Effect of Complexation on Solubility
The solubility of a precipitate can be improved by adding a ligand capable of forming a soluble complex with one of the precipitate’s ions. For example, the solubility of AgI increases in the presence of NH3 due to the formation of the soluble Ag(NH3)2+ complex. As a final illustration of the systematic approach to solving equilibrium problems, let us find the solubility of AgI in 0.10 M NH3.
We begin by writing the equilibria that we need to consider
Counting unknowns, we find that there are seven—[Ag+], [I–], [Ag(NH3)2+], [NH3], [NH4+], [OH–], and [H3O+]. Four of the equations needed to solve this problem are given by the equilibrium constant expressions
Three additional equations are needed. The first of these equations is a mass balance for NH3.
CNH3 = [NH3] + [NH4+]+2 x [Ag(NH3)2+]
Note that in writing this mass balance equation, the concentration of Ag(NH3)2+ must be multiplied by 2 since two moles of NH3 occurs per mole of Ag(NH3) +. The second additional equation is a mass balance on iodide and silver. Since AgI is the only source of I– and Ag+, every iodide in solution must have an associated silver ion; thus
[I–] = [Ag+] + [Ag(NH3) +]
Finally, the last equation is a charge balance equation
[Ag+] + [Ag(NH3)3+]2 + [NH4+]+ [H3O+]= [I–] + [OH–]
Our problem looks challenging, but several assumptions greatly simplify the al- gebra. First, since the formation of the Ag(NH3) + complex is favorable, we will as- sume that
[Ag+] << [Ag(NH3) +]
Second, since NH3 is a base, we will assume that
[H3O+] << [OH–]
[NH4+] << [NH3] + [Ag(NH3) +]
Finally, since Ksp is significantly smaller than β2, it seems likely that the solubility of AgI is small and
[Ag(NH3) +] << [NH3]
Using these assumptions allows us to simplify several equations. The mass bal- ance for NH3 is now
We continue by multiplying together the equations for Ksp and β2, giving
Substituting in the new mass balance equations for NH3 and I–
Before accepting this answer, we first check our assumptions. Using the Ksp equation we calculate the [Ag+] to be
Our first assumption that the [Ag+] is significantly smaller than the [Ag(NH3) +], therefore, is reasonable. Furthermore, our third assumption that the [Ag(NH3)2+] is significantly less than the [NH3] also is reasonable. Our second assumption was
Although the [NH4+] is not significantly smaller than the combined concentrations of NH3 and Ag(NH3)2+, the error is only about 1%. Since this is not an excessively large error, we will accept this approximation as reasonable.
Since one mole of AgI produces one mole of I–, the solubility of AgI is the same as the concentration of iodide, or 3.7 x 10–6 mol/L.
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