1. 1-1 Thermodynamics and kinetics Heats of reaction Thermodynamic laws Electrochemistry Kinetics Acid-Base Equilibrium calculations

2. 1-2 Heats of Reaction 1 cal = 4.184 j Heat Capacity (C p ) §Heat required to raise one gram of substance 1 °C §Al; C p = 0.895 j/gK §What is the heat needed to 40 g Al 10 K §(0.895 j/gK)(40g)(10K)= 358 j Exothermic §Reaction produces heat (at 25 °C) C(s) + O 2 (g) CO 2 (g) + 393.76 kj

3. 1-3 Heats of Reaction Endothermic §Reaction requires energy (at 25 °C) 2 HgO + 181.70 kj 2 Hg + O 2 Enthalpy (∆H) Energy of a system (heat content) §∆H = ∆H products - ∆H reactants §Exothermic reactions have negative ∆H §Negative ∆H tend to be spontaneous

4. 1-4 Enthalpy (∆H) Bond energies §Can be used to estimate ∆H §N 2 + 3 H 2 2 NH 3 §6(351.5)-945.6-3(436.0) = -144.6 kj/2 mole =-72.3 kj/mole (actual -46.1 kj/mol) Aqueous Ions (use ∆H values) §∆H products - ∆H reactants à2 H + + CO 3 2- CO 2 + H 2 O à-393.5 + (-285.8)-(-677.1+2(0)) = -2.2 kj/mol CO 2(g) H 2 O (l) CO 3 2- H+H+

5. 1-5 Bond Energies (kJ/mol) at 298 K HH436.0 HC418.4 HN351.5 HO464.4 HF564.8 HS338.9 HCl431.8 HBr368.2 HI297.1 CC343.1 CN267.8 CO347.3 CF426.8 CS255.2 CCl330.5 CBr276.1 Single Bonds CI217.6 NN133.9 NF234.3 NCl154.8 OO138.1 OF188.3 OSi443.5 OCl209.2 OS518.8 FF158.2 PCl326.4 SS205.0 ClCl242.7 ClI209.2 BrBr192.5 II150.6 Single Bonds Double BondTriple Bond CC610.9836.8 CN615.0891.2 CO740.61071.1 NN418.4945.6 Bond in O 2 498.7

6. 1-6 Heat Capacities Li3.582B1.026 C0.709N1.042 O0.920Al0.900 Cr0.449Cu0.385 Zr0.278Hf0.140 W0.130Au0.128 Hg0.140Pb0.129 Th0.113U0.120 Np0.120Pu0.130 Am0.110 Elements (Jg -1 K -1 ) H 2 28.87 O 2 29.50 N 2 29.04 CO29.16 CO 2 37.49 CH 4 35.98 C 2 H 6 53.18 NH 3 36.11 H 2 O (g) 24.77 HBr27.53 SnO 2 56.61 Al 2 O 3 79.33 Fe 2 O 3 104.35 Compounds (Jmol -1 K -1- )

7. 1-7 Pure Substance Standard ∆H (kJ/mol) Ag(g)284.55 Ag + (g)1021.73 AgCl(c,cerargyrite)-127.068 AgNO 3 (c)-124.39 Al 3+ (g)5483.17 Al(OH) 3 -1276 BaCO 3 (c,witherite)-1216.3 BaC 2 O 4 (c)-1368.6 BaCrO 4 (c)-1446 BaF 2 (c)-1207.1 BaSO 4 (c)-1473.2 BeO(c)-609.6 Bi 2 S 3 (c)-143.1 Br 2 (g)30.907 Br(g)111.88 Br-(g)-219.07 C(c,diamond)1.895 C(g)716.682 CO(g)-110.525 CO 2 (g)-393.509 COCl 2 (g,phosgene)-218.8 CH 4 (g,methane)-74.81 C 2 H 2 (g,ethyne)226.73 C 2 H 2 (g,ethene)52.25 C 2 H 6 (g,ethane)-84.68 CH 3 OCH 3 (g)-184.05 CH 3 OH(g,methanol)-200.66 CH 3 OH(l,methanol)-238.66 C 2 H 5 OH(g,ethanol)-235.1 C 2 H 5 OH(l,ethanol)-277.69 CH 3 COOH(l,acetic acid)-484.51 (CH 3 ) 2 O(g)-184.05 CH 3 CHO(l)-192.3 CH 3 Cl(g)-80.83 CHCl 3 (g)-103.14 CCl 4 (l)-135.44 CaO(c)-635.09 Ca(OH) 2 (c)-986.09 CaCO 3 (c, calcite)-1206.92 CaCO 3 (c, aragonite)-1207.13

