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Chapter 16
Solubility and
Complex Ion Equilibria
Solubility Equilibria
• Many natural processes depend on the
precipitation or dissolving of a slightly
soluble salt.
In the next section, we look at the
equilibria of slightly soluble, or nearly
insoluble, ionic compounds.
Their equilibrium constants can be used to
answer questions regarding solubility and
precipitation.
The Solubility Product
Constant
• When an excess of a slightly soluble ionic
compound is mixed with water, an
equilibrium is established between the solid
and the ions in the saturated solution.
For the salt calcium oxalate, CaC2O4, you
have the following equilibrium.
CaC2O 4 (s )
H2O
2
2
Ca (aq)  C 2O 4 (aq)
The Solubility Product
Constant
• When an excess of a slightly soluble ionic
compound is mixed with water, an
equilibrium is established between the solid
and the ions in the saturated solution.
The equilibrium constant for this process is
called the solubility product constant.
2
2
K sp  [Ca ][C 2O 4 ]
The Solubility Product
Constant
• In general, the solubility product constant is
the equilibrium constant for the solubility
equilibrium of a slightly soluble (or nearly
insoluble) ionic compound.
It equals the product of the equilibrium
concentrations of the ions in the
compound.
Each concentration is raised to a power
equal to the number of such ions in the
formula of the compound.
The Solubility Product
Constant
• In general, the solubility product constant is
the equilibrium constant for the solubility
equilibrium of a slightly soluble (or nearly
insoluble) ionic compound.
For example, lead iodide, PbI2, is another
slightly soluble salt. Its equilibrium is:
PbI 2 (s )
H2O
2

Pb (aq)  2I (aq)
The Solubility Product
Constant
• In general, the solubility product constant is
the equilibrium constant for the solubility
equilibrium of a slightly soluble (or nearly
insoluble) ionic compound.
The expression for the solubility product
constant is:
2
 2
K sp  [Pb ][I ]
Calculating Ksp from the
Solubility
A 1.0-L sample of a saturated calcium oxalate
solution, CaC2O4, contains 0.0061-g of the salt at
25 oC. Calculate the Ksp for this salt at 25 oC.
We must first convert the solubility of
calcium oxalate from 0.0061 g/liter to
moles per liter.
1 mol CaC2O 4
M CaC2O 4  ( 0.0061 g CaC2O 4 / L ) 
128g CaC2O 4
 4.8  10  5 m ol CaC2O 4 / L
Calculating Ksp from the
Solubility
A 1.0-L sample of a saturated calcium oxalate
solution, CaC2O4, contains 0.0061-g of the salt at
25 oC. Calculate the Ksp for this salt at 25 oC.
When 4.8 x 10-5 mol of solid dissolve it
forms 4.8 x 10-5 mol of each ion.
CaC2O 4 (s )
Starting
Change
Equilibrium
H2O
2
2
Ca (aq)  C 2O 4 (aq)
0
0
+4.8 x 10-5 +4.8 x 10-5
4.8 x 10-5 4.8 x 10-5
Calculating Ksp from the
Solubility
A 1.0-L sample of a saturated calcium oxalate
solution, CaC2O4, contains 0.0061-g of the salt at
25 oC. Calculate the Ksp for this salt at 25 oC.
You can now substitute into the
equilibrium-constant expression.
2
2
K sp  [Ca ][C 2O 4 ]
5
5
K sp  ( 4.8  10 )( 4.8  10 )
K sp  2.3  10
9
Calculating Ksp from the
Solubility
By experiment, it is found that 1.2 x 10-3 mol of
lead(II) iodide, PbI2, dissolves in 1.0 L of water at
25 oC. What is the Ksp at this temperature?
Note that in this example, you find that 1.2 x
10-3 mol of the solid dissolves to give 1.2 x 103 mol Pb2+ and 2 x (1.2 x 10-3) mol of I-.
Calculating Ksp from the
Solubility
By experiment, it is found that 1.2 x 10-3 mol of
lead(II) iodide, PbI2, dissolves in 1.0 L of water at
25 oC. What is the Ksp at this temperature?
The following table summarizes.
PbI 2 (s )
Starting
Change
Equilibrium
H2O
2

