Solubility Equilibrium and Complexes PDF

Solubility Equilibrium In saturated solutions dynamic equilibrium exists between undissolved solids and ionic species in solutions Solids continue to dissolve and ion-pairs continue to form solids. The rate of dissolution process is equal to the rate of precipitation.  Solubilit...

Solubility Equilibrium In saturated solutions dynamic equilibrium exists between undissolved solids and ionic species in solutions Solids continue to dissolve and ion-pairs continue to form solids. The rate of dissolution process is equal to the rate of precipitation.  Solubility Product Constant General expression: MmXn(s) ⇄ mMn+(aq) + nXm-(aq) Solubility product, Ksp = [Mn+]m[Xm-]n What is the Ksp expression for AgCl, Cu(NO3)2?  Solubility and Solubility Products Examples: AgCl(s) ⇌ Ag+(aq) + Cl-(aq) Ksp = [Ag+][Cl-] = 1.6 x 10-10 If s is the solubility of AgCl, then: [Ag+] = s and [Cl-] = s Ksp = (s)(s) = s2 = 1.6 x 10-10 s = 1.3 x 10-5 mol/L  Solubility and Solubility Products Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO42-(aq) Ksp = [Ag+]2[CrO42-] = 9.0 x 10-12 If s is the solubility of Ag2CrO4, then: [Ag+] = 2s and [CrO42-] = s Ksp = (2s)2(s) = 4s3 = 9.0 x 10-12 s = 1.3 x 10-4 mol/L  Solubility and Solubility Products More Examples: Ca(IO3)2(s) ⇌ Ca2+(aq) + 2 IO3-(aq) Ksp = [Ca2+][IO3-]2 = 7.1 x 10-7 If the solubility of Ca(IO3)2(s) is s mol/L, then: Ksp = 4s3 = 7.1 x 10-7 s = 5.6 x 10-3 mol/L  Solubility and Solubility Products Mg(OH)2(s) ⇌ Mg2+(aq) + 2 OH-(aq) Ksp = [Mg2+][OH-]2 = 8.9 x 10-12 If the solubility of Mg(OH)2 is s mol/L, then: [Mg2+] = s mol/L and [OH-] = 2s mol/L, Ksp = (s)(2s)2 = 4s3 = 8.9 x 10-12 s = 1.3 x 10-4 mol/L  Solubility and Solubility Products More Examples: Ag3PO4(s) ⇌ 3Ag+(aq) + PO43-(aq) Ksp = [Ag+]3[PO43-] = 1.8 x 10-18 If the solubility of Ag3PO4 is s mol/L, then: Ksp = (3s)3(s) = 27s4 = 1.8 x 10-18 s = 1.6 x 10-5 mol/L Solubility and Solubility Products Cr(OH)3(s) ⇌ Cr3+(aq) + 3 OH-(aq) Ksp = [Cr3+][OH-]3 = 6.7 x 10-31 If the solubility is s mol/L, then: Ksp = [Cr3+][OH-]3 = (s)(3s)3 = 27s4 = 6.7 x 10-31 s = 1.3 x 10-8 mol/L  Solubility and Solubility Products More Examples: Ca3(PO4)2(s) ⇌ 3Ca2+(aq) + 2PO43-(aq) Ksp = [Ca2+]3[PO43-]2 = 1.3 x 10-32 If the solubility is s mol/L, then: [Ca2+] = 3s, and [PO43-] = 2s Ksp = (3s)3(2s)2 = 108s5 = 1.3 x 10-32 s = 1.6 x 10-7 mol/L  Factors that affect solubility Temperature – Solubility generally increases with temperature; Common ion effect – Common ions reduce solubility Salt effect – This slightly increases solubility pH of solution – pH affects the solubility of ionic compounds in which the anions are conjugate bases of weak acids; Formation of complex ion – The formation of complex ion increases solubility  Common Ion Effect Consider the following solubility equilibrium: AgCl(s) ⇌ Ag+(aq) + Cl-(aq); Ksp = 1.6 x 10-10; The solubility of AgCl is 1.3 x 10-5 mol/L at 25oC. If NaCl is added, equilibrium shifts left due to increase in [Cl-] and some AgCl will precipitate out. For example, if [Cl-] = 1.0 x 10-2 M, Solubility of AgCl = (1.6 x 10-10)/(1.0 x 10-2) = 1.6 x 10-8 mol/L  Effect of pH on Solubility Consider the following equilibrium: Ag3PO4(s) ⇌ 3Ag+(aq) + PO43-(aq); If HNO3 is added, the following reaction occurs: H3O+(aq) + PO43-(aq) ⇌ HPO42-(aq) + H2O This reaction reduces PO43- in solution, causing more solid Ag3PO4 to dissolve. In general, the solubility of compounds such as Ag3PO4, which anions are conjugate bases of weak acids, increases as the pH is lowered by adding nitric acid.  