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Analytical Chemistry
Lecture W5
Feb. 05, 2003
Well if you waded through Chapter 13, you saw we are back to equilibrium equations and
calculations.
Most metal ions react with electron-pair donors to form coordination compounds or
coordination complexes.
Back in chapter six we talked about acids and bases as proton donors and acceptors. We
also mentioned the concept of Lewis acids and bases. A Lewis acid accepts a pair of
electrons while a Lewis base donates pairs of electrons. The bond between a Lewis acid
and base is called a coordinate covalent bond and is the basis for complex-ion formation.
The transition elements of the periodic table are all metals, i.e. form cations and many
have empty d orbitals that can accept pairs of electrons from Lewis bases. Because many
of these ions are +2 or +3 they have a high positive charge and are very attractive to the
negative charge of electron pairs.
The donor species is called a ligand and must have at least one pair of unshared electrons
available for coordinate covalent bond formation. Water, ammonia, and halide ions are
common inorganic ligands.
The number of covalent bonds that a cation tends to form with electron donors is its
coordination number. Typical values for coordination numbers are two, four, and six.
The complex formed as a result of coordination can be electrically positive, neutral, or
negative.
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For example, copper(II), which has a coordination number of four, forms a cationic
ammine complex, Cu(NH3)42+; a neutral complex with glycine, Cu(gly)2; and an anionic
complex with chloride ion, CuCl42-.
Complex formation between ligands and metal ions impacts analytical chemistry is
several ways.
First, the anion in a precipitate may form soluble complexes with a metal. This means
that if you try to use the common ion effect to suppress the concentration of cation in
solution, you may redissolve the ppt.
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For example an important method for analyzing silver is by gravimetric analysis (next
week) Precipitates all the silver as AgCl.
AgCl(s) < == > Ag+ + Cl-
Ksp = 1.8 x 10-10
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if we add excess chloride, Le Chatlier's principle tells us that the silver ion concentration
will be further reduced. We want to get all the Ag into the ppt.
But AgCl(s) + Cl- < == > AgCl2-(aq)
K = 3.3 x 10-5
If we add too much chloride we will start to redissolve the ppt in the formation of this
soluble coordination complex.
How do we determine when enough chloride is enough and not too much? By carrying
out the calculations and determining which is the optimal concentration. A good problem
for FREQC.
DEMO
Here is another example. I have in here a solution of copper(II) sulfate – note that it is a
pale blue. If I add base to this solution I get a ppt of the light blue solid copper(II)
hydroxide.
Now if I continue adding base the ppt starts to dissolves and the solution becomes dark
blue. With continued addition, I get a clear dark blue solution. What is the equilibrium
involved?
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CuSO4(s) + 4 H2O == > Cu(H2O)42+ + SO42most of the time we ignore the metal aqua complexes and just write Cu2+
Cu2+ + 2 OH- < == > Cu(OH)2(s)
This solid has copper cations and hydroxide anions. The copper cation is not blue.
Cu2+ + 4 NH3 < == > Cu(NH3)42+
Here we see two roles of complex ions. First if we are to assay copper by gravimetric
hydroxide formation we do not want ammonia in the solution. Complex formation would
be detrimental to analysis because it keeps some of the copper in solution.
Second we could mask the presence of copper by complexing it with ammonia.
The book mentions that aluminum ion is often masked by adding fluoride ion and
forming the AlF63- complex. Complex ions can keep a metal in solution so that reactions
of other analyts can take place.
The third and most important use of complexes in analytical chemistry involves titration
methods based on complex formation (sometimes called complexometric methods).
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Appications of complex ions to analytical chemistry is based primarily on coordination
compounds called chelates. A chelate is a claw. Chelates grab onto cations.
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The copper complex of glycine, is an example. Here, copper bonds to both the oxygen of
the carboxyl group and the nitrogen of the amine group. The copper ion is +2 and each
of the glycine carboxylic acid groups is -1 giving an overall neutral complex.
NH2 O
2+
Cu
+
H C
C OH
H
O
O
C O
H2C N
H2
Cu
O C
+ 2 H+
N C H2
H2
A ligand that has a single donor group, such as ammonia, is called unidentate
(single-toothed), whereas one such as glycine, which has two groups available for
covalent bonding, is called bidentate.
Tridentate, tetradentate, pentadentate, and hexadentate chelating agents are also
known.
As titrants, multidentate ligands, particularly those having four or six donor groups, have
advantages over their unidentate counterparts. They form 1:1 complexes and generally
react more completely with cations and thus provide sharper endpoints.
Fe3+ + L4- < == > FeLFe3+ + Cl- < == > FeCl2+, FeCl2+, FeCl30, FeCl4XXXXXX
Look at this graph representing titrations with a unidentate, a bidentate and a tetradentate
ligand.
First notice the y axis. What does pM mean? Negative log of the M concentration.
As volume of titrant is added, what happens to the concentration of M? Goes down.
Now which curve represents the better titration? Curve A, the titration with the
tetradentate ligand. Why? We want the equivalence/end point to be as sharp as possible.