Pb (aq)  2I (aq)
0
0
+1.2 x 10-3 +2 x (1.2 x 10-3)
1.2 x 10-3 2 x (1.2 x 10-3)
Calculating Ksp from the
Solubility
By experiment, it is found that 1.2 x 10-3 mol of
lead(II) iodide, PbI2, dissolves in 1.0 L of water at
25 oC. What is the Ksp at this temperature?
Substituting into the equilibrium-constant
expression:
2
 2
K sp  [Pb ][I ]
3
3
K sp  (1.2  10 )( 2  (1.2  10 ))
K sp  6.9  10
9
2
Calculating Ksp from the
Solubility
By experiment, it is found that 1.2 x 10-3 mol of
lead(II) iodide, PbI2, dissolves in 1.0 L of water at
25 oC. What is the Ksp at this temperature?
If the solubility product constant is known, the
solubility of the compound can be calculated.
Calculating the Solubility from
Ksp
The mineral fluorite is calcium fluoride, CaF2.
Calculate the solubility (in grams per liter) of
calcium fluoride in water from the Ksp (3.4 x 10-11)
Let x be the molar solubility of CaF2.
CaF2 (s )
Starting
Change
Equilibrium
H2O
0
+x
x
2

Ca (aq)  2F (aq)
0
+2x
2x
Calculating the Solubility from
Ksp
The mineral fluorite is calcium fluoride, CaF2.
Calculate the solubility (in grams per liter) of
calcium fluoride in water from the Ksp (3.4 x 10-11)
You substitute into the equilibrium-constant
equation
2
 2
[Ca ][F ]  K sp
(x)(2x)  3.4  10
2
4x  3.4  10
3
11
11
Calculating the Solubility from
Ksp
The mineral fluorite is calcium fluoride, CaF2.
Calculate the solubility (in grams per liter) of
calcium fluoride in water from the Ksp (3.4 x 10-11)
You now solve for x.
3.4

10
x3
4
-11
 2.0  10
4
Calculating the Solubility from
Ksp
The mineral fluorite is calcium fluoride, CaF2.
Calculate the solubility (in grams per liter) of
calcium fluoride in water from the Ksp (3.4 x 10-11)
Convert to g/L (CaF2 78.1 g/mol).
78.1g CaF2
solubility  2.0  10 mol / L 
1 mol CaF2
4
2
 1.6  10 g CaF2 / L
Solubility Equilibria – Ion Product (reaction quotient)
Q is defined as the ion product (reaction quotient)
Q = same as Ksp only at initial concentration
I.e.
AgCl ↔ Ag+ + Cl-
Q = [Ag+ ]0[Cl- ]0
The subscript 0 reminds us that these are initial
concentrations
Criteria for Precipitation
• To determine whether an equilibrium
system will go in the forward or reverse
direction requires that we evaluate the
reaction quotient, Qc.
To predict the direction of reaction, you
compare Qc with Kc (Chapter 14).
The reaction quotient has the same form
as the Ksp expression, but the
concentrations of products are starting
values.
Criteria for Precipitation
• To determine whether an equilibrium
system will go in the forward or reverse
direction requires that we evaluate the
reaction quotient, Qc.
Consider the following equilibrium.
PbCl 2 (s )
H2O
2

