Solubility and Complex Formation Equilibria
Objectives
1. Write a reaction showing the dissolution of any solid compound.
2. Write the expression for Ksp for any solid compound.
3. Use the value of Ksp to calculate the concentration of one ion knowing the concentration
of the other.
4. Use the value of Ksp to determine whether a precipitate will form.
5. Use the value of Ksp to calculate the solubility of a solid compound in water or in a
solution containing a common ion.
6. Use the values of Ksp to determine which ion will precipitate from solution first from a
mixture.
7. Write reactions and calculate the value of K for dissolving a hydroxide compound in acid
solution.
8. Write reactions and calculate K for dissolving a precipitate in a solution containing a
complexing agent.
9. Write reactions and calculate K for dissolving a precipitate in acid solution when a weak
electrolyte (weak acid) is formed.
10. Discuss and employ qualitative analysis techniques to determine what ions are present
in solutions of several metal ions.
Precipitation Formation: Solubility Product Constant (Ksp)
Background/Terminology
Precipitation is the reverse of dissolution. When studying precipitation/dissolution processes,
reactions are typically represented as follows:
salt(s) <==> cation(aq) + anion(aq)
The presence of water as the solvent is indicated by the subscripted (aq) following the ion
formulas.
As indicated by the double reaction arrows, these are equilibrium processes. Therefore, they can
be treated like other equilibrium systems. That is, the reactions are "governed" by an equilibrium
constant, equilibrium can be approached from both directions, and equilibrium can be shifted
(through the application of Le Chatelier's Principle).
Let's take a look at some examples and learn some terminology.
AgCl(s) <==> Ag+(aq) + Cl-(aq) K(sp) = 1.8 x 10-10
K(sp) is the equilibrium constant for this reaction (at 25o) and it is called the solubility product
constant (hence the "sp").
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Since this is an equilibrium, the equilibrium constant is related to the concentrations as in other
equilibrium expressions . . .
K(sp) = [Ag+][Cl-]
The pure salt and water are not included because they are pure compounds.
You will notice that K(sp) is rather small which indicates that not much of the salt dissolves. This
will be true for all of the salts that are studied in this chapter. The reason for this will be revealed
later.
Another important term that you should be familiar with is solubility (given the symbol "S").
Solubility is the concentration of the salt that is dissolved in solution. It is normally expressed in
terms of molarity (moles/liter) but it can be expressed in other units (g/100 g of water or g/liter or
. . .)
For the reaction above, S = [Ag+] = [Cl-], since the stoichiometry of the compound is one to one.
See if you can determine what "S" is equal to for the following salts"
Ca3(PO4)2
Al(OH)3
Fe2(SO4)3
Ans:
For Ca3(PO4)2, S = 1/3 [Ca2+] = 1/2 [ PO43-]
For Al(OH)3, S = [Al3+] = 1/3 [OH-]
For Fe2(SO4)3, S = 1/2 [Fe3+] = 1/3 [ SO42-]
Do you see the relationship between the compound's stoichiometry and the way that "S" is
related to each ion's concentration?
NOTE: The relationships shown above are true when the only source of the ions in the solution
is the salt itself (i.e. when there is no other species present in the solution that produces either of
the ions produced by the salt in the equilibrium).
Determining if a Precipitate will Form
In order to determine whether a precipitate will form or not, we need merely ask the following
question: "Will the solubility reaction proceed in the forward or reverse direction based on the
ion concentrations present?"
Determining the direction in which an equilibrium will proceed is something we have done
before. In a previous chapter, we learned about the reaction quotient, Q. We learned the
following about Q.
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Q takes the same form as the equilibrium expression. When Q = K, the process being studied is
at equilibrium and there is not net reaction.
When Q < K, the process shifts toward product until Q becomes equal to K.
When Q > K, the process shifts toward reactant until Q becomes equal to K.
