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

Chapter 16
Solubility & Complex Ion Equilibria

Shaun Williams, PhD

Solubility Equilibria

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)

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

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

Addition of hydrogen sulfide to a solution of these ions precipitates copper(II) sulfide and mercury(II) sulfide while leaving the rest in solution. The remaining ions can be precipitated by adding base to get the pH to 8.

Separating the Common Cations by Selective Precipitation

Addition of hydrochloric acid precipitates the group I ions. Addition of hydrogen sulfide to the resulting solution precipitates the group II ions. Adding sodium hydroxide to the resulting solution precipitates the group III ions. Adding sodium carbonate to the resulting solution precipitates the group IV ions. The group V ions and ammonium remain the resulting solution.

Complex Ion Equilibria

$$ \begin{align} &\chem{Be^{2+}(aq)+F^-(aq) \rightleftharpoons BeF^+(aq)} &~~~~& K_1 = 7.9 \times 10^4 \\ &\chem{BeF^{+}(aq)+F^-(aq) \rightleftharpoons BeF_2(aq)} &~~~~& K_2 = 5.8 \times 10^3 \\ &\chem{BeF_2(aq)+F^-(aq) \rightleftharpoons BeF_3^-(aq)} &~~~~& K_3 = 6.1 \times 10^2 \\ &\chem{BeF_3^-(aq)+F^-(aq) \rightleftharpoons BeF_4^{2-}(aq)} &~~~~& K_4 = 2.7 \times 10^1 \end{align} $$

Complex Ions and Solubility

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.

\(9.0\,\mathrm{M}\)

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