1 / 19

Fractions of Dissociating Species in Polyligand Complexes

Fractions of Dissociating Species in Polyligand Complexes When polyligand complexes are dissociated in solution, metal ions, ligand, and intermediates are obtained in equilibrium with the complex. For example, look at the following equilibria

idona-preston
Download Presentation

Fractions of Dissociating Species in Polyligand Complexes

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Fractions of Dissociating Species in Polyligand Complexes When polyligand complexes are dissociated in solution, metal ions, ligand, and intermediates are obtained in equilibrium with the complex. For example, look at the following equilibria Ag+ + NH3D Ag(NH3)+ kf1 = [Ag(NH3)+]/[Ag+][NH3] Ag(NH3)+ + NH3D Ag(NH3)2+ kf2 = [Ag(NH3)2+]/[Ag(NH3)+][NH3] We have Ag+, NH3, Ag(NH3)+, and Ag(NH3)2+ all present in solution at equilibrium where CAg = [Ag+] + [Ag(NH3)+] + [Ag(NH3)2+]

  2. The fraction of each Ag+ species can be defined as: b0 = [Ag+]/ CAg b1 = [Ag(NH3)+]/ CAg b2 = [Ag(NH3)2+]/ CAg As seen for fractions of a polyprotic acid dissociating species, one can look at the b values as b0 for the fraction with zero ligand (free metal ion, Ag+), b1 as the fraction of the species having one ligand (Ag(NH3)+) while b2 as the fraction containing two ligands (Ag(NH3)2+). The sum of all fractions will necessarily add up to unity (b0 + b1 + b2 = 1)

  3. For the case of b0, we make all terms as a function of Ag+ since b0 is a function of Ag+. We use the equilibrium constants of each step: CAg = [Ag+] + [Ag(NH3)+] + [Ag(NH3)2+] kf1 = [Ag(NH3)+]/[Ag+][NH3] [Ag(NH3)+] = kf1 [Ag+][NH3] Kf1 x kf2 = [Ag(NH3)2+]/[Ag+][NH3]2 [Ag(NH3)2+] = Kf1 x kf2 [Ag+][NH3]2 Substitution in the CAg relation gives: CAg = [Ag+] + kf1 [Ag+][NH3] + Kf1 x kf2 [Ag+][NH3]2 CAg = [Ag+]( 1 + kf1 [NH3] + Kf1 kf2 [NH3]2) CAg /[Ag+] = ( 1 + kf1 [NH3] + Kf1 kf2 [NH3]2) The inverse of this equation gives: b0 = 1/ ( 1 + kf1 [NH3] + Kf1 kf2 [NH3]2)

  4. If we use the same procedure for the derivation of relations for other fractions we will get the same denominator but the nominator will change according to the species of interest: b0 = 1/ ( 1 + kf1 [NH3] + Kf1 kf2 [NH3]2) b1 = kf1 [NH3] / ( 1 + kf1 [NH3] + Kf1 kf2 [NH3]2) b2 = Kf1 kf2 [NH3]2/ ( 1 + kf1 [NH3] + Kf1 kf2 [NH3]2)

  5. Example Calculate the concentration of the different ion species of silver for 0.010 M Ag+ in a 0.10 M NH3 solution. Assume that ammonia concentration will not change. Kf1 = 2.5x103, kf2 = 1.0x104 Solution The concentration of silver ion species can be obtained as a function of ammonia concentration and formation constants from the relations above: b0 = 1/ ( 1 + kf1 [NH3] + Kf1 kf2 [NH3]2) Substitution in the above equation yields:

  6. b0 = 1/ ( 1 + 2.5x103 * 0.1 + 2.5x103 * 1.0x104 *( 0.10)2) b0 = 4.0x10-6 b0 = [Ag+]/ CAg 4.0x10-6 = [Ag+]/0.010 [Ag+] = 4.0x10-8 M In the same manner calculations give: b1 = 1.0x10-3 b1 = [Ag(NH3)+]/ CAg 1.0x10-3 = [Ag(NH3)+]/ 0.010 [Ag(NH3)+] = 1.0x10-5 M

  7. b2 = 1.0 b2 = [Ag(NH3)2+]/ CAg 1.0 = [Ag(NH3)2+]/ 0.010 [Ag(NH3)2+] = 0.010 M Therefore, it is clear that most Ag+ will be in the complex form Ag(NH3)2+ since the formation constant is large for the overall reaction: Kf = kf1*kf2 Kf = 2.5x103 * 1.0x104 = 2.5x107 The actual concentration of ammonia is 0.08 M not 0.10 M. The value of 0.08 M NH3 should have been used for the calculation of the fractions.

