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COORDINATION CHEMISTRY COMPLEXATION

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COORDINATION CHEMISTRY COMPLEXATION

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    1. 1 COORDINATION CHEMISTRY (COMPLEXATION)

    2. 2 WHY IS CHEMICAL SPECIATION SO IMPORTANT? The biological availability (bioavailability) of metals and their physiological and toxicological effects depend on the actual species present. Example: CuCO30, Cu(en)20, and Cu2+ all affect the growth of algae differently Example: Methylmercury (CH3Hg+) is readily formed in biological processes, kinetically inert, and readily passes through cell walls. It is far more toxic than inorganic forms. Solubility and mobility depend on speciation.

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    4. 4 DEFINITIONS - I Coordination (complex formation) - any combination of cations with molecules or anions containing free pairs of electrons. Bonding may be electrostatic, covalent or a mix. Central atom (nucleus) - the metal cation. Ligand - anion or molecule with which a cation forms complexes. Multidentate ligand - a ligand with more than one possible binding site. Chelation - complex formation with multidentate ligands. Multi- or poly-nuclear complexes - complexes with more than one central atom or nucleus.

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    6. 6 Chelation

    7. 7 DEFINITIONS - II Species - refers to the actual form in which a molecule or ion is present in solution. Coordination number - total number of ligands surrounding a metal ion. Ligation number - number of a specific type of ligand surrounding a metal ion. Colloid - suspension of particles composed of several units, whereas in true solution we have hydration of a single molecule, atom or ion.

    8. 8 FORMS OF OCCURRENCE OF METAL SPECIES

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    10. 10 STABILITY CONSTANTS MEASURE THE STRENGTH OF COMPLEXATION Stepwise constants MLn-1 + L ? MLn Cumulative constants M + nL ? MLn

    11. 11 For a protonated ligand we have: Stepwise complexation MLn-1 + HL ? MLn + H+ Cumulative complexation M + nHL ? MLn + nH+

    12. 12 STABILITY CONSTANTS FOR POLYNUCLEAR COMPLEXES mM + nL ? MmLn mM + nHL ? MmLn + nH+

    13. 13 METAL-ION TITRATIONS Metal ions can be titrated by ligands in the same way that acids and bases can be titrated. According to the Lewis definition, metal ions are acids because they accept electrons; ligands are bases because they donate electrons.

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    15. 15 HYDROLYSIS The waters surrounding a cation may function as acids. The acidity is expected to increase with decreasing ionic radius and increasing ionic charge. For example: Zn(H2O)62+ ? Zn(H2O)5(OH)+ + H+ Hydrolysis products may range from cationic to anionic. For example: Zn2+ ? ZnOH+ ? Zn(OH)20 (ZnO0) ? Zn(OH)3- (HZnO2-) ? Zn(OH)42- (ZnO22-) May also get polynuclear species. Kinetics of formation of mononuclear hydrolysis products is rather fast, polynuclear formation may be slow.

    16. 16 GENERAL RULES OF HYDROLYSIS The tendency for a metal ion to hydrolyze will increase with dilution and increasing pH (decreasing [H+]) The fraction of polynuclear products will decrease on dilution Compare Cu2+ + H2O ? CuOH+ + H+ log *K1 = -8.0 Mg2+ + H2O ? MgOH+ + H+ log *K1 = -11.4

    17. 17 At infinite dilution, pH ? 7 so ?CuOH+ = (1 + 10-7/10-8)-1 = 1/11 = 0.091 ?MgOH+ = (1 + 10-7/10-11.4)-1 = 1/25119 = 4x10-5 Only salts with p*K1 < (1/2)pKw or p*?n < (n/2)pKw will undergo significant hydrolysis upon dilution. Progressive hydrolysis is the reason some salts precipitate upon dilution. This is why it is necessary to add acid when diluting standards.

