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Chapter 16
Lesson 2
Solubility and
Complex Ion
Equilibria
Chapter 16
Table of Contents
16.1
16.2
16.3
Solubility Equilibria and the Solubility Product
Precipitation and Qualitative Analysis
Equilibria Involving Complex Ions
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2
Precipitation
• Precipitation is merely another way of
looking at solubility equilibrium.
– Rather than considering how much of a
substance will dissolve, we ask: Will
precipitation occur for a given starting
ion concentration?
3
Criteria for Precipitation
• To determine whether an equilibrium
system will go in the forward or reverse
direction requires that we evaluate the
reaction quotient, Q (or IP).
– To predict the direction of reaction, you compare Q
with Ksp.
– The reaction quotient has the same form as the
Ksp expression, but the concentrations of products
are starting values not necessarily saturated conc.
4
Criteria for Precipitation
– Consider the following equilibrium.
H2O
PbCl 2 (s )
2

Pb (aq)  2Cl (aq)
2
 2
eq
K sp  [Pb ]eq [Cl ]
2
 2
i
Q  [Pb ]i [Cl ]
where initial concentration is
denoted by i.
5
Criteria for Precipitation
– If Q = Ksp, the solution is just saturated with ions and any
additional solid will not dissolve in solution but
instead will precipitate out.
solid
ions
Q = [ions]
– If Q < Ksp, the solution is unsaturated and more solid can be
dissolved in the solution; no precipitate forms. Shift
right increase Q
– If Q > Ksp, , the solution is supersaturated meaning the
solution contains a higher concentration of ions than
possible at equilibrium; a precipitate will form. Shift
to left decrease Q
6
• The concentration of calcium ion in blood plasma is 0.0025 M.
If the concentration of oxalate ion is 1.0 x 10-5 M, do you
expect calcium oxalate to precipitate? Ksp for calcium oxalate
is 2.3 x 10-9.
0.0025 M 1.0 x 10-5 M
2
2
Ca (aq)  C 2O 4 (aq)
CaC 2O 4 (s )
– The ion product quotient, Qc, is:
2
i
2
2
Q  [Ca ] [C O 4 ]i
Q  (0.0025)  (1.0  10 )
Q  2.5  10
-5
-8
– This value is larger than the Ksp, so you expect
precipitation to occur past saturation point.
Q  2.5  10  K sp  2 .3  10
-8
-9
7
• The concentration of lead ion is 0.25 M. If the concentration
of chloride ion is 0.0060 M, do you expect lead chloride to
precipitate? Ksp for lead chloride is 1.7 x 10-5.
PbCl2 (s)
Pb2+ (aq) + 2Cl- (aq)
0.25 M
0.0060 M
Note: did not double chloride concentration
Q < Ksp = 1.7 x 10-5, indicating that a no precipitate will form (below
the saturation point).
8
example: A student mixes 0.200 L of 0.0060 M Sr(NO3)2 solution with 0.100 L of
0.015 M K2CrO4 solution to give a final volume of 0.300L. Will a precipitate form
under these conditions? Ksp SrCrO4 = 3.6 x 10-5
Sr(NO3)2 (aq) + K2CrO4 (aq)  SrCrO4 (s) + 2 KNO3 (aq)
(aq)
SrCrO4 (s)
Sr2+ (aq) + CrO42- (aq)
0.0040 M
0.0050 M
we find that Q (2.0 x 10-5) < Ksp (3.6 x 10-5)
indicating that no precipitate will form (below
saturation point)
9
Selective Precipitation
• Selective precipitation is the
technique of separating two or more
ions from a solution by adding a
reactant that precipitates first one ion,
then another, and so forth.
– For example, when you slowly add potassium
chromate, K2CrO4, to a solution containing Ba2+ and
Sr2+, barium chromate precipitates first due to its
lower solubility than SrCrO4.
10
Selective Precipitation
– After most of the Ba2+ ion has precipitated, strontium
chromate begins to precipitate.
– It is therefore possible to separate Ba2+ from Sr2+ by
selective precipitation using K2CrO4.
– Take advantage in qual/quan type analysis.
11
Effect of pH on Solubility
• Sometimes it is necessary to account
for other reactions that aqueous ions
might undergo.
– For example, if the anion is the conjugate base of
a weak acid, it will react with H3O+.
– You should expect the solubility to be affected by
pH. By adding and complexing out ions you can
affect the pH of solution which could affect ppt
reactions.
12
Effect of pH on Solubility
– Consider the following equilibrium.
CaC 2O 4 (s )
H2O
2
2
Ca (aq)  C 2O 4 (aq)
– Because the oxalate ion is conjugate base, it will
react with H3O+ (added acid to lower pH).
2

