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Introduction to Analytical Chemistry

Introduction to Analytical Chemistry. CHAPTER 10 COMPLEXATION AND PRECIPITATION TITRATIONS: TAKING ADVANTAGE OF COMPLEXING AND PRECIPITATING AGENTS. 10A Forming Complexes.

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Introduction to Analytical Chemistry

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  1. Introduction toAnalytical Chemistry CHAPTER 10 COMPLEXATION AND PRECIPITATION TITRATIONS: TAKING ADVANTAGE OF COMPLEXING AND PRECIPITATING AGENTS

  2. 10A Forming Complexes • Most metal ions react with electron-pair donors to form coordination compounds or complexes. The donor species, or ligand, must have at least one pair of unshared electrons. • A chelate is produced when a metal ion coordinates with two or more donor groups of a single ligand to form a five- or six-membered heterocyclic ring.

  3. 10A Forming Complexes

  4. 10A Forming Complexes • A ligand that has a single donor group, is called unidentate, one such as glycine, which has two groups available for covalent bonding, is called bidentate. Tridentate, tetradentate, pentadentate, and hexadentate chelating agents are also known.

  5. 10A-1 Producing Soluble Complexes • Complexation reactions involve a metal ion M reacting with a ligand L to form a complex ML, as shown in Equation 10-1: (10-1)

  6. 10A-1 Producing Soluble Complexes (10-5) (10-6) (10-7)

  7. 10A-2 Forming Insoluble Species (10-8)

  8. Example 10-1 • It is desired to separate Ni and Zn by precipitation as the sulfides NiS and ZnS. The solubility products are (Ksp)NiS= 4.0 × 10¯²⁰ and (Ksp)ZnS = 3.0 × 10¯²⁵. If excess KCN is added to a solution initially containing 0.01 M each of Ni²⁺ and Zn²⁺ and the volume remains approximately constant, both ions form the complex almost exclusively.

  9. Example 10-1 • If KCN is added until the [CN¯] is 1.0 M and the [S2¯] is maintained at 0.5 M, can we precipitate ZnS while leaving most of the Ni in solution? Adding CN¯ causes the formation of the Zn²⁺ and Ni²⁺ complexes according to

  10. Example 10-1 • From mass balance, Since the formation constants of the complexes are so large, we can estimate that most of the metals will be in the form of the complex ion.

  11. Example 10-1 • Thus, We can then estimate the free [Ni²⁺] and [Zn²⁺] con-centrations from the β4 values:

  12. Example 10-1 • Now the product of the ion concentrations for NiS ([Ni²⁺][S²¯] = 6.25×10¯³³ × 0.5 = 3.1 ×10¯³³) is seen to be much smaller than Ksp (4.0 × 10¯²⁰) so that the Ni²⁺ remains in solution. The product of the ion concentrations for ZnS ([Zn²⁺][S²¯] = 2.38 × 10¯²² × 0.5 = 1.2 ×10¯²²), however, exceeds Ksp (3.0 × 10¯²⁵) so that ZnS readily precipitates, which then allows a good separation of Zn from Ni.

  13. 10A-3 Ligands That Can Protonate • Complexation with Protonating Ligands • For a diprotic acid, like oxalic acid, the fraction of the total oxalate containing species in any given form, Ox²¯, HOx¯, and H₂Ox, is given by an alpha value . (10-9)

  14. 10A-3 Ligands That Can Protonate • Since we are interested in the free oxalate concentration, (10-10) (10-11) (10-12) (10-13)

  15. 10A-3 Ligands That Can Protonate • Accounting for pH Effects with Conditional Formation Constants • At a particular pH value, α2 is constant, and we can combine K1 and α2 to yield a new conditional constant K’1 : (10-14) (10-15)

  16. 10B-1 Complexation Titrations • As titrants, multidentate ligands, particularly those having four or six donor groups, have two advantages over their unidentate counterparts. • First, they generally react more completely with cations and thus provide sharper end. • Second, they ordinarily react with metal ions in a single-step process.

