chapter 18 solubility and complex ion equilibria
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Chapter 18: Solubility and Complex-Ion Equilibria

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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
Chapter 18: Solubility andComplex-Ion Equilibria

Chemistry 1062: Principles of Chemistry II

Andy Aspaas, Instructor

solubility product constant
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
solubility and k sp
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
solubility and k sp1
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
calculating solubility
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
solubility and the common ion effect
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
common ion effect calculation
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
predicting precipitation
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
precipitation prediction
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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