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What happens to the absorbed energy?

What happens to the absorbed energy?. s 1. t 1. Energy. s o. EDTA Titrations. Outline. What is EDTA? Metal-Chelate Complexes ATP 4- with Mg 2+ Fe(NTA) 2 3- Fe(DTPA) 2- Chelate Effect EDTA Acid Base Properties a Y nomenclature Conditional Formation Constants EDTA Titration.

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What happens to the absorbed energy?

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  1. What happens to the absorbed energy?

  2. s1 t1 Energy so

  3. EDTA Titrations

  4. Outline • What is EDTA? • Metal-Chelate Complexes • ATP4- with Mg2+ • Fe(NTA)23- • Fe(DTPA)2- • Chelate Effect • EDTA • Acid Base Properties • aY nomenclature • Conditional Formation Constants • EDTA Titration

  5. Metal-Chelate Complexes Lewis Acid/Base Chemistry Monodentate Multidentate and Chelates

  6. Review: • What is a Lewis Acid? Examples? • And a Lewis Base? Examples?

  7. Transition Metal with ligand Central Metal ion is a Lewis Acid Ligand – All ligands are Lewis Bases

  8. Multidentate Multidentate or chelating ligand attaches to a metal ion through more than one atom is said to be multidentate, or a chelating ligand. Examples?

  9. ATP4- can also form complexes with metals

  10. Complex of Iron and NTA Fe3+ + 2 Fe(NTA)23-

  11. Medical Applications The Thalassemia Story

  12. The Chelate Effect Question: Describe in your own words, the “chelate effect”.

  13. The Chelate Effect! Cd(H2O)62+ + 2 Cd(H2O)62+ + 4CH3NH2 + 4 H2O K =B2 = 8 x 109 + 4 H2O K =B2 = 4 x 106

  14. 13-2 EDTA “EDTA is by far, the most widely used chelator in analytical chemistry. By direct titration or through indirect series of reactions, virtually every element of the periodic table can be measured with EDTA.” - Daniel Harris

  15. Acid/Base Properties H H (H6Y2+)

  16. Acid/Base Properties pKa = 0.0 H (H5Y+)

  17. Acid/Base Properties pKa = 0.0 pKa = 1.5 (H4Y)

  18. Acid/Base Properties pKa = 0.0 - pKa = 2.0 pKa = 1.5 (H3Y-)

  19. Acid/Base Properties pKa = 0.0 - pKa = 2.0 - pKa = 1.5 pKa = 2.7 (H2Y-2)

  20. Acid/Base Properties pKa = 6.16 pKa = 0.0 - pKa = 2.0 - pKa = 1.5 pKa = 2.7 (HY-3)

  21. Acid/Base Properties pKa = 10.24 pKa = 6.16 pKa = 0.0 - pKa = 2.0 - pKa = 1.5 pKa = 2.7 (Y-4)

  22. Fraction as Y4- The fraction of EDTA in form Y4- is given as a4- Fraction of EDTA ion the form Y4- (13-3) Concentration in the form Y4- Total Concentration of EDTA

  23. Fraction as Y4- Equation 13-4 in text

  24. Example You make a solution of 0.10 M EDTA and you buffer the pH to (a) 10.0. What is aY4-? (b) What is aY4- if the pH of the solution is buffered to 11.0?

  25. EDTA reactions with Metals Silver – Ag+ Mercury - Hg2+ Iron (III) – Fe3+

  26. EDTA ethylenediaminetetraacetate anion => EDTA-4 => Y-4 +1 cation Ag+ + Y-4D AgY-3

  27. EDTA ethylenediaminetetraacetate anion => EDTA-4 => Y-4 +2 cation Hg+2 + Y-4D HgY-2

  28. EDTA ethylenediaminetetraacetate anion => EDTA-4 => Y-4 +3 cation Fe+3 + Y-4D FeY-1

  29. EDTA ethylenediaminetetraacetate anion => EDTA-4 => Y-4 +n ion M+n + Y-4D MY(n-4)+

  30. EDTA [MY(n-4)+] KMY = -------------- [M][Y-4] [MY(n-4)+] KMY = ------------------- [M+n] * a4 * [EDTA] Conditional formation constant! [MY(n-4)+] K'MY = KMY x a4 = ------------------- [M+n] [EDTA]

  31. Example • Calculate the concentration of Ni2+ in a solution that was prepared by mixing 50.0 mL of 0.0300 M Ni2+ with 50.0 mL of 0.0500 M EDTA. The solution was buffered to pH of 3.00. Two Parts 1. Reaction 2. Then equilibrium is established

  32. EDTA Titrations

  33. Figure 13-10 Theoretical titration curves

  34. EXAMPLE: Derive a curve (pCa as a function of volume of EDTA) for the titration of 50.0 mL of 0.0500 M Ca+2 with 0.1000 M EDTA in a solution buffered to a constant pH of 10.0. Calculate the conditional constant: Equivalence Volume pCa at Initial Point pCa at Equivalence pCa at Pre-Equivalence Point pCa at Post-Equivalence Point

  35. Example Derive a curve (pCa as a function of volume of EDTA) for the titration of 50.0 mL of 0.0500 M Ca+2 with 0.1000 M EDTA in a solution buffered to a constant pH of 10.0. [CaY-2] K'CaY = KCaY * a4 = ---------------- [Ca+2] * [EDTA] where aY4- = 0.36 at pH = 10.0 KCaY = 4.9 x 1010 K'CaY = KCaY * a4 = 4.9 x 1010 * 0.36 = 1.8 x 1010

  36. EXAMPLE: Derive a curve (pCa as a function of volume of EDTA) for the titration of 50.0 mL of 0.0500 M Ca+2 with 0.1000 M EDTA in a solution buffered to a constant pH of 10.0. Equivalence Volume 1 Mole of EDTA = 1 Mole of Metal M1V1 = M2V2 (Careful of Stoichiometry) 50.0 mL (0.0500 M) = 0.1000 M (V2) V2 = 25.0 mL

  37. EXAMPLE: Derive a curve (pCa as a function of volume of EDTA) for the titration of 50.0 mL of 0.0500 M Ca+2 with 0.1000 M EDTA in a solution buffered to a constant pH of 10.0. K'CaY = 1.8 x 1010 0.00 mL EDTA added Initial Point pCa = - log[Ca+2] = - log(0.00500 M) = 2.301

  38. EXAMPLE: Derive a curve (pCa as a function of volume of EDTA) for the titration of 50.0 mL of 0.0500 M Ca+2 with 0.1000 M EDTA in a solution buffered to a constant pH of 10.0. At 25.0 mL (Equivalence Point) What can contribute to Ca2+ “after” reaction?

  39. EXAMPLE: Derive a curve (pCa as a function of volume of EDTA) for the titration of 50.0 mL of 0.0500 M Ca+2 with 0.1000 M EDTA in a solution buffered to a constant pH of 10.0. Ca2+ + Y4-DCaY2- I - - 0.0025 moles/V C +x +x -x E +x + x 0.0333 –x 0.0025moles/0.075 L X = [Ca2+] = 1.4 x10-6 pX = p[Ca2+] = 5.866

  40. Pre-Equivalence Point Let’s try 15 mL

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