8. 1-8 Pure Substance Standard ∆H (kJ/mol) CaC 2 O 4 (c)-1360.6 CaF 2 (c)-1219.6 Ca 3 (PO 4 ) 2 (c)-4109.9 CaSO4(c,anhydrite)-1434.1 Cd(g)2623.54 Cd 2+ (g)112.01 Cd(OH) 2 (c)-560.7 CdS(c)-161.9 Cl(g)121.679 Cl-(g)-233.13 ClO 2 (g)102.5 Cu(g)338.32 Cu 2 O(c,cuprite)-168.6 CuO(c,tenorite)-157.3 Cu(OH) 2 (c)-449.8 Cu2S(c,chalcocite)-79.5 CuS(c,covellite)-53.1 F(g)78.99 F-(g)-255.39 Fe(g)416.3 Fe 3+ (g)5712.8 Fe 2 O 3 (c,hematite)-824.2 H + (g)1536.202 H 2 O(g)-241.818 H 2 O(l)-285.83 H 2 O 2 (g)-136.31 H 2 O 2 (l)-187.78 H 2 SO 4 (l)-813.989 HF(g)-271.1 HCl(g)-92.307 HBr(g)-36.4 HI(g)26.48 HCN(g)135.1 PbO(c,red)-218.99 PbO 2 (c)-277.4 Pb 3 O 4 (c)-718.4 PbS(c,galena)-100.4 PbSO 4 (c)-919.94 ThO 2 (c)-1226.4 UO2(c)-1084.9

9. 1-9 Solution Standard ∆H (kJ/mol) Ag+105.579 AgCl 2 - -245.2 Ag(NH 3 ) 2+ -111.29 Ag(S 2 O 3 ) 2 - -1285.7 Al 3+ -531 Br--121.55 BrO 3 --67.07 Ca 2+ -542.83 Cd 2+ -75.9 Cd(CN) 4 2- 428 Cd(NH 3 ) 4 2+ -450.2 Ce 3+ -696.2 Ce 4+ -537.2 CH 3 COO--486.01 CH 3 COOH-485.76 CN-150.6 CNS-76.44 Cl--167.15 ClO 4 --129.33 CO 2 -413.8 CO 3 2 --677.14 H+0 H 2 O 2 -191.17 I--55.19 I 3 --51.5 IO 3 --221.3 K+-252.38 NH 3 -80.29 NH 4 +-132.51 NO 3 --205 Na+-240.12 OH--229.99 O 2 -11.7 SO 4 2 --909.27 Sn 2+ -8.8 Sr 2+ -545.8 Tl 3+ 196.6 U 4+ -591.2 UO 2 2 +-1019.6

10. 1-10 ∆H determination from other data Reactants at 298 K ∆H reactants =(  C p )(T 2 -298) Reactants at T 2 ∆H T2 Products at T 2 ∆H products =(  C p )(298-T 2 ) Products at 298 K ∆H 298  C p is the sum of the heat capacities

11. 1-11 Entropy (∆S) Randomness of a system §increase in ∆S tends to be spontaneous Enthalpy and Entropy can be used for evaluating the free energy of a system Gibbs Free Energy §∆G = ∆H -T∆S §∆G=-RTlnK àK is equilibrium constant àActivity at unity

12. 1-12 Calculations Compound∆G° (kJ/mol) at 298.15 K H 2 O-237.129 OH - (aq) -157.244 H + (aq) 0 H 2 O  H + +OH - §What is the constant for the reaction? At 298.15 K ∆G(rxn) = 0 + -157.244 - (-273.129) = 79.9 kJ/mol lnK= (79.9E3/(-8.314*298.15))=-32.2 K=1E-14

