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AT Chemistry
2012
Solubilty Product, Complex Ion Equilibria
And Qualitative Analysis
(Chapter 16 Notes)
Solubility Equilibria and the Solubility Product
Solubility refers to the amount of a salt that can be dissolved in water. Your text
points out that the solubility of a salt is variable but the solubility product ("Ksp") is
constant (at a given temperature).
The solubility equilibrium expression is set up as any other. For example, the
equilibrium expression for the dissolution of Ag2S in water is
Ag2S(s) = 2Ag+(aq) + S2-(aq)
Ksp = 1.6 X 10-49
Ksp = [Ag+]2[S2-]
Remember the pure solid, Ag2S, is not included in the equilibrium expression.
One type of problem you will need to solve is to interconvert between solubility and Ksp.
The keys to solving solubility problems are to


properly define solubility, and
properly use the reaction stoichiometry
Problem: Barium sulfate has a solubility of 3.9 X 10-5 M at 25oC. Calculate Ksp for
BaSO4.
Problem: Calculate the solubility of NiCO3 (Ksp = 1.4 X 10-7) in moles per liter and grams
per liter.
Note that relative solubilities among different compounds cannot be measured simply by
comparing Ksp values. You must take the composition of the salt into account.
Problem: Which of the following compounds is the most soluble?
AgCl;
Ag2CrO4
Ag3PO4
Ksp = 1.5 X 10-10
Ksp = 9.0 X 10-12
Ksp = 1.8 X 10-18
Another type of problem you will encounter involves the common ion effect. Recall that
LeChatelier's Principle predicts that adding a common ion to the solution shifts the
equilibrium to the left in solubility equations. In other words adding a common ion (that
does not react with other species in the solution) reduces the solubility.
Problem: Calculate the solubility of SrF2 (Ksp = 7.9 X 10-10) in
(a) pure water
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(b) 0.100 M Sr(NO3)2
(c) 0.200 M NaF
Keep in mind that substances (such as acids or bases) that can combine with one of
the ions of the salt will increase the solubility of the salt by removing product, thus forcing
the reaction to the right.
For example,
Mg(OH)2(s) = Mg2+(aq) + 2OH-(aq)
If we add acid (H3O+) to the above saturated solution, the equilibrium will shift
right as [OH-] decreases because it reacts with H3O+ to form water.
Precipitation
In precipitation reactions we answer two questions:
1. If two solutions are mixed, will a precipitate form?
2. If it forms, what will be the concentrations of each ion at equilibrium?
Recall the idea of "Q" the calculated equilibrium constant for initial conditions. In this
case if Q > Ksp then precipitation will occur.
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Problem: A 200.0 mL solution of 1.3 X 10-3 M AgNO3 is mixed with 100.0 mL of a 4.5 X
10-5 M Na2S. Will precipitation occur? (The Ksp for silver sulfide is 1.6 X 10-49).
Once we determine that precipitation will occur, we are faced with the problem of
determining the equilibrium concentrations of each of our ions of interest. The general
strategy involves assuming that because Ksp is so low, we can assume that if a precipitate
forms, it will do so quantitatively. We can then use the equilibrium expression involving
the solubility of the salt (with a common ion) to solve for the equilibrium concentration of
each ion.
Problem: Calculate the equilibrium concentration of each ion in a solution obtained by
mixing 50.0 mL of 6.0 X 10-3 M CaCl2 with 30.0 mL of 0.040 M NaF. (Ksp for CaF2 = 4.0
X 10-11).
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Complex Ion Equilibria
The main idea here is that introducing a Lewis base into a solution enhances the
solubility of an otherwise insoluble salt. You should know the following terms:
complex ion: a charged species consisting of a metal ion surrounded by ligands.
ligand: typically an anion or neutral molecule that has an unshared electron pair (lone
pair) that can be shared with an empt metal ion orbital to form a metal-ligand bond.
Ligands are Lewis bases and some common examples are H2O, NH3, l- and CN- .
coordination number: the number of ligands attached to a metal ion. The most common
coordination numbers are six (as in Co(OH2)62+ and Ni(NH3)62+), four (as in CoCl42- and
Cu(NH3)42+), and two (as in Ag(NH3)2+, although others are known.
formation constant: metal ions add ligands one at a time in steps characterized by
equilibrium constants called formation constants
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Note: formation of a complex ion increases the solubility of an "insoluble" salt. The
addition of a Lewis base will cause it to form a comlex ion with the cation (metal ion) in
equilibrium with the salt. This will decrease the concentration of the cation, and via Le
Chatelier’s principle, the equilibrium will shift right to replace the cation. As a result, the
solid will have to dissolve.