  8. Precipitation Reactions and Titrations

  9. We had previously looked at precipitation equilibria in solution and we are familiar with calculations of solubility of sparingly soluble salts in pure water, in presence of a common ion, and in presence of diverse ions. However, we should remember that a salt is formed from a metal ion (a very weak conjugate Lewis acid) that will not react with water, and a conjugate base that may be weak and will not react with water (like Cl-, Br-, I-) or a strong conjugate base that will react with water (all conjugate bases of weak acids). Calculations presented so far deal with the first situation where the conjugate base is a very weak base. Let us now look at sparingly soluble salts of weak acids:

  10. Effect of Acidity on the Solubility of Precipitates The conjugate base of a weak acid is strong enough to react with water upon dissociation of the precipitate. At low pH values, excessive amounts of the conjugate base will be converted to the weak acid and thus forces the precipitate to further dissociate to produce more anions to satisfy the equilibrium constant (ksp). Look at the general example assuming A-is the conjugate base of a weak acid and in presence of acid solution (low pH):

  11. MA(s)DM++ A-ksp= [M+][A-] A-+ H+DHAka= [HA]/[A-][H+] Since some of the formed A-is converted to HA, [M+] no longer equals [A-]. However, we can write: CT= [A-] + [HA], where CT= [M+] [A-] =a1CT Recall EDTA equilibria in solution when its chelate dissociates in water. Therefore, for the equilibrium reaction: MA(s)DM++ A-

  12. We can write: Ksp= s *a1s S = (ksp/a1)1/2

  13. Example Calculate the solubility of CaC2O4(ka1= 6.5x10-2, ka2= 6.1x10-5, ksp= 2.6x10-9) in a 0.001 M HCl solution. Solution CaC2O4DCa2++ C2O42- The released oxalate will form H2C2O4, HC2O4-, and some will remain as C2O42-where: CC2O42-= s = [Ca2+] = [H2C2O4] + [HC2O4-] + [C2O42-] and: [C2O42-] =a2s

  14. Ksp= s *a2s S = (ksp/a2)1/2 Therefore, we must calculatea2 a2= ka1ka2/ ([H+]2+ ka1[H+] + ka1ka2) a2= (6.5x10-2* 6.1x10-5)/( (0.001)2+ (6.5x10-2* 0.001) + (6.5x10-2* 6.1x10-5) ) a2= 5.7x10-2 s = (2.6x10-9/5.7x10-2)1/2= 2.1x10-4M Compare with solubility in absence of acidity!!

  15. Example Find the solubility of Ag3PO4at pH 3.a3at pH 3=3.3x10-14,ksp= 1.3x10-20 Solution Ag3PO4D3 Ag++ PO43- Solid 3s a3s Ksp= (3 s)3*a3s S = (ksp/27a3)1/4 S = (1.3x10-20/27x3.3x10-14)1/4 S = 1.1x10-2M

  16. Effect of Complexation on Solubility We have seen earlier that acidity affects the solubility of precipitates that contains the conjugate base of a weak acid. In the same manner, if a sparingly soluble precipitate was placed in a solution that contains a ligand which can form a complex with the metal ion of the precipitate, solubility of the precipitate will increase due to complex formation. For example, if solid AgCl is placed in an ammonia solution, the following equilibria will take place:

  17. AgCl(s)D Ag+ + Cl- Ag+ + NH3D Ag(NH3)+ Ag(NH3)+ + NH3D Ag(NH3)2+ The dissociated Ag+ ions will form complexes with ammonia, forcing AgCl to further dissolve. Now, the solubility of AgCl will increase and will equal [Cl-]. However, CAg = [Cl-] = s CAg = [Ag+] + [Ag(NH3)+] + [Ag(NH3)2+] Therefore, [Ag+] = bo s One can then write: AgCl(s)D Ag+ + Cl- Ksp = [Ag+][Cl-] Ksp = bo s * s S = (ksp/bo)1/2

  18. Example Find the solubility of AgBr (ksp = 4x10-13) in pure water and in 0.10 M NH3 (bo in 0.10 M ammonia = 4.0x10-6). Solution a. In pure water AgBr(s)D Ag+ + Br- Ksp = s * s S = (4.0x10-13)1/2 , S = 6.3x10-7 M

  19. b. In presence of 0.10 M NH3 Ksp = bo s * s S = (ksp/bo)1/2 S = (4.0x10-13/4.0x10-6)1/2 S = 3.2x10-4 M % increase in solubility = {(3.2x10-4 – 6.3x10-7)/6.3x10-7} x 100 = 5.1X104 % Huge increase in solubility is expected as calculation above suggests.

More Related