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Chemical
Equilibrium
Equilibrium 2007-2008
1
Chemical Equilibrium
• Few chemical reactions proceed in only one
direction. Most are reversible, at least to
some extent. At the start of a reversible
process, the reaction proceeds toward the
formation of products. As soon as some
product molecules are formed, the reverse
process begins to take place and reactant
molecules are formed from product molecules.
Chemical equilibrium is achieved when the
rates of the forward and reverse reactions
are equal and the concentrations of the
reactants and products remain constant.
Equilibrium 2007-2008
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Equilibrium is a dynamic process.
• Physical equilibrium
H2O(s)
H2O(l)
At equilibrium, liquid water is reforming ice as
fast as ice is melting.
Equilibrium 2007-2008
3
Chemical equilibrium
N2O4(g)
colorless
2NO2(g)
brown
Equilibrium is a state in which there are no observable
changes as time goes by. When a chemical reaction
has reached the equilibrium state, the concentrations
of reactants and products remain constant over time,
and there are no visible changes in the system.
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The Law of Mass Action
For a reversible reaction at equilibrium
and at a constant temperature, the ratio
of reactant and product concentrations
has a constant value, k (the equilibrium
constant).
• Note that although the concentrations
may vary, as long as a given reaction is
at equilibrium and the temperature does
not change, according to the law of mass
action, the value of k remains constant.
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We can generalize this phenomenon with the following
reversible reaction:
aA + bB
cC + dD
where a, b, c and d are the stoichiometric
coefficients for the reacting species A, B, C, and D.
k=
[C]c[D]d
[A]a[B]b
Law of Mass
Action
•kc indicates that the concentrations of the
reacting species are expressed in molarity or
moles per liter.
•kp is used when equilibrium concentrations are
expressed in terms of pressure.
6
For the reaction,
N2O4(g)
at 25 oC
k=
2NO2(g)
[NO2]2
[N2O4]
Note that the exponent 2 for [NO2] in this
expression is the same as the stoichiometric
coefficient for NO2 in the balanced chemical
equation.
•k is defined as having no units.
7
Changes in the concentrations of NO2 and N2O4 with time, in three situations.
(a) Initially only NO2 is present
(b) Initially only N2O4 is present
(c) Initially a mixture of NO2 and N2O4 is present
Note that even though equilibrium is reached in all cases, the equilibrium
concentrations of NO2 and N2O4 are not the same as in the previous trials.
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Solving for k, the equilibrium constant
The following table shows some
experimental data for the NO2-N2O4
system at 25 oC. The gas
concentrations are expressed in
molarity, which can be calculated from
the number of moles of the gases
present initially and at equilibrium and
the volume of the flask in liters.
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The NO2-N2O4 System at 25 oC
Initial
Concentrations
(M)
Equilibrium
Concentrations
(M)
Ratio of
Concentrations at
Equilibrium
[NO2]
[N2O4]
[NO2]
[N2O4]
[NO2]
[N2O4]
0.000
0.670
0.0547
0.643
0.0851
0.0500
0.446
0.0457
0.448
0.102
0.0300
0.500
0.0475
0.491
0.0967
0.0400
0.600
0.0523
0.594
0.0880
0.200
0.000
0.0204
0.0898
0.227
[NO2]2
[N2O4]
Analysis of the equilibrium data shows that the ratio [NO2]/[N2O4] gives scattered values. 10
The NO2-N2O4 System at 25 oC
Initial
Concentrations
(M)
Equilibrium
Concentrations
(M)
Ratio of
Concentrations at
Equilibrium
[NO2]
[N2O4]
[NO2]
[N2O4]
[NO2]
[N2O4]
0.000
0.670
0.0547
0.643
0.0851
0.0500
0.446
0.0457
0.448
0.102
0.0300
0.500
0.0475
0.491
0.0967
0.0400
0.600
0.0523
0.594
0.0880
0.200
0.000
0.0204
0.0898
0.227
[NO2]2
[N2O4]
Analysis of the equilibrium data shows that the ratio [NO2]2/[N2O4] gives a
nearly constant value that averages 4.63 x 10-3
11
The magnitude of the equilibrium constant tells
us whether an equilibrium reaction favors the
products or reactants.
