Chapter 13 EDTA titrations
Problems 1-6, 8, 13, 16, 18, 19, 20
Overhead of EDTA
Overhead of table 13-1
12-1 Metal Chelate Complexes
We are now going to talk about Metal Complexes. These are compounds in
which there is a Central Metal atom surrounded by Ligands. The central metal atom
is a Lewis acid (accepts electron pairs) and the ligands are Lewis bases (Donate Lewis
pairs). In this chapter we will focus on the chemistry of one particular ligand,
ethylenediaminetetraacetic acid (EDTA).
The strength of coordinate-covalent bond between the ligand and the metal ion
is somewhere between that of ionic bonds and covalent bonds. What happens to the
ionic bond between Na and Cl when you dissolve salt in water? It’s as if there were no
bond at all - the Na drifts off from the Cl. On the other hand, what happens to the
covalent bond between C and H in a methyl group in water? They stay together. So
Bonds between metal and their chelates may be strong, and or weak depending on the
metal and the ligand.
A metal ions can form complexes with 4, 6, or even 8 ligands at the same time.
On the other hand the ligand itself may have one place where it can bond to the metal
(Monodentate ) or more (multidentate).
EDTA the focus of this chapter is a multidentate ligand. That is, it can
coordinate to a central metal using up to 6 electron pairs (2 on N and 4 on COO- ) See
picture Figure 13-1.
We will focus on EDTA because it is an extremely useful analytical tool. It can
be bind to virtually any metal ion (See Table 13-1) . It is also a very well studied
system and can serve as a model for understanding other metal complexes (like Fe in
Hemoglobin or Mg in ATP.
What does the equation for complex formation look like? Its really quite simple
M + nL 6MLN
We measure this equilibrium using a Formation Constant Kf
Kf = [MLn]/[M][L]n
Now if we are dealing with a complex ion system that has several different states
lie ML1 and ML2 and ML3...ML6 you have lots of equilibria that are all linked together
and things get relatively ugly pretty quickly.
However for a multidentate like EDTA that forms a 1:1 complex it is quite simple
M + EDTA WM@EDTA
Kf = [M@EDTA]/[M][EDTA]
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The Chelate Effect
It is a general property of multi-dentate ligands like EDTA, that they will form a
1:1 complex with the metal ion, thus greatly simplifying the above equilibrium
equations. Further, their binding constants are much stronger than those of unidentate
ions. There are a several of reasons for this.
One reason that multidentate ions have strong interactions is that while an
individual interaction with the central metal may be weak, the total interaction ends up
multiplying all the small interaction together to become a large number.
K1=100
M + Ligand site 1
M + Ligand site 2
K2=100
M + Ligand site 3
K3=100
M + Ligand site 4
K4=100
M + Ligand site 5
K5=100
M + Ligand site 6
K6=100
M + Ligandall sites
K1*K2*K3*K4*K5*K6 =1012
A second reason would be cooperativity. Once the EDTA is bound at one site,
the metal and the EDTA are now physically tied together so the second site is
physically close and can bind almost instantly.
A third, more subtle argument effect is based on the entropy of the reaction.
Entropy is a measure of randomness, and thing that more ordered are generally less
favored than thing that are more random. If we have a complex between 1 metal and 6
monodentate ligands, you enforce order on 7 different compounds. If you have a
complex between a metal and a single 6-dentate ligand, you only enforce order on two
molecules, so this is much more favored.
13-2 EDTA
While there are a host of biologically relevant complexes we might study, we will
focus on complexes between EDTA and metals because the turn out to be excellent
complexes for quantifying metals in solutions.
Here is the Structure of EDTA Figure 13-1
Notice that EDTA has 4 COOH functionalities and 2 NH2 functionalities. Each of
these groups can complex with a metal, so this one molecule usually acts as a
multidentate molecules to form a 1:1 complex with metal thus our Formation Constant
Kf usually looks very simple.
Kf = [MEDTA]/[M][EDTA]
The Kf for most multivalent metals and EDTA is VERY large. 1010-1030 (See
table 13-1)
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5
However you should appreciate the fact that each of the functional groups is and
acid or a base. You can see that this compound is going to have some interesting
Acids-base properties , and you might wonder which of the acid/base forms of EDTA
the [EDTA] in the above equation refers to.
