remember that $K_{f}$ , known as the formation constant, describes the formation of a complex ion

This introduction of the ammonia ligand upsets the equilibrium of the saturated solution. To make up for the $Cu^{2+}$ ions that are pulled out of solution, Le Chatelier's principle tells us that we can drive the reaction to the right by removing one of the products. By doing so, more $CuS_{s}$ must go into solution.

Let's sum together the two equilibriums.

Original: $CuS_{s} \Leftrightarrow Cu^{2+}_{(aq)} + S^{2-}_{aq}$ $K_{sp}=6.3x10^{-36}$

New: $Cu^{2+}_{aq} + 4NH_{3(aq)} \Leftrightarrow Cu(NH_{3})_{4(aq)}^{2+} K_{f}=1.1x10^{13}$

Sum: $CuS_{s} + 4NH_{3(aq)} \Leftrightarrow Cu(NH_{3})_{4(aq)}^{2+} + S^{2-}$

This equilibrium constant for this overall equilibrium expression is:

$K_{c}= \frac{[Cu(NH_{3})_{4}^{2+}][S^{2-}]}{[NH_{3}]}$

This equation shows that as ammonia forms a complex with the $CuS$ , more of the solid will dissolve as the reaction proceeds towards the products, thus increasing the solubility of $CuS$ in solution.

This process of changing one of the salts ions to a complex ion will increase the solubility

The value of $K_{c}$ can be found my multiplying $K_{f}$ and $K_{sp}$ together.

$K_{c}=K_{f} x K_{sp}$

$K_{c}= (6.3x10^{-36})x(1.1x10^{13)} = 6.99x10^{-23}$

Comparing $K_{c}$ of the overall reaction with the $K_{sp}$ of the original saturated solution shows that the formation of a complex ion increases the solubility of $CuS$ by approximately $1 x 10^{13}$

So there you have it! The complex ion formation can substantially increase the solubility of sparingly soluble salts IF the complex ion formed has a large formation constant, $K_{f}$ .

Complex Ion Formation Effects on Solubility — Overview

(Video) Increasing Solubility via Complex Ion Formation

Keep going