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 CaC 2 O 4 (s) = Ca 2+ (aq) + C 2 O 4 2- (aq) Equilibrium constant for this process is called solubility product constant, K sp K sp = [Ca 2+ ][C 2 O 4 2- ] (since only aqueous components are included in an equilibrium expression) In the K sp equation, raise an ion’s concentration to the power of its stoichiometric number in the chemical equation

Solubility and K sp The solubility of silver chloride is 1.9 x g/L. What is K sp ? –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 K sp expression

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

Calculating solubility What is the solubility of calcium phosphate, in g/L? K sp = 1 x 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 K sp expression Convert mol/L to g/L

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 Ca 2+ to this equilibrium: CaC 2 O 4 (s) = Ca 2+ (aq) + C 2 O 4 2+ (aq) This will cause the equilibrium to shift to the left, and the solid will become less soluble

Common-ion effect calculation Compare the molar solubilities of BaF 2 in pure water and in 0.15 M NaF. K sp = 1.0 k 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 << x

Predicting precipitation Recall reaction quotient, Q c –Same calc as equilibrium constant, but system is not necessarily at equilibrium If Q c < K c, reaction goes forward If Q c = K c, reaction is at equilibrium If Q c > K c, reaction goes reverse Ion product is Q c for a solubility reaction –Reverse means the mixture will precipitate, since the solid dissociates in the forward direction

Precipitation prediction [Ca 2+ ] = M [C 2 O 4 2- ] = 1.0 x M Will calcium oxalate precipitate? K sp = 2.3 x Calculate ion product Compare to solubility product constant