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19 Heterogeneous and complex equilibria

19 Heterogeneous and complex equilibria. Formation of crystals such as AgCl in a solution is a heterogeneous equilibrium , because there are more than one phase , AgCl (s) = Ag + (aq) + Cl – (aq) Species such as Ag(NH 3 ) 2 + & Ag(CN) 2 – are complexes ( or complex ions ).

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19 Heterogeneous and complex equilibria

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  1. 19 Heterogeneous and complex equilibria Formation of crystals such as AgCl in a solution is a heterogeneous equilibrium, because there are more than one phase, AgCl (s) = Ag+ (aq) + Cl– (aq) Species such as Ag(NH3)2+& Ag(CN)2– are complexes (or complex ions). Ag+ (aq) + 2 NH3 (aq) = Ag(NH3)2+ (aq) CaCO3 exists as aragonite  calcite 19 Heterogeneous & complex equilibria

  2. Beauty due to heterogeneous equilibria There are many natural heterogeneous equilibria. Please think of some! 19 Heterogeneous & complex equilibria

  3. The solubility product For the dissolution, CaCO3 (s) = Ca2+ (aq) + CO32- (aq) Ksp = [Ca2+] [CO32-] = 3.8e-9 (a constant)solubility product Ksp = [Al3+] [OH-]3 Ksp = [Ca2+]3 [PO4]2 Which is the most and least soluble? 19 Heterogeneous & complex equilibria

  4. Compound Ksp Compound Ksp AgBr 5.0 x 10-13 Fe(OH)3 4.0 x 10-38 AgCl 1.8 x 10-10 FeS 6.3 x 10-18 AgI 8.3 x 10-17 HgS 1.6 x 10-52 AgIO3 3.1 x 10-8 Mg(OH)2 1.8 x 10-11 Ag3PO4 1.3 x 10-20 MgC2O4 8.6 x 10-5 Al(OH)3 2.0 x 10-32 Mn(OH)2 1.9 x 10-13 Ba(OH)2 5.0 x 10-3 MnS 2.5 x 10-13 BaSO4 1.1 x 10-10 NiS 1.0 x 10-24 Bi2S3 1.0 x 10-97 PbCl2 1.6 x 10-5 CaCO3 4.8 x 10-9 PbSO4 1.6 x 10-8 CaC2O4 4.0 x 10-9 PbS 8.0 x 10-28 CaSO4 1.2 x 10-6 SrSO4 3.2 x 10-7 CdS 8.0 x 10-27 Zn(OH)2 3.3 x 10-17 CoS 2.0 x 10-25 ZnS 1.6 x 10-23 CuS 6.3 x 10-36 Solubility product constants Table like this is available in handbooks or data bases. Know where to find them when you need them. WriteKsp expressions. 19 Heterogeneous & complex equilibria

  5. Ksp and solubility 19-2 The Ksp = 1.0e-6 for BaF2, what are [Ba2+] and [F¯]? Solution: BaF2 = Ba2+ + 2 F¯assume  x 2 x Ksp = x (2x)2 = 1.0e-6 x = 3 (1.0e-6 / 4) = 6.3e-3 M [Ba2+] = x = 6.3e-3 M  molar solubility = 6.3e-3 *175 = 1.1g/L [F¯] = 2 x = 0.013 M Checking: 6.3e-3 *0.0132 = 1e-6 = Ksp Calculate solubility of BaCrO3, Ksp = 1.2e-10 Molar mass of BaF2=175 g/mol 19 Heterogeneous & complex equilibria

  6. Concentrations of ions in solution CaF2(s) = Ca2+(aq) + 2 F- (aq), Ksp know where to find Ksp = 5.3e-9 19 Heterogeneous & complex equilibria