Pb (aq)  2Cl (aq)
Criteria for Precipitation
• To determine whether an equilibrium
system will go in the forward or reverse
direction requires that we evaluate the
reaction quotient, Qc.
The Qc expression is
Q c  [Pb
2
 2
]i [Cl ]i
where initial concentration is
denoted by i.
Criteria for Precipitation
• To determine whether an equilibrium
system will go in the forward or reverse
direction requires that we evaluate the
reaction quotient, Qc.
If Qc exceeds the Ksp, precipitation occurs.
If Qc is less than Ksp, more solute can
dissolve.
If Qc equals the Ksp, the solution is
saturated.
Predicting Whether
Precipitation Will Occur
The concentration of calcium ion in blood plasma
is 0.0025 M. If the concentration of oxalate ion is
1.0 x 10-7 M, do you expect calcium oxalate to
precipitate? Ksp for calcium oxalate is 2.3 x 10-9.
The ion product quotient, Qc, is:
2
2
Q c  [Ca ]i [C 2O 4 ]i
-7
Q c  (0.0025)  (1.0  10 )
Q c  2.5  10
-10
Predicting Whether
Precipitation Will Occur
The concentration of calcium ion in blood plasma
is 0.0025 M. If the concentration of oxalate ion is
1.0 x 10-7 M, do you expect calcium oxalate to
precipitate? Ksp for calcium oxalate is 2.3 x 10-9.
This value is smaller than the Ksp, so you
do not expect precipitation to occur.
Q c  2.5  10
-10
 K sp
Solubility Equilibria
AgCl (s)
Ksp = [Ag+][Cl-]
MgF2 (s)
Ag2CO3 (s)
Ca3(PO4)2 (s)
Ag+ (aq) + Cl- (aq)
Ksp is the solubility product constant
Mg2+ (aq) + 2F- (aq)
Ksp = [Mg2+][F-]2
2Ag+ (aq) + CO32- (aq)
Ksp = [Ag+]2[CO32-]
3Ca2+ (aq) + 2PO43- (aq)
Ksp = [Ca2+]3[PO43-]2
Dissolution of an ionic solid in aqueous solution:
Q < Ksp
Unsaturated solution
Q = Ksp
Saturated solution
Q > Ksp
Supersaturated solution
No precipitate
Precipitate will form
Molar solubility (mol/L) is the number of moles of solute
dissolved in 1 L of a saturated solution.
Solubility (g/L) is the number of grams of solute dissolved in
1 L of a saturated solution.
What is the solubility of silver chloride in g/L ?
AgCl (s)
Initial (M)
Change (M)
Equilibrium (M)
[Ag+] = 1.3 x 10-5 M
Ag+ (aq) + Cl- (aq)
0.00
0.00
+s
+s
s
s
[Cl-] = 1.3 x 10-5 M
Ksp = 1.6 x 10-10
Ksp = [Ag+][Cl-]
Ksp = s2
s = Ksp
s = 1.3 x 10-5
1.3 x 10-5 mol AgCl 143.35 g AgCl
Solubility of AgCl =
x
= 1.9 x 10-3 g/L
1 L soln
1 mol AgCl
If 2.00 mL of 0.200 M NaOH are added to 1.00 L of
0.100 M CaCl2, will a precipitate form?
The ions present in solution are Na+, OH-, Ca2+, Cl-.
Only possible precipitate is Ca(OH)2 (solubility rules).
Is Q > Ksp for Ca(OH)2?
[Ca2+]0 = 0.100 M
[OH-]0 = 4.0 x 10-4 M
Q = [Ca2+]0[OH-]02 = 0.10 x (4.0 x 10-4)2 = 1.6 x 10-8
Ksp = [Ca2+][OH-]2 = 8.0 x 10-6
Q < Ksp
No precipitate will form
Fractional Precipitation
Selective Precipitation (Mixtures of Metal Ions)
• Fractional 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.
Fractional Precipitation
• Fractional 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.
After most of the Ba2+ ion has precipitated,
strontium chromate begins to precipitate.
It is therefore possible to separate Ba2+ from
Sr2+ by fractional precipitation using
K2CrO4.
What concentration of Ag is required to precipitate ONLY
AgBr in a solution that contains both Br- and Cl- at a
concentration of 0.02 M?
AgBr (s)
Ag+ (aq) + Br- (aq)
Ksp = 7.7 x 10-13
Ksp = [Ag+][Br-]
-13
K
7.7
x
10
sp
-11 M
=
=
3.9
x
10
[Ag+] =
0.020
[Br-]
AgCl (s)
[Ag+]
Ag+ (aq) + Cl- (aq)
Ksp = 1.6 x 10-10
Ksp = [Ag+][Cl-]
Ksp
1.6 x 10-10
-9 M
=
=
8.0
x
10
=
0.020
[Cl-]
3.9 x 10-11 M < [Ag+] < 8.0 x 10-9 M
Solubility and the Common-Ion
Effect
• In this section we will look at calculating
solubilities in the presence of other ions.
The importance of the Ksp becomes
apparent when you consider the solubility
of one salt in the solution of another
having the same cation.
Solubility and the Common-Ion
Effect
• In this section we will look at calculating
solubilities in the presence of other ions.
For example, suppose you wish to know
the solubility of calcium oxalate in a
solution of calcium chloride.
Each salt contributes the same cation
(Ca2+)
The effect is to make calcium oxalate less
soluble than it would be in pure water.
A Problem To Consider
What is the molar solubility of calcium oxalate in
0.15 M calcium chloride? The Ksp for calcium
oxalate is 2.3 x 10-9.
Note that before the calcium oxalate
dissolves, there is already 0.15 M Ca2+ in
the solution.
H2O
2
2
CaC2O 4 (s )
Ca (aq)  C 2O 4 (aq)
Starting
Change
Equilibrium
0.15
+x
0.15+x
0
+x
x
A Problem To Consider
What is the molar solubility of calcium oxalate in
0.15 M calcium chloride? The Ksp for calcium
oxalate is 2.3 x 10-9.
You substitute into the equilibrium-constant
equation
2
2
[Ca ][C 2O 4 ]  K sp
( 0.15  x )( x )  2.3  10
9
A Problem To Consider
What is the molar solubility of calcium oxalate in
0.15 M calcium chloride? The Ksp for calcium
oxalate is 2.3 x 10-9.
Now rearrange this equation to give
9
2.3  10
2.3  10
x