The various pKas are
pK1 0.0
1.5
2.0
2.69
6.13
10.37
Now you might expect that dealing with 1 Kf and 6 Ka’s might make thing pretty
complicated, but we can simplify the math very easily, so we don’t have to deal with
these complications. (Will look at in more details in section 13-5)
There are actually two different pH effects you have to worry about. AS you will
see, the binding constants are determined for the EDTA-4 form, because this is the form
that binds metals the best. This form only occurs when the pH of the solution is 11 or
so. As the pH gets lower, the % of EDTA in this form gets lower, so the binding of
metal and EDTA also gets lower
In the other direction you should remember from Gen Chen, that metal hydroxide
complexs are only slightly soluble, so they tend to precipitate out of solution. As the pH
of a solution goes up, the [OH] goes up, and this tends to make many metals ppt out
has hydroxides. Once the metal has ppt’s out, it is not in solution, so it cannot react,
and no chemistry can occur.
The effect of hydroxides can be ameliorated through the use of auxiliary
complexing agents. These are others anions that form metal complexes, like
ammonia, tartarate, citrate, or ethanolamine, however thee complexes are soluble, so
the metal stays in solution and can continue to react.
The use of auxiliary complexing agents is a little trick, because you have a
competition between the metal-complexing agent and Metal-EDTA complex. The Kf for
the metal-EDTA complex must be many orders of magnitude larger than the Kf for the
metal-auxiliary complex agent so that the EDTA can remove the metal from the
complexing agent.
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13-3 Metal Ion Indicators
The most common way to follow complexometric titration is either with a
potentiometer and electrode designed to sense the metal involved, or by using a metal
ion indicator.
Like Acid-Base indicators that were simply another acid or base, that had a color
change between forms, Metal Ion indicators are metal complexing agents that have one
when it binds a metal ion but it may have several different colors when it is not bound to
the metal. Why several different colors? All of the indicators have several acid base
functionalities and the ionization state of these functional groups affects the color of the
unbound indicator. Thus the color of the unbound indicator will vary with the pH of the
solution. This book shows only 2 indicators in Table 13-2, Calmagite and Xylenol
orange, but there are many others as seen in Figure 13-4. Note the different colors
and pKa’s in the first table, and the proper indicator for a given ion and pH in the other
table.
The choice of proper indicator at the right pH is tied to the Kass of both indicator
and EDTA. Since we are doing our titration in EDTA, K of the metal ion complex must
be large enough that the complex will form, but it must bind the metal more weakly than
EDTA. Why?
At the start of the titration the metal and the indicator form the color complex.
The titration proceeds until all the free metal is complexed with EDTA and only the
small amount complexed with the indicator is left. Now with one more drop of EDTA,
the EDTA must remove the metal from the indicator complex, so it changes back to its
free form and you can see the endpoint. If a metal and the indicator complex is too
strong the metal is said to block the indicator.
The color of the indicators is frequently a function of pH as well, so you have to
be careful of pH of the indicator as well as the EDTA.
13-4 EDTA Techniques.
EDTA titrations can be performed in many ways, let’s look at a few
Since several of these techniques are pretty specialized and you won’t see them
again I will just talk about the first and last, Direct titrations and Masking
Direct Titrations
In direct titrations you simply add an indicator to a solution of the metal ion and
titrate with EDTA. Before you start the titration yo need to check that the pH of the
solution gives a good Kf’ and that the pH is consistent with you indicator color change
as well. auxiliary complexing agents like ammonia, tartarate, or citrate may be added
to block formation of insoluble OH complexes
Back Titrations (Skip)
In a back titration an excess of EDTA is added to the metal ion solution, and the
excess EDTA is titrated with a known concentration of a second metal ion. The second
metal ion must form a weaker complex with EDTA than the analyte ion so the second
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metal does not displace the analyte ion from its complex with EDTA.
Back titration are used when the metal ion blocks the indicator (see above),
when the metal-EDTA complex forms too slowly, or when the metal precipitates in the
absence of EDTA.
Displacement Titrations (Skip)
For metal ions that do not have a good indicator a second titration method is the
displacement titration. Here the analyte is treated with an excess of a second metal
bound to EDTA. The analyte ion displaces the second metal from the EDTA complex,
and then the second metal is titrated with EDTA. A typical displacement titration
involves Hg2+ as the analyte and MgEDTA at the displacement titrant.
