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EDTA Titrations Metal Chelate Complexes – complexes formed between a Lewis acid and a Lewis base. Metals ions are Lewis acids, because they accept electrons from Lewis bases. When metal cations combine with Lewis bases, the resulting species is called a complex ion, and the base is called a ligand. The coordination number is the number of covalent bonds that the metal cation tends to form with the electron donor. Ligands: monodentate, bidentate, multidentate [dentate = “toothed”] Tetradentate ligand: ATP (adesonine triphosphate) Chelate effect - ability of multidentate ligands to form more stable metal compounds than those formed by monodentate ligands. K ≡ β2 = 8 x 109 1 EDTA, H6Y2+, is a hexaprotic system; it has six acidic hydrogens that are lost upon metal-complex formation. The first four pK values apply to carboxyl protons, and the last two are for the ammonium protons 2+ Hexaprotic system: H6Y COOH H6Y2+ H+ + H5Y+ pK1 = 0.0 COOH H5Y+ H+ + H4Y pK2 = 1.5 Tetraprotic system: H4Y COOH H4Y H+ + H3Y − pK3 = 2.0 COOH H3Y − H+ + H2Y 2− pK4 =2.69 + Commonly used: Na2H2Y• 2H2O NH H2Y 2− H+ + HY 3− pK5 = 6.13 NH+ HY 3− H+ + Y 4− pK6 = 10.37 o Note: pK values are at 25 C and µ = 0.1 M except pK1 applies at µ = 1.0 M Complexometric titration: a titration based n a complex formation. The pH of an EDTA solution affects the equilibrium constant of complex formation. Solutions of high pH used in analytical procedures do not significantly effect the stability of a complex. The fraction of EDTA in each of its protonated forms is shown in the figure below: 2 We define α for each species as the fraction of EDTA at that form: [EDTA] = [H6Y2+]+ [H5Y+]+ [H4Y]+ [H3Y − ]+ [H2Y 2− ]+ [HY 3− ]+ [Y 4− ] [Y 4− ] αY 4− = [ EDTA] ← this is the concentration of “free” EDTA, not complexed to metal ions. The α fraction is a function of pH. (Equation 12-4 of Harris) αY = 4− K1 K 2 K 3 K 4 K 5 K 6 {[ H ] + [ H ] K1 + [ H ] K1 K 2 + [ H ] K1 K 2 K 3 + [ H + ]2 K1 K 2 K 3 K 4 + [ H + ]K1 K 2 K 3 K 4 K 5 + K1 K 2 K 3 K 4 K 5 K 6 } + 6 + 5 + 4 + 3 (Equation 12-4 of Harris) Reaction with a Metal Ion This Kf is defined only for Y 4− This is a problem!!!! We saw from the fraction plot and Table 12-1 that most of the EDTA is not in the form of Y 4− below a pH ~10. 3 Solution to the problem: Conditional formation constant, K 'f • We can derive a more useful equilibrium equation by rearranging the fraction relationship: αY = 4− [Y ] 4- [EDTA ] [ ] ⇒ Y 4- = α Y 4− [EDTA ] [MY ] = [MY ] = [M ][Y ] [M ]α [EDTA] n −4 Kf n+ 4− n −4 n+ Y 4- αY 4− • If we fix the pH of the titration with a buffer, then is a constant that can be combined with Kf K = α Y 4- K f ' f [MY ] = [M ][EDTA ] n −4 n+ K f' = α Y 4− K f M n + + EDTA ⇔ MY n − 4 Example Calculate the concentration of free Ca2+ in a solution of 0.10 M CaY 2− at pH 10 and pH 6. Kf for CaY 2− is 4.5x1010 (Table 12-2) K f' = α Y 4− K f Ca 2+ + EDTA ↔ CaY 2− at pH = 10.00, K f' = α Y 4− K f = (0.30)( 4.5 × 1010 ) = 1.34 × 1010 at pH = 6.00, K f' = α Y 4− K f = (1.8 × 10− 5 )( 4.5 × 1010 ) = 8.0 × 105 Ca 2+ + EDTA ↔ CaY 2− Conci 0 0 0.1 Concf x x 0.1 - x K ' f [CaY ] = 0.1 − x = [Ca ][EDTA] x 2− 2+ 2 4 We solve w/r/t x and at pH 10: [Ca2+] = 2.7 x 10-6 and at pH 6: [Ca2+] = 3.5 