  7. Perception of a saturate solution Ag2S = 2 Ag+ + S2– Ksp = [Ag+]2[ S2–] 19 Heterogeneous & complex equilibria

  8. Calculate Ksp from Solubility When 0.50 L saturated CaC2O4 solution was dried, 0.0030 g dry salt was obtained. Evaluate the Ksp. Solution: 3.0e-3 g————0.50 L 1 mol———128 g = 4.7e-5 mol / L = [Ca2+] CaC2O4 = Ca2+ + C2O42- 4.7e-5 4.7e-5 Ksp = 4.7e-5 * 4.7e-5 = 2.2e-9 Figure out how to calculate solubility from Ksp . How many grams of HgS will dissolve in 1 L? 19 Heterogeneous & complex equilibria

  9. Precipitation of AgCl Goto URL: http:// dl.clackamas.cc.or.us/ch105-05/solubili.htm for a demonstration if you have not seen the experiment Ag+ (aq) + Cl- (aq) = AgCl(s) 19 Heterogeneous & complex equilibria

  10. Common-ion effect on solubility 19-3 Like acid-base equilibria, presence of common ions from more than one electrolyte affects the solubility, since Ksp remains constant. For example, the Ksp = 1.8e-10 for AgCl. The maximum [Ag+] is governed by the condition, NaCl = Na+ + Cl- 0.10 0.10 AgCl = Ag+ + Cl- [Ag+] 0.10Thus [Ag+] * 0.10 = 1.8e-10 [Ag+] = 1.8e-9 MSolubility of AgCl = 1.8e-10 = 1.3e-5 M in pure water is 7,454 times more. ? Should we considerAgCl = Ag+ + Cl- x 0.10+x 19 Heterogeneous & complex equilibria

  11. Graph the common ion effect AgCl = Ag+ + Cl- NaCl = Na+ + Cl- 19 Heterogeneous & complex equilibria

  12. Condition for precipitation 19-5 Recall: Predicting reaction directions by comparing Qc and Kc. Same principle applies to precipitation (ppt) Q < Ksp, unsaturated (solution)Q = Ksp, saturated (usually two phases are present)Q > Ksp, super-saturated (unstable, often needs a seed to start the ppt) 19 Heterogeneous & complex equilibria

  13. Separation by precipitation 19-6 A 10-mL solution contains 0.10 M each of Cl–, Br–, and I– ions. Micro amounts of 0.10 M AgNO3 solution is added to the system. What are the Ag+ concentrations before AgI, AgBr, and AgCl precipitate?Solution: Data sheet Ksp:AgCl 1.8e-10, AgBr 5.0e-13, and AgI 8.3e-17. [Ag+] for AgI, AgBr, & AgCl solids to form: [I–] = 0.10 M [Ag+] = 8.3e-17/ 0.10 = 8.3e-16 AgI(s) appears [Br–] = 0.10 M [Ag+] = 5.0e-13/ 0.10 = 5.0e-12 AgBr(s) appears [Cl–] = 0.10 M [Ag+] = 1.8e-10 / 0.10 = 1.8e-9 AgCl(s) appears As [Ag+] increases, how do [Cl-], [Br-] and [I-] vary ? [Ag+] [Cl–] [Br–] [I–] 0  8.3e-16 0.1 0.1 0.10 AgI(s) 5.0e-12 0.1 0.10 AgBr(s) 1.7e-5  1.8e-90.10 AgCl(s)  2.8e-4 4.6e-8 1e-7  1.8e-3  5.0e-6  8.3e-10 0.1 ____? ____? ____? 19 Heterogeneous & complex equilibria

