The solubility of a precipitate can be improved by adding a ligand capable of forming a soluble complex with one of the precipitateâ€™s ions.

**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 precipitateâ€™s ions. For example, the solubility
of AgI increases in the presence of NH_{3} due to the formation of the soluble
Ag(NH_{3})_{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 NH_{3}.

We begin by writing the equilibria that we need to consider

Counting
unknowns, we find that there are sevenâ€”[Ag+], [Iâ€“], [Ag(NH_{3})_{2}+], [NH_{3}], [NH_{4}+], [OHâ€“],
and [H_{3}O+]. 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 NH_{3}.

*C*_{NH}3 = [NH_{3}] + [NH_{4}+]+2 x [Ag(NH_{3})_{2}+]

Note that in writing this
mass balance equation, the concentration of Ag(NH_{3})_{2}+ must be multiplied by 2 since
two moles of NH_{3} occurs
per mole of Ag(NH_{3}) +.
The second additional equation
is a mass balance on iodide and silver. Since AgI is the
only source of Iâ€“ and Ag+, every iodide
in solution must
have an associated silver ion; thus

[Iâ€“]
= [Ag+]
+ [Ag(NH_{3}) +]

Finally, the last equation is a charge balance equation

[Ag+] + [Ag(NH_{3})_{3}+]_{2}
+ [NH_{4}+]+
[H_{3}O+]=
[Iâ€“] +
[OHâ€“]

Our problem looks
challenging, but several
assumptions greatly simplify
the al- gebra. First,
since the formation of the Ag(NH_{3}) + complex is favorable, we will as- sume
that

[Ag+] << [Ag(NH_{3}) +]

Second, since NH_{3} is a base, we will assume that

[H_{3}O+] << [OHâ€“]

[NH_{4}+] << [NH_{3}] + [Ag(NH_{3})
+]

Finally, since *K*_{sp} is significantly smaller
than Î²_{2}, it seems likely
that the solubility of AgI is small and

[Ag(NH_{3}) +] << [NH_{3}]

Using these assumptions allows us to simplify several
equations. The mass
bal- ance for NH_{3} is now

We continue by multiplying together the equations for *K*_{sp} and Î²_{2}, giving

Substituting in the new mass balance equations for NH_{3}
and Iâ€“

Before accepting this answer, we first check our assumptions.
Using the *K*_{sp} equation we calculate the [Ag+] to be

Our first assumption that the [Ag+] is significantly smaller than the [Ag(NH_{3}) +], therefore, is reasonable. Furthermore, our third assumption that the [Ag(NH_{3})_{2}+] is
significantly less than
the [NH_{3}] also
is reasonable. Our
second assumption was

Although the [NH_{4}+] is not significantly smaller than the combined concentrations of NH_{3} and Ag(NH_{3})_{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 Iâ€“, the solubility of AgI is the same as
the concentration of iodide, or 3.7 x 10â€“6 mol/L.

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Modern Analytical Chemistry: Equilibrium Chemistry : Effect of Complexation on Solubility |

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