• If k is much greater than 1 (in this context,
any number greater than 10 is considered to be
much greater than 1) the equilibrium will lie to
the right and favors the products.
• Conversely, if the equilibrium constant is much
smaller than 1 (any number less than 0.1 is much
less than 1), the equilibrium will lie to the left
and favor the reactants.
Equilibrium 2007-2008
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The concentrations of reactants and products in
gaseous reactions can also be expressed in
terms of their partial pressures.
For the equilibrium process
N2O4(g)
we can write
kp =
2NO2(g)
(PNO )2
2
(PN O )
2
4
Where PNO2 and PN2O4 are the equilibrium
partial pressures (in atm) of NO2 and
N2O4, respectively.
13
Example – Use the Law of Mass Action to write the
equilibrium constant expression kc and kp if applicable,
for the following system:
(NH4)2Se(s)
k=
2NH3(g) + H2Se(g)
[NH3]2[H2Se]
[(NH4)2Se]
• The “concentration” of a solid or a liquid is a constant, (an
intensive property that does not depend on how much of
the substance is present) and can be omitted when
writing equilibrium expressions.
kc = [NH3]2[H2Se]
kp = (PNH3)2(PH Se)
2
14
Use the Law of Mass Action to write the equilibrium
expressions for the following reactions.
CH3COOH(aq)
H+(aq) + CH3COOH+(aq)
kc =
N2O4(g)
2NO2(g)
kp =
Fe(OH)3(s)
Fe+3(aq) + 3OH-(aq)
kc =
Equilibrium 2007-2008
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In general, kc is not equal to kp, since the partial
pressures of reactants and products are not
equal to their concentrations expressed in
moles per liter.
• A simple relationship exists between kp and kc
kp = kc(RT)Dn
where
Dn = moles of gaseous product – moles of gaseous reactants
R = the gas constant 0.0821 L.atm/mol. K
and T = temperature in Kelvins
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Consider the following equilibrium at 395 K
NH4HS(s)
NH3(g) + H2S(g)
The partial pressure of each gas is .265 atm. Calculate
kp and kc for the reaction
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Multiple Equilibria
The reactions we have considered so far are all
relatively simple. A more complicated
situation is one in which the product
molecules in one equilibrium system are
involved in a second equilibrium process:
aA + bB
cC + dD
k’c
cC + dD
eE + fF
k”c
aA + bB
eE + fF
kc
kc = k’c . k”c
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A typical operating temperature of an auto
engine (~100 oC), N2 and O2 form nitric oxide,
which then combines with more O2 to form
nitrogen dioxide, a toxic pollutant:
(1) N2(g) + O2(g)
(2) 2NO(g) + O2(g)
2NO(g)
2NO2(g)
kc1 = 4.3 x 10-25
kc2 = 6.4 x 109
Calculate kc for the overall reaction.
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The balanced chemical equation must be
written when quoting the numerical value of an
equilibrium constant.
• If a reversible reaction is written in the
opposite direction, the equilibrium constant
becomes the reciprocal of the original
equilibrium constant.
• The value of k also depends on how the
equilibrium equation is balanced. (If you
double the chemical equation throughout, the
corresponding equilibrium constant will be the
square of the original value.)
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Q, the Reaction Quotient
The equilibrium constant kc for the formation
of hydrogen iodide from molecular hydrogen
and molecular iodine in the gas phase
H2(g) + I2(g)
2HI(g)
is 54.3 at 430 oC. Suppose that in a certain
experiment we place 0.243 mole of H2, 0.146
mole of I2, and 1.98 moles of HI all in a 1.00 L
container at 430 oC. Will there be a net
reaction to form more H2 and I2 or more HI?
[HI]02
[1.98]2
Q= [H ] [I ] = [0.243][0.146] = 111
2 0 2 0
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Because the quotient [HI]02/[H2]0[I2]0 is
greater than kc, this system is not at
equilibrium. Consequently, some of the
HI will react to form more H2 and I2
(decreasing the value of the quotient).
Thus the net reaction proceeds from
right to left to reach equilibrium.