  14. pH H2S and solubility of metal ions 19-7 The pH affects the equilibrium of many species, for example: H2S = H+ + HS–Ka1 = 1e-9 HS– = H+ + S2–Ka2 = 1e-14 (doubtful but adopt) H2S = 2 H+ + S2–Koverall = 1e-23 [S2–] = [H2S] *1e-23 / [H+]2 = [H2S] *1e(2*pH-23), strongly affected by pH If [H2S] = 1.0 M pH [S2–] Ksp of MS [S2–] for [M2+] = 0.001 1 1e-21 1.6e-52 HgS 1.6e-49 (ppt) 2 1e-19 2.5e-27 PbS 2.5e-24 (ppt) 2.52 1.1e-18 1.1e-21 ZnS 1.1e-18 (no ppt pH < 2.5) 4.39 6e-15 6e-18 FeS 6e-15 (no ppt pH < 4.4) 7 1e-9 2.5e-13 MnS _____ (no ppt pH < ____) 10 1e-3 11 0.1 Recalculate [s2-] at various pH if [H2S] = 0.10 M 19 Heterogeneous & complex equilibria

  15. pH, and CO2 on CaCO3 solubility The [CO32–] in a 0.0010 M H2CO3 solution is determined by, H2CO3 = H+ + HCO3–Ka1 = 4.3e-7 HCO3– = H+ + CO32– Ka2 = 5.6e-11 H2CO3 = 2 H+ + CO32–Koverall = Ka1*Ka2 = 2.4e-17 = [H+]2 [CO32–] / [H2CO3][CO32–] = (0.0010*2.4e-17) [H+]–2 = 2.4e-20 * 1e(2*pH) = 2.4e(2*pH-20) [CO32–] affects [Ca2+] due to equilibrium CaCO3 = Ca2+ + CO32–Ksp = 8.7e–9[Ca2+] = 8.7e–9 [CO32–]–1 = 8.7e–9 / 2.4e(2*pH–20) = 3.6e(20 – 9–2*pH) = 3.6e(11-2*pH) Decrease pH by 1 increases [Ca2+] by 2 order of magnitude pH [Ca2+] 8 3.6e-5 7 3.6e-3 6 0.36 19 Heterogeneous & complex equilibria

  16. Stalactites and stalagmites stalactites Rain dissolves limestone and when water drops form stalactites and stalagmites in caves. They grew about 2 cm per 1000 years. The slightly acidic rain dissolves lime stone:CaCO3 (s) + H+ = Ca2+ (aq) + HCO3- (aq) When acidity is reduced, solid forms:Ca2+ (aq) + HCO3- (aq) + OH- (aq) = CaCO3(s) + H2O stalagmites 19 Heterogeneous & complex equilibria

  17. Stalactites hanging down from the ceiling formed over hundreds of years in Mercer Cavern 19 Heterogeneous & complex equilibria

  18. The "Angel Wings" are two delicate and translucent crystalline formations, over 9 feet long and 2.5 feet wide in Mercer Caverns Aragonite formations found at Mercer Caverns, California 19 Heterogeneous & complex equilibria

  19. Equilibrium of complexes 19-8 Metal ions tend to attract Lewis bases forming coordinated complexes, or complex ions. These formation is governed by equilibrium, metal ionligandStep-wise formation constant Ag+ (aq) + NH3 (aq) = Ag(NH3)+ (aq) K1 Ag(NH3)+ (aq) + NH3 (aq) = Ag(NH3)2+ (aq) K2formation constant Ag+ (aq) + 2 NH3 (aq) = Ag(NH3)2+ (aq) Kf = K1 K2 = 1.7e7 Ag+ (aq) + 2 S2O32– (aq) = Ag(S2O3)23– (aq) Kf = 2.9e13 Ag+ (aq) + 2 CN– (aq) = Ag(CN)2– (aq) Kf = 5.6e18Dissociation constant Ag(NH3)2+ (aq) = Ag+ (aq) + 2 NH3 (aq) Kd = 1 / Kf = 5.9e-8 Ag(S2O3)23– (aq) = Ag+ (aq) + 2 S2O32– (aq) Kd = 1/2.9e13 = 3.4e-14 19 Heterogeneous & complex equilibria

