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Environmental Analysis
Analytical Chemistry Lecture 2
Oct. 2, 2002
Environmental Chemistry involves many chemical species that are soluble in water as ions.
If you encounter some you are not familiar with be sure and look them up and learn their
properties.
For example Jim said carbonate and expect that you knew this is CO32-, an ion with two
negative charges. There are only a small number of common inorganic ions so be sure you
review these and learn their formulae and charge. Go over these in your study group.
Now let’s turn to the topic of Equilibrium already introduced by Jim. You are going to
find that topics in this program come up over and over and over again. Jim talked about
least-squares regression yesterday. We will encounter it again in Chapter 5 of Harris and
again in the statistics coverage during winter quarter. You will use it all year long.
As I said Monday that chemical equilibrium plays a major role in analytical chemistry.
Many of the separation steps in analytic procedures remove interferents by establishing
situtations where the interference reacts and the analyte does not, or visa versa.
In order to determine the amount of an analyte involved in a chemical reaction, we must
be sure that essentially all of the substance present reacts.
However, the reactions used in analytical chemistry never result in complete conversion
of reactants to products. Instead, they proceed to a state of chemical equilibrium where
the concentrations of reactants and products remain constant.
Equilibrium-constant expressions are algebraic equations that describe the
relationships that exist among reactants and products at equilibrium.
Among other things, equilibrium equations permit calculation of the error resulting from
the quantity of unreacted analyte that remains when a steady state have been reached.
Suppose we have a chemical reaction involving the precipitation of CaCO3.
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We write the chemical reaction as
Ca2+(aq) + CO32-(aq) < == > CaCO3(s)
Minerals are? Calcite and aragonite
Rocks? Limestone and marble
When calcium ions and carbonate ions are mixed in solution calcium carbonate may
precipitate.
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Here we see that at the start of the reaction the mass of Ca2+ decreases and the mass of
CaCO3 increases. Eventually, however, the reaction reaches a point after which no
further changes occur in the amounts of these species. This is the condition of
equilibrium.
Although a system at equilibrium appears static on a macroscopic level, it is important to
remember that the forward and reverse reactions still occur. A reaction at equilibrium
exists in a “steady state” in which the forward and reverse reaction rates are equal.
The time required for equilibrium to be attained varies from one reaction to another. The
time may be anything from a fraction of a second to many days to many years.
Most of the reactions that we will consider in analytical chemistry reach equilibrium
within a few minutes. However many chemical reactions of geological interests take
long time intervals to come to equilibrium.
The book points out that there is a relationship between equilibrium and thermodynamics.
Suppose we have a generalized equilibrium reaction that involves the solutes A, B, C,
and with stoichiometric coefficients a, b, c, and d.
a A + b B < == > c C + d D
We say a moles of A react with b moles of B to produce c moles of C and d moles of D.
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You all remember.
By convention, species to the left of the arrows are called reactants, and those on the
right side are called products. Simply writing a reaction in this fashion does not
guarantee that the reaction of A and B to produce C and D is favorable.
Depending on the initial conditions, the reaction may move to the left, to the right or be
in a state of equilibrium and not change at all.
Chemical systems spontaneously react in a fashion that lowers their overall free energy.
At a constant temperature and pressure, conditions typical to the laboratory and a good
approximation to the conditions on the earth’s surface, the free energy of a chemical
reaction is given by the Gibb’s free energy function.
Stability in chemistry involves reactions that give off free energy.
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G = H - TS
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where T is the temperature in kelvins, and G, H, and S are the differences in the
Gibb’s free energy, the enthalpy and the entropy between the products and reactants.
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Enthalpy is a measure of the net flow of energy, as heat, during a chemical reaction.
Reactions in which heat is produced have a negative H and are called exothermic.
Endothermic reactions absorb heat from their surroundings and have a positive H.
Entropy is a measure of randomness or disorder.
The sign of G can be used to predict the direction in which a reaction moves to reach its
equilibrium position.
As a system moves from a nonequilibrium to an equilibrium position, G must change
from its initial value to zero. When G is zero we are at a point of equilibrium.
At the same time, the species involved in the reaction undergo a change in their
concentrations. The Gibb’s free energy change, therefore, iselated to the concentrations
of the reactants and products.
The Gibb’s free energy can be divided into two terms.
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G = Go + RT ln Q
The first term, Go, is the change in Gibb’s free energy under standard state conditions
which are defined as
Temperature 298K
All gases with partial pressures of 1 atm.