These curves demonstrate that a much sharper end point is obtained with a reaction that
takes place in a single step.
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Ethylenediaminetetraacetic acid, which is commonly shortened to EDTA, is the most
widely used complexometric titrant. EDTA has the structure
HOOC-CH2
CH2-COOH
\
/
:N-CH2-CH2-N:
/
\
HOOC-CH2
CH2-COOH
The EDTA molecule has six potential sites for bonding a metal ion: the four carboxyl
groups and the two amino groups, each of the latter with an unshared pair of electrons.
Thus EDTA can be a tetra- or hexadentate ligand.
But the advantage of EDTA as a chelating agent is that we can mask the binding sites.
By changing the pH of a solution we can change carboxylic acid groups, which are weak
binders to carboxlate anion groups which are strong binders.
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Thus in solution EDTA is a conjugate acid with 6 different acid dissociations. Here you
can see five of the seven ionic forms of EDTA. As we go from acidic to basic solutions
four carboxylic acid hydrogens are lost and then the two amine hydrogens.
It is the Y4- form that is the strongly binding form.
Solutions of EDTA are particularly valuable as titrants because the reagent combines
with metal ions in a 1:1 ratio regardless of the charge on the cation. For example,
formation of the silver and aluminum complexes is described by the equations:
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Ag+ + Y4- < == > AgY3Al3+ + Y4- < == > AlYEDTA is a remarkable reagent not only because it forms chelates with nearly every cation
of every metal, but also because most of these chelates are sufficiently stable to form the
basis for a titrimetric method.
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One form of the complex is depicted on this transparency. Note that six donor atoms are
involved in bonding the metal ion and that in the cage-like structure, the cation is
effectively surrounded and isolated from solvent molecules.
The equilibrium for complex ions is summarized as formation constants. This involves
the ligand and cation as reactants that form the complex as the product.
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Mn+ + Y4- < == > MY(n-4)+
n+
Kf
= [MY(n-4)+]
4-
[M ][Y ]
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This transparency lists the formation constants KMY for common EDTA complexes. Note
that the constant refers to the equilibrium involving the deprotonated species Y4- with the
metal ion.
Table 13-2 in Harris on p.264 gives formation constant values for 63 cations.
You can see that all the numbers are very large. This means that equilibrium lies far to
the right in the equation. So EDTA binds tightly and does not let go.
The problem with these numbers is that, since EDTA is a multiprotic acid, both the
concentration of MY(n-4)+ and Y4- are pH dependent as we have noted.
Fortunately, EDTA titrations are always performed in solutions that are acid/base
buffered to a known pH to avoid interference by other cations and to ensure satisfactory
ligand behavior, i.e. that the ligand stays in one ionic form.
Calculating [Mn+] in a buffered solution containing EDTA is a relatively straightforward
procedure provided the pH is known. The secret to simplifying the equilibrium
calculations is in determining the fraction of the EDTA that is in the Y4- form.
Here is where all the calculations in the chapter come in.
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The quantity that is used is the fraction of the total uncomplexed EDTA ( cT or [EDTA])
in the Y4- form. This is called the alpha value or the free EDTA. [Note that this alpha is
different from activity which is sometimes given the symbol alpha as well.]
4 = [Y4-] =
[EDTA]
.
[Y4-]
.
43[Y ] + [HY ] + [H2Y2-] + [H3Y-] + [H4Y] + [H5Y+] + [H6Y2+]
Notice that we can also define  functions for each of the other 6 forms of EDTA.
The point is that the value of each of the concentration terms is dependent on [H+] or the
pH, and therefore the value of 4 is thus fixed when the pH is fixed.
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This transparency shows the log  term for each of the EDTA forms as a function of pH.
Notice that it is just at very high pH that the Y4- form becomes important.
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We can see the relative concentration of Y4- in this table. Now how do we take
advantage of the pH dependence of the Y4- concentration?
If we have a metal ion which a very strong formation constant for its EDTA complex,
then it will react completely even if the Y4- concentration is low, i.e. we can use a lower
pH for its titration. A metal ion with a weaker formation constant will require high Y4concentration which means higher pH.
Thus by varying the pH we can selectively react EDTA with different analytes.
The way these calculations are usually applied is through conditional or effective
formation constants as described in the chapter.
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since [Y4-] = 4 [EDTA]
where [EDTA] is the concentration of all the EDTA (also called cT) in solution not bound
to metal ion.
the equilibrium constant can be rewritten as
Kf
= [MY(n-4)+]
[Mn+][Y4-]
=
[MY(n-4)+]
[Mn+] 4 [EDTA]
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and now we define a conditional formation constant
Kf' = Kf4 =
[MY(n-4)+]
[Mn+][EDTA]
The conditional formation constant allows us to look at EDTA complex formation as
using the measurable quantity of total EDTA concentration
Mn+ + EDTA < == > MYn-4
and we can calculate the value of K' from the value of 4 at the pH of the experiment.
As an example let’s look at a titration of calcium ion with EDTA.