• For reactions that have not reached
equilibrium, such as the formation of HI
considered above, we obtain the reaction
quotient (Qc), instead of the equilibrium
constant by substituting the initial
concentrations into the equilibrium
constant expression.
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To determine the direction in which the net
reaction will proceed to achieve equilibrium,
we compare the values of Qc and kc.
• If Qc > kc, the reaction proceeds from right to
left (consuming products, forming reactants)
to reach equilibrium
• If Qc < kc, the reaction proceeds from left to
right (consuming reactants, forming products)
to reach equilibrium.
• If Qc = kc, the initial concentrations are
equilibrium concentrations. The system is at
equilibrium.
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Calculating concentrations in an equilibrium system
If we know the equilibrium constant for a
particular reaction, we can calculate the
concentrations in the equilibrium mixture from
the initial concentrations.
Let’s consider the following system involving two
organic compounds, cis-stilbene and
trans-stilbene
cis-stilbene
trans-stilbene
The equilibrium constant for this system is 24.0 at
200 oC.
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cis-stilbene
Suppose that initially only Initial
cis-stilbene is present
at a concentration of
0.850 mol/L.
Change
How do we calculate the
concentrations of cis
and trans-stilbene at
equilibrium?
trans-stilbene
0.850
0.000
-x
+x
Equilbrium (0.850 – x)
Equilibrium 2007-2008
x
25
Next we set up the equilibrium constant
expression
kc =
[trans-stilbene]
[cis-stilbene]
24.0 =
x
0.850 - x
x = 0.816 M
[cis-stilbene] = (0.850- 0.816) M = 0.034 M
[trans-stilbene] = 0.816 M
26
Factors that Affect Equilibrium
Chemical equilibrium represents a balance
between forward and reverse reactions. In
most cases, this balance is quite delicate.
Changes in experimental conditions
(concentration, pressure, volume and
temperature) may disturb the balance and
shift the equilibrium position so that more or
less of the desired product is formed. When
we say that an equilibrium position shifts to
the right, for example, we mean that the net
reaction proceeds from left to right.
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Le Chatelier’s Principle
There is a general rule that helps us to predict
the direction in which an equilibrium reaction
will move when a change in concentration,
pressure, volume, or temperature occurs. The
rule, known as Le Chatelier’s principle, states
that if an external stress is applied to a
system at equilibrium, the system adjusts in
such a way that the stress is partially offset.
The word “stress” here means a change in
concentration, pressure, volume, or
temperature that removes a system from the
equilibrium state.
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Changes in concentration
An aqueous
solution of
Fe(SCN)3
FeSCN+2(aq)
red
Fe+3(aq) + SCN-(aq)
pale yellow
Equilibrium 2007-2008
colorless
29
What happens if we add some sodium thiocyanate,
NaSCN, to the solution?
In this case, the stress applied to
the equilibrium system is an
increase in the concentration of
SCN- (from the dissociation of
NaSCN). To offset this stress,
some Fe+3 ions react with the
added SCN- ions, and the
equilibrium shifts from right to
left, and the red color of the
solution deepens.
FeSCN+2(aq)
red
Fe+3(aq) + SCN-(aq)
pale yellow
Equilibrium 2007-2008
colorless
30
Similarly, if we added iron
(III) nitrate to the
original solution, the red
color would also deepen
because the additional
Fe+3 ions (from the
Fe(NO3)3 solution) would
shift the equilibrium
from right to left.
FeSCN+2(aq)
red
Fe+3(aq) + SCN-(aq)
pale yellow
Equilibrium 2007-2008
colorless
31
Now suppose we add some oxalic
acid, H2C2O4, to the original
solution. Oxalic acid ionizes
in water to form the oxalate
ion, C2O4-2, which binds
strongly to the Fe+3 ions.
This formation of Fe(C2O4)3-3
removes free Fe+3 ions in
solution. Consequently, more
FeSCN+2 units dissociate and
the equilibrium shifts from
left to right, and the solution
will turn yellow.
