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Chapter 18: Solubility and Complex-Ion Equilibria

Chapter 18: Solubility and Complex-Ion Equilibria. Chemistry 1062: Principles of Chemistry II Andy Aspaas, Instructor. Solubility product constant. Excess of a slightly-soluble ionic compound is mixed with water An equilibrium occurs between the solid ionic compound and the dissociated ions

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Chapter 18: Solubility and Complex-Ion Equilibria

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  1. Chapter 18: Solubility andComplex-Ion Equilibria Chemistry 1062: Principles of Chemistry II Andy Aspaas, Instructor

  2. Solubility product constant • Excess of a slightly-soluble ionic compound is mixed with water • An equilibrium occurs between the solid ionic compound and the dissociated ions CaC2O4(s) = Ca2+(aq) + C2O42-(aq) • Equilibrium constant for this process is called solubility product constant, Ksp Ksp = [Ca2+][C2O42-] (since only aqueous components are included in an equilibrium expression) • In the Kspequation, raise an ion’s concentration to the power of its stoichiometric number in the chemical equation

  3. Solubility and Ksp • The solubility of silver chloride is 1.9 x 10-3 g/L. What is Ksp? • First convert the solubility to molar solubility (mol/L) • Use an ICE table to find equilibrium molar concentrations of the ions (using 0 as initial) – ignore the solid • Substitute concentrations into a Ksp expression

  4. Solubility and Ksp • Remember to account for the correct number of ions forming in their molar concentrations • The solubility of Pb3(AsO4)2 is 3.0 x 10-5 g/L. What is Ksp? • Since a single formula unit of lead arsenate forms 3 Pb2+ ions when dissolved, be sure to multiply 3 by the molar concentration in the C row of the ICE table

  5. Calculating solubility • What is the solubility of calcium phosphate, in g/L? Ksp = 1 x 10-26 • Create an equilibrium expression with correct stoichiometry • In an ICE table, use x as the unknown change of molar concentrations, multiplying stoichiometric numbers by x • Solve for x in Ksp expression • Convert mol/L to g/L

  6. Solubility and the common-ion effect • Addition of extra ion to a solubility equilibrium solution will shift the equilibrium according to Le Chatelier’s principle • Ex. Adding additional Ca2+ to this equilibrium: CaC2O4(s) = Ca2+(aq) + C2O42+(aq) • This will cause the equilibrium to shift to the left, and the solid will become less soluble

  7. Common-ion effect calculation • Compare the molar solubilities of BaF2 in pure water and in 0.15 M NaF. Ksp = 1.0 k 10-6. • Set up the water solution as before and solve for molar solubility (x) • In the NaF solution, you have an initial concentration of 0.15 M of F- • You can most likely assume that x << 0.15 + x

  8. Predicting precipitation • Recall reaction quotient, Qc • Same calc as equilibrium constant, but system is not necessarily at equilibrium • If Qc < Kc, reaction goes forward • If Qc = Kc, reaction is at equilibrium • If Qc > Kc, reaction goes reverse • Ion product is Qcfor a solubility reaction • Reverse means the mixture will precipitate, since the solid dissociates in the forward direction

  9. Precipitation prediction • [Ca2+] = 0.0025 M • [C2O42-] = 1.0 x 10-7M • Will calcium oxalate precipitate? Ksp = 2.3 x 10-9 • Calculate ion product • Compare to solubility product constant

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