Chapter 13 EDTA titrations

Problems 1-6, 8, 13, 16, 18, 19, 20Overhead of EDTAOverhead of table 13-1

12-1 Metal Chelate ComplexesWe 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 atomis a Lewis acid (accepts electron pairs) and the ligands are Lewis bases (Donate Lewispairs). 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 ionis somewhere between that of ionic bonds and covalent bonds. What happens to theionic bond between Na and Cl when you dissolve salt in water? It’s as if there were nobond at all - the Na drifts off from the Cl. On the other hand, what happens to thecovalent bond between C and H in a methyl group in water? They stay together. SoBonds between metal and their chelates may be strong, and or weak depending on themetal 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 cancoordinate to a central metal using up to 6 electron pairs (2 on N and 4 on COO ) - Seepicture Figure 13-1.

We will focus on EDTA because it is an extremely useful analytical tool. It canbe bind to virtually any metal ion (See Table 13-1) . It is also a very well studiedsystem and can serve as a model for understanding other metal complexes (like Fe inHemoglobin or Mg in ATP.

What does the equation for complex formation look like? Its really quite simple

N M + nL 6ML

fWe measure this equilibrium using a Formation Constant K

f nK = [ML ]/[M][L]n

Now if we are dealing with a complex ion system that has several different states

1 2 3 6lie ML and ML and ML ...ML you have lots of equilibria that are all linked togetherand 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

fK = [M@EDTA]/[M][EDTA]

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The Chelate EffectIt 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 equilibriumequations. Further, their binding constants are much stronger than those of unidentateions. There are a several of reasons for this.

One reason that multidentate ions have strong interactions is that while anindividual interaction with the central metal may be weak, the total interaction ends upmultiplying all the small interaction together to become a large number.

site 1M + Ligand 1K =100

site 2 2M + Ligand K =100

site 3 3M + Ligand K =100

site 4 4M + Ligand K =100

site 5 5M + Ligand K =100

site 6 6M + Ligand K =100

all sites 1 2 3 4 5 6M + Ligand K *K *K *K *K *K =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 isphysically 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 lessfavored than thing that are more random. If we have a complex between 1 metal and 6monodentate ligands, you enforce order on 7 different compounds. If you have acomplex between a metal and a single 6-dentate ligand, you only enforce order on twomolecules, so this is much more favored.

13-2 EDTAWhile 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 excellentcomplexes for quantifying metals in solutions.

Here is the Structure of EDTA Figure 13-1

2Notice that EDTA has 4 COOH functionalities and 2 NH functionalities. Each ofthese groups can complex with a metal, so this one molecule usually acts as amultidentate molecules to form a 1:1 complex with metal thus our Formation Constant

fK usually looks very simple.

fK = [MEDTA]/[M][EDTA]

fThe K for most multivalent metals and EDTA is VERY large. 10 -10 10 30 (Seetable 13-1)

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However you should appreciate the fact that each of the functional groups is andacid or a base. You can see that this compound is going to have some interestingAcids-base properties , and you might wonder which of the acid/base forms of EDTAthe [EDTA] in the above equation refers to.

a The various pK s arepK1 0.0

1.52.02.696.1310.37

f aNow you might expect that dealing with 1 K and 6 K ’s might make thing prettycomplicated, but we can simplify the math very easily, so we don’t have to deal withthese 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 willsee, the binding constants are determined for the EDTA form, because this is the form-4

that binds metals the best. This form only occurs when the pH of the solution is 11 orso. As the pH gets lower, the % of EDTA in this form gets lower, so the binding ofmetal and EDTA also gets lower

In the other direction you should remember from Gen Chen, that metal hydroxidecomplexs are only slightly soluble, so they tend to precipitate out of solution. As the pHof a solution goes up, the [OH] goes up, and this tends to make many metals ppt outhas 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 auxiliarycomplexing agents. These are others anions that form metal complexes, likeammonia, tartarate, citrate, or ethanolamine, however thee complexes are soluble, sothe metal stays in solution and can continue to react.

The use of auxiliary complexing agents is a little trick, because you have a

fcompetition between the metal-complexing agent and Metal-EDTA complex. The K for

fthe metal-EDTA complex must be many orders of magnitude larger than the K for themetal-auxiliary complex agent so that the EDTA can remove the metal from thecomplexing agent.

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13-3 Metal Ion IndicatorsThe most common way to follow complexometric titration is either with a

potentiometer and electrode designed to sense the metal involved, or by using a metalion indicator.