x 10-4 Note : At low pH, the metal-complex is less stable Calcium/EDTA Titration Curve For calcium, the end point becomes hard to detect below ~pH=8. The formation constant is too small below this point. This can be used to separate metals. At pH=4, Ca2+ does not perform significant complexaion with EDTA. Generic Titration Curve We’ll consider a titration where we have 50.0 mL of 0.040 M Ca2+ (buffered at pH=10) with 0.080 M EDTA Ca 2+ + EDTA ↔ CaY 2− K f' = α Y 4− K f (50.0 mL)(0.040 M) = Ve (0.080 M) ⇒ Ve= 25.0 mL From the previous example at pH 10.00 K 'f = 1.34 x1010 • Before the Equivalence Point: What is pCa2+ when we have added 5.0 mL of EDTA? 5 - 5.0 50.0 [Ca ] = 25.0 (0.040) = 0.0291 M 25.0 55.0 2+ ↑ Fraction remaining ↑ ↑ initial dilution factor concentration pCa 2+ = − log(1.2 × 10−6 ) = 5.91 • At the Equivalence Point: What is pCa2+ when we have added 25.0 mL of EDTA? At the equivalence point almost all the metal is in the form CaY 2− [CaY ] = (0.040) 50.0 = 0.0267 M 75.0 2- ↑ ↑ initial dilution factor concentration Free Ca2+ is small and can be found: Ca 2+ + EDTA ↔ CaY 2− K ' f Conci 0 0 0.0267 Concf x x 0.0267 - x [CaY ] = 0.0267 − x = 1.34 × 10 = [Ca ][EDTA] x 2− 2+ 10 2 x = 1.4 × 10−6 M pCa 2+ = − log(1.4 × 10 −6 ) = 5.85 • After the Equivalence Point: What is pCa2+ when we have added 26.0 mL of EDTA? We have 1 mL excess of EDTA. [EDTA ] = (0.080) 1.0 = 1.05x10-3 M 76.0 ↑ ↑ Initial [EDTA] dilution factor 6 [CaY ] = (0.040) 50.0 = 2.63 × 10 76.0 2- ↑ 2+ −2 M ↑ Initial [Ca ] dilution factor K ' f [CaY ] = 2.63 × 10 = [Ca ][EDTA ] [Ca ](1.05 × 10 2− −2 2+ 2+ −3 ) = 1.34 × 1010 [Ca2+] = 1.9 x 10-9 M pCa2+ = 8.73 Auxiliary Complexing Agent This is a ligand that binds strongly enough to the metal to prevent hydroxide precipitation, but weak enough to be displaced by EDTA. Ammonia is a common auxiliary complex for transition metals like Zn. Metal Ion Indicators To detect the end point of EDTA titrations, we usually use a metal ion indicator or an ion-selective electrode (chapter in Electrochemistry). Metal ion indicators change color when the metal ion is bound to EDTA: MgEbT + EDTA ↔ MgEDTA + EbT (Red) (Clear) (Clear) (Blue) The indicator must bind less strongly than EDTA. EDTA Titration Techniques • Direct titration: analyte is titrated with standard EDTA with solution buffered at a pH where Kf’ is large. • Back titration: known excess of EDTA is added to analyte. Excess EDTA is titrated with 2nd metal ion. 7 • Displacement titration: For metals without a good indicator ion, the analyte can be treated with excess Mg(EDTA)2-. The analyte displaces Mg2+, and than Mg2+ can be titrated with standard EDTA. • Indirect titration: Anions can be analyzed by precipitation with excess metal ion and then titration of the metal in the dissolved precipitate with EDTA. Example 25.0 mL of an unknown Ni2+ solution was treated with 25.00 mL of 0.05283 M Na2EDTA. The pH of the solution was buffered to 5.5 and than back-titrated with 17.61 mL of 0.02299 M Zn2+. What was the unknown Ni2+ M? Zn 2+ + Y 4- ↔ ZnY 2mol EDTA = (25.00 mL)(0.05283 M) = 1.32 mmol EDTA mol Zn 2+ = (17.61 mL)(0.02299 M) = 0.4049 mmol Zn 2+ Ni 2+ + Y 4- ↔ NiY 2mol Ni 2+ = 1.321 mmol EDTA - 0.4049 mmol Zn 2+ = 0.916 mmol M Ni 2+ = (0.916 mmol)/(25.00 mL) = 0.0366 M 8