ERT207 Analytical Chemistry Complexometric Titration Pn Syazni Zainul Kamal PPK Bioproses

CO4: ABILITY TO DIFFERENTIATE AND CALCULATE CONCENTRATION OF ANALYTES OF VARIOUS TITRIMETRIC METHODS (ACID-BASE, COMPLEXATION, REDOX, PRECIPITATION)

Types of Titrimetric Methods Classified into four groups based on type of reaction involve; Acid-base titrations Complexometric titrations Redox titrations Precipitation titrations

Complexes Most metal ions (kation) can react with lone pair electron from a molecule or anion to form covalent bonds and produce coordination compound or complexes. Molecule or ion with at least 1 pair of unshared electron can form covalent bond with metal ion = ligands The bonding between metal and ligand generally involves formal donation of one or more of the ligand's electron pairs Eg of ligands = ammonia, cyanide ions, halide ions, water (neutral/-ve charge mols or ions)

Metal ion – lewis acid (electron pair acceptor) Compound form between ligand and metal ion = complexes or coordination compound (has charges or neutral). Examples of complex formation : Ag+ + 2CN- Ag(CN)2- Ag+ + 2NH3 Ag (NH3)2+ Complex/coordination compound Metal ion ligand Metal ion – lewis acid (electron pair acceptor) Ligand – lewis base (electron pair donor) Coordination number – number of covalent bond formed between metal ion and ligand

A ligand that has one pair of unshared electron such as NH3, is called unidentate. Glycine (NH2CH2COOH) and ethylenediamine (NH2CH2CH2NH2) which has two pairs of unshared electron available for covalent bonding, is called bidentate. EDTA, has 6 pairs of unshared electrons = hexadentate

EDTA Equilibrium Chelating agent - An organic agent that has two or more group capable of complexing with a metal ion (also called ligand) Chelate – complex formed Titration with chelating agent = chelometric titration , a type of complexometric titration Most widely used chelating agent in titration – ethylenediaminetetraacetic acid (EDTA).

EDTA = Ethylenediaminetetraacetic acid Hexadentate ligand Has six bonding sites (the four carboxyl groups and the two nitrogen providing six lone pairs electrons) Tetraprotic acid (H4Y), can exist in many forms H3Y-, H2Y2-, HY3- and Y4- only unprotonated ligand (Y4-) can complex with metal ion

H4Y H+ + H3Y- Ka1 = [H+][H3Y-] = 1.0 x10-2 Since EDTA is a tetraprotic acid, the dissociation of EDTA as follows : H4Y H+ + H3Y- Ka1 = [H+][H3Y-] = 1.0 x10-2 [H4Y] H3Y- H+ + H2Y2- Ka2 = [H+][H2Y2-] = 2.1 x10-3 [H3Y-] H2Y2- H+ + HY3- Ka3 = [H+][HY3-] = 6.9 x10-7 [H2Y2-] HY3- H+ + Y4- Ka4 = [H+][Y4-] = 5.5 x10-11 [HY3-]

Complex formation constant of EDTA EDTA can form complex with Ca2+ as the following equilibrium : Ca2+ + Y4- CaY2- (1) The complex formation constant is : Kf = KCaY2- = [CaY2-] (2) [Ca2+][Y4-]

Effect of pH on EDTA equilibria If H+ concentration increases, equilibrium in equation 1 will shift to the left. Chelating anion (Y4-) will react with H+. Dissociation of CaY2- in presence of acid From the overall equilibrium

Let us consider that CH4Y represent the total uncomplexed EDTA If we substitute the values of [HY3-], [H2Y2-], [H3Y-] and [H4Y] derived from the Ka values to this equation and divide each term with [Y4-], we will get the following equation:- Where α4 is the fraction of the total EDTA exists as Y4- .

Conditional formation constant The equation for the complex formation between Ca and EDTA is : Ca2+ + Y4- CaY2- (3) Kf = KCaY2- = [CaY2-] (4) [Ca2+][Y4-] α4 = [Y4-] , hence [Y4-] = α4 CH4Y (5) CH4Y

Replacing [Y4-] into equation (4) : Kf = KCaY2- = [CaY2-] (6) [Ca2+] α4CH4Y Kfα4 = [CaY2-] = K’f (7) [Ca2+]CH4Y K’f is the conditional formation constant which depends on α4 therefore K’f depends on pH. We can use equation (7) to calculate the value of equilibirum concentration for EDTA species at specific pH to replace equation (4)

Metal-EDTA titration curves Titration is perform by adding the chelating agent (EDTA) to the sample (metal). Titration curve – plotting the changes in metal ion concentration (pM) versus volume of titrant (EDTA) Example of complexometric titration is by adding 0.100 M EDTA to 100 ml 0.100 M Ca2+ solution buffered at pH 11 Ca2+ + Y4- CaY2-

Before titration started – only have Ca2+ solution. pCa = - log [Ca2+] Titration proceed – part of Ca2+ is reacted with EDTA to form chelate. [Ca2+] gradually decrease. pCa= -log [remaining Ca2+] At equivalence point – have convert all Ca2+ to CaY2- So pCa can be determined from the dissociation of chelate at a given pH using Kf. K’f = Kf α4 = [CaY2-] [Ca2+] CH4Y Excess titrant added – pCa can be determined from the dissociation of chelate at a given pH using Kf.

Exercise Calculate pCa in 100 ml of a solution of 0.100 M Ca2+ at pH10 after addition of 0, 50, 100, 150 ml of 0.100 M EDTA. Kf for CaY2- is 5.0x1010 and α4 is 0.35. K’f = Kf x α4 = 5.0x1010 x 0.35 = 1.75x1010

a) Addition of 0.00 ml EDTA [Ca2+] = 0.100 M pCa = - log 0.100 = 1.00

b) Addition of 50.00 ml EDTA Initial mmol Ca2+ =100ml x 0.100 M =10 mmol mmol EDTA added = 50ml x 0.100 M = 5 mmol mmol Ca2+ left = 5 mmol [Ca2+] = 5 mmol = 0.0333 M (100+50)ml pCa = - log 0.0333 = 1.48

c) Addition of 100 ml EDTA Initial mmol Ca2+ =100ml x 0.100 M =10 mmol mmol EDTA added =100ml x 0.100 M = 10 mmol Equivalence point is reached. We have convert all Ca2+ to CaY2-. mmol CaY2- = mmol initial Ca2+ [CaY2-] = 10 mmol = 0.05 M (100+100)ml K’f = Kf α4 = [CaY2-] [Ca2+] CH4Y K’f = [CaY2-] = 1.75 x 1010 0.05 = 1.75 x 1010 (x)(x) x = 1.7 x 10-6 so pCa = - log 1.7x10-6 = 5.77

d) Addition of 150 ml EDTA Initial mmol Ca2+ =100ml x 0.100 M =10 mmol mmol EDTA added =150ml x 0.100 M =15 mmol mmol EDTA excess = 5 mmol CH4Y = 5 mmol = 0.02M [CaY2-] = 10 = 0.04M (100+150)ml (100+150)ml K’f = [CaY2-] = 1.75 x 1010 [Ca2+] (0.02) 0.04 = 1.75 x 1010 (x)(0.02) x = 1.14 x 10-10 so pCa = - log 1.14x10-10 = 9.94