C 2O 4 (aq )  H 3O (aq)
H2O
s
by
pH

HC 2O 4 (aq)  H 2O(l)
– According to Le Chatelier’s principle, as C2O42- ion is
removed by the reaction with H3O+, more calcium oxalate
dissolves (increase solubility).
– Therefore, you expect calcium oxalate to be more soluble in acidic
solution (lower pH) than in pure water. The acidity will react with
the oxalate and shift the equil toward the right and allow more
calcium oxalate to dissolve.
13
Separation of Metal Ions by Sulfide
Precipitation
• Many metal sulfides are insoluble in
water but dissolve in acidic solution.
– Qualitative analysis uses this change in solubility
of the metal sulfides with pH to separate a mixture
of metal ions.
– By adjusting the pH in an aqueous solution of H2S,
you adjust the sulfide concentration to precipitate the
least soluble metal sulfide first.
– Typically do some qual experiment similar in lab.
14
Qualitative Analysis
• Qualitative analysis involves the
determination of the identity of
substances present in a mixture.
– In the qualitative analysis scheme for metal
ions, a cation is usually detected by the
presence of a characteristic precipitate.
– Next slide shows a figure that summarizes
how metal ions in an aqueous solution are
separated into five analytical groups.
15
Ebbing, D. D.; Gammon, S. D. General
Chemistry, 8th ed., Houghton Mifflin, New
York, NY, 2005.
16
Section 16.2
Separating the Common Cations by Selective Precipitation
Atomic Masses
Precipitation
and Qualitative Analysis
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17
Section 16.2
Atomic Masses
Precipitation
and Qualitative Analysis
Separation of Cu2+ and Hg2+ from Ni2+ and Mn2+ using H2S
• At a low pH, [S2–] is relatively low and only the
very insoluble HgS and CuS precipitate.
• When OH– is added to lower [H+], the value of
[S2–] increases, and MnS and NiS precipitate.
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18
Section 16.2
Atomic Masses
Precipitation
and Qualitative Analysis
Separation of Cu2+ and Hg2+ from Ni2+ and Mn2+ using H2S
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19
Section 16.3
The Mole Involving Complex Ions
Equilibria
Complex Ion Equilibria
• Charged species consisting of a metal ion
surrounded by ligands.
 Ligand: Lewis base
• Formation (stability) constant.
 Equilibrium constant for each step of the
formation of a complex ion by the addition of
an individual ligand to a metal ion or complex
ion in aqueous solution.
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20
Section 16.3
The Mole Involving Complex Ions
Equilibria
Complex Ion Equilibria
Be2+(aq) + F–(aq)
BeF+(aq)
K1 = 7.9 x 104
BeF+(aq) + F–(aq)
BeF2(aq)
K2 = 5.8 x 103
BeF2(aq) + F–(aq)
BeF3– (aq)
K3 = 6.1 x 102
BeF3– (aq) + F–(aq)
BeF42– (aq)
K4 = 2.7 x 101
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21
Section 16.3
The Mole Involving Complex Ions
Equilibria
Complex Ions and Solubility
• Two strategies for dissolving a water–insoluble
ionic solid.
 If the anion of the solid is a good base, the
solubility is greatly increased by acidifying
the solution.
 In cases where the anion is not sufficiently
basic, the ionic solid often can be dissolved
in a solution containing a ligand that forms
stable complex ions with its cation.
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22
Section 16.3
The Mole Involving Complex Ions
Equilibria
How can one dissolve silver chloride and pull the rxn
to the right even though chloride is a weak base?
AgCl(s)
Ag+(aq) + Cl-(aq)
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23
Section 16.3
The Mole Involving Complex Ions
Equilibria
Concept Check
Calculate the solubility of silver chloride in 10.0
M ammonia given the following information:
Ksp (AgCl) = 1.6 x 10–10
Ag+ + NH3
AgNH3+
AgNH3+ + NH3
Ag(NH3)2+
0.48 M
K = 2.1 x 103
K = 8.2 x 103
Calculate the concentration of NH3 in the final
equilibrium mixture.
9.0 M
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24
Formation Constants
for Complex Ions
The solution of a slightly soluble salt
increase when one of its ions can be
changed to a soluble complex ion.
AgBr (s)
Add NH3
Ag+ + 2NH3
Ag+ + Br -
Ksp = 5.0 x 10-13
Ag(NH3)2+ kform = 1.6 x 107
Formation Constants
for Complex Ions
• The very soluble silver complex ion
removes Ag+ from the solution and shifts
the equilibrium to the right increasing the
solubility of AgCl.
AgBr + 2NH3
Ag(NH3)2+ + Br Kc = 8.0 x 10-6
= [Ag(NH3)2+][Br -]
[NH3]2
Kc = kform x ksp = (1.6 x 107)(5.0x10-13)
= 8.0 x 10-6
Example
• How many moles of AgBr can dissolve
in 1 L of 1.0 M NH3?
AgBr + 2NH3 Ag(NH3)2+ + Br –
1.0
0
0
-2x
+x
+x
1.0-2x
x
x
Kc =
x2
(1.0-2x)2 = 8.0x10-6
X = 2.8x10-3, 2.8 x 10-3 mol of AgBr dissolves
in 1L of NH3
Relative Solubility
Comparing solubilities: Which is most soluble in water?
CaCO3 (s)
Ca2+ (aq) + CO32CaCO3
Ksp = 3.8 x 10-9
Ksp = [Ca2+] [CO32-] =s2 = 3.8 x 10-9
s = 6.2 x 10-5 M
AgBr
Ksp = 5.0 x 10-13
AgBr (s)
Ksp = [Ag+] [Br-] = s2 = 5.0 x 10-13
s = 7.1 x 10-7 M
CaF2
Ksp = 3.4 x 10-11
CaF2 (s)
s
(aq)
s
Ag+ (aq) Br- (aq)
s
s
Ca2+ (aq) + 2 F- (aq)
s
2s
Ksp = [Ca2+] [F-] 2 =(s)(2s)2 = 4s3 = 3.4 x 10-11
s = 2.0 x 10-4 M
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