Effect of pH on Solubility Consider the following equilibrium: Mg(OH)2(s) ⇌ Mg2+(aq) + 2 OH-(aq); Increasing the pH means increasing [OH-] and equilibrium will shift to the left, causing some of Mg(OH)2 to precipitate out. If the pH is lowered, [OH-] decreases and equilibrium shifts to the right, causing solid Mg(OH)2 to dissolve. The solubility of compounds of the type M(OH)n decreases as pH is increased, and increases as pH is decreased. Formation of Complex Ions on Solubility Many transition metals ions have strong affinity for ligands to form complex ions. Ligands are molecules, such as H2O, NH3 and CO, or anions, such as F-, CN- and S2O32-. Complex ions are soluble – thus, the formation of complex ions increases solubility of slightly soluble ionic compounds. Effect of complex ion formation on solubility Consider the following equilibria: AgCl(s) ⇌ Ag+(aq) + Cl-(aq); Ksp = 1.6 x 10-10 Ag+(aq) + 2NH3(aq) ⇌ Ag(NH3)2+(aq) ; Kf = 1.7 x 107 Combining the two equations yields: AgCl(s) + 2NH3(aq) ⇌ Ag(NH3)2+(aq) + Cl-(aq); Knet = Ksp x Kf = (1.6 x 10-10) x (1.7 x 107) = 2.7 x 10-3 Knet > Ksp implies that AgCl is more soluble in aqueous NH3 than in water.  Solubility Exercise #1 Calculate the solubility of AgCl in water and in 1.0 M NH3 solution at 25oC. Solutions: Solubility in water = (Ksp) = (1.6 x 10-10) = 1.3 x 10-5 mol/L  Solubility Exercise #1 Solubility of AgCl in 1.0 NH3: AgCl(s) + 2NH3(aq) ⇌ Ag(NH3)2+(aq) + Cl-(aq) ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ [Initial], M - 1.0 0.0 0.0 [Change] - -2S +S +S [Equilm.] - (1 – 2S) S S ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ + [Ag(NH 3 ) 2 ][Cl - ] S2 K net = = = 2.7 x 10 -3 [ NH 3 ]2 (1 - 2 S ) 2  Solubility Exercise #1 Solubility of AgCl in 1.0 NH3 (continued): S = 2.7 x 10 -3 = 0.052 (1 - 2 S ) S = 0.052 – 0.104S; S = 0.052/1.104 = 0.047 mol/L AgCl is much more soluble in NH3 solution than in water.  Predicting Formation of Precipitate Qsp = Ksp ➔ saturated solution, but no precipitate Qsp > Ksp ➔ saturated solution, with precipitate Qsp < Ksp ➔ unsaturated solution, Qsp is ion product expressed in the same way as Ksp for a particular system.  Predicting Precipitation Consider the following case: 20.0 mL of 0.025 M Pb(NO3)2 is added to 30.0 mL of 0.10 M NaCl. Predict if precipitate of PbCl2 will form. (Ksp for PbCl2 = 1.6 x 10-5) Calculation: [Pb2+] = (20.0 mL x 0.025 M)/(50.0 mL) = 0.010 M [Cl-] = (30.0 mL x 0.10 M)/(50.0 mL) = 0.060 M Qsp = [Pb2+][Cl-]2 = (0.010 M)(0.060 M)2 = 3.6 x 10-5 Qsp > Ksp ➔ precipitate of PbCl2 will form.  Complex Ion Equilibria Complex ions are ions consisting central metal ions and ligands covalently bonded to the metal ions; Ligands can be neutral molecules such as H2O, CO, and NH3, or anions such as Cl-, F-, OH-, and CN-; For example, in the complex ion [Cu(NH3)4]2+, four NH3 molecules are covalently bonded to Cu2+.  Formation of Complex Ions In aqueous solutions, metal ions form complex ions with water molecules as ligands. If stronger ligands are present, ligand exchanges occur, and equilibrium is established. For example: Cu2+(aq) + 4NH3(aq) ⇌ [Cu(NH3)4]2+(aq) 2+ [Cu(NH 3 ) 4 ] Kf = = 1.1 x 1013 [Cu 2+ ][ NH 3 ]4 Stepwise Formation of Complex Ion At molecular level, ligand molecules or ions combine with metal ions in stepwise manner; Each step has its equilibrium and equilibrium constant; For example: (1) Ag+(aq) + NH3(aq) ⇌ Ag(NH3)+(aq) [Ag(NH 3 ) + ] K f1 = = 2.1 x 10 3 [Ag + ][NH 3 ] (2) Ag(NH3)+(aq) + NH3(aq) ⇌ Ag(NH3)2+(aq); + [Ag(NH 3 ) 2 ] Kf 2 = = 8.2 x 10 3 [Ag(NH 3 ) + ][NH 3 ]  Stepwise Formation of Complex Ion Individual equilibrium steps: (1) Ag+(aq) + NH3(aq) ⇌ Ag(NH3)+(aq); Kf1 = 2.1 x 103 (2) Ag(NH3)+(aq) + NH3(aq) ⇌ Ag(NH3)2+(aq); Kf2 = 8.2 x 103 Combining (1) and (2) yields: Ag+(aq) + 2NH3(aq) ⇌ Ag(NH3)2+(aq); + [Ag(NH 3 ) 2 ] Kf = = K x K = 1.7 x 10 7 [Ag + ][NH 3 ]2 f1 f2 Stepwise complex ion formation for Cu(NH3)42+ Individual equilibrium steps: 1. Cu2+(aq) + NH3(aq) ⇌ Cu(NH3)2+(aq); K1 = 1.9 x 104 2. Cu(NH3)2+(aq) + NH3(aq) ⇌ Cu(NH3)22+(aq); K2 = 3.9 x 103 3. Cu(NH3)22+(aq) + NH3(aq) ⇌ Cu(NH3)32+(aq); K3 = 1.0 x 103 4. Cu(NH3)32+(aq) + NH3(aq) ⇌ Cu(NH3)42+(aq); K4 = 1.5 x 102 Combining equilibrium: Cu2+(aq) + 4NH3(aq) ⇌ Cu(NH3)42+(aq); 2+ [Cu(NH 3 ) 4 ] Kf = [Cu 2+ ][NH 3 ]4 Kf = K1 x K2 x K3 x K4 = 1.1 x 1013