All solids and liquids pure
All solutes present in 1 M concentrations.
The second term, which includes the reaction quotient Q, accounts for nonstandard state
concentrations and pressures.
For the general reaction we started with
Q = [C]c [D]d
[A]a [B]b
where the terms in brackets are the molar concentrations of the solutes. Tis is products
over reactants and that each concentration terms is raised to the power equal to its
stoichiometric coefficient in the balanced chemical equation.
Note Jim would write activities here instead of concentrations. Activities are more
correct but we will assume concentration = activity at least until we complete chapter 8.
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Partial pressures in bars are substituted for concentrations when the reactant or product is
a gas. Also the concentrations of pure solids and pure liquids do not change during a
reaction and are excluded from the reaction quotient. Page 13 has pressure units
1 bar = 105 Pa
1 atm = 101325 Pa
1 torr = 133.3 Pa
760 torr = 1 atm
At equilibrium the Gibb’s free energy is zero so the equation simplifies
0 = Go + RT ln Q

Go = - RT ln Qeq = - RT ln Keq
where K is called the equilibrium constant and that defines the reaction’s equilibrium
position and R is the gas constant 8.3145 J/K mol. It is important to remember that the
value of K is determined by the concentrations of solutes at equilibrium.
The last equation can be rearrange to give
Keq = e-G/RT
So we can calculate equilibrium constants from tables of thermodynamic data the give
G for reactants and products.
The equilibrium constant for a given reaction has a specific value at a stated temperature.
Unless otherwise indicated, the value given for an equilibrium constant in a table is for
equilibrium at room temperature, 25oC and 1 atmosphere pressure. We have tables of
equilibrium constants in Appendices F and G in Harris.
It is also true that these equations are idealized and are actually true only for infinitely
dilute solutions where activity coefficients are 1. We will correct for real systems soon
enough.
If you think about it, the reaction quotient Q can be used to test whether a chemical
system is at equilibrium and, if not, in which direction equilibrium lies.
If Q is smaller than Keq then the concentration of reactants compared to products must be
too high and the reaction must continue from left to right. If Q is larger then the reaction
goes from right to left, and if Q = K then the systems is at equilibrium.
The equilibrium point of a chemical reaction may be shifted by varying the
concentrations of reactants or products. If the concentration of A or B is increased or the
concentration of C or D is decreased, the equilibrium point is shifted farther to the right.
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This situation is covered by le Chatelier’s Principle, which states that when a stress is
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applied to a chemical system at equilibrium, the system acts so as to counteract the stress.
In particular if a concentration is increased, the reaction goes in the direction that
decreases the concentration. And if a concentration is decreased the reaction goes in the
direction that produces that substance.
This is a very useful principle because it helps you predict what will happen when
conditions change in a system.
We are going to encounter equilibria of the following types:
Solubility equilibrium - extremely important to groundwater chemistry as it can describe
the concentrations of ions in water when minerals dissolve or precipitate.
Acid/base equilibrium - important to groundwater chemistry, particularly in its
relationship to minerals containing carbonates and silicates has major effects on
oxidation/reduction reactions.
Complex ion equilibria are important in geology and analytical chemistry as we will see.
Redox equilibria involved in oxidation/reduction reactions.
Finally the water ionization reaction HOH < == > H+ + OH- is important because so
many chemical analyses are carried out in water.
Now lets add a few rules for equilibrium constants:
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When a stoichiometric equation is reversed, the equilibrium constant for the reversed
reaction is the reciprocal of that for the forward reaction.
Kw = [H+][OH-] but for
1
.
[H+][OH-]
=
H+ + OH- < == > HOH
1
Kw
When stoichiometric equations are added to give an overall equilibrium equation, the
associated equilibrium constants must be multiplied together to obtain the equilibrium
constant for the overall equation.
H2CO3 < ====>
H+ + HCO3-
HCO3- < ====> H+ + CO32-
K1 = [H+][HCO3-]
[H2CO3]
K2 = [H+][CO32-]
[HCO3-]
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+
H2CO3 < ====> 2 H + CO3
2-
Ka = [H ] [CO3 ] = K1∙ K2
[H2CO3]
+ 2
2-
Most equilibrium expressions relating to real systems can be evaluated with the
operations of addition and reversal of chemical equations and with the appropriate
manipulation of the equilibrium expressions.
