Equilibria of Other Reaction Classes

Systems of Multiple Equilibria

If a solution contains multiple solutes, the solutes can affect each other's equilibria, causing a substance that is insoluble on its own to dissolve or causing a substance that is soluble to precipitate out of solution at a concentration below its saturation point.

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(NH3)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 NH3 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.

Consider a sample of AgCl in an aqueous solution of 1.0 M NH3. In this system, there are two relevant equilibrium equations.
AgCl(s)Ag+(aq)+Cl(aq)Ksp=1.77×1010Ag+(aq)+2NH3(aq)[Ag(NH3)2]+(aq)Kf=1.6×107\begin{alignedat}{2}{\rm{AgCl}}(s) &\rightleftharpoons{\rm{Ag^+}}(aq)+{\rm{Cl^-}}(aq)\quad &&K_{\rm{sp}}=1.77\times10^{-10}\\{\rm{Ag^+}}(aq)+{2\rm{NH_3}}(aq) &\rightleftharpoons{\left[\rm{Ag(NH_3)_2}\right]^+}(aq) &&K_{\rm{f}}=1.6\times10^{7}\hspace{25pt}\end{alignedat}
AgCl is a white powder that is essentially insoluble in water (Ksp=1.77×1010K_{\rm {sp}}=1.77\times10^{-10}). When NH3 is introduced into the solution, the solid AgCl dissolves because NH3 forms a complex ion with Ag+.
NH3 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:
AgCl(s)+2NH3(aq)[Ag(NH3)2]+(aq)+Cl(aq),K=?{\rm{AgCl}}(s)+{2\rm{NH}_{3}}(aq)\rightleftharpoons \left[\rm{Ag}(\rm{NH}_{3})_{2}\right]^{+}\!(aq)+{\rm{Cl}^{-}}(aq),\hspace{15pt}K=?
The expression for K follows the usual rules of equilibrium constants. Writing out the expression for K, however, reveals something interesting:
K=[[Ag(NH3)2]+][Cl][NH3]2=[[Ag(NH3)2]+][Ag+][NH3]2×[Ag+][Cl]=Kf×KspK=\frac{\left[\left[\rm{Ag}(\rm{NH}_{3})_{2}\right]^{+}\right][\rm{Cl}^{-}]}{[\rm{NH}_{3}]^{2}}=\frac{\left[\left[\rm{Ag}(\rm{NH}_{3})_{2}\right]^{+}\right]}{[\rm{Ag}^{+}][\rm{NH}_{3}]^{2}}\times{[\rm{Ag}^{+}]}{[\rm{Cl}^{-}]}=K_{\rm{f}}\times K_{\rm{sp}}
K is simply equal to the product of Kf and Ksp. Recall that Kf is the formation constant and Ksp is the solubility constant. Both are calculated in a similar way, but Kf describes the formation of a complex ion and Ksp describes the formation of a saturated solution. Use the known values of Kf and Ksp to calculate the value of K. Note that extra digits are included since this is an intermediate calculation.
K=Kf×Ksp=(1.6×107)(1.77×1010)=2.832×103\begin{aligned}K&={K_{\rm{f}}} \times {K_{{\rm{sp}}}}\\& = \left( {1.6 \times {{10}^7}} \right)\left( {1.77 \times {{10}^{-10}}} \right)\\&= 2.832 \times {10^{-3}}\end{aligned}
Now let 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(NH3)2]+ and Cl are also s mol/L. Then calculate the final concentrations of all species.
Let s represent the number of moles of AgCl dissolved. The balanced equation then indicates that 2s moles of NH3 react, producing s moles of [Ag(NH3)2]+ and s moles of Cl-. The final concentrations can be determined by subtracting or adding.
Next, use the final concentrations and the value of the equilibrium constant, K, to solve for s:
K=[[Ag(NH3)2]+][Cl][NH3]22.832×103=s2(1.02s)22.832×103=s1.02ss=(1.02s)2.832×103s=4.8×102\begin{aligned}K &= \frac{\left[\left[\rm{Ag(NH}_3\rm{)}_2\right]^+\right]\![\rm{Cl}^-]}{\left[\rm{NH}_3\right]^2}\\2.832\times10^{-3}&= \frac{{{s^2}}}{{{{(1.0-2s)}^2}}}\\\sqrt {2.832 \times {{10}^{-3}}} &= \frac{s}{{1.0-2s}}\\s &= \left( {1.0-2s} \right)\sqrt {2.832 \times {{10}^{-3}}} \\s&= 4.8 \times {10^{-2}}\end{aligned}
Remember that the variable s is defined to be the molar solubility of AgCl, so the molar solubility of solid AgCl in 0.1 M aqueous NH3 is 4.8×102M4.8\times10^{-2}\,\rm{M}. Compare that to the molar solubility of AgCl without NH3, which is a simpler calculation that requires only Ksp. Once again let s represent the molar solubility so that s=[Ag+]=[Cl]s=\rm{[Ag}^+\rm{]}=\rm{[Cl}^-\rm{]}, and use the Ksp expression to solve for s:
Ksp=[Ag+][Cl]=ss=1.77×1010s=1.77×1010=1.33×105\begin{aligned}{K_{{\rm{sp}}}} &= [ {{\rm{Ag}}^ +}][ {{\rm{Cl}}^-} ]\\&= s \cdot s\\&= 1.77 \times {10^{-10}}\\s &= \sqrt {1.77 \times {{10}^{-10}}} \\&= 1.33 \times {10^{-5}}\end{aligned}
Without NH3, the molar solubility of AgCl drops from 4.8×102M4.8 \times {10^{-2}}{\,\rm{M}} to 1.33×105M1.33 \times {10^{-5}}{\,\rm{M}}. So a 1.0 M NH3 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(NH3)2]+ reduces the concentration of [Ag+] and shifts the equilibrium of the reaction to the right.
AgCl(s)Ag+(aq)+Cl(aq){\rm{AgCl}}(s)\rightleftharpoons {\rm{Ag}^{+}}(aq)+{\rm{Cl}^{-}}(aq)