Complex Ion Formation
Ag+ + 2NH3 = Ag(NH3)2+
Ag+ + 2CN- = Ag(CN)2Cu2+ + 4NH3 = Cu(NH3)42+
Zn2+ + 4NH3 = Zn(NH3)42+
Zn2+ + 4OH- = Zn(OH)42Fe3+ + 6CN- = Fe(CN)63-
Amphoterism
Many metal hydroxides and oxides that are relatively insoluble in neutral water dissolve in
strongly acidic and strongly basic media. These subbstances because are soluble in strong
acids and bases because they themselves are capable of beahving as either an acid or a
base; they are amphoteric. Examples of amphoteric substances include the hydroxides
and oxides of Al3+, Cr3+, Zn2+ and Sn2+.
This behavior results from the formation of complex anions containing several (typically
four) hydroxides bound to the metal ion:
Al(OH)3(s) + OH-(aq) = Al(OH)4-(aq)
Amphoterism is often interpreted in terms of the behaqvior of the water molecules that
surround the metal ion and that are bonded to it by Lewis acid-base interactions. For
example, Al3+(aq) is more accurately represented as Al(H2O)63+(aq); six water molecules
are bonded to the Al3+ in aqueous solution. This species acts as a weak acid. As a strong
base is added, Al(H2O)63+ loses protons in a tepwise fashion, eventually forming neutral
and water-insoluble Al(H2O)3(OH)3. This substance then dissolves upon removal of an
additional proton to form the anion Al(H2O)2(OH)4-. The reactions that occur are as
follows:
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Al(H2O)63+(aq) + OH-(aq) = Al(H2O)5(OH)2+(aq) + H2O(l)
Al(H2O)5(OH)2+(aq) + OH-(aq) = Al(H2O)4(OH)2+(aq) + H2O(l)
Al(H2O)4(OH)2+(aq) + OH-(aq) = Al(H2O)3(OH)3(s) + H2O(l)
Al(H2O)3(OH)3(s) + OH-(aq) = Al(H2O)2(OH)4-(aq) + H2O(l)
Further proton removals are possible, but each successive reaction occurs less readily than
the one before (as the charge on the ion becomes more negative it is increasingly more
difficult to remove a positively charged proton.
Addition of acid reverses the reaction. The proton adds in a stepwise fashion to convert the
OH- groups to H2O, eventually reforming Al(H2O)63+(aq).
The common practice is to simplify the equations for these reactions by excluding the
bound H2O molecules. Thus, we usually write Al3+ instead of Al(H2O)63+, Al(OH)3
instead of Al(H2O)3(OH)3, Al(OH)4- instead of Al(H2O)2(OH)4-, and so forth.
The extent to which an insoluble metal hydroxide reacts with either acid or base varies
with the particular metal ion involved. Many metal hydroxides – for example, Ca(OH)2,
Fe(OH)2, and Fe(OH)3 – are capable of dissolving in acidic solution but do not react with
excess base. These hydroxides are not amphoteric.
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Qualitative Analysis Flowchart
Group I cations = insoluble chlorides
Ag+(aq)
Hg22+(aq)
Pb2+(aq)
add HCl(aq)
AgCl(s)
Hg2Cl2(s)
PbCl2(s)
heat (hot water)
Pb2+(aq)
AgCl(s)
Hg2Cl2(s)
Add CrO42-(aq)
PbCrO4(s)
(yellow ppt)
add NH3(aq)
Ag(NH3)2+(aq)
and Cl-(aq)
Hg(l) and
HgNH2Cl2(s)
(white ppt)
HNO3(aq)
AgCl(s)
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Group II and Group III Separation
Hg2+, Cd2+, Bi3+, Cu2+, Sn4+, Co2+, Zn2+, Mn2+, Ni2+, Fe2+, Cr3+, Al3+
(in acid)
H2S
HgS, CdS, Bi2S3, CuS, SnS2
Co2+, Zn2+, Mn2+, Ni2+, Fe2+, Cr3+
(Group II insoluble sulfides
in acidic solution)
NaOH
CoS, ZnS, MnS, NiS, FeS, Cr(OH)3, Al(OH)3
(Group III insoluble sulfides in basic solution)
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