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Acids and Bases Chapter 12 EDTA Titrations • Definitions: Lewis – Electrons (acid: electron pair acceptor); BrØnsted-Lowry (acid: proton donor) Lewis acid-base concept in Metal-Chelate Complexes Metal ions (electron pair acceptor) Lewis acid Ligand (electron pair donor) Lewis base Coordination Number Complex Formation Formation of coordinate bonds between Lewis Acids/Bases K1 Ag ( aq ) + NH (aq) + • The atom of the ligand that supplies the nonbonding electrons for the metal-ligand bond is the donor atom. • The number of these atoms is the coordination number. 3 Ag ( NH ) ( aq ) + NH (aq) 3 Ag ( NH ) ( aq ) + K2 + 3 1 [ Ag ( NH ) ] [ Ag ][ NH ] + Ag ( NH ) ( aq ) 3 + K = 3 2 + K = 3 2 + [ Ag ( NH ) ] [ Ag ( NH ) ][ NH ] 3 2 + 3 3 3 Formation constants (Kf) are the equilibrium constants for complex ion formation. The overall, or cumulative, formation constants are denoted i Ag (aq) + 2 NH (aq) + Ag ( NH ) (aq) Kf + 3 3 2 + K = f [ Ag ( NH ) ] = β = K ⋅K [ Ag ][ NH ] 3 2 + 2 2 1 2 3 Geometries There are two common geometries for metals with a coordination number of four: Tetrahedral & Square planar By far the mostencountered geometry, when the coordination number is six, is octahedral. Ligands • Monodentate ligand: binds to a metal ion through only one atom, e.g., CN• Multidentate ligand or chelating ligand: has more than one ligand donor atoms. • In ethylenediamine, NH2CH2CH2NH2 (i.e., en), each N is a donor atom. en is bidentate. 1 Chelating Effect A multidentate ligand to form more stable metal complexes than those formed by similar monodentate ligand Chelating Agents • Porphyrins (tetradentate ligands, in heme and chlorophyll) • Adenosine triphosphate (ATP) Some Metals Form 7 or 8 Coordinate Complexes EDTA Ethylenediaminetetraacetic acid (H4EDTA or H4Y) Ethylenediaminetetraacetate anion (EDTA-4 or Y-4) Ethylenediaminetetraacetate, mercifully abbreviated EDTA, has six donor atoms. EDTA is a primary standard material. EDTA Complexes • EDTA forms 1:1 complexes with most metals (Not with Group 1A metals) • EDTA complexes are usually stable water soluble complexes with high formation constants • Formation constant, Kf, (or stability constant): M n + + Y 4− ↔ MY n −4 Kf = [MY ] [M ][Y ] n−4 n+ 4− • Kf could have been defined for any form of EDTA 2 pH affects EDTA titration Acid-Base Properties of EDTA: CH2CO2H HO2CH2C EDTA is a hexaprotic system (H6Y2+) with 4 carboxylic acids and 2 ammoniums HNCH2CH2NH CH2CO2H HO2CH2C H6Y2+ pK1 = 0.0 pK 2 = 1.5 pK 5 = 6.16 pK 6 = 10.24 pK 3 = 2.0 pK 4 = 2.66 Mn+ + Y4− ↔ MYn−4 pH: Hn Yn−4 Fraction of EDTA in the form Y4αY = 4− [H Y ]+ [H Y 4− 44−Y − Y4- [Y ] ] + [H Y ] + [H Y ] + [H Y ]+ [HY ]+ [Y ] 4− αY = ααY == pH Dependence of 6 2+ 5 + 4 3 − 2 2− 3− 4− [Y 4− ] [Y 4− ] = [ EDTA] C EDTA K1K2K3K4 K5K6 {[H ] +[H ] K +[H ] K K +[H ] K K K +[H ] K K K K +[H ]K K K K K + K K K K K K } + 6 + 5 1 + 4 1 2 + 3 1 2 3 + 2 1 2 3 4 + 1 2 3 4 5 1 2 3 4 5 6 Conditional Formation Constant Fractional Composition Diagram for EDTA αY = 4− [Y ] 4- [EDTA] [Y ] = α [EDTA] 4- Kf = Y 4− Conditional formation constant: [MY ] [MY ] [M ][Y ] = [M ]α [EDTA] n−4 n+ n−4 4− n+ Y K f' = α Y 4- K f = Fixing the pH by buffering, then α Y At any fixed pH, find α Y [ MY n − 4 ] [ M n + ][Y 4 − ] Kf = M n + + Y 4− ⇔ MY n− 4 4− 4− 4- [MY ] [M ][EDTA ] n−4 n+ is a constant. and evaluate Kf’ 3 EDTA Titration Curve Example: Calculate the concentration of free Ca2+ in a solution of 0.10 M CaY2- at pH 10 and pH 6. Kf for CaY2- is 1010.65 (Table 12-2) Ca 2+ + EDTA ↔ CaY 1. Excess Mn+ left after each addition of EDTA. Conc. of free metal equal to conc. of unreacted Mn+. K = α Y 4− K f 2− ' f at pH = 10.00, K = α K = (0.30)(10 ) = 1.3 × 10 ' 10.65 Y 4− f 10 f at pH = 6.00, K = α K = (1.8 ×10 )(10 ) = 8.0 × 10 −5 ' Y 4− f 10.65 5 f 2. Equivalence point: [Mn+] = [EDTA] Some free Mn+ generated by MYn-4 ⇔ Mn+ + EDTA Ca 2+ + EDTA ↔ CaY2− K f' = Conci 0 0 0.1 Concf x x 0.1 - x [CaY ] 0.1 − x [Ca ][EDTA ] = x 2− 2+ 2 [ 3. Excess EDTA. Virtually all metal in MYn-4 form. ] x = Ca 2+ = 2.7 ×10−6 M at pH = 10 = 3.5 ×10-4 M at pH = 6 EDTA Titration Curve EDTA Titration Curve 1. Titration reaction: ' M n+ + EDTA⇔ MYn−4 Kf =αY4− Kf EXAMPLE: Derive a titration curve for the titration of 50.0 mL of 0.040 M Ca2+ (buffered at pH=10) with 5.00, 25.00, and 26.00 mL of 0.080 M EDTA. Ca 2+ + EDTA → CaY 2 − K f' = α Y 4− K f = 0.30 *1010.65 = 1.3 × 1010 2. Reaction completes at each point in the titration if Kf’ is large. At equivalence point, Vol. of EDTA = 25.0 mL 5.00 ml, 25.00 ml, 26.00 ml, 3. Plot pM (= -log[Mn+]) vs. volume of EDTA added EDTA Titration Curve EXAMPLE: Derive a titration curve for the titration of 50.0 mL of 0.040 M Ca2+ (buffered at pH=10) with 5.00, 25.00, and 26.00 mL of 0.080 M EDTA. Before the equivalence point At the equivalence point After the equivalence point EDTA Titration Curve EXAMPLE: Derive a titration curve for the titration of 50.0 mL of 0.040 M Ca2+ (buffered at pH=10) with 5.00, 25.00, and 26.00 mL of 0.080 M EDTA. At equivalence point, Vol. of EDTA = 25.0 mL At equivalence point, Vol. of EDTA = 25.0 mL 5mL before the equivalence point almost all the metal is in the form, CaY2- [CaY ] = (0.040) 2- [Ca ] 2+ 25.0 - 5.0 50.0 = (0.040) = 0.0291 M 25.0 5.00 + 50.0 Fraction Remaining pCa 2+ = − log(0.0291) = 1.54 50.0 = 0.0267 M 25.00 + 50.0 Ca 2+ + EDTA ↔ CaY 2− Conci 0 0 0.0267 Concf x x 0.0267 - x K f' = [CaY ] 0.0267 − x [Ca ][EDTA] = x = 1.8 ×10 2− 2+ 10 2 x = 1.2 ×10 −6 M pCa 2+ = − log(1.2 × 10−6 ) = 5.91 4 EDTA Titration Curve EXAMPLE: Derive a titration curve for the titration of 50.0 mL of 0.040 M Ca2+ (buffered at pH=10) with 5.00, 25.00, and 26.00 mL of 0.080 M EDTA. At equivalence point, Vol. of EDTA = 25.0 mL 26.00 mL 1.0 mL excess EDTA, after the equivalence point [EDTA ] = (0.080) [CaY ] = (0.040) 2- K f' = 1.00 = 1.05 × 10 −3 M 50.0 + 26.00 50.0 = 2.63 ×10 −2 M 50.0 + 26.00 [CaY ] 2.63 ×10 [Ca ][EDTA] = [Ca ](1.05 ×10 2− 2+ −2 2+ −3 ) = 1.8 ×1010 Ca 2+ = 1.4 ×10 −9 M pCa 2+ = 8.86 pH affects the titration of Ca2+ with EDTA Auxiliary Complexing Agents • A ligand that binds strongly enough to the metal to prevent hydroxide precipitation, but weak enough to be displaced by EDTA (e.g., ammonia, tartrate, citrate, or trithanolamine) Ammonia is a common auxiliary complex for transition metals like zinc (p. 239) K 'f' = α Zn 2+ α Y 4− K f Kf’’ is the effective formation constant at a fixed concentration of auxiliary complexing agent. 5 Metal Ion Indicators Metal Ion Indicators • To detect the end point of EDTA titrations, we usually use a metal ion indicator or an ion-selective electrode (Ch. 15) • Metal ion indicators change color when the metal ion is bound to EDTA: MgEbT + EDTA ↔ MgEDTA + EbT (Red) (Colorless ) (Blue) – Eriochrome black T is an organic ion • 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. 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+ in M? mol EDTA = (25.00 mL)(0.05283 M) = 1.32 mmol EDTA Zn 2+ + Y 4- ↔ ZnY 2- 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 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. • Displacement titration: For metals without a good indicator ion, the analyte can be treated with excess Mg(EDTA)2-. The analyte displaces Mg, and than Mg 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. • Masking agent: protects some componet of the analyte from reaction with EDTA (render metal ions inactive without actually removing them from solution). Demasking: releasing metal ion from a masking agent. 6