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Chapter 19: Solubility and Simultaneous Equilibria

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Title: Chapter 19: Solubility and Simultaneous Equilibria


1
Chapter 19 Solubility and Simultaneous
Equilibria
  • None of the salts describe as insoluble in
    Chapter 5 are totally insoluble
  • For example, if solid AgCl is added to water a
    small amount dissolves
  • The ion product is the product of the molar
    concentrations of the dissolved ions of the solute

2
  • The solubility product constants for a number of
    ionic compounds are given in Table 19.1 (and
    Appendix C)
  • Ksp can be calculated by determining the molar
    solubility or the number of moles of salt
    dissolved in one liter of the saturated solution
  • We can assume that the salt that dissolves is
    100 dissociated (see Facets of Chemistry 19.1
    for a discussion of the validity of this
    assumption)

3
  • Example The molar solubility of PbF2 in a 0.10 M
    Pb(NO3)2 solution at 25oC is 3.1x10-4 mol L-1.
    What is Ksp for PbF2?
  • ANALYSIS Write the solubility equation, setup
    and evaluate the ICE table, and calculated the
    requested quantity.
  • SOLUTION

4
  • Often, a value for Ksp is available and the molar
    solubility needs to be calculated
  • This can be done with or without the presence of
    a common ion
  • An ion in solution that has been supplied from
    more than one solute is called a common ion
  • When a common ion is present, the molar
    solubility of an ionic compound decreases
  • This is called the common ion effect
  • Care must be taken to use all the information
    correctly

5
  • Example What is the molar solubility of PbI2 in
    0.10 M NaI?
  • ANALYSIS This is a common ion problem. NaI is
    soluble in water and provides the common ion
    (I-).
  • SOLUTION

6
  • This is a common ion problem, and Ksp is small
  • Ksp can be used to determine if a precipitation
    will occur
  • If the ion product gt Ksp, the solution is
    supersaturated and a precipitate will form
  • If the ion product Ksp, the solution is
    saturated and no precipitate will form
  • If the ion product lt Ksp, the solution is
    unsaturated and no precipitate will form

7
  • The pH dependence of metal carbonate solubility
    is similar to the pH dependence of metal sulfide
    solubility (see Example 19.10) and can provide a
    basis for selective precipitation
  • Metal ions are Lewis acids and so can become
    covalently bonded with ions or neutral molecules
    in solution
  • The resulting species is called a complex ion

8
  • Copper(II) ion is a typical example
  • The Lewis base that attaches itself to the metal
    ion is called a ligand (H2O is the ligand)
  • The atom in the ligand that actually provides the
    electron pair is called the donor atom (O is the
    donor atom)
  • The metal ion is called the acceptor (Cu2 is the
    acceptor)
  • Compounds that contain complex ions are called
    coordination compounds and the complex itself is
    sometimes called a coordination complex

9
  • Formation or stability constants, Kform, can be
    used to describe complex ion formation
  • For example, for copper(II) ions in solutions of
    ammonia
  • The inverse of the formation constant is called
    the instability constant, Kinst, which are
    preferred by some

10
  • Table 19.3 lists both formation and instability
    constants for some ions
  • Appendix C lists only formation constants
  • Complex ion formation can affect the solubility
    of salts

11
  • The solubility of a slightly soluble salt
    increases when one of its ions can be changed
    into soluble complex ion
  • Example Calculate the solubility of silver
    chloride in 0.10 M NH3.
  • ANALYSIS The equations describing the solubility
    and complex ion formation must be combined into a
    single equation and solved.

12
  • SOLUTION

13
  • Setup the ICE table and solve for the molar
    solubility x

14
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