EDTA Titration EDTA = Ethylenediaminetetraacetic acid Lewis acid : electron pair acceptor eg metal ions Lewis base : electron pair donor eg ligands A ligand that attaches to a metal ion through more than one ligand atom is called a chelating ligand or multidentate ligand

A titration based on complex formation is called complexometic titration The equilibrium constant for the reaction of a metal with a ligand is called the formation constant, Kf , or the stability constant Consider : Ag+ + NH3 Ag(NH3)+ Ag(NH3)+ + NH3 Ag(NH3)2+ Kf1 = Kf2 = [Ag(NH3)+] [Ag+][NH3] [Ag(NH3)2+] [Ag(NH3)+][NH3]

and the overall formation constant : Kf =Kf1 Kf2 = The overall reaction : Ag+ + 2NH3 = Ag(NH3)2+ and the overall formation constant : Kf =Kf1 Kf2 = [Ag(NH3)2+] [Ag+][NH3]2 Example : A divalent metal M2+ reacts with a ligand L to form a 1:1 complex : M2+ + L ML2+ Calculate the concentration of M2+ in a solution prepared by mixing equal volumes of 0.20 M M2+ and 0.20 M L. Given Kf = 1.0 x 108 Given Kf = 1.0 x 108 complex is sufficiently strong such that the reaction is virtually complete Since equal volumes were added initial concentration is halved

Let x = residual concentration of M2+ M2+ + L ML2+ x x 0.1 – x Kf = = 1.0 x 108 x = 3.2 x 10-5 M x2 0.1 - x Similarly, if L is a multidentate ligand and M + nL MLn then Kf = [MLn] [M][L]n

EDTA Complexes It can be represented as having four Ka values : EDTA is a hexaprotic system – H6Y2+ Neutral acid is tetraprotic – H4Y It can be represented as having four Ka values : H4Y H+ + H3Y- Ka1 = 1.0 x 10-2 H3Y- H+ + H2Y2- Ka2 = 2.2 x 10-3 H2Y2- H+ + HY3- Ka3 = 6.9 x 10-7 HY2- H+ + Y4- Ka4 = 5.5 x 10-11

CH Y = [Y4-] + [HY3-] + [H2Y2-] + [H3Y-] + [H4Y] 4 Hence in complexing with - +1 cation : Ag+ + Y4- = AgY3- Kf = +2 cation : Hg2+ + Y4- = HgY2- Kf = +3 cation : Fe3+ + Y4- = FeY- Kf = +n cation : Mn+ + Y4- = MYn-4 Kf = [MYn-4] [Mn+] [Y4-] [HgY2-] [Hg2+][Y4-] [FeY-] [Fe3+][Y4-] [AgY3-] [Ag+][Y4-] Fraction of the total EDTA species that exists as Y4- = a4 = [Y4-]/CH Y where CH Y = [Y4-] + [HY3-] + [H2Y2-] + [H3Y-] + [H4Y] 4

Substituting [Y4- ] = a4CH Y Kf = 4 a4Kf = = Kf’ 4 [MYn-4] [Mn+] a4CH Y [Mn+]CH Y Conditional formation constant – holds only for a particular pH – describes the formation of MYn-4 at any particular pH – allows us to look at EDTA complex formation as if the uncomplexed EDTA were all in one form Example : Consider the reaction: Fe3+ + EDTA FeY- (Kf = 1.3 x 1025; aY (at pH1) = 1.9 x 10-18 and aY (at pH4) = 3.8 x 10-9). Calculate the concentration of free Fe3+ in a solution of 0.10 M FeY- at pH4 and at pH1. 4-

Let x = [Fe3+] = [EDTA], thus = = 4.9 x 1016 (at pH4) Using Kf’ = a4Kf , at pH = 4 : Kf’ = (3.8 x 10-9)(1.3 x 1025) = 4.9 x 1016 at pH = 1 : Kf’ = (1.9 x 10-18)(1.3 x 1025) = 2.5 x 107 Let x = [Fe3+] = [EDTA], thus = = 4.9 x 1016 (at pH4) x = 1.4 x 10-9 M Similarly, at pH1: x = 6.4 x 10-5 M pH affects the stability of the complex. The Kf’ values show that the metal-EDTA complex becomes less stable at lower pH [FeY-] [Fe3+][EDTA] 0.1 - x x2

For a titration reaction to be effective, the equilibrium constant must be large the analyte and titrant should be completely reacted at the equivalence point Titration of Ca2+ with EDTA as a function of pH :

