Analytical Chemistry Lecture W5 Feb. 05, 2003

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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 electronpair 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

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+ + SO42-

most 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).

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.

A ligand that has a single donor group, such as ammonia, is called unidentate

(singletoothed), 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- < == > FeL-

Fe3+ + Cl- < == > FeCl2+, FeCl2+, FeCl30, FeCl4-

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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.

Ethylenediaminetetraacetic acid, which is commonly shortened to EDTA, is the most widely used complexometric titrant. EDTA has the structure

HOOCCH2 CH2COOH

\ /

:NCH2CH2N:

/ \

HOOCCH2 CH2COOH

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- < == > AgY3-

Al3+ + Y4 < == > AlY-

EDTA 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.

Mn+ + Y4- < == > MY(n-4)+ Kf = [MY(n-4)+]

[Mn+][Y4-]

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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.]

a4 = [Y4-] = . [Y4-] .

[EDTA] [Y4-] + [HY3-] + [H2Y2-] + [H3Y-] + [H4Y] + [H5Y+] + [H6Y2+]

Notice that we can also define a 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 a4 is thus fixed when the pH is fixed.

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This transparency shows the log a 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 Y4- concentration 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-] = a4 [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)+] = [MY(n-4)+]

[Mn+][Y4-] [Mn+] a4 [EDTA]

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and now we define a conditional formation constant

Kf' = Kfa4 = [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 a4 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-

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 a4 value for EDTA at pH 10.0 is 0.36 as given in Table 13-1.

K'CaY = a4 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 preequivalencepoint 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.

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

before eq .00333 0 0

after eq .00333 –x x x