Like Acid-Base indicators that were simply another acid or base, that had a colorchange between forms, Metal Ion indicators are metal complexing agents that have onewhen it binds a metal ion but it may have several different colors when it is not bound tothe metal. Why several different colors? All of the indicators have several acid basefunctionalities and the ionization state of these functional groups affects the color of theunbound indicator. Thus the color of the unbound indicator will vary with the pH of thesolution. This book shows only 2 indicators in Table 13-2, Calmagite and Xylenolorange, but there are many others as seen in Figure 13-4. Note the different colors

aand pK ’s in the first table, and the proper indicator for a given ion and pH in the othertable.

assThe choice of proper indicator at the right pH is tied to the K of both indicatorand EDTA. Since we are doing our titration in EDTA, K of the metal ion complex mustbe large enough that the complex will form, but it must bind the metal more weakly thanEDTA. 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 thesmall 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 itsfree form and you can see the endpoint. If a metal and the indicator complex is toostrong the metal is said to block the indicator.

The color of the indicators is frequently a function of pH as well, so you have tobe 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 fewSince 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 TitrationsIn 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

fsolution gives a good K ’ and that the pH is consistent with you indicator color changeas well. auxiliary complexing agents like ammonia, tartarate, or citrate may be addedto 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 secondmetal 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 theabsence 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 metalbound 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 titrationinvolves Hg as the analyte and MgEDTA at the displacement titrant.2+

Indirect Titrations (Skip)

4With a little clever thought EDTA can be used as a titrant for anions like SO 2-

4 4BaSO is insoluble to one way to determine SO is to precipitate with Ba , filter and2- 2+

wash the ppt, then boil in excess EDTA to complex all the Ba. Back titrate to determine

4how much Ba you had, and that, in turn, tells you how much SO you had.2-

MaskingI 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 metalion that it will not titrate with EDTA. These can be used to prevent other ions frominterfering in a given titration. For instance CN (cyanide) will form strong complexs-

with Zn , Hg Co ,Cu , Ag , Ni Pd Pt Fe and Fe , but not Mg ,Ca , Mn or2+ 2+ 2+ + + 2+ 2+ 2+ 2+ 3+ 2+ 2+ 2+

Pb so you can titrate any into in the second set in the presence of an ion from the first2+

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 sein the lab.

In the lab that you so, however, we do something like making. We will titration amixture that contains both Mg and Ca . In our first titration the indicator won’t2+ 2+

change color until both metals are bound by the EDTA, so what we wil determine is thetotal metal in the sample. In the second titration we will add more OH . This additional-

2OH makes the Mg precipitate out of the solution as Mg(OH) so only Ca remains in- 2+ 2+

solution to be titrated.

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13-5 The pH-Dependent Metal-EDTA EquilibriumFrom Section 12-5 you will remember that we can calculate the fraction of any species

fin solution using " equations. Since we have K ’s for EDTA in the EDTA form, will will-4

Y-4focus on the " fraction where Y is short for EDTA-4 -4

If you think back to the previous chapter you should remember that the " value

-4for a particular ionization state is a function of the K’s and the [H ] of the solution for "+

the equation is:

Don’t worry about memorizing this equation or making calculations on it. To save

-4people math errors and memorization, the values for " have already been calculatedand tabulated in table 13-3 of your text. (Overhead)

-4Why are we worried about the " value?? Because, as shown table 13-1, the Kof complex formation between metals and EDTA are usually given in terms of thecomplex of the metal and the Y form of EDTA. Yet, if you look at table 13-1, at any pH-4

less that 13, less than 100% of the EDTA in solution is in the Y form.-4

So what do we do if we aren’t at pH 13 or above where EDTA is ion the Y form-4

to match our table??We calculate a

Conditional Formation Constant

fIf K = [My ]/[M ][Y ]n-4 +n -4

y-4and we know that [Y ] = " F(EDTA)-4

Then we can combine these to get the equation

f y-4K = [My ]/[M ] " F(EDTA)n-4 +n

Then rearrange to get

f y-4K " = [My ]/[M ]F(EDTA)n-4 +n

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fThis allows us to calculate the CONDITIONAL FORMATION Constant (K ’) at any pH.

