1 / 9

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

livana
Download Presentation

Chapter 18: Solubility and Complex-Ion Equilibria

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related