  17. Figure 10-1 Figure 10-1 Titration curves for complexometric titrations. Titration of 60.0 mL of a solution that is 0.020 M in M with (curve A) a 0.020-M solution of the tetradentate ligand D to give MD as the product; (curve B) a 0.040-M solution of the bidentate ligand B to give MB2 ; and (curve C) a 0.080- M solution of the unidentate ligand A to give MA4 . The overall formation constant for each product is 1020.

  18. 10B-2 Precipitation Titrations • Titrations with silver nitrate are sometimes called argentometric titrations.

  19. Example 10-2 • Calculate the pAg of the solution during the titration of 50.00 mL of 0.0500 M NaCl with 0.1000 M AgNO₃ after the addition of the following volumes of reagent: (a) 0.00 mL, (b) 24.50 mL, (c) 25.00 mL, (d) 25.50 mL. • (a) Because no AgNO₃ has been added, [Ag⁺] 0 and pAg is indeterminate. • (b) At 24.5 mL, [Ag⁺] is very small and cannot be computed from stoichiometric considerations, but [Cl¯] can be obtained readily.

  20. Example 10-2

  21. Example 10-2 • (c) This volume corresponds to the equivalence point where [Ag⁺]=[Cl¯] and • (d)

  22. Figure 10-2 Figure 10-2 Effect of titrant concentration on precipitation titration curves. Curve A shows 50.00 mL of 0.0500 M NaCl with 0.1000 M AgNO3 , and curve B shows 50.00 mL of 0.00500 M NaCl with 0.01000 M AgNO3 . Note the increased sharpness of the break for the more concentrated solution, A.

  23. Figure 10-3 Figure 10-3 Effect of reaction completeness on precipitation titration curves. For each curve, 50.00 mL of a 0.0500 M solution of the anion was titrated with 0.1000 M AgNO3 . Note that smaller values of Ksp give much sharper breaks at the end point.

  24. 10B-2 Precipitation Titrations • Signaling the End Point for Argentometric Titrations • Three types of end points are encountered in titrations with silver nitrate: (1) chemical, (2) potentiometric, and (3) amperometric.

  25. 10B-2 Precipitation Titrations • Formation of Colored Precipitate: The Mohr Method.

  26. 10B-2 Precipitation Titrations • The high chromate ion concentration imparts such an intense yellow color to the solution. • Lower concentrations of chromate ion are generally used, and, as a consequence, excess silver nitrate is required before precipitation begins. • An additional excess of the reagent must also be added to produce enough silver chromate to be seen.

  27. 10B-2 Precipitation Titrations • The Mohr titration must be carried out at a pH of 7 to 10 because chromate ion is the conjugate base of the weak chromic acid. Consequently, in acidic solutions, where the pH is less than 7, the chromate ion concentration is too low to produce the precipitate.

  28. 10B-2 Precipitation Titrations • Adsorption Indicators: The Fajans Method. • Fluorescein is a typical adsorption indicator • In the early stages of the titration of chloride ion with silver nitrate, the colloidal silver chloride particles are negatively charged because of adsorption of excess chloride ions. The dye anions are repelled from this surface by electrostatic repulsion.

  29. 10B-2 Precipitation Titrations • Beyond the equivalence point, the silver chloride particles strongly adsorb silver ions and thereby acquire a positive charge. Fluoresceinate anions are now attracted into the counter-ion layer. • The net result is the appearance of the red color of silver fluoresceinate in the surface layer of the solution surrounding the solid.

  30. 10B-2 Precipitation Titrations • Forming a Colored Complex: The Volhard Method. • Iron(III) serves as the indicator. • The titration must be carried out in acidic solution to prevent precipitation ofiron(III) as the hydrated oxide.

  31. 10B-2 Precipitation Titrations • The most important application of the Volhard method is for the indirect determination of halide ions. A measured excess of standard silver nitrate solution is added to the sample, and the excess silver ion is determined by back-titration with a standard thiocyanate solution.