FeSCN+2(aq)
red
Fe+3(aq) + SCN-(aq)
pale yellow
Equilibrium 2007-2008
colorless
32
Changes in Volume and Pressure
Changes in pressure do not affect the
concentrations of solids, liquids or
aqueous solutions because they are
virtually incompressible. On the other
hand, concentrations of gases are
greatly affected by changes in
pressure.
Equilibrium 2007-2008
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Suppose that the equilibrium system
N2O4(g)
2NO2(g)
is in a cylinder fitted with a movable
piston.
What happens if we increase the
pressure on the gases by pushing
down on the piston at constant
temperature?
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Since the volume decreases, the
concentration of both NO2 and N2O4
increases. Because the concentration of
NO2 is squared in the equilibrium
constant expression, the increase in
pressure increases the numerator more
than the denominator. The system is no
longer at equilibrium, so we write
Qc =
[NO2]02
[N2O4]0
Equilibrium 2007-2008
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Qc > Kc, and the net
reaction will shift to
the left until Qc = kc.
(a)
(b)
(c)
(a)Brown NO2(g) and
colorless N2O4(g) in
equilibrium.
(b)When the volume in the
syringe is suddenly
decreased, the
concentration of NO2 and
N2O4 are both increased,
(indicated by the darker
brown color).
(c)A few seconds after the
sudden volume decrease,
the color is much lighter
brown as the equilibrium
shifts the brown NO2(g)
to colorless N2O4(g).
36
In general, an increase in pressure
(decrease in volume) favors the net
reaction that decreases the total
number of moles of gases (the reverse
reaction, in this case), and a decrease in
pressure (increase in volume) favors the
reaction that increases the total
number of moles of gases.
For the reactions in which there is no
change in the number of moles of gases,
a pressure (or volume) change has no
effect on the position of equilibrium.
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We can also increase the pressure of a system
without changing its volume by adding an inert
gas (helium, for example) to the equilibrium
system.
Adding helium to the equilibrium mixture at
constant volume increases the total gas
pressure and decreases the mole fractions of
both NO2 and N2O4; but the partial pressure
of each gas, given by the product of its mole
fraction and total pressure, does not change.
Thus the presence of an inert gas does not
affect the equilibrium.
Equilibrium 2007-2008
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Changes in Temperature
A change in concentration, pressure, or
volume may alter the equilibrium
position, but it DOES NOT change the
value of the equilibrium constant (k).
Only a change in temperature can alter
the equilibrium constant.
Equilibrium 2007-2008
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N2O4(g)
2NO2(g)
DH = 58.0 kJ
The formation of NO2 from N2O4 is an endothermic
process. Another way to write this reaction would be
N2O4(g) + heat
2NO2(g)
The formation of N2O4 from NO2 is an exothermic
process. This process could be written
2NO2(g)
2NO2(g)
N2O4(g)
or
DH = -58.0 kJ
N2O4(g) + heat
Equilibrium 2007-2008
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What will happen if the following system
N2O4(g)
2NO2(g)
is heated at constant volume?
Equilibrium 2007-2008
41
Since endothermic
processes absorb
heat from the
surroundings, heating
favors the
dissociation of
colorless N2O4 into
brown NO2 molecules,
(an endothermic
process).
N2O4(g)
2NO2(g)
Equilibrium 2007-2008
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Cooling favors the
exothermic
process (the
formation of
colorless N2O4).
N2O4(g)
2NO2(g)
Equilibrium 2007-2008
43
The Effect of a Catalyst
A catalyst enhances the
rate of a reaction by
lowering the reaction’s
activation energy of the
forward and reverse
reaction to the same
extent. We can
therefore conclude that
the presence of a
catalyst does not alter
the equilibrium constant,
nor does it shift the
position of an equilibrium
system.
Equilibrium 2007-2008
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Adding a catalyst to a reaction mixture
that is not at equilibrium will simply
cause the mixture to reach equilibrium
sooner. The same equilibrium mixture
could be obtained without the catalyst,
but we might have to wait much longer
for it to happen.
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ksp, Solubility Equilibria
Precipitation reactions are important in
industry, medicine, and everyday life.
Although useful, the solubility rules do
not allow us to make quantitative
predictions about how much of a given
ionic compound will dissolve in water.
