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Analytical Chemistry
Lecture 5
October 20, 2002
Schedule Revision:
Week 4: Chapter 9
Monday: Systematic Treatment of Equilibrium
Wednesday: Review & Workshop
Week 5: Chapter 7 Titrations (particularly 7-1,7-2,7-4,7-6)
Monday: Open-book Exam
Wednesday: Titrations
Week 6: Chapters 6:7-9, Chapter 10 Acids and Bases
Week 7: Chapter 11 Polyprotic Acids and Bases
Week 8: Chapter 12 Acid-Base Titrations
Chapter 27:1-3 Gravimetric Analysis
Week 9: Chapter 13 EDTA Titrations and Review
Week 10: Exam II & Chapter 5
The exam will be a week from today. I have not written the exam yet, but it will be
open-book, but closed for all notes, homework etc. That is, you do not need to memorize
equations, but instead need to know how to solve problems.
Looking over the sample problems in the chapters and homework should give you a good
review of problem solving methods and approaches. In addition, I will give you a
practice exam on Wednesday that you can used to practice for the test. Solutions to the
practice exam will be in a file in the CAL.
By now you should be pretty comfortable with solubility equilibrium. Consider the
solubility of silver chloride.
AgCl(s) < === > Ag+ + Cl-
Ksp = [Ag+][Cl-] = 1.8 x 10-10
You should be able to write the chemical equation, set up a concentration table and
calculate the solubility of silver chloride in water in g/L.
As you have see, this kind of equilibrium is extremely important to geology since it is the
basis for rocks dissolving into ground and surface waters.
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We will see in Chapter 27 that we can use this equation to determine silver or chloride
ion concentrations in solution. What happens if we add a solution of sodium chloride to a
silver ion solution?
Le Chatlier's principle tells us that the silver ion concentration will be reduced and we
will form solid silver chloride. Which we could dry and weight. The mass of AgCl
could be used to calculate the original concentration of silver ions in solution.
If we add an excess of chloride ion we will be able to remove virtually all of the silver
ions in solution.
But there is another equilibrium that comes into play at high chloride concentrations
AgCl(s) + Cl- < == > AgCl2-(aq)
K = 3.3 x 10-5
This is an example of complex ion equilibrium and is the subject of section 6 in Chapter
6. It is also the basis for the analytical titrations described in Chapter 13.
In our example above, if we add too much chloride ion we will start to redissolve the ppt
through the formation of the soluble complex. The result would be to throw off the
analysis method based on the mass of precipitate.
Note that we have not changed the silver chloride equilibrium. Silver ion concentration
continues to go down, but silver goes back into solution as a new ionic species.
Both Jim and I have talked about carbonic acid equilibrium.
H2CO3(aq) < === > H+ + HCO3HCO3- (aq) < === > H+ + CO32The reverse of these reactions are the basis for your alkalinity titrations.
These equations are examples of a third type of equilibrium called acid/base equilibrium.
Finally, both of these chemical systems involve several competing chemical reactions. In
order to do these analyses successfully and to understand water samples containing ions
which may interact with each other, we need to be able to consider several chemical
equilibria at the same time.
This is the focus of Chapter 9 which presents a systematic method for solving complex
equilibria.
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Today I will go over this method for solving simultaneous equilibria and work through an
example.
Suppose we have a a solution that is saturated with slightly soluble barium sulfate, barite.
XXXXX
By now you should be able to write the solubility equation, but there are also two other
equilibria involved:
BaSO4(s)
< == > Ba2+ + SO42-
The first is should be easy to remember and it is normally not involved in the solubility
reaction.
H2O < == > H+ + OHHowever, in this case, sulfate ions will react with hydrogen ions to form the hydrogen
sulfate ion
SO42- + H+ < == > HSO4If hydrogen ions are added to this system, the third equilibrium is shifted to the right by
the Le Chatelier’s principle. The resulting decrease in sulfate concentration causes the
first equilibrium to shift to the right as well, thus increasing the solubility of the barium
sulfate.
This effect is the basis for the common environmental effect of acid rainfall and acid
mine waste. Many minerals are more soluble in slightly acidic water.
XXXXX
The solubility of barium sulfate is also increased when acetate ions are added to an
aqueous suspension of the solid because acetate ions tend to form a soluble complex with
barium ions as shown by the reaction
Ba2+ + OAc- < == > BaOAc+
but acetate is also a base so we also must include the equilibrium
H+ + OAc- < == > HOAc
Again, Le Chatelier’s principle causes both this equilibrium and the solubility
equilibrium to shift to the right, thus causing a solubility increase.
If we need to calculate the solubility of barium sulfate in a system containing hydrogen
and acetate ions, we must take into account not only the solubility equilibrium, but the
other three equilibria as well.