  32. 10B-2 Precipitation Titrations • A distinct advantage over other titrimetric methods of halide analysis because such ions as carbonate, oxalate, and arsenate (which form slightly soluble silver salts in neutral media but not in acidic media) do not interfere. • Silver chloride is more soluble than silver thiocyanate.

  33. 10B-2 Precipitation Titrations • This reaction causes the end point to fade. • This error can be circumvented by filtering the silver chloride before undertaking the back-titration.

  34. 10C-1 Reagents for Precipitating Metals • One important type of reaction involving an organic complexing agent is that in which an insoluble, uncharged complex is formed. • Precipitation occurs when the solubility of the species MXn has been exceeded. (10-16) (10-17)

  35. 10C-1 Reagents for Precipitating Metals • and the solubility product expression as (10-18)

  36. 10C-2 Forming Soluble Complexes for Extractions and Other Uses • Many organic reagents are useful in converting metal ions into forms that can be readily extracted from water into an immiscible organic phase. • Organic complexing agents for extractions are listed in Table 10-4.

  37. 10D-1 Versatile Ethylenediaminetetraacetic Acid(EDTA) • EDTA, is the most widely used complexometric titrant. • EDTA is a hexadentate ligand. • The various EDTA species are often abbreviated H₄Y, H₃Y¯, H₂Y²¯, HY³¯, and Y⁴¯.

  38. Figure 10-5 Figure 10-5 Structure of H4Y and its dissociation products. Note that the fully protonated species H4Y exist as the double zwitterion with the amine nitrogens and two of the carboxylic acid groups protonated. The first two protons dissociate from the carboxyl groups, whereas the last two come from the amine groups

  39. 10D-2 The Nature of EDTA Complexes with Metal Ions • Solutions of EDTA are particularly valuable as titrants because the reagent combines with metal ions in a 1 : 1 ratio regardless of the charge on the cation. • One of the common structures for metal/EDTA complexes is shown in Figure 10-6.

  40. Figure 10-6 Figure 10-6 Structure of a metal /EDTA complex. Note that EDTA behaves here as a hexadentate ligand in that six donor atoms are involved in bonding the divalent metal cation.

  41. 10D-2 The Nature of EDTA Complexes with Metal Ions • Formation constants KMY (10-20)

  42. 10D-3 Equilibrium Calculations Involving EDTA • where cT is the total molar concentration of uncomplexed EDTA: (10-21)

  43. 10D-3 Equilibrium Calculations Involving EDTA • Conditional Formation Constants • Conditional formation constant K’MY: (10-22) (10-23)

  44. 10D-3 Equilibrium Calculations Involving EDTA • Computing α4 Values for EDTA Solutions • where K₁ , K₂ , K₃ , and K₄ are the four dissociation constants for H₄Y and D is the denominator of Equation 10-24. (10-24) (10-25)

  45. Example 10-4 • Calculate the molar Y⁴¯ concentration in a 0.0200 M EDTA solution buffered to a pH of 10.00. At pH 10.00, α4 is 0.35 (Figure 10-7). Thus,

  46. Example 10-5 • Calculate the equilibrium concentration of Ni²⁺ in a solution with an analytical NiY²¯ concentration of 0.0150 M at pH (a) 3.0 and (b) 8.0. 10-46

  47. Example 10-5 • If we assume that [Ni²⁺] << 0.0150, an assumption that is almost certainly valid in light of the large formation constant of the complex, • Since the complex is the only source of both Ni²⁺ and the EDTA species,

  48. Example 10-5 • Substitution of this equality into Equation 10-23 gives • α4 is 2.5 × 1011 at pH 3.0.

  49. Example 10-5 • (b) At pH 8.0,

  50. Example 10-6 • 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 mixture was buffered to a pH of 3.0. • The solution has an excess of EDTA,

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