$$\newcommand{\xrightleftharpoons}[2]{\overset{#1}{\underset{#2}{\rightleftharpoons}}}$$ $$\newcommand{\conc}[1]{\left[\mathrm{#1}\right]}$$ $$\newcommand{\chem}[1]{\mathrm{#1}}$$

# Chapter 16 Solubility & Complex Ion Equilibria

Shaun Williams, PhD

### Solubility Equilibria

• Solubility product ($$K_{sp}$$) – equilibrium constant; has only one value for a given solid at a given temperature.
• Solubility – an equilibrium position.
• $$\chem{Bi_2S_3(s) \rightleftharpoons 2Bi^{3+}(aq) + 3S^{2-}(aq)}$$ $$K_{sp}=\conc{Bi^{3+}}^2\conc{S^{2-}}^3$$

### Exercise

Calculate the solubility of silver chloride in water. $$K_{sp} = 1.6 \times 10^{–10}$$

$$1.3 \times 10^{-5}\,\mathrm{M}$$

Calculate the solubility of silver phosphate in water. $$K_{sp} = 1.8 \times 10^{–18}$$

$$1.6 \times 10^{-5}\,\mathrm{M}$$

### Exercise

Calculate the solubility of $$\chem{AgCl}$$ in: $$K_{sp}=1.6 \times 10^{-10}$$

1. $$100.0\,\mathrm{mL}$$ of $$4.00 \times 10^{-3}\,\mathrm{M}$$ calcium chloride.
2. $$2.0 \times 10^{-8}\,\mathrm{M}$$

3. $$100.0\,\mathrm{mL}$$ of $$4.00 \times 10^{-3}\,\mathrm{M}$$ calcium nitrate.
4. $$1.3 \times 10^{-5}\,\mathrm{M}$$

### Precipitation (Mixing Two Solutions on Ions)

• $$Q > K_{sp}$$: precipitation occurs and will continue until the concentrations are reduced to the point that they satisfy $$K_{sp}$$.
• $$Q < K_{sp}$$: no precipitation occurs.
• Selective Precipitation (Mixtures of Metal Ions)
• Use a reagent whose anion forms a precipitate with only one or a few of the metal ions in the mixture.
• Example:
• Solution contains $$\chem{Ba^{2+}}$$ and $$\chem{Ag^+}$$ ions.
• Adding $$\chem{NaCl}$$ will form a precipitate with $$\chem{Ag^+}$$ ($$\chem{AgCl}$$), while still leaving $$\chem{Ba^{2+}}$$ in solution.

### Separation of $$\chem{Cu^{2+}}$$ and $$\chem{Hg^{2+}}$$ from $$\chem{Ni^{2+}}$$ and $$\chem{Mn^{2+}}$$ using $$\chem{H_2S}$$

• At a low $$\mathrm{pH}$$, $$\conc{S^{2-}}$$ is relatively low and only the very insoluble $$\chem{HgS}$$ and $$\chem{CuS}$$ precipitate.
• When $$\chem{OH^–}$$ is added to lower $$\conc{H^+}$$, the value of $$\conc{S^{2-}}$$ increases, and $$\chem{MnS}$$ and $$\chem{NiS}$$ precipitate.

### Complex Ion Equilibria

• Charged species consisting of a metal ion surrounded by ligands.
• Ligand: Lewis base
• Formation (stability) constant.
• Equilibrium constant for each step of the formation of a complex ion by the addition of an individual ligand to a metal ion or complex ion in aqueous solution.
\begin{align} &\chem{Be^{2+}(aq)+F^-(aq) \rightleftharpoons BeF^+(aq)} &~~~~& K_1 = 7.9 \times 10^4 \\ &\chem{BeF^{+}(aq)+F^-(aq) \rightleftharpoons BeF_2(aq)} &~~~~& K_2 = 5.8 \times 10^3 \\ &\chem{BeF_2(aq)+F^-(aq) \rightleftharpoons BeF_3^-(aq)} &~~~~& K_3 = 6.1 \times 10^2 \\ &\chem{BeF_3^-(aq)+F^-(aq) \rightleftharpoons BeF_4^{2-}(aq)} &~~~~& K_4 = 2.7 \times 10^1 \end{align}

### Complex Ions and Solubility

• Two strategies for dissolving a water–insoluble ionic solid.
• If the anion of the solid is a good base, the solubility is greatly increased by acidifying the solution.
• In cases where the anion is not sufficiently basic, the ionic solid often can be dissolved in a solution containing a ligand that forms stable complex ions with its cation.

### Concept Check

Calculate the solubility of silver chloride in 10.0 M ammonia given the following information: $$K_{sp}(\chem{AgCl}) = 1.6 \times 10^{-10}$$ \begin{align} &\chem{Ag^++NH_3 \rightleftharpoons AgNH_3^+} &~~~~& K = 2.1 \times 10^3 \\ &\chem{AgNH_3^++NH_3 \rightleftharpoons Ag\left( NH_3\right)_2^+} &~~~~& K = 8.2 \times 10^3 \end{align}

$$0.48\,\mathrm{M}$$

Calculate the concentration of $$\chem{NH_3}$$ in the final equilibrium mixture.

$$9.0\,\mathrm{M}$$

/