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Consider a saturated solution of silver
chloride that is in contact with solid
silver chloride. The solubility equilibrium
can be represented as
AgCl(s)
Ag+(aq) + Cl-(aq)
Because salts are considered strong
electrolytes, all the AgCl that dissolves
in water is assumed to dissociate
completely into Ag+ and Cl- ions.
Equilibrium 2007-2008
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Because the concentration of a solid is
constant, the equilibrium constant for
the dissolution of AgCl can be written as
ksp = [Ag+][Cl-]
where ksp is called the solubility product.
In general, the solubility product of a
compound is the product of the molar
concentrations of the constituent ions,
each raised to the power of its
stoichiometric coefficient in the
equilibrium expression.
Equilibrium 2007-2008
48
Write the equilibrium expressions for the
dissolution of
• magnesium fluoride
MgF2(s)
Mg+2(aq) + 2F-(aq)
ksp = [Mg+2][F-]2
•silver carbonate
Ag2CO3(s)
2Ag+(aq) + CO3-2(aq)
ksp = [Ag+]2[CO3-2]
•calcium phosphate
Ca3(PO4)2(s)
3Ca+2(aq) + 2PO4-3(aq)
Equilibrium 2007-2008
ksp = [Ca+2]3[PO4-3]2
49
The value of ksp indicates the solubility of
an ionic compound-the smaller the value,
the less soluble the compound in water.
However, in using ksp values to compare
solubilities, you should choose
compounds that have similar formulas,
such as AgCl and ZnS, or CaF2 and
Fe(OH)2.
Equilibrium 2007-2008
50
For equilibrium reactions involving an ionic solid in
aqueous solution, any one of the following
conditions may exist:
•
•
•
the solution is unsaturated, (a solution that
contains less solute than it has the capacity
to dissolve)
the solution is saturated, (at a given
temperature, the solution that results when
the maximum amount of a substance has
dissolved in a solvent), or
the solution is supersaturated, (a solution
that contains more of the solute than is
present in a saturated solution).
Equilibrium 2007-2008
51
Q, the ion product
For concentrations that do not correspond to
equilibrium conditions, we use the reaction
quotient which in this case is called the ion
product (Q), to predict whether a precipitate
will form.
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52
• For example, if we mix a solution
containing Ag+ ions with one containing
Cl- ions, then the ion product is given by
Q = [Ag+]0[Cl-]0
• The subscript 0 reminds us that these
are initial concentrations, and do not
necessarily correspond to those at
equilibrium.
Equilibrium 2007-2008
53
The possible relationships between Q and ksp
are
Q< ksp
[Ag+]0[Cl-]0 < 1.6 x 10-10
Unsaturated solution
Q = ksp
[Ag+]0[Cl-]0 = 1.6 x 10-10
Saturated solution
Q > ksp
[Ag+]0[Cl-]0 > 1.6 x 10-10
Supersaturated solution; AgCl will precipitate
out until the product of the ionic
concentrations is equal to 1.6 x 10-10
Equilibrium 2007-2008
54
There are two other ways to express a
substance’s solubility:
• molar solubility, which is the number of
moles of solute in one liter of a
saturated solution (mol/L), and
• solubility, which is the number of grams
in one liter of a saturated solution (g/L).
• Both solubility and molar solubility can
be used to determine ksp.
Equilibrium 2007-2008
55
The solubility of calcium sulfate is found
experimentally to be 0.67 g/L. Calculate
the value of ksp for calcium sulfate.
Equilibrium 2007-2008
56
Sometimes we are given the value of ksp
for a compound and are asked to
calculate the compound’s molar
solubility.
For example, the ksp of silver bromide is
7.7 x 10-13. Calculate the molar
solubility of silver bromide.
Equilibrium 2007-2008
57
Common Ion Effect
The solubility product is an equilibrium
constant; precipitation of an ionic
compound from solution occurs
whenever the ion product exceeds ksp
for that substance.
In a saturated solution of AgCl, for
example, the ion product [Ag+][Cl-] is, of
course, equal to ksp. Furthermore,
simple stoichiometry tells us that
[Ag+]=[Cl-]. But this equality does not
hold true in all situations.