You must keep in mind, however, that an equilibrium-constant expression yields no
information concerning the rate at which equilibrium is approached.
In fact, we sometimes encounter reactions that have highly favorable equilibrium
constants but are of little analytical use because their rates are low. Sometimes this
limitation can be overcome by use of a catalyst.
C6H12O6(s) + 6 O2(g) < ==== > 6 CO2(g) + 6 H2O (l)
G = - 2,870 kJ/mol
But reaction is very slow without catalyst, such as a hungry EA student.
Mercury(I) or mercurous chloride has the formula Hg2Cl2. The chemical equation for
mercurous chloride dissolving in water is
Hg2Cl2(s) < === > Hg22+ + 2 ClThis is a dissociation reaction.
The mercurous cation is a dimer as discussed by Harris in chapter 6 and so consists of
two mercury ions with a total of +2 charge. Thus each atom of mercury is in the +1
oxidation state.
In a saturated aqueous solution of mercurous chloride the above equation describes an
equilibrium condition. What does saturated mean?
We can write an equilibrium equation
K = [Hg22+] [Cl- ]2
[Hg2Cl2(s)]
But the concentration of the solid in the solid doesn’t change, so we rewrite this as
Ksp = [Hg22+] [Cl-]2 = 1.2 x 10-18
where the new constant is called the solubility-product constant or simply the solubility 7
product. It is important to remember that this equation applies only to equilibrium/
saturated solutions and that they are written as dissociations.
Remember that the square-bracketed, concentration terms are molar concentration if the
species is a dissolved solute partial pressure in bars if the species is a gas.
The physical meaning of the solubility product is that if an aqueous solution is left in
contact with excess solid [Hg2Cl2(s)] , the solid will dissolve until the Ksp equilibrium is
achieved.
After that, the amount of undissolved solid [Hg2Cl2(s)] will remain constant. Unless
excess solid remains, there is no guarantee that equilibrium has been obtained.
Also, if Hg22+ and Cl- are mixed together such that the product [Hg22+][Cl-]2 exceeds Ksp,
then Hg2Cl2 will precipitate.
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What will be the concentration of mercurous ions in a solution saturated with Hg2Cl2(s)?
We see from the equation that two chloride ions are produced for every mercurous dimer
ion produced. We can construct a concentration table to display this information:
Hg2Cl2(s) < === > Hg22+ + 2 ClInitial concentration
Final concentration
solid
solid
0
x
0
2x
Let x = [Hg22+]
So
[Hg22+] [Cl-]2 = x(2x)2 = Ksp = 1.2 x 10-18
4x3
= 1.2 x 10-18
x = [Hg22+] = [(1.2 x 10-18)/4]1/3
(to find the cube root of a number, raise the
number to the 0.333333333 power.)
[Hg22+] = 6.7 x 10-7 M
The next step in developing solubility is to point out what happens when there is a
common ion present.
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What will be the concentration of Hg22+ in a solution containing 0.030 M NaCl and
saturated with Hg2Cl2?
First tell me what Le Chateleir's principle would say about this change in conditions?
This is called the common ion effect – A salt will be less soluble if one of its constituent
ions is already present in the solution.
Let us show this.
The concentration table will now be:
Initial concentration
Final concentration
Hg2Cl2(s) < === > Hg22+ + 2 Clsolid
0
0.030
solid
x
2x + 0.030
[Hg22+] [Cl-]2 = x(2x + 0.030)2 = Ksp = 1.2 x 10-18
Think about the size of x for a minute. In pure water it was less than 10-6 M and it is
going to be even smaller. In particular 2x is vanishingly small compared to 0.030. So we
can ignore the 2x in comparison with 0.030 and solve.
x (0.030)2 = 1.2 x 10-18
x = 1.3 x 10-15
We can see that x is indeed small compared to 0.030. There is a section in your book on
the logic of approximations. Read it carefully because knowing when to make
approximations and how to test if the approximations are good, can save you a lot of
time. [just watch out for the paranhas in the lake.]
Notice that the common ion decreased the solubility by a factor of 5 x 108.
If we can get away with it (not be affected by another equilibrium reaction) then we can
remove more of an ion from solution and force it into the solid form that can be weighed.
We get better assay when more of the analyte is in the precipitate.
Homework due next Wednesday. Must be legible.
Solutions Manual is available in the library.
In you have questions you want to ask, write in margins or at top or bottom of page.
Think about answers and questions.
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