If a solution contains many different solutes, the solution must also contain many different equilibria. The different solutes can affect each other's equilibria by changing their solubilities and causing something that is insoluble on its own to dissolve or by causing something that is soluble to precipitate out of solution at a concentration below its saturation point.

Both the common ion effect (observed when KI is added to a solution of AgI) and the example of the effect of the complex ion [Ag(NH_{3})_{2}]^{+} can be considered systems of multiple equilibria. In each case, the ions formed by one equilibrium reaction caused the equilibrium of a second reaction to shift one way or another. In the first case, the addition of KI to a solution of AgI causes a precipitate to form, and in the second case, the introduction of NH_{3} causes the previously insoluble AgCl to dissolve into the solution. So the equilibrium of one reaction can affect the equilibrium of another. To put it another way, introducing additional equilibria to a system can change the concentrations of species involved.

_{3}. In this system, there are two relevant equilibrium equations.

_{3}increases the molar solubility of AgCl, and the equilibrium constants can be used to calculate the new solubility. The first step is to calculate the overall equilibrium constant for the entire system by combining the equilibrium equations for AgCl into one overall equation:

*K*follows the usual rules of equilibrium constants. Writing out the expression for

*K*, however, reveals something interesting:

*K*is simply equal to the product of

*K*

_{f}and

*K*

_{sp}. Recall that

*K*

_{f}is the formation constant and

*K*

_{sp}is the solubility constant. Both are calculated in a similar way, but

*K*

_{f}describes the formation of a complex ion and

*K*

_{sp}describes the formation of a saturated solution. Use the known values of

*K*

_{f}and

*K*

_{sp}to calculate the value of

*K*. Note that extra digits are included since this is an intermediate calculation.

*s*equal the molar solubility of AgCl. Remember that the molar solubility is equal to the concentration of a solute at equilibrium. If

*s*mol of AgCl/L dissolves, the equilibrium equation requires that the concentrations of [Ag(NH

_{3})

_{2}]

^{+}and Cl

^{–}are also

*s*mol/L. Then calculate the final concentrations of all species. Next, use the final concentrations and the value of the equilibrium constant,

*K,*to solve for

*s*:

*s*is defined to be the molar solubility of AgCl, so the molar solubility of solid AgCl in 0.1 M aqueous NH

_{3}is $4.8\times10^{-2}\,\rm{M}$. Compare that to the molar solubility of AgCl without NH

_{3}, which is a simpler calculation that requires only

*K*

_{sp}. Once again let

*s*represent the molar solubility so that $s=\rm{[Ag}^+\rm{]}=\rm{[Cl}^-\rm{]}$, and use the

*K*

_{sp}expression to solve for

*s:*

_{3}, the molar solubility of AgCl drops from $4.8 \times {10^{-2}}{\,\rm{M}}$ to $1.33 \times {10^{-5}}{\,\rm{M}}$. So a 1.0 M NH

_{3}solution increases the molar solubility of AgCl by three orders of magnitude. This increase in solubility occurs because the formation of the complex ion [Ag(NH

_{3})

_{2}]

^{+}reduces the concentration of [Ag

^{+}] and shifts the equilibrium of the reaction to the right.