For precipitation/dissolution equilbria, we change the name from reaction quotient to "ion
product" and symbol for Q to either “QIP”or "P".
example If a solution contains [Ag+] = [Cl-] = 0.10 M, will a precipitate of AgCl form? If so,
what will [Ag+] and [Cl-] be after AgCl has completed precipitation?
Ans. The equilibrium is: AgCl(s) <==> Ag+(aq) + Cl-(aq) K(sp) = 1.8 x 10-10
Precipitation will occur if QIP > K(sp).
QIP = [Ag+][Cl-] = (0.10)(0.10) = 0.010
QIP > K(sp), so AgCl will precipitate.
Now, how do we determine what [Ag+] and [Cl-] will be at equilibrium?
Let's set up a reaction chart!
AgCl(s)<==> Ag+(aq) + Cl-(aq)
initial --0.10 M
0.10 M
change ---S
-S
equil. --0.10 - S M 0.10 - S M
K(sp) = 1.8 x 10-10 = [Ag+][Cl-]
1.8 x 10-10 = [0.10 - S][0.10 - S]
We cannot assume S is small because most of the ions will precipitate out of solution.
However, solving the above equation can be done explicitly by taking the square root of each
side and solving for S.
Doing so yields S = 0.099987 M . . . which is many sig figs. more than we should have.
Note that [Ag+] = [Cl-] = 1.3 x 10-5 M
IMPORTANT NOTE: The answer to the second part of the example problem was solved in a
very general way. i.e. the method would work for any equilibrium system. However, in the
sample above there is a much easier solution.
Since AgCl has one to one compound stoichiometry AND since [Ag+] = [Cl-] in the starting
solution, the final concentrations of both must be equal. That means that
K(sp) = 1.8 x 10-10 = [Ag+][Cl-]
and [Ag+] = [Cl-] square root of K(sp) = 1.3 x 10-5 M.
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Common Ion Effect
The solubility of a salt in the presence of another source of ions common to the salt will be lower
than in pure water.
For example:
Find the solubility of lead (II) chloride in pure water and in aqueous 0.10 M NaCl.
In pure water the equilibrium table would look like:
PbCl2 (s)<==>Pb2+(aq)+2 Cl-(aq)
initial --0M
0M
change --+S
+2S
equil. --SM
2S M
K(sp) = 1.7 x 10-5 = [Pb2+][Cl-]2
1.7 x 10-5 = [S][2S]2 = 4S3
S = [Pb2+] = 1.6 x 10-2 M, and 2S = [Cl-] = 3.2 x 10-2 M
In aqueous 0.10 M NaCl the equilibrium will be altered to look like this:
PbCl2 (s)<==>Pb2+(aq)+ 2 Cl-(aq)
initial --0M
0.10 M
change --+S
+2S
equil. --S M 0.10 + 2S M
The initial concentration of chloride is due to the presence of dissolved NaCl.
K(sp) = 1.7 x 10-5 = [Pb2+][Cl-]2
1.7 x 10-5 = [S][0.10 + 2S]2
assume 2S is small compared to 0.10 M, then
1.7 x 10-5 = [S][0.10]2
S = [Pb2+] = 1.7 x 10-3 M, and [Cl-] = 0.10 + 2S = 0.10 M
The assumption was valid. Also note that the solubility of the lead (II) chloride is lower in the
aqueous NaCl solution than in pure water.
The Uncommon Ion Effect
Salts will be more soluble in a solution that contains other dissolved salts with no ions in
common with the salt in the equilibrium. The reasons are discussed below.
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example: lead (II) chloride would have a higher solubility in an aqueous solution of sodium
nitrate than in pure water. However, there is no "easy" mathematical method to account for the
change in solubility caused by the uncommon ions. The treatment of these types of systems is
covered in higher level chemistry courses.