  20. Logical, but impractical method!!! Evaluate [Ag+] in a solution containing 1.0 M NH3 and 0.10 M AgNO3. Solution: Ag+ (aq) + 2 NH3 (aq) = Ag(NH3)2+ (aq) Kf = 1.7e7 (find data) 0.1-y 1-2y y (think  way) y ——————— = 1.7e7 (1-2y)2(0.10-y) y ——— = (1-2y)2 (0.1-y) By trial an error, y = 0.10 1.7e7 [Ag+] = 0.10 – y ~ 0 (cannot evaluate [Ag+]) 5.9e-8y = 0.1 –0.8y –2y2 0.10-y = 6e-9, too small, impractical! 19 Heterogeneous & complex equilibria

  21. Concentration of ions in presence of ligands Evaluate [Ag+] in a solution containing 1.0 M NH3 and 0.1 M AgNO3. Solution: Ag+ (aq) + 2 NH3 (aq) = Ag(NH3)2+ (aq) Kf = 1.7e7 (find data) x 0.8+2x 0.1-x (think  way) 0.1-x (treated as dissociation) ————— = 1.7e7 (0.8+2x)2 x 0.1 - x ~ 0.1; 0.8+2x ~ 0.8 (x is very small) then 0.1 0.1 ——— = 1.7e7 x = ————— = 9.2e-9 = [Ag+] 0.82 x 0.82 *1.7e7 3.4e7 x2 + 1.7e7 x – 0.1 = 0x =[ -1.7e7+{(1.7e7)2 + 4*0.1}] /2 = 0 (cannot solve) Small indeed 19 Heterogeneous & complex equilibria

  22. Complex and precipitate formation-1 What is the maximum [Cl-] before AgCl(s) forms in a solution containing 1.0 M NH3 and 0.1 M AgNO3? Solution: Previous slide showed [Ag+] = 9.2e-9 M in a solution when [NH3] = 1.0 MKsp = 1.8e-10 for AgCl (know where to look up) Thus, the max. [Cl-] = 1.8e-10 / 9.2e-9 = 0.02 MWhen no NH3 is present, max. [Cl-] = 1.8e-10/0.1 = 1.8e-9 What is the maximum [Br-] before AgBr(s) forms in a solution containing 1.0 M NH3 and 0.1 M AgNO3? Ksp = 5e-13 for AgBr,Thus, the max. [Br-] = 5e-13 / 9.2e-9 = 5.4e-5 M. 0.02-------- = 368 times5.4e-5 19 Heterogeneous & complex equilibria

  23. Complex and precipitate formation-2 What is the maximum [Br-] in a solution suppose to containing 0.10 M AgNO3 and 0.20 M Na2S2O3 before AgBr(s) (Ksp = 5e-13) forms? Solution: Kf = 2.9e13 for Ag(S2O3)23–; Ag(S2O3)23– = Ag+ + 2 S2O32– , K = 1/2.9e13 = 3.4e-14 0.1-x x 2x 4x3 ——— = 3.4e-14; x = (3.4e-14*0.1)1 / 3 = 1.5e-5 = [Ag+] (0.1–x) small max [Br–] = Ksp/[Ag+] = 5e-13/1.5e-5 = 3.3e-8 (very low) What is [Br–] = ? If [Na2S2O3] is increased to 1.0 M? See next page 19 Heterogeneous & complex equilibria

  24. Complex and precipitate formation-3 What is the maximum [Br-] in a solution suppose to containing 0.10 M AgNO3 and 1.0 M Na2S2O3 before AgBr(s) (Ksp = 5e-13) forms? Solution: Kf = 2.9e13 for Ag(S2O3)23–K = 1/2.9e13 = 3.4e-14 Ag(S2O3)23– = Ag+ + 2 S2O32– 0.1-x x 2x+0.8 (0.8+2x)2x ———— = 3.4e-14, x = 3.4e-14*0.1/0.82 = 5.3e-14 = [Ag+] (0.1-x) small max [Br –] = 5e-13/5.3e-14 =94 M,unrealistically large What is the maximum [I – ] in a solution containing 0.10 M AgNO3 and 1.0 M Na2S2O3 before AgI(s) (Ksp = 8e-17) forms? [I-] = 8e-17 / 5.3e-14 =0.0015 M;small compare to 0.1 M 19 Heterogeneous & complex equilibria