EDTA Titration Curves Titration is carried out by adding the chelating agent to the sample : Mn+ + EDTA MYn-4 Kf’ = a4Kf The titration curve is a graph of pM (= -log[M]) versus the volume of added EDTA Titration curve consists of 3 regions- Before the equivalence point : There is excess Mn+ in the solution after the EDTA has been consumed. Concentration of free metal ion = concentration of excess unreacted Mn+

At the equivalence point : [Mn+] = [EDTA] Free Mn+ is from the dissociation of MYn-4 : MYn-4 Mn+ + EDTA After the equivalence point : There is excess EDTA and virtually all the metal ion is in the MYn-4 form Example : Calculate the shape of the titration curve for the reaction of 50.0 ml of 0.0500 M Mg2+ (buffered to pH10.0) with 0.0500 M EDTA (Kf = 6.2 x 108 ; at pH10.0 : aY = 0.36) Mg2+ + EDTA MgY2- Using Kf’ = a4Kf = (0.36)(6.2 x 108) = 2.2 x 108

Since Kf’ is large the reaction goes to completion with each addition of titrant Before equivalence point : Consider the addition of 5.0 ml EDTA to the solution Moles of EDTA added = (0.005 l)(0.0500 M) = 2.5 x 10-4 Moles of Mg2+ present initially= (0.050 l)(0.050M) = 2.5 x 10-3 Moles of Mg2+ present after the addition of EDTA = (2.5 x 10-3 ) – (2.5 x 10-4) = 0.00225 [Mg2+] = = 0.0409 M pMg2+ = -log [Mg2+] =1.39 At equivalence point : [Mn+] = [EDTA] Volume of EDTA added = (2.5 x 10-3)/(0.0500M) = 50.0 ml 0.00225 0.055

Since there is negligible dissociation, [MgY2-] = = 0.025 M From: 0.0025 mol 0.100 l [MgY2-] [Mg2+][EDTA] 0.025 - x x2 Since there is negligible dissociation, [MgY2-] = = 0.025 M From: Mg2+ + EDTA MgY2- Let x =[Mg2+] = [EDTA], then Kf’ = = = 2.2 x 108 x = 1.07 x 10-5 M pMg2+ = -log [Mg2+] =4.97 After equivalence point : If 51.0 ml of EDTA is added there will be 1.0 ml excess EDTA in the solution

[EDTA] ={(0.001)(0.05M)}/(0.101 l) =4.95 x 10-4 M [MgY2-] =(2.5 x 10-3)/(0.101 l) =2.48 x 10-2 M Using Kf’ = = = 2.2 x 108 [Mg2+] = 2.3 x 10-7 M p[Mg2+] = -log [Mg2+] = 6.64 [MgY2-] [Mg2+][EDTA] 2.48 x 10-2 [Mg2+](4.95 x 10-4)

Metal Ion Indicators Methods to detect the end point in EDTA titrations are : metal ion indicators A metal ion indicator is a compound whose color changes when it binds to a metal ion (eg Eriochrome black T) This compound must bind metal less strongly than EDTA MgIn + EDTA MgEDTA + In (red) (colorless) (colorless) (blue) mercury electrode : measurement to potential glass pH electrode ion-selective electrode

EDTA Titration Techniques Direct Titration analyte is titrated with standard EDTA analyte is buffered to an appropriate pH at which the conditional formation constant for the metal-EDTA complex is large and the color of the free indicator is distinctly different from that of the metal-indicator complex Back titration a known excess of EDTA is added to the analyte the excess EDTA is titrated with a standard solution of a second metal this method is useful if - analyte precipitates in the absence of EDTA - the analyte reacts too slowly with EDTA under titration conditions - analyte blocks the indicator - metal used in back titration must not displace the analyte metal ion from its EDTA complex

Displacement titration Method is useful is the metal ions do not have a satisfactory indicator analyte is treated with excess Mg(EDTA)2- to displace Mg2+ Mn+ + MgY2- MYn-4 + Mg2+ - the displaced Mg2+ is titrated with standard EDTA Example : 2Ag+ + Ni(CN)42- 2Ag(CN)2- + Ni2+ (iv) Indirect titration Anions (such as SO42-, CrO42-, CO32- and S2-) that precipitate with certain metal ions can be analyzed with EDTA through indirect titration For example, SO42 can be precipitated with excess Ba2+. The BaSO4(s) is washed and boiled with excess EDTA at pH10 to bring Ba2+ back into solution as Ba(EDTA)2-. The excess EDTA is back titrated with Mg2+