Chapter 18: Solubility and Complex-Ion Equilibria

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Chapter 18 Solubility andComplex-Ion Equilibria
Chemistry 1062 Principles of Chemistry II Andy
Aspaas, Instructor
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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 Ca2C2O42-
• (since only aqueous components are included in
  an equilibrium expression)
• In the Ksp equation, raise an ions concentration
  to the power of its stoichiometric number in the
  chemical equation

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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

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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

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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

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Solubility and the common-ion effect

• Addition of extra ion to a solubility equilibrium
  solution will shift the equilibrium according to
  Le Chateliers 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

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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 ltlt 0.15 x

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Predicting precipitation

• Recall reaction quotient, Qc
• Same calc as equilibrium constant, but system is
  not necessarily at equilibrium
• If Qc lt Kc, reaction goes forward
• If Qc Kc, reaction is at equilibrium
• If Qc gt Kc, reaction goes reverse
• Ion product is Qc for a solubility reaction
• Reverse means the mixture will precipitate, since
  the solid dissociates in the forward direction

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Precipitation prediction

• Ca2 0.0025 M
• C2O42- 1.0 x 10-7 M
• Will calcium oxalate precipitate? Ksp 2.3 x
  10-9
• Calculate ion product
• Compare to solubility product constant