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In order to solve this kind of problem, most analytical chemistry books put forward the
systematic approach given in Chapter 9. The order and number of steps is not always the
same, but the approaches are identical.
Solving a multiple-equilibrium problem requires us to develop as many independent
algebraic equations as there are participants in the system being studied. For example, if
we wish to compute the solubility of barium sulfate in a solution of acid we need to be
able to calculate the concentration of all of the species present in the solution.
XXXXX
In the first three chemical equations there are 5 such species
Ba2+; SO42-; H+; HSO4-; and OHif acetate were present there would be three more
OAc-; BaOAc+; and HOAc
In order to calculate the solubility of barium sulfate in the first case of an acid solution
rigorously, it is necessary to develop five independent algebraic equations containing the
five unknowns and solve them simultaneously.
Three types of algebraic equations are used in solving multiple-equilibrium problems:
(1) equilibrium-constant expressions,
(2) mass-balance equations, and [also called material-balance or atom-balance]
(3) a single charge-balance equation.
We already know how to generate equilibrium-constant expressions, so we will go on to
Mass-Balance Equations.
Mass-balance equations relate the equilibrium concentrations of various species in a
solution to one another and to the analytical concentrations of the various solutes.
They are derived from information about how the solution was prepared and from a
knowledge of the kinds of equilibria established in the solution. As the book points out
this is really an atom-balance approach instead of mass-balance.
This is best understood by example.
XXXXX
Write mass-balance expressions for a 0.0100 M solution of HCl that is in equilibrium
with an excess of solid BaSO4.
We look first at barium sulfate. We don’t know how much is in solution, but for every
barium ion in solution there will be one sulfate ion, however some of the sulfate will have
moved on to the hydrogen sulfate form.
So
[Ba2+] = [SO42-] + [HSO4-]
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This is called a mass-balance equation.
We may write a second mass-balance expression based upon the fact that the hydrogen
ions in the solution have two sources, namely, from the HCl and from the dissociation of
the solvent water.
[H+] = cHCl + cdis water
=
0.0100 M + [OH-]
Here, the second term on the right side of the equation accounts for the concentration of
hydrogen ions that result from the dissociation of water.
Let's try another example.
XXXX
Write mass-balance expressions for the system formed when a 0.010 M NH3 solution is
saturated with AgBr.
Here are the equations for the pertinent equilibria in the solution are
solubility
AgBr(s) < == > Ag+ + Br-
complex ion
Ag+ + 2 NH3 < == > Ag(NH3)2+
acid/base
NH3 + H2O < == > NH4+ + OH-
acid/base
H2O < == > H+ + OH-
AgBr is the only source of Br-, Ag+, and Ag(NH3)2+, and because silver and bromide ions
exist in a 1:1 ratio in the starting material, it follows that one mass-balance equation is
[Ag+] + [Ag(NH3)2+] = [Br-]
Also, we know, that the only source of ammonia containing species is the 0.010 M NH3
Therefore,
cNH3 = [NH3] + [NH4+] + 2 [Ag(NH3)2+] = 0.010 M
Notice the 2 in front of the silver ammonia complex concentration. Does that make
sense?
From the last two equilibria, we see also that one hydroxide ion is formed for each NH4+
and each hydrogen ion. Therefore,
[OH-] = [NH4+] + [H+]
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The other new equation, the charge-balance equation is much easier to write. There is
only one such equation for each solution.
Solutions are neutral because the molar concentration of positive charge in an electrolyte
solution always equals the molar concentration of negative charge. That is, for any
solution containing electrolytes, we may write
no. mol/L positive charge = no. mol/L negative charge
This equation represents the condition of charge balance and is called the charge-balance
equation. To be useful for equilibrium calculations, this equality must be expressed in
terms of the molar concentrations of the species that carry the charge in the solution.
How much charge is contributed to a solution by 1 mol of Na+? 1 mol of Ca2+?
The concentration of charge contributed to a solution by an ion is equal to the molar
concentration of that ion multiplied by its charge.
XXXXX
For our solubility of silver bromide in ammonia problem, the cations are
[Ag+] + [Ag(NH3)2+] + [NH4+] + [H+] =
[Br-] + [OH-]
What is the charge-balance equation for our first system?
2 [Ba2+] + [H+]
= 2 [SO42-] + [HSO4-] + [OH-]
Now lets go through the steps involved in solving problems involving several equilibria.
Different set than in the book, but exactly the same method.
XXXXX
For our first example we will calculate the solubility of BaCO3 in water.