0.15  x
0.15
9
We expect x to be negligible compared to
0.15.
A Problem To Consider
What is the molar solubility of calcium oxalate in
0.15 M calcium chloride? The Ksp for calcium
oxalate is 2.3 x 10-9.
Now rearrange this equation to give
9
2.3  10
2.3  10
x

0.15  x
0.15
x  1.5  10
8
9
A Problem To Consider
What is the molar solubility of calcium oxalate in
0.15 M calcium chloride? The Ksp for calcium
oxalate is 2.3 x 10-9.
Therefore, the molar solubility of calcium
oxalate in 0.15 M CaCl2 is 1.5 x 10-8 M.
In pure water, the molarity was 4.8 x 10-5
M, which is over 3000 times greater.
The Common Ion Effect and Solubility
The presence of a common ion decreases
the solubility of the salt.
What is the molar solubility of AgBr in (a) pure water
and (b) 0.0010 M NaBr?
AgBr (s)
Ag+ (aq) + Br- (aq)
Ksp = 7.7 x 10-13
s2 = Ksp
s = 8.8 x 10-7
NaBr (s)
Na+ (aq) + Br- (aq)
[Br-] = 0.0010 M
AgBr (s)
Ag+ (aq) + Br- (aq)
[Ag+] = s
[Br-] = 0.0010 + s  0.0010
Ksp = 0.0010 x s
s = 7.7 x 10-10
Effect of pH on Solubility
• Sometimes it is necessary to account for other
reactions 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.
Effect of pH on Solubility
• Sometimes it is necessary to account for other
reactions aqueous ions might undergo.
Consider the following equilibrium.
CaC2O 4 (s )
H2O
2
2
Ca (aq)  C 2O 4 (aq)
Because the oxalate ion is conjugate to a
weak acid (HC2O4-), it will react with H3O+.
2

C 2O 4 (aq )  H 3O (aq)
H2O

HC 2O 4 (aq)  H 2O(l)
Effect of pH on Solubility
• Sometimes it is necessary to account for other
reactions aqueous ions might undergo.
According to Le Chatelier’s principle, as
C2O42- ion is removed by the reaction with
H3O+, more calcium oxalate dissolves.
Therefore, you expect calcium oxalate to
be more soluble in acidic solution (low
pH) than in pure water.
pH and Solubility
•
•
•
The presence of a common ion decreases the solubility.
Insoluble bases dissolve in acidic solutions
Insoluble acids dissolve in basic solutions
add
Mg(OH)2 (s)
remove
Mg2+ (aq) + 2OH- (aq)
Ksp = [Mg2+][OH-]2 = 1.2 x 10-11
Ksp = (s)(2s)2 = 4s3
4s3 = 1.2 x 10-11
s = 1.4 x 10-4 M
[OH-] = 2s = 2.8 x 10-4 M
pOH = 3.55 pH = 10.45
At pH less than 10.45
Lower [OH-]
OH- (aq) + H+ (aq)
H2O (l)
Increase solubility of Mg(OH)2
At pH greater than 10.45
Raise [OH-]
Decrease solubility of Mg(OH)2
Complex Ion Equilibria and Solubility
A complex ion is an ion containing a central metal cation
bonded to one or more molecules or ions.
CoCl42- (aq)
Co2+ (aq) + 4Cl- (aq)
The formation constant or stability constant (Kf) is the
equilibrium constant for the complex ion formation.
Co(H2O)2+
6
CoCl24
Kf =
[CoCl42- ]
[Co2+][Cl-]4
Kf
stability of
complex
Complex-Ion Equilibria
• Many metal ions, especially transition
metals, form coordinate covalent bonds
with molecules or anions having a lone pair
of electrons.
This type of bond formation is essentially a
Lewis acid-base reaction.
Complex-Ion Equilibria
• Many metal ions, especially transition
metals, form coordinate covalent bonds
with molecules or anions having a lone pair
of electrons.
For example, the silver ion, Ag+, can react
with ammonia to form the Ag(NH3)2+ ion.

