1. ERT207 Analytical Chemistry Complexometric Titration Pn Syazni Zainul Kamal PPK Bioproses

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

3. Types of Titrimetric Methods Classified into four groups based on type of reaction involve: Classified into four groups based on type of reaction involve: 1. Acid-base titrations 2. Complexometric titrations 3. Redox titrations 4. Precipitation titrations

4. 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)

5. 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 + + 2NH 3 Ag(NH 3 ) 2 + Metal ion ligand Complex/coordination compound Metal ion – lewis acid (electron pair acceptor) – lewis base (electron pair donor) Ligand – lewis base (electron pair donor) Coordination number – number of covalent bond formed between metal ion and ligand

6. A ligand that has one pair of unshared electron such as NH 3, is called unidentate. A ligand that has one pair of unshared electron such as NH 3, is called unidentate. Glycine (NH 2 CH 2 COOH) and ethylenediamine (NH 2 CH 2 CH 2 NH 2 ) which has two pairs of unshared electron available for covalent bonding, is called bidentate. Glycine (NH 2 CH 2 COOH) and ethylenediamine (NH 2 CH 2 CH 2 NH 2 ) which has two pairs of unshared electron available for covalent bonding, is called bidentate. EDTA, has 6 pairs of unshared electrons = hexadentate EDTA, has 6 pairs of unshared electrons = hexadentate

7. Complex: Formation Constant NH 3 forms complex with Ag + through 2 steps: NH 3 forms complex with Ag + through 2 steps:

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

10.  EDTA = Ethylenediaminetetraacetic acid  Hexadentate ligand  Has six bonding sites (the four carboxyl groups and the two nitrogen providing six lone pairs electrons)  Tetraprotic acid (H 4 Y), can exist in many forms H 3 Y -, H 2 Y 2-, HY 3- and Y 4-  only unprotonated ligand (Y 4- ) can complex with metal ion

11. Since EDTA is a tetraprotic acid, the dissociation of EDTA as follows : Since EDTA is a tetraprotic acid, the dissociation of EDTA as follows : H 4 Y H + + H 3 Y - K a1 = [H + ][H 3 Y - ] = 1.0 x10 -2 [H 4 Y] H 3 Y - H + + H 2 Y 2- K a2 = [H + ][H 2 Y 2- ] = 2.1 x10 -3 [H 3 Y - ] H 2 Y 2- H + + HY 3- K a3 = [H + ][HY 3- ] = 6.9 x10 -7 [H 2 Y 2- ] HY 3- H + + Y 4- K a4 = [H + ][Y 4- ] = 5.5 x10 -11 [HY 3- ]

12. Complex formation constant of EDTA EDTA can form complex with Ca 2+ as the following equilibrium : EDTA can form complex with Ca 2+ as the following equilibrium : Ca 2+ + Y 4- CaY 2- (1) The complex formation constant is : K f = K CaY 2- = [CaY 2- ] (2) [Ca 2+ ][Y 4- ] [Ca 2+ ][Y 4- ]

14. Effect of pH on EDTA equilibria From the overall equilibrium If H + concentration increases, equilibrium in equation 1 will shift to the left. Chelating anion (Y 4- ) will react with H +. Dissociation of CaY 2- in presence of acid

15. If we substitute the values of [HY 3- ], [H 2 Y 2- ], [H 3 Y - ] and [H 4 Y] derived from the Ka values to this equation and divide each term with [Y 4- ], we will get the following equation:- Where α 4 is the fraction of the total EDTA exists as Y 4-. Let us consider that C H 4 Y represent the total uncomplexed EDTA

16. Conditional formation constant The equation for the complex formation between Ca and EDTA is : Ca 2+ + Y 4- CaY 2- (3) K f = K CaY 2- = [CaY 2- ] (4) [Ca 2+ ][Y 4- ] [Ca 2+ ][Y 4- ] α 4 = [Y 4- ], hence [Y 4- ] = α 4 C H 4 Y (5) C H 4 Y C H 4 Y