13. 1-13 Thermodynamic Laws 1st law of thermodynamics §If the state of a system is changed by applying work or heat or both, then the change in the energy of the system must equal the energy applied. à∆E = q (heat absorbed) + w (work) §Conservation of energy àEnergy can be transferred, but no direction is given

14. 1-14 Thermodynamic Laws 2nd law of thermodynamics §Reactions tend towards equilibrium àIncrease in entropy of a system §Spontaneous reaction for -∆G §∆G = 0, system at equilibrium 3rd law of thermodynamics §Entropies of pure crystalline solids are zero at 0K

15. 1-15 Faraday Laws In 1834 Faraday demonstrated that the quantities of chemicals which react at electrodes are directly proportional to the quantity of charge passed through the cell 96487 C is the charge on 1 mole of electrons = 1F (faraday)

16. 1-16 Faraday Laws Cu(II) is electrolyzed by a current of 10A for 1 hr between Cu electrode §anode: Cu Cu 2+ + 2e - §cathode: Cu 2+ + 2e - Cu §Number of electrons à(10A)(3600 sec)/(96487 C/mol) = 0.373 F à0.373 mole e - (1 mole Cu/2 mole e - ) = 0.186 mole Cu

17. 1-17 Half-cell potentials Standard potential  Defined as  °=0.00V àH 2 (atm) 2 H + (1.000M) + 2e - Cell reaction for §Zn and Fe 3+/2+ at 1.0 M §Write as reduction potentials  Fe 3+ + e - Fe 2+  °=0.77 V  Zn 2+ + 2e - Zn  °=-0.76 V §Fe 3+ is reduced, Zn is oxidized

18. 1-18 Half-Cell Potentials Overall  2Fe 3+ +Zn 2Fe 2+ + Zn 2+  °=0.77+0.76=1.53 V Half cell potential values are not multiplied Application of Gibbs If work is done by a system  ∆G = -  °nF (n= e - ) Find ∆G for Zn/Cu cell at 1.0 M  Cu 2+ + Zn Cu + Zn 2+   °=1.10 V §2 moles of electrons (n=2) à∆G =-2(96487C/mole e - )(1.10V) à∆G = -212 kJ/mol

19. 1-19 Reduction Potentials Electrode Couple"E0, V" Na+ + e- --> Na-2.7144 Mg2+ + 2e- --> Mg-2.3568 Al3+ + 3e- --> Al-1.676 Zn2+ + 2e- --> Zn-0.7621 Fe2+ + 2e- --> Fe-0.4089 Cd2+ + 2e- --> Cd-0.4022 Tl+ + e- --> Tl-0.3358 Sn2+ + 2e- --> Sn-0.141 Pb2+ + 2e- --> Pb-0.1266 2H+ + 2e- --> H2(SHE)0 S4O62- + 2e- --> 2S2O32-0.0238 Sn4+ + 2e- --> Sn2+0.1539 SO42- + 4H+ + 2e- --> H2O + H2SO3(aq)0.1576 Cu2+ + e- --> Cu+0.1607 S + 2H+ + 2e- --> H2S0.1739 AgCl + e- --> Ag + Cl-0.2221 Saturated Calomel (SCE)0.2412 UO22+ + 4H+ + 2e- --> U4+ + 4H2O0.2682

20. 1-20 Reduction Potentials Hg2Cl2 + 2e- --> 2Cl- + 2Hg0.268 Bi3+ + 3e- --> Bi0.286 Cu2+ + 2e- --> Cu0.3394 Fe(CN)63- + e- --> Fe(CN)64-0.3557 Cu+ + e- --> Cu0.518 I2 + 2e- --> 2I-0.5345 I3- + 2e- --> 3I-0.5354 H3AsO4(aq) + 2H+ + 2e- -->H3AsO3(aq) + H2O0.5748 2HgCl2 + 4H+ + 2e- -->Hg2Cl2 + 2Cl-0.6011 Hg2SO4 + 2e- --> 2Hg + SO42-0.6152 I2(aq) + 2e- --> 2I-0.6195 O2 + 2H+ + 2e- --> H2O2(l)0.6237 O2 + 2H+ + 2e- --> H2O2(aq)0.6945 Fe3+ + e- --> Fe2+0.769 Hg22+ + 2e- --> Hg0.7955 Ag+ + e- --> Ag0.7991 Hg2+ + 2e- --> Hg0.8519 2Hg2+ + 2e- --> Hg22+0.9083 NO3- + 3H+ + 2e- -->HNO2(aq) + H2O0.9275

