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COMPLEXOMETRIC TITRATION

COMPLEXOMETRIC TITRATION. Lecturer : Budi Hastuti, S.Pd, M.Si. Complex. Anhydrous CuSO 4 (no crystal water) → white CuSO 4 5 H 2 O → blue oldish Cu 2+ ion changes color? → The occurrence of complexes between Cu 2+ ions and other species in solution (H 2 O). The Notice of Complex.

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COMPLEXOMETRIC TITRATION

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  1. COMPLEXOMETRICTITRATION Lecturer : Budi Hastuti, S.Pd, M.Si

  2. Complex • Anhydrous CuSO4 (no crystal water) → whiteCuSO4 5 H2O → blue oldish Cu2+ ion changes color?→ The occurrence of complexes between Cu2+ ions and other species in solution (H2O)

  3. The Notice of Complex • Complex:→ a new unit formed from units that can stand alone, but form a new bond. • Atomic Centre:→ metal ions in the complex • Ligand→ The group is bound to the central atom • Coordination Numbers:→ The number of bonds formed by the central metal

  4. Reaction of Complex Formation • Reaction Ag+ + 2CN- Ag(CN)2- • The reactions that form the complex → lewis acid-base reaction. • The ligands works as a base by provids a pair of electrons to the cation which is an acid - Bond formed between the central of metal atom and ligand → covalent bond - Some of the complex establish a substitution reaction very quicktly → labile complex

  5. Next Example : Cu(H2O)42+ + 4NH3  Cu(NH3)42+ + 4H2O Light Blue Dark Blue • Many complexes provided from the first row transition metals

  6. Complex Reaction • Some of the complex establish a substitution reaction very slowly → stabile or inert • For example: all the complex formed from cobalt and chromium in oxidation state +3, such as Co(NH3)63+, Cr (CN) 63+, Fe (CN) 63 +

  7. Ligand • Monodentateligands: → ligands containing only one electron pair donor atomsfor example: I-, NH3, CN- • - PolydentateLigand:→ ligands have donor atoms more than one For example: bidentat and hexadentat

  8. N OH Ligand (next) • Bidentate Ligands - ethylenediamine (H2N ─ CH2CH2 ─ NH2) The donor are both of N atoms - 8 hidroxiquinolin (oxin)    The donor are N and O atoms • Hexadentat Ligans • EDTA (ethylenediamine tetra acetic acid)→ The donors: both of N atoms and four O atoms (from OH)

  9. CH2COOH HOOCCH2 • N ─ CH2CH2 ─ N CH2COOH HOOCCH2 Ligand (next) • HexadentateLigands EDTA (ethylenediamine tetra acetic acid) → The donors: both of N atoms and four O atoms (from OH)

  10. NH2 NH2 H2C CH2 Zn H2C CH2 NH2 NH2 Chelate • If Zn2+ make Complex with ethylene diamine: Zn2++2H2N ─ CH2CH2 ─ NH2 • The circle called the chelate circle • The complex containing the chelate circle → called chelate • It’s ligand called forming chelat → (chelating or chelating agent) • In the volumetric, chelat reagent as titrant

  11. The Stability of complex • In the reaction :M + L ML M = cation L = Ligand ML = complex • Complex stability constants: K

  12. Gradual formation constant • For example, the formation of complex Cu (NH3) 2+Held in 4 stages: Cu2+ + NH3 CuNH32+ K1 = 1,9x104 CuNH32+ + NH3 Cu(NH3)22+ K2 = 3,6x103 Cu(NH3)22+ + NH3 Cu(NH3)32+ K3 = 7,9x103 Cu(NH3)32+ + NH3Cu(NH3)42+ K4 = 1,5x103 • By review of the overall reaction,Cu2+ + 4NH3Cu(NH3)42+

  13. The balance involve EDTA Titration • In the reaction : Cu2+ + Y4- CuY2- • Analogous to the regular neutralization reaction Where: Y4-is an anion of EDTA → electron donor (base) Cu2+ is a Cation of EDTA → acceptor proton (acid) So pCu = - log [Cu]

  14. Absolute stability or formation constant • In the reaction : Mn+ + Y4- MY-(4-n) So : Kabs = absolute stability constant / absolute formation constant

  15. Effect of the pH The Reaction : H4Y + H2O H3O+ + H3Y- Ka1 = 1,02 x 10-2 H3Y- + H2O H3O+ + H2Y2- Ka1 = 8,14 x 10-3 H2Y2- + H2O H3O+ + HY3- Ka1 = 6,92 x 10-7 HY3- + H2O H3O+ + Y4 Ka1 = 5,50 x 10-11 if Cy = total concentration of EDTA isn’t complexed Cy = [Y4-] + [HY3-] + [H2Y2-] + [H3Y-] + [H4Y]

  16. Effect of the pH Keff = Effective Constant or conditional stability Constant

  17. Effect of the pH • Not like Kabs, so Keff fluctuate with the pH because the pH dependence on 4 • Keff more useful than Kabs , because it shows the actual tendency to form metal chelonat on the pH value • If the pH falls, 4becomessmaller → Keffsmaller

  18. List Price 4 for EDTA in the solution of varian pH

  19. Example 1 • 50 ml 0.01 M Ca2+ buffered solution at pH 10, titrated with 0.01 M EDTA solution. Calculate the price of pCa on the varians titration phase!

  20. Answer Kabs for CaY-2 = 5,0.1010 4 on pH 10 = 0,35 so Keff = 5.1010 x 0,35 = 1,8.1010 • The beginning of the titration [Ca2+] = 0,01 M pCa = - log [Ca2+] = - log 10-2 = 2 • After adding 10 ml of titran

  21. d. After addition 60 ml of titrant. Excess EDTA = 0.1 mmol c. Equivalence point

  22. 12 pH 12 pH 10 10 8 pH 8 6 4 2 0 10 20 30 40 50 60

  23. The suitable of complexometric titration • Example 2 50 ml M2+of 0.010 M titrated with EDTA. Calculate the price of Keff so the reaction is complete if 49, 95 ml of titrant was added, and pM change as much as 2 units on the addition of 2 drops (0.10 ml) of the titrant again !

  24. Answer • 1 drop before the equivalence point = 49.95 x 0.01 = 0.4995 mmole at first M2+= 50 x 0.01 = 0.5 mmol • Reaction: Mn+ + Y4- MY-(4-n)b : 0.5 0.4995rx: 0.4995 0.4995 0.4995r : 0.0005 0 0.4995

  25. If ΔpM = 2 units, so pM = 7.3 and [M2+] = anti log - 7.3 = [M2+] = 5.10-8 MWhen the titrant is added 50.05 ml (50.05 -50) ml = 0.05 ml

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