Indirect Titrations (Skip)
With a little clever thought EDTA can be used as a titrant for anions like SO42BaSO4 is insoluble to one way to determine SO42- is to precipitate with Ba2+, filter and
wash the ppt, then boil in excess EDTA to complex all the Ba. Back titrate to determine
how much Ba you had, and that, in turn, tells you how much SO42- you had.
Masking
I mentioned earlier that you can add auxiliary complexing agents to keep metal
ions in solution. You can also add masking agent that will bind so tightly to a metal
ion that it will not titrate with EDTA. These can be used to prevent other ions from
interfering in a given titration. For instance CN- (cyanide) will form strong complexs
with Zn2+, Hg2+ Co2+,Cu+, Ag+, Ni2+ Pd2+ Pt2+ Fe2+ and Fe3+, but not Mg2+,Ca2+, Mn2+ or
Pb2+ so you can titrate any into in the second set in the presence of an ion from the first
set by adding CN- to the solution (Note CN- is extremely toxic -don’t do this at home)
CN- is especially nice because you can demask it with formaldehyde
F- is an example of another masking agent, if you read some of the warnings in
the text you find that it too, is fairly nasty stuff, so we won’t use masking agents, per se
in the lab.
In the lab that you so, however, we do something like making. We will titration a
mixture that contains both Mg2+ and Ca2+. In our first titration the indicator won’t
change color until both metals are bound by the EDTA, so what we wil determine is the
total metal in the sample. In the second titration we will add more OH-. This additional
OH- makes the Mg2+ precipitate out of the solution as Mg(OH)2 so only Ca2+ remains in
solution to be titrated.
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13-5 The pH-Dependent Metal-EDTA Equilibrium
From Section 12-5 you will remember that we can calculate the fraction of any species
in solution using " equations. Since we have Kf’s for EDTA in the EDTA-4 form, will will
focus on the "Y-4 fraction where Y-4 is short for EDTA-4
If you think back to the previous chapter you should remember that the " value
for a particular ionization state is a function of the K’s and the [H+] of the solution for "-4
the equation is:
Don’t worry about memorizing this equation or making calculations on it. To save
people math errors and memorization, the values for "-4 have already been calculated
and tabulated in table 13-3 of your text. (Overhead)
Why are we worried about the "-4 value?? Because, as shown table 13-1, the K
of complex formation between metals and EDTA are usually given in terms of the
complex of the metal and the Y-4 form of EDTA. Yet, if you look at table 13-1, at any pH
less that 13, less than 100% of the EDTA in solution is in the Y-4 form.
So what do we do if we aren’t at pH 13 or above where EDTA is ion the Y-4 form
to match our table??
We calculate a
Conditional Formation Constant
If Kf = [Myn-4]/[M+n][Y-4]
and we know that [Y-4] = "y-4 F(EDTA)
Then we can combine these to get the equation
Kf = [Myn-4]/[M+n] "y-4 F(EDTA)
Then rearrange to get
Kf "y-4= [Myn-4]/[M+n]F(EDTA)
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This allows us to calculate the CONDITIONAL FORMATION Constant (Kf’) at any pH.
Let’s try an example. Say we wish to titrate Ca+2 with EDTA at pH 10, what is our
Conditional Formation Constant?
Log K f for the Ca EDTA complex is 10.65 according to table 13-1
K=10 10.65 = 4.47x10 10.
If pH=10 then "y-4 = 0.30 according to table 13-3
Therefore (K f’= .30(4.47x10 +10)
=1.34x10 +10
One general trend you should see is that as pH gets lower so does the Kf. At what point
does it get so low that we can no longer do the titration?? As a general rule of thumb
we use 108 as a cutoff. Thus Kf’ must be >=108 for an EDTA titration to work.
Let’s try another one. What is the lowest pH at which you can successfully titrate
+2
Fe ?
Kf’ = Kf x "Y4- $108
Kf = 1014.30 = 2.0x1014
2.0x1014 × "Y4- $ 108
"Y4-$108/2x1014
"Y4-$5x10-7
"Y4-=2.9x10-7 at pH 5, and this is just a touch low, so we could do the
titration at pH 6 or above
13-6 EDTA Titration Curves
Let’s see if we can do a titration curve for an EDTA titration
For an Acid-Base reaction our titration curve was pH vs ml of titrant. What shall
we use here?
pMetal vs ml of titrant
The curve has three region to worry about, before the EP, at the equiv. point and
after the EP.