17. Replacing [Y 4- ] into equation (4) : K f = K CaY 2- = [CaY 2- ] (6) [Ca 2+ ] α 4 C H 4 Y [Ca 2+ ] α 4 C H 4 Y K f α 4 = [CaY 2- ] = K’ f (7) [Ca 2+ ]C H 4 Y [Ca 2+ ]C H 4 Y K’ f is the conditional formation constant which depends on α 4 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)

18. Metal-EDTA titration curves Titration is perform by adding the chelating agent (EDTA) to the sample (metal). 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) 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 Ca 2+ solution buffered at pH 11 Example of complexometric titration is by adding 0.100 M EDTA to 100 ml 0.100 M Ca 2+ solution buffered at pH 11 Ca 2+ + Y 4- CaY 2-

19. Before titration started – only have Ca 2+ solution. Before titration started – only have Ca 2+ solution. pCa = - log [Ca 2+ ] pCa = - log [Ca 2+ ] Titration proceed – part of Ca 2+ is reacted with EDTA to form chelate. [Ca 2+ ] gradually decrease. Titration proceed – part of Ca 2+ is reacted with EDTA to form chelate. [Ca 2+ ] gradually decrease. pCa= -log [remaining Ca 2+ ] pCa= -log [remaining Ca 2+ ] At equivalence point – have convert all At equivalence point – have convert all Ca 2+ to CaY 2- So pCa can be determined from the dissociation of chelate at a given pH using K f. K’ f = K f α 4 = [CaY 2- ] [Ca 2+ ] C H 4 Y Excess titrant added – Excess titrant added – pCa can be determined from the dissociation of chelate at a given pH using K f.

21. Exercise Calculate pCa in 100 ml of a solution of 0.100 M Ca 2+ at pH10 after addition of 0, 50, 100, 150 ml of 0.100 M EDTA. K f for CaY 2- is 5.0x10 10 and is 0.35. Calculate pCa in 100 ml of a solution of 0.100 M Ca 2+ at pH10 after addition of 0, 50, 100, 150 ml of 0.100 M EDTA. K f for CaY 2- is 5.0x10 10 and α 4 is 0.35. K’ f = K f x α 4 = 5.0x10 10 x 0.35 = 5.0x10 10 x 0.35 = 1.75x10 10 = 1.75x10 10

22. a) Addition of 0.00 ml EDTA [Ca 2+ ] = 0.100 M pCa = - log 0.100 = 1.00 = 1.00

23. b) Addition of 50.00 ml EDTA Initial mmol Ca 2+ =100ml x 0.100 M =10 mmol mmol EDTA added = 50ml x 0.100 M = 5 mmol mmol Ca 2+ left = 5 mmol [Ca 2+ ] = 5 mmol = 0.0333 M (100+50)ml (100+50)ml pCa = - log 0.0333 = 1.48

24. c) Addition of 100 ml EDTA Initial mmol Ca 2+ =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 Ca 2+ to CaY 2-. mmol CaY 2- = mmol initial Ca 2+ [CaY 2- ] = 10 mmol = 0.05 M (100+100)ml (100+100)ml K’ f = K f α 4 = [CaY 2- ] [Ca 2+ ] C H 4 Y K’ f = [CaY 2- ] = 1.75 x 10 10 [Ca 2+ ] C H 4 Y 0.05 = 1.75 x 10 10 (x)(x) 5.77 x = 1.7 x 10 -6 so pCa = - log 1.7x10 -6 = 5.77

25. d) Addition of 150 ml EDTA Initial mmol Ca 2+ =100ml x 0.100 M =10 mmol mmol EDTA added =150ml x 0.100 M =15 mmol mmol EDTA excess = 5 mmol C H 4 Y = 5 mmol = 0.02M [CaY 2- ] = 10 = 0.04M (100+150)ml (100+150)ml (100+150)ml (100+150)ml K’ f = [CaY 2- ] = 1.75 x 10 10 [Ca 2+ ] (0.02) 0.04 = 1.75 x 10 10 (x)(0.02) 9.94 x = 1.14 x 10 -10 so pCa = - log 1.14x10 -10 = 9.94