21. 1-21 Reduction Potentials VO2+ + 2H+ + e- --> VO2+ + H2O1.0004 HNO2(aq) + H+ + e- --> NO + H2O1.0362 Br2(l) + 2e- --> 2Br-1.0775 Br2(aq) + 2e- --> 2Br-1.0978 2IO3- + 12H+ + 10e- -->6H2O + I21.2093 O2 + 4H+ + 4e- --> 2H2O1.2288 MnO2 + 4H+ + 2e- -->Mn2+ + 2H2O1.1406 Cl2 + 2e- --> 2Cl-1.3601 MnO4- + 8H+ + 5e- -->4H2O + Mn2+1.5119 2BrO3- + 12H+ + 10e- -->6H2O + Br21.5131

22. 1-22 Nernst Equation Compensated for non unit activity (not 1 M) Relationship between cell potential and activities aA + bB +ne - cC + dD At 298K 2.3RT/F = 0.0592 What is potential of an electrode of Zn(s) and 0.01 M Zn 2+ Zn 2+ +2e - Zn  °= -0.763 V activity of metal is 1

23. 1-23 Kinetics and Equilibrium Kinetics and equilibrium are important concepts in examining and describing chemistry §Identify factors which determine rates of reactions àTemperature, pressure, reactants, mixing §Describe how to control reactions §Explain why reactions fail to go to completion §Identify conditions which prevail at equilibrium

24. 1-24 Kinetics Rate of reaction §Can depend upon conditions àPaper reacts slowly with oxygen, but increases with temperature Free energy does not dictate kinetics 16 H + + 2 MnO 4 - + 5 Sn 2+ 5 Sn 4+ + 2 Mn 2+ + 8 H 2 O  °= 1.39 V, ∆G = -1390 kcal/mol 8 H + + MnO 4 - + 5 Fe 2+ 5 Fe 3+ + Mn 2+ + 4 H 2 O  °= 0.78 V, ∆G = -376 kcal/mol §The reaction with Fe is much faster

25. 1-25 Kinetics Rate can depend upon reaction path §C 6 H 12 O 6 (s) + 6 O 2 (g) 6 CO 2 (g) + 6 H 2 O(l) §∆G = -791 kcal/mol §Glucose can be stored indefinitely at room temperature, but is quickly metabolized in a cell Thermodynamics is only concerned with difference between initial and final state Kinetics account for reaction rates and describe the conditions and mechanisms of reactions

26. 1-26 Kinetics Kinetics are very difficult to describe from first principles §structure, elements, behavior General factors effecting kinetics §Nature of reactants §Effective concentrations §Temperature §Presence of catalysts §Number of steps

27. 1-27 Nature of Reactants Ions react rapidly §Ag + + Cl - AgCl(s) Very fast Reactions which involve bond breaking are slower §NH 4 + + OCN - OC(NH 2 ) 2 Redox reactions in solutions are slow §Transfer of electrons are faster than those of atomic transfer Reactions between covalently bonded molecules are slow §2 HI(g) H 2 (g) + I 2 (g)

28. 1-28 Nature of Reactants Structure §P 4 Same formula Concentration §Reaction can occur when molecules contact P P P P P P P P P P P P White Phosphorus Red Phosphorus

29. 1-29 Concentration Surface area §larger surface area increases reaction Mixing increases interaction Need to minimized precipitation or colloid formation Rate Law Concentration of reactant or product per unit time Effect of initial concentration on rate can be examined

30. 1-30 Rate Law rate = k[A] x [B] y rate order = x + y knowledge of order can help control reaction rate must be experimentally determined Injection mixing detector Flow meter

31. 1-31 Rates Rate=k[A] n ; A=conc. at time t, A o =initial conc., X=product conc. Orderrate equationk 0[A 0 ]-[A]=kt, [X]=ktmole/L sec 1 ln[A 0 ]-ln[A]=kt, ln[A 0 ]-ln([A o ]-[X])=kt 1/sec 2L/mole sec 3 L 2 /mole 2 sec