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Ca++ + Y4- < == > CaY2-
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Suppose we want to calculate the concentration of calcium at various points during the
titration of 50.0 mL of 0.00500 M Ca2+with 0.0100 M EDTA in a solution buffered to a
constant pH of 10.0.
What we want is points on the titration curve of pCa as a function of volume of EDTA
added.
First we calculated the conditional formation constant for the calcium/EDTA complex at
pH 10.0. The value for Kf for calcium is given in Table 13-2 as log Kf = 10.69 and the 4
value for EDTA at pH 10.0 is 0.36 as given in Table 13-1.
K'CaY = 4 Kf = 0.36 x 1010.69 = 1.76 x 1010
Now to calculate point along the titration curve we break the curve up into three different
regions.
Before the equivalence point is reached, the equilibrium concentration of Ca2+ is equal to
the sum of the contributions from the untitrated excess calcium still in solution and any
cation due to the dissociation of the complex.
CaY2- < == > Ca2+ + EDTA
The dissociation contribution is thus equal to [EDTA] or cT.
It is ordinarily reasonable to assume that cT is small relative to the analytical
concentration of the uncomplexed calcium ion.
Now look at the point in the titration when we have added 10.0 mL of reagent,
we started out with 50.0 mL x 0.00500 mmol/mL = 0.250 mmol Ca2+ and have reacted
with
10.0 mL x 0.0100 mmol/mL EDTA = 0.100 mmol EDTA.
Since 1 mol Ca reacts with 1 mol EDTA there should be left
0.250 – 0.100 mmol Ca/60.0 mL of solution. [Ca2+] = 2.50 x 10-3 M;
pCa = 2.60
Other preequivalence-point data are derived in this same way. This approach works fine
until we get very close to the equivalence point.
It is generally easy to calculate the value of pCa at the equivalence point itself.
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First we assume that all of the calcium has reacted with EDTA which is our definition of
an equivalence point. Thus we now have only CaY2- in the solution.
Since the EDTA solution is twice as concentrated as the Ca solution it will take ½ the
volume of titrant as the original volume of Ca. That is 50 mL of 0.00500 M will react
completely with 25 mL of 0.0100 M EDTA.
[CaY2-] = 50.0 x 0.00500 = 3.33 x 10-3 M
50.0 + 25.0
The only source of Ca2+ ions is the dissociation of this complex. It also follows that the
calcium ion concentration must be identical to the sum of the concentrations of the
uncomplexed EDTA ions.
CaY2- < == > Ca2+ + EDTA
.00333
0
0
.00333 –x
x
x
before eq
after eq
Let x = [Ca2+] = [EDTA]
[CaY2-] = 3.33 x 10-3 M – x
 0.00333 M
Substituting into the conditional formation-constant expression gives
[CaY2-]
= 0.00333 M
= 1.76 x 1010
2
[Ca ][EDTA]
x
_________________
x =  (0.00333/1.76 x 1010) = 4.35 x 10-7 M
K'f =
2+
was our assumption good?
pCa = log (4.35 x 10-7) = 6.36
Now lets look at the last or postequivalence-point region.
Beyond the equivalence point, the analytical concentrations of CaY2- and EDTA are
obtained directly from the stoichiometry.
Let's look at the point of adding 35.0 mL of EDTA titrant.
[CaY2-] = (50.0 x 0.00500) mmol = 2.94 x 10-3 M - [Ca2+] (assume small)
(50.0 + 35.0) mL
[EDTA] = 35.0 mL x 0.0100 mmol/mL - (50.0 x 0.00500) mmol = 1.18 x 10-3 M + x
(35.0 + 50.0 )mL
assume small
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before eq
after eq
K'f =
CaY2- < == > Ca2+ + EDTA
.00294
0
0.00118
.00294 –x
x
0.00118 + x
[CaY2-]
=
2+
[Ca ][EDTA]
0.00294 M
x (0.00118)
= 1.76 x 1010
x = [Ca2+] = 1.42 x 10-10
pCa = - log (1.42 x 10-10) = 9.85
You can see we have used the same principles to calculate EDTA titrations points that we
used with acid/base titrations.
Before we discuss these calcium results, lets look at some points about EDTA titration
curves.
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This transparency shows titration curves for calcium ion in solutions buffered to various
pH levels. Remember 4 and thus the conditional formation constant become smaller as
the pH decreases.
The less favorable equilibrium constant leads to a smaller change in pCa in the
equivalence-point region. It is apparent from the figure that an adequate end point in the
titration of calcium requires a pH of about 8 or greater.
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This transparency shows, however, that cations with larger formation constants provide
good end points even in acidic media of pH 6.
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Finally this transparency shows the minimum permissible pH for a satisfactory end point
in the titration of various metal ions in the absence of competing complexing agents.
Note that a moderately acidic environment is satisfactory for many divalent heavy-metal
cations and that a strongly acidic medium can be tolerated in the titration of such ions as
iron(III) and indium(III).
Also note the location of Ca2+ and Mg2+ on the curve.