Equilibrium 2007-2008
58
Suppose we study a solution containing
two dissolved substances that share a
common ion, say, AgCl and AgNO3. In
addition to the dissociation of AgCl, the
following process also contributes to
the total concentration of the common
silver ions in solution:
AgNO3(s)
Ag+(aq) + NO3-(aq)
Equilibrium 2007-2008
59
If AgNO3 is added to a saturated AgCl solution,
the increase in [Ag+] will make the ion product
greater than the solubility product:
Q = [Ag+]0[Cl-]0> ksp
To reestablish equilibrium, some AgCl will
precipitate out of the solution, as LeChatelier’s
principle would predict, until the ion product is
one again equal to ksp.
The effect of adding a common ion, then, is a
decrease in the solubility of the salt (AgCl) in
solution. Note that in this case [Ag+] is no
longer equal to [Cl-] at equilibrium; rather,
[Ag+]>[Cl-].
Equilibrium 2007-2008
60
pH and Solubility
The solubilities of many substances also
depend on the pH of the solution.
Consider the solubility equilibrium of
magnesium hydroxide:
Mg(OH)2(s)
Mg+2(aq) + 2OH-(aq)
Equilibrium 2007-2008
61
Adding OH- ions (increasing the pH)
shifts the equilibrium from right to
left, thereby decreasing the solubility
of Mg(OH)2.
On the other hand, adding H+ ions
(decreasing the pH) shifts the
equilibrium from left to right, (because
H+ and OH- will form water, thus
removing the OH- from solution) and
the solubility of Mg(OH)2 increases.
Thus, insoluble bases tend to dissolve in
acidic solutions.
Equilibrium 2007-2008
62
Lewis acids and base reactions in which a metal
cation (electron pair acceptor) combines with a
Lewis base (electron pair donor) result in the
formation of complex ions:
Ag+(aq) + 2NH3(aq)
Ag(NH3)2+
Lewis acid
Lewis base
Transition metals have a particular tendency to form
complex ions. For example, a solution of cobalt (II)
chloride is pink because of the presence of the
Co(H2O)6+2 ions. When HCl is added, the solution turns
blue as a result of the formation of the complex ion
CoCl4-2
Co+2(aq) + 4Cl-(aq)
CoCl4-2
Equilibrium 2007-2008
63
Separation of Ions by Fractional
Precipitation
In chemical analysis, it is sometimes desirable
to remove one type of ion from solution by
precipitation while leaving other ions in
solution. For instance, the addition of sulfate
ions to a solution containing both potassium
and barium ions causes BaSO4 to precipitate
out, thereby removing most of the Ba+2 ions
from the solution. The other “product”
K2SO4, is soluble and will remain in solution.
The BaSO4 precipitate can be separated from
the solution by filtration or decanting.
Equilibrium 2007-2008
64
Application of the Solubility Product
Principle to Qualitative Analysis
Qualitative analysis is the determination
of the types of ions present in a solution.
(We will focus on the cations.)
• There are some twenty cations that can
be analyzed readily in aqueous solution.
• These cations can be divided into groups
according to the solubility products of
their insoluble salts.
Equilibrium 2007-2008
65
The separation of cations into groups is
made as selective as possible; that is,
the anions that are added as reagents
must be such that they will precipitate
the fewest types of cations.
The separation of cations at each step
must be carried out as completely as
possible.
Since an unknown solution may contain
from one to all twenty ions, any analysis
must be carried out systematically from
Group I through group III.
Equilibrium 2007-2008
66
A flow chart for the separation of cations in qualitative
analysis
Solution containing ions of
all cation groups
Group I precipitates
HCl
decanting
AgCl, Hg2Cl2, and PbCl2
Solution containing ions of
remaining groups
decanting
NH4OH and (NH4)2S
Group II precipitates
CoS, FeS, MnS, NiS, ZnS,
Al(OH)3 and Cr(OH)3
Solution containing ions of
remaining groups
decanting
(NH4)2CO3
Group III precipitates
BaCO3, CaCO3 MgCO3 and
SrCO3
Solution contains
Na+, K+ and NH4+ ions
Equilibrium 2007-2008
67
The End
Equilibrium 2007-2008
68