32. 1-32 Rate Law Half-life  A =A o e - t  = ln2/t 1/2 §If a rate half life is known, fraction reacted or remaining can be calculated (CH 3 ) 2 N 2 (g) N 2 (g) + C 2 H 6 (g) TimePressure (torr) 036.2 3046.5

33. 1-33 Rate Law Temperature §Reactions tend to double for every 10 °C Catalysts §Accelerate reaction but are not used àPt surface §Thermodynamically drive, catalysts drive kinetics §If not thermodynamically favored, catalysts will not drive reaction Autocatalytic reactions form products which act as catalysts Catalytic amount is moles of catalysts needed to cause reaction

34. 1-34 Complexation Kinetics Uranium and cobalt with pyridine based ligands 111Py12111Py14222Py14 Examine complexation by UV-Visible spectroscopy

35. 1-35 Complexation of U with 111Py12

36. 1-36 Absorbance Kinetics Absorbance sum from 250 nm to 325 nm for 111Py12 and uranium at pH 4

37. 1-37 Kinetic Data Evaluation Evaluation of absorbance and kinetic data for 111Py12 and 111Py14 with uranium at pH 4. The concentration of ligand and uranium is 50x10 -6 mol/L. LigandAbs o ∆Abs eq k (min -1 )95% Equilibrium Time (min) 111Py127.86±0.825.66±1.284.65±0.47x10 -5 6.44±0.65x10 4 111Py144.82±1.707.06±5.764.24±0.80x10 -5 7.07±1.33x10 4 Evaluation of change in absorbance

38. 1-38 Acid-Base Equilibria Water can act as solvent and reactant Brønsted Theory of Acids and Bases §Acid àSubstance which donates a proton §Base àAccepts proton from another substance NH 3 + HCl NH 4 + + Cl - H 2 O + HCl H 3 O + + Cl - NH 3 + H 2 O NH 4 + + OH - Remainder of acid is base Complete reaction is proton exchange between sets Extent of exchange based on strength

39. 1-39 Acid Strengths Strong acids tend towards completion HCl + H 2 O H 3 O + + Cl - §Strong acids tend to have weak conjugate bases Weak acid forms only slightly ionized species CH 3 COOH + H 2 O CH 3 COO - + H 3 O + §Stronger conjugate base Relative strengths of acids can be compared §Can be rated relative to water Some salts can have acid-base properties §NH 4 Cl

40. 1-40 Relative Strengths of Acids and Bases Conjugate AcidConjugate Base HClO 4 ClO 4 - H 2 SO 4 SO 4 2- HClCl - H 3 O + H 2 O H 2 SO 3 HSO 3 - HFF - HC 2 H 3 O 2 C 2 H 3 O 2 - HSO 3 - SO 3 2- H 2 SHS - NH 4 + NH 3 HCO 3 CO 3 2- H 2 OOH - HS - S 2- OHO 2- H 2 H - Acid Strength Base Strength

41. 1-41 Dissociation Constants Equilibrium expression for the behavior of acid HA + H 2 O A - + H 3 O + Water concentration is constant pK a =-logK a Can also be measured for base Constants are characteristic of the particular acid or base

42. 1-42 Dissociation Constants for Acids at 25°C AcidFormulaK a AceticHC 2 H 3 O 2 1.8E-5 HydrocyanicHCN7.2E-10 CarbonicH 2 CO 3 3.5E-7 HCO 3 - 5E-11 NitrousHNO 2 4.5E-4 HydrosulfuricH 2 S1E-7 HS - 1E-14 PhosphoricH 3 PO 4 7.5E-3 H 2 PO 4 - 6.2E-8 HPO 4 2- 4.8E-13 OxalicH 2 C 2 O 4 5.9E-2 HC 2 O 4 - 6.4E-5

43. 1-43 Dissociation Constants for Bases at 25°C BaseFormulaK b AmmoniaNH 3 1.8E-5 PyridineC 5 H 5 N2.3E-9 MethylamineCH 3 NH 2 4.4E-4 Protonation of amine group