  25. Chemical casserole 19-9 Experiment AgNO3 solution |  Cl – AgCl (s) |  6.0 M NH3 Ag(NH3)2+ + Cl – |  Br– AgBr (s) + NH3, Cl – |  S2O32– Ag(S2O3)23– + Br – + NH3, Cl– |  I– AgI (s) Data Solid Ksp AgCl 1.8e-10 AgBr 5.0e-13 AgI 8.0e-17 Complex Kf Ag(NH3)2+ 1.7e7 Ag(S2O3 )23– 2.9e13 white ivory brown June 29 19 Heterogeneous & complex equilibria

  26. Some complexes Ag(NH3)2+, Cu(CN)32- (Cu(I)), Cu(PPh3)2Br, (NH3)2PtCl2 (cis & trans), Ni(CN)53-, Ligands (Lewis bases): Negative ions, F-, Cl-, Br-, I-, OH-, CN-, SO42-, RCOO-, … Neutral molecules: H2O, NH3, NR3, ROR, CO, C5H5, PR3, C5H5N, …Multidentate ligands: H2NCH2CH2NH2, H2NCH2CH2NHCH2CH2NH2, 19 Heterogeneous & complex equilibria

  27. The heme and hemoglobin (a) The structure of heme is a planar porphoryin ring with iron at the center. (b) Four heme units and four coiled polypeptide chains are bonded together in a molecule of hemoglobin. 19 Heterogeneous & complex equilibria

  28. Amphoteric hydroxides Amphoteric metal hydroxides react with both acids and bases. Examples: (M = Fe, Zn) M(H2O)4(OH)2 + H+ = M(H2O)5(OH)+ M(H2O)5(OH)+ + H+ = M(H2O)62+ M(H2O)4(OH)2 + OH– = M(H2O)3(OH)3 – M(H2O)3(OH)3 – + OH– = M(H2O)2(OH)42 – Aluminum hydroxide behave similarly: Al(H2O)3(OH)3 + H+ = Al(H2O)4(OH)2+ Al(H2O)4(OH)2+ + H+ = Al(H2O)5(OH)2+ Al(H2O)5(OH)2+ + H+ = Al(H2O)63+ Al(H2O)3(OH)3 + OH – = Al(H2O)2(OH)4 – Al(H2O)2(OH)4– + OH – = Al(H2O)(OH)52– Al(H2O)(OH)52– + OH – = Al(OH)63 – 19 Heterogeneous & complex equilibria

  29. Qualitative analysis of metal ions Mixture of metal ions add HClgroup I Ag+, Hg22+, Pb2+filtrate of soluble chloride chloride |  add 0.3 M H+ & H2Sgroup II Cu2+, Cd2+, Hg2+, Pb2+filtrate of soluble metal sulfide As3+, Sb3+, Bi3+, Sn4+ | | add OH- & H2Sacid insoluble sulfide |  (NH4OH) group III Mn2+, Fe2+, Co2+, Ni2+, Zn2+ filtrate of soluble metal ionsbase insoluble sulfide| | add CO32- or PO43-Al(OH)3, Cr (OH)3| |hydroxide |  group IV Mg2+, Ca2+, Sr2+, Ba2+ filtrate of soluble metal ions ppt as carbonates or phosphates | K+, Na+ group V 19 Heterogeneous & complex equilibria

  30. Heterogeneous and complex ion equilibriaSummary Calculate Ksp Evaluate molar solubility (or in other units) from Ksp Discuss concentrations of species in solution with two or more electrolytes (common ion effect, pH effect, separation by ppt) Predicts ppt (heterogeneous equilibria) Describe ligands, metal ions, complexes (ions), and formation constants, and dissociation constant Apply complex formation to explain solubility Be able to get out of traps (think in both directions) 19 Heterogeneous & complex equilibria

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