Let’s try an example. Say we wish to titrate Ca with EDTA at pH 10, what is our+2

Conditional Formation Constant?

fLog K for the Ca EDTA complex is 10.65 according to table 13-1K=10 = 4.47x10 .10.65 10

y-4If pH=10 then " = 0.30 according to table 13-3

fTherefore (K ’= .30(4.47x10 )+10

=1.34x10+10

fOne general trend you should see is that as pH gets lower so does the K . At what pointdoes it get so low that we can no longer do the titration?? As a general rule of thumb

fwe use 10 as a cutoff. Thus K ’ must be >=10 for an EDTA titration to work. 8 8

Let’s try another one. What is the lowest pH at which you can successfully titrateFe ?+2

f f Y4- K ’ = K x " $108

fK = 10 = 2.0x1014.30 14

Y4-2.0x10 × " $ 1014 8

Y4-" $10 /2x108 14

Y4-" $5x10-7

Y4-" =2.9x10 at pH 5, and this is just a touch low, so we could do the-7

titration at pH 6 or above

13-6 EDTA Titration CurvesLet’s see if we can do a titration curve for an EDTA titrationFor an Acid-Base reaction our titration curve was pH vs ml of titrant. What shall

we use here?

pMetal vs ml of titrantThe 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,

fsince we have already calculated that K ’= 1.7610+10

Lets define the rest of the parameters, we are titrating a 50 ml solution of0.01M 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 EQUIVALENCEInitial point, and points up to equivalence point are determined directly by

calculating free Ca in solution1. 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 mlsAt 30 mls

What are the moles of EDTA and metal at this point?Mole Ca = 50 ml x .01M = .5 mmoles2+

Mole EDTA = 30 mls x .01M = .3 mmoles

Reaction table:Ca + EDTA 6 Ca@EDTA2+

Before reaction .5 .3 0Reaction -.3 -.3 +.3After reaction .2 0 .3

So [Ca ] .2 mmole/(50ml+30ml) = 2.5x10 ; pCa = 2.602+ -2 2+

2. Equivalence pointAt the equivalence point we have

moles Ca = 50 ml x .01 M = .5 mmole2+

moles EDTA = 50 ml x .01 M = .5 mmole

Ca + EDTA 6 Ca@EDTA2+

Before reaction .5 .5 0Reaction -.5 -.5 +.5After reaction 0 0 .5

And your first guess might be that [Ca ] = 0, pCa = 42+ 2+

This is wrong, can you figure out why?

(K is not infinite, so there is always a little back reaction, so there is alwaysa little free Ca .)2+

It is our task then, to use the K at this point to figure out how much free Ca2+

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

effSo K =.005-X/X2

Now, you could multiply through by X and use the quadratic equation to2

solve the above equation, but it is a waste of your time. With K so large, X isgoing to be small compared to .005, so the -X term can be neglected. Thus:

f fK ’=.005/x ; X =.005/K ’; X=sqrt(.005/1.76x10 )=sqrt(2.84x10 );2 2 +10 -13

X=5.33x10 ;pCa=6.27-7

Let's double check our assumption; is 5.33x10 <<5x10 ?-7 -3

3. After the Equivalence pointAfter the equivalence point we continue with the same logic; since the only

Ca can come from the equilibrium, we need to use the equilibrium constant2+

equation. Let’s try the point 10 mL.

moles Ca = 50 ml x .01 M = .5 mmole2+

moles EDTA = 60 ml x .01 M = .6 mmole

Ca + EDTA 6 Ca@EDTA2+

Before reaction .5 .6 0Reaction -.5 -.5 +.5After reaction 0 .1 .5

So our nominal concentrations are:[Ca ] = 02+

[EDTA] = .1/(50+60) = .1/110 = 9.1x10 M-4

[Ca@EDTA] = .5/(50+60) = .5/110 = 4.55x10 M-3

Now let’s throw in the back reaction[Ca ] = 0+x2+

[EDTA] = 9.1x10 M +X-4

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[Ca@EDTA] = 4.55x10 M - x-3

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And finally go to the equilibrium equation:K = [CaEDTA]/[EDTA][Ca]

fK ’=4.55x10 -X/X(9.1x10 +X)-3 -3

Again we assume that X is small (if it was small before, it will be evensmaller now) and simplify the equation and avoid the quadratic:

fK ’=4.55x10 /9.1x10 (X);X=4.55/.91(1.76x10 )=2.84x10-3 -3 +10 -10

pCa=9.55

BEWAREIn 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 conditionstakes a bit more work and a few more pieces of scratch paper. We will have to skipover this for this class.