44. 1-44 Calculations 1 L of 0.1 M acetic acid has pH = 2.87 What is the pK a for acetic acid CH 3 COOH + H 2 O CH 3 COO - + H 3 O + [CH 3 COO - ] = [H 3 O + ] =10 -2.87 pK a =4.73

45. 1-45 Calculations What is pH of 0.1 M NH 3 K b =1.8E-5 NH 3 + H 2 O NH 4 + + OH - [NH 4 + ] = [OH - ] = x [NH 3 ] = 0.1 - x x 2 + 1.8E-5x -1.8E-6 = 0 [OH - ] = 1.33E-3 M, pOH = 2.87, pH ≈ 14-pOH ≈11.12

46. 1-46 Acid-Base Reactions For Water 2 H 2 O H 3 O + + OH - Water concentration remains constant, so for water: K w = [H 3 O + ][OH - ]= 1E-14 at 25°C

47. 1-47 Common ion effect Same products from different acids (hydronium ion) 0.2 M CH 3 COOH in 0.1 M HCl What is the free acetic acid concentration HCl is totally dissociated x=[CH 3 COO - ], [CH 3 COOH]=0.2-x, [H 3 O + ] = 0.1 + x K a = 1.8E-5 Small K a, x is much smaller than 0.1 x=3.6E-5 M

48. 1-48 Hydrolysis Constants Reaction of water with metal ion §Common reaction §Environmentally important §Dependent upon metal ion oxidation state M z+ + H 2 O MOH z-1+ + H + Constants are listed for many metal ion with different hydroxide amounts

49. 1-49 Buffers Weak acid or weak base with conjugate salt Acetate as example §Acetic acid, CH 3 COONa §CH 3 COOH + H 2 O CH 3 COO - + H 3 O + §If acid is added àhydronium reacts with acetate ion, forming undissociated acetic acid §If base is added àHydroxide reacts with hydronium, acetic acid dissociates to removed hydronium ion large quantityhuge quantitylarge quantitysmall quantity

50. 1-50 Buffer Solutions Buffers can be made over a large pH range Can be useful in controlling reactions and separations §Buffer range àEffective range of buffer àDetermined by pK a of acid or pK b of base HA + H 2 O A - + H 3 O - Write as pH

51. 1-51 Buffer Solutions The best buffer is when [HA]=[A - ] §largest buffer range for the conditions §pH = pK a - log1 For a buffer the range is determined by [HA]/[A - ] §[HA]/[A - ] from 0.1 to 10 §Buffer pH range = pK a ± 1 §Higher buffer concentration increase durability

52. 1-52 Buffers What is a good buffer system for maintaining a solution at pH 4? Want 0.1 M total in 1L total volume For acetic acid, pK a =4.75 We have 0.1 M each Volume CH 3 COO - =1/6.62 L = 0.15 L Acid =1-0.15 =0.85 L

53. 1-53 Buffers You have a buffer with 0.1 M acetic acid and 0.1 M sodium acetate in 1L. 0.01 moles of NaOH(s) is added. What is the pH? NaOH reacts with 0.01 moles of proton from the acid. At equilibrium, we have 0.11 M sodium acetate and 0.09 M acetic acid [H 3 O + ]=1.5E-5, pH =4.8 §(pK a =4.75)

54. 1-54 Equilibrium Reactions proceed in the forward and reverse direction simultaneously § N 2 + 3 H 2 2 NH 3 §Initially contains nitrogen and hydrogen àForward rate decreases as concentration (pressure) decreases àAmmonia production increase reverse rate àEventually, forward rate is equal to reverse rate àNo net change in concentration àReaction still occurring

55. 1-55 Equilibrium Some reactions have a negligible reverse rate §Proceeds in forward direction §Reaction is said to go to completion Le Châtelier’s Principle At equilibrium, no further change as long as external conditions are constant Change in external conditions can change equilibrium §A stressed system at equilibrium will shift to reduce stress àconcentration, pressure, temperature

56. 1-56 Equilibrium N 2 + 3 H 2 2 NH 3 + 22 kcal § What is the shift due to àIncreased temperature? àIncreased N 2 ? àReduction of reactor vessel volume?