    18. 18 POLYNUCLEAR SPECIES DECREASE IN IMPORTANCE WITH DILUTION Consider the dimerization of CuOH+: 2CuOH+ ? Cu2(OH)22+ log *K22 = 1.5 Assuming we have a system where: CuT = [Cu2+] + [Cu(OH)+] + 2[Cu2(OH)22+] we can write:

    19. 19 HYDROLYSIS OF IRON(III) Example 1: Compute the equilibrium composition of a homogeneous solution to which 10-4 (10-2) M of iron(III) has been added and the pH adjusted in the range 1 to 4.5 with acid or base. The following equilibrium constants are available at I = 3 M (NaClO4) and 25°C: Fe3+ + H2O ? FeOH2+ + H+ log *K1 = -3.05 Fe3+ + 2H2O ? Fe(OH)2+ + 2H+ log *?2 = -6.31 2Fe3+ + 2H2O ? Fe2(OH)24+ + 2H+ log *?22 = -2.91 FeT = [Fe3+] + [FeOH2+] + [Fe(OH)2+] + 2[Fe2(OH)24+]

    20. 20 Now we define: ?0 = [Fe3+]/FeT; ?1= [FeOH2+]/FeT; ?2= [Fe(OH)2+]/FeT; and ?22= 2[Fe2(OH)24+]/FeT.

    21. 21 These equations can then be employed to calculate the speciation diagrams on the next slide.

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    23. 23 Example 2: Compute the composition of a Fe(III) solution in equilibrium with amorphous ferric hydroxide given the additional equilibrium constants: Fe(OH)3(s) + 3H+ ? Fe3+ + 3H2O log *Ks0 = 3.96 Fe(OH)3(s) + H2O ? Fe(OH)4- + H+ log *Ks4 = -18.7 Fe3+ log [Fe3+] = log *Ks0 - 3pH Fe(OH)4- log [Fe(OH)4-] = log *Ks4 + pH

    24. 24 FeOH+ Fe(OH)3(s) + 3H+ ? Fe3+ + 3H2O log *Ks0 = 3.96 Fe3+ + H2O ? FeOH2+ + H+ log *K1 = -3.05 Fe(OH)3(s) + 2H+ ? FeOH2+ + 2H2O log *Ks1 = 0.91 log [FeOH2+] = log *Ks1 - 2pH Fe(OH)2+ Fe(OH)3(s) + 3H+ ? Fe3+ + 3H2O log *Ks0 = 3.96 Fe3+ + 2H2O ? Fe(OH)2+ + 2H+ log *?2 = -6.31 Fe(OH)3(s) + H+ ? Fe(OH)2+ + H2O log *Ks2 = -2.35 log [Fe(OH)2+] = log *Ks2 - pH

    25. 25 Fe2(OH)24+ 2Fe(OH)3(s) + 6H+ ? 2Fe3+ + 6H2O 2log *Ks0 = 7.92 2Fe3+ + 2H2O ? Fe2(OH)24+ + 2H+ log *?22 = -2.91 2Fe(OH)3(s) + 4H+ ? Fe2(OH)24+ + 4H2O log *Ks22 = 5.01 log [Fe2(OH)24+] = log *Ks22 - 4pH These equations can be used to obtain the concentration of each of the Fe(III) species as a function of pH. They can all be summed to give the total solubility.

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    30. 30 PEARSON HARD-SOFT ACID-BASE (HSAB) THEORY Hard ions (class A) small highly charged d0 electron configuration electron clouds not easily deformed prefer to form ionic bonds Soft ions (class B) large low charge d10 electron configuration electron clouds easily deformed prefer to form covalent bonds

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    33. 33 ION PAIRS VS. COORDINATION COMPLEXES ION PAIRS formed solely by electrostatic attraction ions often separated by coordinated waters short-lived association no definite geometry also called outer-sphere complexes COORDINATION COMPLEXES large covalent component to bonding ligand and metal joined directly longer-lived species definite geometry also called inner-sphere complexes

    34. 34 STABILITY CONSTANTS OF ION PAIRS CAN BE ESTIMATED FROM ELECTROSTATIC MODELS For 1:1 pairs (e.g., NaCl0, LiF0, etc.) log K ? 0 - 1 (I = 0) For 2:2 pairs (e.g., CaSO40, MgCO30, etc.) log K ? 1.5 - 2.4 (I = 0) For 3:3 pairs (e.g., LaPO40, AlPO40, etc.) log K ? 2.8 - 4.0 (I = 0) Stability constants for covalently bound coordination complexes cannot be estimated as easily.