Solubility is often higher than expected for two reasons. First ions can combine in solution to
form "ion pairs". For example, when studying the solubility of magnesium fluoride, the reaction
of interest is:
MgF2 (s) <==> Mg2+(aq) + 2 F-(aq)
However, a Mg2+(aq) and a F-(aq) can combine to form an ion pair MgF+(aq). This allows more
magnesium fluoride to dissolve to replace these ions.
The second reason for finding increased solubilities is a little more nebulous. The "ionic
atmosphere" of the solution is affected by the presence of high (> 0.010 M) concentrations of
ions. You may think about this in the following way. In very dilute solutions, ions are separated
by large distances (on the atomic scale) and numerous solvent molecules "insulate" the ions from
the charges on other ions. However, in more concentrated solutions, ions "feel" the presence of
other ions and the solvent is made more “ionic” by the presence of a large amount of ions. This
increases the solubility of the sparingly soluble salts.
Dissolving Precipitates
In the section above you learned that the common ion effect acts to shift the solubility equilibria
in a way that decreases the solubility of salts. In this section we will investigate how to enhance
the solubility of salts by coupling reactions with the solubility equilibria.
Formation of water:
Hydroxides generally have low solubilities in water. However, we can make the salts more
soluble by dissolving them in an acidic solution. Consider the following example:
Aluminum hydroxide has K(sp) = 2 x 10-32 = [Al3+][OH-]3
Clearly this is not a highly soluble compound. However, if we dissolve the hydroxide in a buffer
with a pH = 4.00 the salt will dissolve much more readily. (neutral is pH = 7.00, a pH of 4.00 is
1000 times more acidic). It is easy to understand how the presence of additional acid helps
dissolve the hydroxide if you examine the following reactions.
solubility
Al(OH)3(s) <==> Al3+(aq) +3 OH-(aq)
acid/base neut.3 OH-(aq)+ 3 H+ (aq) <==>3 H2O(l)
overall
3 H+ (aq) +Al(OH)3 (s)<==> Al3+(aq) + 3 H2O(l)
The first reaction is simply the solubility equilibria. The second reaction is an acid/base
neutralization. Look closely at the second reaction and you will recognize that it it 3 times the
reverse of the water self-ionization reaction. When these two reactions are added together, the
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overall reaction is found. To find the equilibrium constant for the overall reaction, we simply
find the product of the equilibrium constants of the first two reactions:
overall K = K(sp)(1/Kw3)
overall K = (2 x 10-32)(1.0 x 10+42) = 2 x 10+10
The overall K is a large number - this indicates that the overall reaction proceeds toward product
to a relatively large extent.
Now that we have determined the equilibrium constant for the reaction, we can calculate the
equilibrium concentrations. Proceed as follows:
overall 3 H+ (aq) +Al(OH)3 (s)<==>Al3+(aq)+3 H2O(l)
initial 1.0 x 10-4 M
--0M
--change
~0
--+S
---4
equil. 1.0 x 10 M
--SM
--The initial concentration of H+(aq) is determined from the pH of the buffer. The change in the
concentration of H+(aq) is ~ 0 because we choose to use a BUFFER in which to dissolve our
compound. (buffers resist change in pH). This greatly simplifies the math we must do.
Now we can solve for S using the following:
overall K = 2 x 10+10 = [Al3+]/[H+]3
S = [Al3+] = 0.020 M
which is a much greater solubility than we would find for aluminum hydroxide in water.
Formation of a weak electrolyte:
Consider the following example: dissolve calcium fluoride in pH = 2.00 buffer. Fluoride ions are
weak bases (they are the conjugate of a weak acid, HF).