Ag  2(: NH 3 )  ( H 3 N : Ag : NH 3 )

Complex-Ion Equilibria
• A complex ion is an ion formed from a
metal ion with a Lewis base attached to it
by a coordinate covalent bond.
A complex is defined as a compound
containing complex ions.
A ligand is a Lewis base (an electron pair
donor) that bonds to a metal ion to form a
complex ion.
Complex-Ion Formation
The aqueous silver ion forms a complex ion with
ammonia in steps.

Ag (aq )  NH 3 (aq)

Ag ( NH 3 ) (aq )  NH 3 (aq)

Ag ( NH 3 ) (aq )

Ag ( NH 3 ) 2 (aq )
When you add these equations, you get the
overall equation for the formation of Ag(NH3)2+.

Ag (aq )  2 NH 3 (aq)

Ag ( NH 3 ) 2 (aq )
Complex-Ion Formation
The formation constant, Kf , is the equilibrium
constant for the formation of a complex ion from the
aqueous metal ion and the ligands.
The formation constant for Ag(NH3)2+ is:

[ Ag( NH 3 )2 ]
Kf 

2
[ Ag ][NH 3 ]
The value of Kf for Ag(NH3)2+ is 1.7 x 107.
Complex-Ion Formation
The formation constant, Kf, is the equilibrium
constant for the formation of a complex ion from the
aqueous metal ion and the ligands.
The large value means that the complex ion
is quite stable.
When a large amount of NH3 is added to a
solution of Ag+, you expect most of the Ag+
ion to react to form the complex ion.
Complex-Ion Formation
The dissociation constant, Kd , is the reciprocal, or
inverse, value of Kf.
The equation for the dissociation of Ag(NH3)2+
is

Ag ( NH 3 ) 2 (aq )
Ag  (aq )  2 NH 3 (aq)
The equilibrium constant equation is

2
1 [ Ag ][NH 3 ]
Kd 


K f [ Ag( NH 3 )2 ]
Equilibrium Calculations with
Kf
What is the concentration of Ag+(aq) ion in
0.010 M AgNO3 that is also 1.00 M NH3?
The Kf for Ag(NH3)2+ is 1.7 x 107.
In 1.0 L of solution, you initially have 0.010
mol Ag+(aq) from AgNO3.
This reacts to give 0.010 mol Ag(NH3)2+,
leaving (1.00- (2 x 0.010)) = 0.98 mol NH3.
You now look at the dissociation of
Ag(NH3)2+.
Equilibrium Calculations with
Kf
What is the concentration of Ag+(aq) ion in
0.010 M AgNO3 that is also 1.00 M NH3?
The Kf for Ag(NH3)2+ is 1.7 x 107.
The following table summarizes:

Ag ( NH 3 ) 2 (aq )
Starting
Change
Equilibrium
0.010
-x
0.010-x
Ag  (aq )  2 NH 3 (aq)
0
+x
x
0.98
+2x
0.98+2x
Equilibrium Calculations with
Kf
What is the concentration of Ag+(aq) ion in
0.010 M AgNO3 that is also 1.00 M NH3?
The Kf for Ag(NH3)2+ is 1.7 x 107.
The dissociation constant equation is:

2
[ Ag ][NH 3 ]
1
 Kd 

Kf
[ Ag( NH 3 )2 ]
Equilibrium Calculations with
Kf
What is the concentration of Ag+(aq) ion in
0.010 M AgNO3 that is also 1.00 M NH3?
The Kf for Ag(NH3)2+ is 1.7 x 107.
Substituting into this equation gives:
( x )(0.98  2x )
1

7
( 0.010  x )
1.7  10
2
Equilibrium Calculations with
Kf
What is the concentration of Ag+(aq) ion in
0.010 M AgNO3 that is also 1.00 M NH3?
The Kf for Ag(NH3)2+ is 1.7 x 107.
If we assume x is small compared with
0.010 and 0.98, then
2
( x )(0.98)
8
 5.9  10
( 0.010)
Equilibrium Calculations with
Kf
What is the concentration of Ag+(aq) ion in
0.010 M AgNO3 that is also 1.00 M NH3?
The Kf for Ag(NH3)2+ is 1.7 x 107.
and
8
x  5.9  10 
( 0.010 )
( 0.98 ) 2
 6.1  10
10
The silver ion concentration is 6.1 x 10-10
M.
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.
Separating the Common
Cations by Selective
Precipitation
Copyright © Cengage Learning. All rights reserved
63
WORKED
EXAMPLES
Worked Example 16.8
Worked Example 16.9
Worked Example 16.10
Worked Example 16.11
Worked Example 16.12
Worked Example 16.13
Worked Example 16.14
Worked Example 16.15
Worked Example 16.16