Chapter 13 EDTA titrations

Transcription

1 Chapter 13 EDTA titrations Problems 1-6, 8, 13, 16, 18, 19, 20 Overhead of EDTA Overhead of table 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 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 ML N We measure this equilibrium using a Formation Constant K f K = [ML ]/[M][L] n f n Now if we are dealing with a complex ion system that has several different states lie ML 1 and ML 2 and ML 3...ML 6 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 M EDTA K f = [M EDTA]/[M][EDTA]

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3 3 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. M + Ligand K site 1 1=100 M + Ligand site 2 K 2=100 M + Ligand site 3 K 3=100 M + Ligand site 4 K 4=100 M + Ligand site 5 K 5=100 M + Ligand K =100 site 6 6 M + Ligandall sites K 1*K 2*K 3*K 4*K 5*K 6 =10 12 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 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 NH 2 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 K usually looks very simple. f K f = [MEDTA]/[M][EDTA] The K f for most multivalent metals and EDTA is VERY large table 13-1) (See

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5 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 pk Now you might expect that dealing with 1 K f and 6 K a 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 -4 see, the binding constants are determined for the EDTA 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 K f for the metal-edta complex must be many orders of magnitude larger than the K f for the metal-auxiliary complex agent so that the EDTA can remove the metal from the complexing agent.

6 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 Note the different colors and pk a 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 K ass 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 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 K f 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 6

7 7 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 Hg as the analyte and MgEDTA at the displacement titrant. Indirect Titrations (Skip) 2- With a little clever thought EDTA can be used as a titrant for anions like SO 4 2- BaSO 4 is insoluble to one way to determine SO 4 is to precipitate with Ba, filter and wash the ppt, then boil in excess EDTA to complex all the Ba. Back titrate to determine 2- how much Ba you had, and that, in turn, tells you how much SO 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 Zn, Hg Co,Cu, Ag, Ni Pd Pt Fe and Fe, but not Mg,Ca, Mn or Pb 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 Mg and Ca. 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 Mg precipitate out of the solution as Mg(OH) 2 so only Ca remains in solution to be titrated. 4

8 13-5 The ph-dependent Metal-EDTA Equilibrium From Section 12-5 you will remember that we can calculate the fraction of any species -4 in solution using equations. Since we have K f s for EDTA in the EDTA form, will will -4-4 focus on the fraction where Y is short for EDTA Y-4 8 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 -4 complex 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 to match our table?? We calculate a Conditional Formation Constant n-4 +n -4 If K f = [My ]/[M ][Y ] -4 and we know that [Y ] = y-4 F(EDTA) Then we can combine these to get the equation n-4 +n K f = [My ]/[M ] y-4 F(EDTA) Then rearrange to get n-4 +n K = [My ]/[M ]F(EDTA) f y-4

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10 10 This allows us to calculate the CONDITIONAL FORMATION Constant (K f ) at any ph. +2 Let s try an example. Say we wish to titrate Ca with EDTA at ph 10, what is our Conditional Formation Constant? Log K f for the Ca EDTA complex is according to table K=10 = 4.47x10. If ph=10 then y-4 = 0.30 according to table Therefore (K f =.30(4.47x10 ) =1.34x One general trend you should see is that as ph gets lower so does the K f. At what point does it get so low that we can no longer do the titration?? As a general rule of thumb 8 8 we use 10 as a cutoff. Thus K must be >=10 for an EDTA titration to work. f +2 Fe? Let s try another one. What is the lowest ph at which you can successfully titrate K f = K f x Y K = 10 = 2.0x10 f Y x10 10 Y4-10 /2x10 5x10-7 Y4- -7 Y4-=2.9x10 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 K = f 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

11 11 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 = 50 ml x.01m =.5 mmoles Mole EDTA = 30 mls x.01m =.3 mmoles Reaction table: Ca + EDTA Ca EDTA Before reaction Reaction After reaction So [Ca ].2 mmole/(50ml+30ml) = 2.5x10 ; pca = Equivalence point At the equivalence point we have moles Ca = 50 ml x.01 M =.5 mmole moles EDTA = 50 ml x.01 M =.5 mmole Ca + EDTA Ca EDTA Before reaction Reaction After reaction And your first guess might be that [Ca ] = 0, pca = 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.) It is our task then, to use the K at this point to figure out how much free Ca is left in the solution from the back reaction. Let s look at the equilibrium calculation:

12 12 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 2 Now, you could multiply through by X 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 ; X =.005/K f ; X=sqrt(.005/1.76x10 )=sqrt(2.84x10 ); -7 X=5.33x10 ;pca= Let's double check our assumption; is 5.33x10 <<5x10? 3. After the Equivalence point After the equivalence point we continue with the same logic; since the only Ca can come from the equilibrium, we need to use the equilibrium constant equation. Let s try the point 10 ml. moles Ca = 50 ml x.01 M =.5 mmole moles EDTA = 60 ml x.01 M =.6 mmole Ca + EDTA Ca EDTA Before reaction Reaction After reaction So our nominal concentrations are: [Ca ] = 0 [EDTA] =.1/(50+60) =.1/110 = 9.1x10-4 M -3 [Ca EDTA] =.5/(50+60) =.5/110 = 4.55x10 M Now let s throw in the back reaction [Ca ] = 0+x -4 [EDTA] = 9.1x10 M +X

13 -3 [Ca EDTA] = 4.55x10 M - x 13

14 14 And finally go to the equilibrium equation: K = [CaEDTA]/[EDTA][Ca] -3-3 K f =4.55x10 -X/X(9.1x10 +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 /9.1x10 (X);X=4.55/.91(1.76x10 )=2.84x10 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.