Calcium fluoride has K(sp) = 1.5 x 10-10 = [Cas+][F-]2
Clearly this is not a highly soluble compound. However, if we dissolve the fluoride in a buffer
with a pH = 2.00 the salt will dissolve much more readily. (neutral is pH = 7.00, a pH of 4.00 is
1000 times more acidic). It is easy to understand how the presence of additional acid helps
dissolve the fluoride if you examine the following reactions.
solubility
CaF2(s) <==> Ca2+(aq) + 2 F-(aq)
rev. acid dissoc. 2 F (aq) +2 H+ (aq)<==>2 HF(aq)
overall
2 H+ (aq)+ CaF2 (s) <==> Ca2+(aq) +2 HF(aq)
The first reaction is simply the solubility equilibria. The second reaction is 2 times the reverse of
an acid dissociation. When these two reactions are added together, the overall reaction is found.
To find the equilibrium constant for the overall reaction, we simply find the product of the
equilibrium constants of the first two reactions:
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overall K = K(sp)(1/Ka2)
overall K = (1.5 x 10-10)/(6.9 x 10-4)2 = 3.2 x 10-4
The overall K is a larger number than Ksp - this indicates that the overall reaction proceeds
toward product to a larger extent.
Now that we have determined the equilibrium constant for the reaction, we can calculate the
equilibrium concentrations. Proceed as follows:
overall 2 H+ (aq) +CaF2 (s)<==>Ca2+(aq)+2 HF(aq)
initial 1.0 x 10-2 M --0M
0M
change
~0
--+S
+ 2S
equil. 1.0 x 10-2 M --SM
2S M
The initial concentration of H+(aq) is determined from the pH of the buffer. The change in the
concentration of H+(aq) is ~ 0 because we choose to use a BUFFER in which to dissolve our
compound. (buffers resist change in pH). This greatly simplifies the math we must do.
Now we can solve for S using the following:
overall K = 3.2 x 10-4 = [Ca2+][HF]2/[H+]2
S = [Ca2+] = 0.0020 M
which is a much greater solubility than we would find for calcium fluoride in water.
Formation of a complex ion:
Salts which contain an ion that can form a complex ion can be dissolved to a greater extent
through the formation of that complex ion. Consider the following . . . AgCl is relatively
insoluble (Ksp = 1.8 x 10-10). However, silver can form a complex with ammonia. You can see
how this would change the position of the solubility equilibria by examining the following
reactions and calculating the solubility of Ag Cl in 0.100 M ammonia.
solubility
AgCl(s) <==>
Ag+(aq)
+
Cl-(aq)
complex form. Ag+(aq) +2 NH3 (aq)<==>[Ag(NH3)2]+(aq)
overall
2 NH3 (aq)+ AgCl (s) <==>
Cl-(aq)
+[Ag(NH3)2]+(aq)
The first reaction is simply the solubility equilibria. The second reaction is a complex formation
equilibria. When these two reactions are added together, the overall reaction is found. To find the
equilibrium constant for the overall reaction, we simply find the product of the equilibrium
constants of the first two reactions:
overall K = (K(sp))(Kf)
overall K = (1.8 x 10-10)/(1.7 x 10+7) = 3.1 x 10-3
The overall K is a larger number than Ksp - this indicates that the overall reaction proceeds
toward product to a relatively large extent.
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Now that we have determined the equilibrium constant for the reaction, we can calculate the
equilibrium concentrations. Proceed as follows:
overall 2 NH3 (aq) +AgCl (s)<==>Cl-(aq)+[Ag(NH3)2]+(aq)
initial
0.10 M
--0M
0M
change
- 2S
--+S
+S
equil. 0.10 M - 2S M
--SM
SM
Now we can solve for S using the following:
overall K = 3.1 x 10-3 = [Ag(NH3)2+] [Cl-]/[NH3]2
S = [Cl-] = 5.0 x 10-3 M
which is a much greater solubility than we would find for silver chloride in water.
Keywords
common ion
complex ion
equilibrium rules
-coefficent
-multiple
-reciprocal
formation constant, Kf
ion product, P
net ionic equations
-spectator ions
-adding reactions
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pH
precipitate
reaction quotient, Q
solubility, S
solubility product constant,
Ksp