Step 1. Pertinent Equilibria
BaCO3(s)
< == > Ba2+ + CO32-
CO32- + H2O < == >
HCO3- + OH-
HCO3- + H2O < == > H2CO3 + OH-
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< == > H+ + OH-
H2 O
# of species = 6
Step 2. Definition of the Unknown
x = solubility = [Ba2+] = [CO32-] + [HCO3-] + [H2CO3]
Step 3. Equillbrium-Constant Expressions
[Ba2+] [CO32-] = Ksp = 5.0 x 10-9
and
(1)
[HCO3-][OH-]
[CO32-]
=
2.13 x 10-4
(2)
[H2CO3][OH-]
[HCO3-]
=
2.25 x 10-8
(3)
[H+] [OH-]
= 1.00 x 10-14
(4)
Step 4. Mass-Balance Expression
[Ba2+] = [CO32-] + [HCO3-] + [H2CO3]
(5)
XXXXXX
Step 5. Charge-Balance Expression
2 [Ba2+] + [H+] = 2 [CO32-] + [HCO3-] + [OH-]
(6)
Step 6. Number of Equations and Unknowns
We have developed six equations which are sufficient to solve for the six unknowns.
Step 7. Approximations
Can we eliminate one or more terms on the ground that their elimination does not create a
significant error? One candidate is [H+]. The solution should be basic because the
carbonate reactions release hydroxide into the solution. So [H+] must be smaller than
10-7. Also, [Ba2+] must be considerably larger than 10-7 because, it is close to the square
root of Ksp so must be greater than 10-5.
So we consider [H+] small compared to 2 [Ba2+] in equation (6) and can therefore
eliminate equation (4) entirely. Now we have 5 equations and 5 unknowns.
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A second possible assumption is that [H2CO3] is small and can be deleted from Equation
(6). We base this assumption on our knowledge that the solution is basic and therefore
likely to support bicarbonate and carbonate ions and not carbonic acid.
This assumptions allows us to eliminate equation (3) and equations (5) and (6) have
become
[Ba2+] = [CO32-] + [HCO3-]
(5)
2 [Ba2+] = 2 [CO32-] + [HCO3-] + [OH-]
(6)
Step 8. Solution of Equations
If we multiply Equation (5) by 2 and subtract -the product from Equation (6), we obtain,
[HCO3-] = [OH-]
Substitution of [HCO3-] for [OH-] in Equation (2) gives
- 2
[HCO3 ] = 2.13 x 10
[CO32-]
-4
or
_________________
[HCO3 ] =  (2.13 x 10-4)*[CO32-]
-
From Equation (1) we get [CO32-] = Ksp/[Ba2+] so
_______________
[HCO3 ] =  (2.13 x 10-4)*Ksp
[Ba2+]
Notice we are trying to get everything in terms of the barium ion concentration. Why are
we doing that? Because we know from step 2 that that is the concentration that is related
to solubility. We do not have to solve for every variable, just barium.
Now Equation (5) becomes
XXXXX
______________
2+
[Ba ] = Ksp/[Ba2+] +  (2.13 x 10-4)*Ksp
[Ba2+]
2+
if we multiply both sides by [Ba ]
______________
2+ 2
[Ba ] = Ksp + [Ba2+]1/2 * (2.13 x 10-4)*Ksp
[Ba2+]
[Ba2+]2 = Ksp + 1.03 x 10-6 * [Ba2+]1/2
_______________________________
=  {(1.03 x 10-6 *[Ba2+]1/2) + 5.0 x 10-9 }
It is best to do this by approximation and we can start by assuming [Ba2+] = x1  Ksp1/2.
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1st approx
[Ba2+]
=  {(1.03 x 10-6 *[7.07 x 10-5] 1/2) + 5.0 x 10-9 }
2nd approx
[Ba2+]
x1 = Ksp1/2 = 7.07 x 10-5
x2 = 1.17 x 10-4
=  {(1.03 x 10-6 *[1.17 x 10-4] 1/2) + 5.0 x 10-9 }
3rd approx
x3 = 1.27 x 10-4
4th approx
x4 = 1.29 x 10-4
5th approx
x5 = 1.29 x 10-4 M
Thus,
solubility = [Ba2+] = 1.29 x 10-4 M = 1.3 x 10-4 M
Step 9. Check of Approximations
To check the two approximations that were made, we must calculate the concentrations
of some of the other ions in the solution.
Here are the values I obtained:
[CO32-] = 3.9 x 10-5 M
[HCO3-] = 9.0 x 10-5 M
[H+]
= 1.1 x 10-10 M
[OH-]
= 9.0 x 10-5 M
[H2CO3] = 2.2 x 10-8 M
The hydrogen ion concentration is indeed small compared to the barium ion
concentration. 10-10 compared to 10-4
And the bicarbonate concentration is about 4000 times larger than carbonic acid
concentration, so this assumption was also good.