Let’s carry on with the problem we started above, a titration of Ca with EDTA at pH 10,
since we have already calculated that Kf’=
1.7610 +10
Lets define the rest of the parameters, we are titrating a 50 ml solution of
0.01M Ca with 0.01 EDTA. Since EDTA and Ca react to form a 1:1 complex, you
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should immediately recognize that the equivalence point is at 50 mls
INITIAL POINT AND.5 ON TO EQUIVALENCE
Initial point, and points up to equivalence point are determined directly by
calculating free Ca in solution
1.
Initial point
[Ca]=0.01M, pCa = -log(.01); pCa=2
2. Another point This is easy let's pick the initial point and, say 30 mls
At 30 mls
What are the moles of EDTA and metal at this point?
Mole Ca 2+ = 50 ml x .01M = .5 mmoles
Mole EDTA = 30 mls x .01M = .3 mmoles
Reaction table:
Before reaction
Reaction
After reaction
Ca 2+
.5
-.3
.2
+ EDTA 6 Ca@EDTA
.3
0
-.3
+.3
0
.3
So [Ca 2+] .2 mmole/(50ml+30ml) = 2.5x10 -2; pCa2+ = 2.60
2.
Equivalence point
At the equivalence point we have
moles Ca 2+ = 50 ml x .01 M = .5 mmole
moles EDTA = 50 ml x .01 M = .5 mmole
Before reaction
Reaction
After reaction
Ca 2+
.5
-.5
0
+ EDTA 6 Ca@EDTA
.5
0
-.5
+.5
0
.5
And your first guess might be that [Ca 2+] = 0, pCa 2+ = 4
This is wrong, can you figure out why?
(K is not infinite, so there is always a little back reaction, so there is always
a little free Ca 2+.)
It is our task then, to use the K at this point to figure out how much free Ca 2+
is left in the solution from the back reaction.
Let’s look at the equilibrium calculation:
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From our reaction table we would expect
[Ca@EDTA] = .5mmole/(50ml+50ml) =.005M
[Ca]=[EDTA]=X;
Taking into account the back reaction
[CaEDTA]=.005-X
So
K eff=.005-X/X 2
Now, you could multiply through by X 2 and use the quadratic equation to
solve the above equation, but it is a waste of your time. With K so large, X is
going to be small compared to .005, so the -X term can be neglected. Thus:
K f’=.005/x 2; X 2=.005/K f’; X=sqrt(.005/1.76x10 +10)=sqrt(2.84x10 -13);
X=5.33x10 -7 ;pCa=6.27
Let's double check our assumption; is 5.33x10 -7 <<5x10 -3?
3. After the Equivalence point
After the equivalence point we continue with the same logic; since the only
Ca 2+ can come from the equilibrium, we need to use the equilibrium constant
equation. Let’s try the point 10 mL.
moles Ca 2+ = 50 ml x .01 M = .5 mmole
moles EDTA = 60 ml x .01 M = .6 mmole
Before reaction
Reaction
After reaction
Ca 2+
.5
-.5
0
+ EDTA 6 Ca@EDTA
.6
0
-.5
+.5
.1
.5
So our nominal concentrations are:
[Ca 2+] = 0
[EDTA] = .1/(50+60) = .1/110 = 9.1x10 -4M
[Ca@EDTA] = .5/(50+60) = .5/110 = 4.55x10 -3 M
Now let’s throw in the back reaction
[Ca 2+] = 0+x
[EDTA] = 9.1x10 -4M +X
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[Ca@EDTA] = 4.55x10 -3 M - x
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And finally go to the equilibrium equation:
K = [CaEDTA]/[EDTA][Ca]
K f’=4.55x10 -3-X/X(9.1x10 -3+X)
Again we assume that X is small (if it was small before, it will be even
smaller now) and simplify the equation and avoid the quadratic:
K f’=4.55x10 -3/9.1x10 -3(X);X=4.55/.91(1.76x10 +10)=2.84x10 -10
pCa=9.55
BEWARE
In the above treatment I have totally ignored any other equilibrium like Metal-OH
or Metal-Complexing agent. To calculate a true titration curve under these conditions
takes a bit more work and a few more pieces of scratch paper. We will have to skip
over this for this class.