1 / 30

Chapter 12

Chapter 12. EDTA Titrations. Antibiotic chelate captures its prey. 15A Forming Metal-Chelate Complexes. M n+ + nL  ML n n+ M n+ : central metal ion (Lewis acid) L: ligand (Lewis base) Complexometric titrations are useful for the determination of metal.

hien
Download Presentation

Chapter 12

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 12 EDTA Titrations

  2. Antibiotic chelate captures its prey

  3. 15A Forming Metal-Chelate Complexes Mn+ + nL  MLnn+ Mn+: central metal ion (Lewis acid) L: ligand (Lewis base) Complexometric titrations are useful for the determination of metal.

  4. Complexes: Formation Constants ligand: have a pair of unshared electrons Ex: N, O, S

  5. 15D EDTA: (1) A hexadentate The most widely used chelating agent in titration is ethylenediaminetetraacetic acid (EDTA)

  6. 15D EDTA (2) The chelon effect or chelate effect: Multidentate chelating agents form stronger complexes (Kf ) with metal ions than bidentate or monodentate ligands. 指 multidentate 之 Kf 比 monodentate 之Kf為大，即較穩定。

  7. (3) Neutral EDTA is a tetrabasic acid

  8. (5) Five species EDTA as a function of pH

  9. Effect of pH on EDTA equilibria

  10. 15D-2 EDTA Complex with Metal Ions (1) Forms strong 1:1 complexes regardless of the charge on the cation (2) Chelate with all cations

  11. (3) Since the anion Y4- is the ligand species in complex formation, the complexation equilibria are affected markedly by the pH. (4) The formation constant (Table 15-5)

  12. 15D-3 Equilibrium: pH dependent M-Y

  13. Ex • Calculate the fraction of EDTA that exists as Y-4 at pH 10, and from this calculate pCa in 100mL of solution of 0.100 M Ca+2 at pH 10 after adding 100 mL of 0.100 M EDTA.

  14. Calculate the fraction of EDTA that exists as Y-4 at pH 10, and from this calculate pCa in 100mL of solution of 0.100 M Ca+2 at pH 10 after adding 100 mL of 0.100 M EDTA. ∵ pH = 10 ∴ [H+] = 1.0 × 10-10 1/α4 = 1 + [H+]/Ka4 + [H+]2/Ka3Ka4 + …… = 1 + [1.0 × 10-10 /5.5 × 10-11]+[(1.0 × 10-10)2/(6.9 × 10-7)(5.5 × 10-11)] +…… = 1 + 1.82 + 2.6 × 10-4 + …...  = 2.82 ∴ α4 = 0.35 Ca+2 + EDTA [or Y-4] = CaY-2 X a4[EDTA} [(0.1 × 100)/200] – X ≒ 0.05 Kf = [CaY-2]/[Ca+2] α4 [CH4Y] = 0.05/(X)(0.35)(X) = 5.0 × 1010 X = 1.7 × 10-6 M pCa = -log (1.7 × 10-6) = 5.77

  15. The conditional formation constant (ex:15.5) • We can use K’f to calculate the equilibrium concentrations of the different species at a given pH. • K’f value holds for only a specified pH. • HgY-2之Kf ＞PbY-2之Kf＞CaY-2之Kf； Kf不受pH值之影響，Kf’則受pH值之影響，上述三者在pH值≦9.0時， Kf’開始變小，也就是EDTA的滴定需在(pH＞9.0)之鹼性溶液中進行

  16. 15D-4 EDTA Titration Curves • The end point break depends upon • [Mn+] • [L1] • [pH]  selectivity • Kf The smaller Kf, the more alkaline the solution must be to obtain a k’f of 106.

  17. 15D-4 • The titration rxn: Mn++ EDTA  MYn-4 K’f = a4Kf Three regions: • Before • At • After equivalence point

  18. The Kf for CaY-2 is 5.0 × 1010. At pH 10, α4 is calculated to be 0.35 to give a Kf` of 1.8 × 1010. Calculate the pCa in 100 mL of a solution of 0.100 M Ca+2 at pH 10 after addition of (a) 0 mL (b) 50 mL (c) 100 mL (d) 150 mL of 0.100 M EDTA.

  19. The Kf for CaY-2 is 5.0 × 1010. At pH 10, α4 is calculated to be 0.35 to give a Kf` of 1.8 × 1010. Calculate the pCa in 100 mL of a solution of 0.100 M Ca+2 at pH 10 after addition of (a) 0 mL (b) 50 mL (c) 100 mL (d) 150 mL of 0.100 M EDTA. • (a).pCa = -log [Ca+2] = -log 0.1 = 1.00 • (b).pCa = -log [0.1 × 100 – 0.1 × 50]/150 = 1.48 • (c).[0.05 - X]/[X][X] = 1.8 × 1010 X = 1.7 × 10-6 M = [Ca+2] pCa = -log [1.7 × 10-6] = 5.77 • (d).[CaY-2] = 10 mmol/250 mL = 0.04 M [CH4Y] = [0.1 × 150 – 0.1 × 100]/250 = 0.02 M 0.04/[Ca+2](0.02) = 1.8 × 1010 = Kf` [Ca+2] = 1.1 × 10-10 M; pCa = -log 1.1 × 10-10 = 9.95

  20. Various pH: K’ becomes smaller as the pH decreases

  21. Kf: cation with larger formation const provide good end point even in acidic media.

  22. Minimum permissible pH for a satisfactory end point • 以EDTA滴定一系列金屬離子，當各自之Kf`≒106且能給出一個sharp end point之最小pH值。 • 整條曲線依金屬之Kf可約略分為三區，於pH＜3可滴定者，3＜pH＜7可滴定者，pH＞7可滴定者。 • 當pH＜3時只有第一區之metals能以EDTA滴定來分析，3＜pH＜7時，則第一、二區之metals理論上都可以EDTA滴定來分析。但當pH為強鹼時，雖然理論上所有的金屬都可被EDTA來滴定，但卻無法區分不同之金屬，此外也有不少金屬在強鹼時會產生氫氧化物的沈澱，也無法以EDTA滴定來分析

  23. 15D-5 How do other complexing agents affect EDTA titration • At the highest pH range, all the metals will react, but not all can be titrated directly due to precipitation of hydroxides. Ex: Fe3+ or Th4+ • [ammonia] influence

  24. 15D-6 Detection of the End Point: Indicators O,O’-dihydroxy azo type (metallochromic indicators) EX: Eriochrome Black T: (EBT)

  25. H2In- HIn2- + H+ pKa=6.3 redblue HIn2-  In3- pKa=11.6 orange pH > 7 MIn + HY3-  HIn2- + MY2- redblue

  26. In general, the metal-indicator complex should be 10 to 100 times less stable than the metal-titrant complex’ The formation constants of the EDTA complexes of calcium and magnesium are too close to differentiate between them in an EDTA titration, so they will titrate together.

  27. Eriochrome Black T cannot be used to indicate the direct titration of calculium alone with EDTA, because the indicator forms too weak a complex with calcium to give a sharp end point.

  28. Calcium can actually be titrated in the presence of magnesium by raising the pH to 12 with strong alkali; Mg(OH)2 precipitates and does not titrate. EBT (pH<6)  H2In- red (pH 6~12)  HIn2- blue (pH>12)  In3- orange

More Related