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Chapter 12 EDTA Titrations. Antibiotic chelate captures its prey.

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Presentation on theme: "Chapter 12 EDTA Titrations. Antibiotic chelate captures its prey."— Presentation transcript:

1 Chapter 12 EDTA Titrations

2 Antibiotic chelate captures its prey

3 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. 15A Forming Metal-Chelate Complexes

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

5

6

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

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

9 (3) Neutral EDTA is a tetrabasic acid

10 (5) Five species EDTA as a function of pH

11 Effect of pH on EDTA equilibria

12 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

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

14 15D-3 Equilibrium: pH dependent M-Y

15 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.

16 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/α 4 = 1 + [H + ]/K a4 + [H + ] 2 /K a3 K a4 + …… = 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 ∴ α 4 = 0.35 Ca +2 + EDTA [or Y -4 ] = CaY -2 X  4 [EDTA} [(0.1 × 100)/200] – X ≒ 0.05 K f = [CaY -2 ]/[Ca +2 ] α 4 [C H4Y ] = 0.05/(X)(0.35)(X) = 5.0 × 10 10 X = 1.7 × 10 -6 M pCa = -log (1.7 × 10 -6 ) = 5.77

17 The conditional formation constant (ex:15.5) (1) We can use K ’ f to calculate the equilibrium concentrations of the different species at a given pH. (2) K ’ f value holds for only a specified pH. (3) HgY -2 之 K f > PbY -2 之 K f > CaY -2 之 K f ; K f 不受 pH 值之影響, K f ’ 則 受 pH 值之影響,上述三者在 pH 值≦ 9.0 時, K f ’ 開始變小,也就是 EDTA 的滴定需在 (pH > 9.0) 之鹼性溶液中進行

18 15D-4 EDTA Titration Curves  The end point break depends upon 1) [M n+ ] 2) [L 1 ] 3) [pH]  selectivity 4) K f The smaller K f, the more alkaline the solution must be to obtain a k ’ f of 10 6.

19 15D-4  The titration rxn: M n+ + EDTA  MY n-4 K ’ f =  4 K f Three regions: (1) Before (2) At (3) After equivalence point

20  The K f for CaY -2 is 5.0 × 10 10. At pH 10, α 4 is calculated to be 0.35 to give a K f ` of 1.8 × 10 10. 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.

21 The K f for CaY -2 is 5.0 × 10 10. At pH 10, α 4 is calculated to be 0.35 to give a K f ` of 1.8 × 10 10. 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 × 10 10 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 [C H4Y ] = [0.1 × 150 – 0.1 × 100]/250 = 0.02 M 0.04/[Ca +2 ](0.02) = 1.8 × 10 10 = K f ` [Ca +2 ] = 1.1 × 10 -10 M; pCa = -log 1.1 × 10 -10 = 9.95

22  Various pH: K ’ becomes smaller as the pH decreases

23 K f : cation with larger formation const provide good end point even in acidic media.

24 Minimum permissible pH for a satisfactory end point  以 EDTA 滴定一系列金屬離子,當各自 之 K f ` ≒ 10 6 且能給出一個 sharp end point 之最小 pH 值。  整條曲線依金屬之 K f 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

25 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: Fe 3+ or Th 4+  [ammonia] influence

26 15D-6 Detection of the End Point: Indicators O,O ’ -dihydroxy azo type (metallochromic indicators) EX: Eriochrome Black T: (EBT)

27 H 2 In -  HIn 2- + H + pK a =6.3 red blue HIn 2-  In 3- pK a =11.6 orange pH > 7 MIn + HY 3-  HIn 2- + MY 2- red blue

28 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.

29 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.

30 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)  H 2 In -  red (pH 6~12)  HIn 2-  blue (pH>12)  In 3-  orange


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