    35. 35 COMPLEX FORMATION AND SOLUBILITY Total solubility of a system is given by: [Me]T = [Me]free + ?[MemHkLn(OH)i] Solubilities of relatively “insoluble” phases such as: Ag2S (pKs0 = 50); HgS (pKs0 = 52); FeOOH (pKs0 = 38); CuO (pKs0 = 20); Al2O3 (pKs0 = 34) are probably not determined by simple ions and solubility products alone, but by complexes such as: AgHS0, HgS22- or HgS2H-, Fe(OH)+, CuCO30 and Al(OH)4-.

    36. 36 Calculate the concentration of Ag+ in a solution in equilibrium with Ag2S with pH = 13 and ST = 0.1 M (20°C, 1 atm., I = 0.1 M NaClO4). Ks0 = 10-49.7 = [Ag+]2[S2-] At pH = 13, [H2S0] << [HS-] because pK1 = 6.68 and pK2 = 14.0 so ST = [HS-] + [S2-] = 0.1 M

    37. 37 [S2-] = 9.1x10-3 M [Ag+]2 = 10-49.7/10-2.04 = 10-47.66 [Ag+] = 10-23.85 = 1.41x10-24 M Obviously, in the absence of complexation, the solubility of Ag2S is exceedingly low under these conditions. The concentration obtained corresponds to ~1 Ag ion per liter. What happens if we take 100 mL of such a solution? Do we then have 1/10 of an Ag ion? No, the physical interpretation of concentration does not make sense here. However, an Ag+ ion-selective electrode would read [Ag+] = 10-23.85 nevertheless.

    38. 38 Estimate the concentration of all species in a solution of ST = 0.02 M and saturated with respect to Ag2S as a function of pH (in other words, calculate a solubility diagram). [Ag]T = [Ag+] + [AgHS0] + [Ag(HS)2-] + 2[Ag2S3H22-] Ks0 = [Ag+]2[S2-], but [S2-] = ?2ST so Ks0 = [Ag+]2 ?2ST

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    43. 43 Region 1: AgHS0 and H2S0 are predominant Ag2S(s) + H2S0 ? 2AgHS0 log [AgHS0] = 1/2log [H2S0] + 1/2log K

    44. 44 Region 3: Ag(HS)2- and HS- are predominant Ag2S(s) + 3HS- + H+ ? 2Ag(HS)2- log [Ag(HS)2-] = 3/2log [HS-] + 1/2log K - 1/2pH

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    46. 46 THE CHELATE EFFECT Multidentate ligands are much stronger complex formers than monodentate ligands. Chelates remain stable even at very dilute concentrations, whereas monodentate complexes tend to dissociate.

    47. 47 WHAT IS THE CAUSE OF THE CHELATE EFFECT? ?Gro = ?Hro - T?Sr0 For many ligands, ?Hro is about the same in multi- and mono-dentate complexes, but there is a larger entropy increase upon chelation! Cu(H2O)42+ + 4NH30 ? Cu(NH3)42+ + 4H2O Cu(H2O)42+ + N4 ? Cu(N4)2+ + 4H2O The second reaction results in a greater increase in ?Sr0.

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    51. 51 METAL-ION BUFFERS Analogous to pH buffers. Consider: Me + L ? MeL

    52. 52 Example: Calculate [Ca2+] of a solution with the composition - EDTA = YT = 1.95x10-2 M, CaT = 9.82x10-3 M, pH = 5.13 and I = 0.1 M (20°C). For EDTA, pK1 = 2.0; pK2 = 2.67; pK3 = 6.16; and pK4 = 10.26.

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    54. 54 MIXED COMPLEXES Examples: Zn(OH)2Cl22-, Hg(OH)(HS)0, PdCl3Br2-, etc. Generalized complexation reaction: M + mA + nB ? MAmBn

    55. 55 In general, mixed complexes usually only predominate under a very restricted set of conditions.

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    57. 57 COMPETITION FOR LIGANDS The ratio of inorganic to organic substances in most natural waters are usually very high. Does a large excess of, say, Ca2+ or Mg2+, decrease the potential of organic ligands to complex trace metals? Example: Fe3+, Ca2+ and EDTA Fe3+ + Y4- ? FeY- log KFeY = 25.1 Ca2+ + Y4- ? CaY2- log KCaY = 10.7 These data suggest that Fe3+ should be complexed by EDTA.

    58. 58 But, let us combine the two above expressions to get: CaY2- + Fe3+ ? FeY- + Ca2+ log Kexchange = 14.4

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