57. 1-57 Equilibrium Constants For a reaction §aA + bB cC + dD At equilibrium the ratio of the product to reactants is a constant §The constant can change with conditions §By convention, constants are expressed as products over reactants Conditions under which the constant is measured should be listed §Temperature, ionic strength

58. 1-58 Activities Strictly speaking, activities, not concentrations should be used At low concentration, activities are assumed to be 1 The constant can be evaluated at a number of ionic strengths and the overall activities fit to equations

59. 1-59 Activities Debye-Hückel (Physik Z., 24, 185 (1923)) Z A = charge of species A µ = molal ionic strength R A = hydrated ionic radius in Å (from 3 to 11) First estimation of activity

60. 1-60 Debye-Hückel term can be written as: Specific ion interaction theory §Uses and extends Debye-Hückel àlong range Debye-Hückel àShort range ion interaction term  ij = specific ion interaction term Activities

61. 1-61 Activities Pitzer §Binary (3) and Ternary (2) interaction parameters Ca 2+ K+K+ Al 3+ Fe(CN) 6 4-

62. 1-62 Cm-Humate at pH 6 Experimental Data shows change in stability constant with ionic strength Ion Specific Interaction Theory used

63. 1-63 Constants Constants can be listed by different names §Equilibrium constants (K) àReactions involving bond breaking *2 HY H 2 + 2Y §Stability constants (ß), Formation constants (K) àMetal-ligand complexation *Pu 4+ + CO 3 2- PuCO 3 2+ *Ligand is written in deprotonated form

64. 1-64 Constants §Conditional Constants àAn experimental condition is written into equation *Pu 4+ + H 2 CO 3 PuCO 3 2+ +2H + ËConstant can vary with concentration, pH Must look at equation! Often the equation is not explicitly written

65. 1-65 Using Equilibrium Constants Constants and balanced equation can be used to evaluate concentrations at equilibrium §2 HY H 2 + 2Y, §K=4E-15 §If you have one mole of HY initially, what are the concentration of the species at equilibrium? §Try to write species in terms of one unknown àStart will species of lowest concentration à[H 2 ] =x, [Y]=2x, [HY]=1-2x §Since K is small, x must be small, 1-2x ≈ 1 §K=4x 3, x =1E-5, 2x=2E-5

66. 1-66 Realistic Case Consider uranium in an aquifer §Species to consider include àfree metal ion: UO 2 2+ àhydroxides: (UO 2 ) x (OH) y àcarbonates: UO 2 CO 3 àhumates: UO 2 HA(II), UO 2 OHHA(I) §Need to get stability constants for all species àUO 2 2+ + CO 3 2- UO 2 CO 3 §Know or find conditions àTotal uranium, total carbonate, pH, total humic concentration

67. 1-67 Stability constants for selected uranium species at 0.1 M ionic strength Specieslogß UO 2 OH + 8.5 UO 2 (OH) 2 17.3 UO 2 (OH) 3 - 22.6 UO 2 (OH) 4 2- 23.1 (UO 2 ) 2 OH 3+ 11.0 (UO 2 ) 2 (OH) 2+ 22.0 UO 2 CO 3 8.87 UO 2 (CO 3 ) 2 2- 16.07 UO 2 (CO 3 ) 3 4- 21.60 UO 2 HA(II)6.16 UO 2 (OH)HA(I)14.7±0.5 Other species may need to be considered. If the total uranium concentration is low enough, binary or tertiary species can be excluded.

68. 1-68 Equations Write concentrations in terms of species §[UO 2 ] tot = UO 2free +U-carb+U-hydroxide+U-humate §[CO 3 2- ] free =f(pH) §[OH - ] = f(pH) §[HA] tot = UO 2 HA + UO 2 OHHA+ HA free Write the species in terms of metal, ligands, and constants §[(UO 2 ) x A a B b ] = 10 -(xpUO2+apA+bpB-log(UO2)xAaBb) §pX = -log[X] free à[(UO 2 ) 2 (OH) 2 2+ ]=10 -(2pUO2+2pOH-22.0) Set up equations and solve for known terms

69. 1-69 U speciation with different CO 2 partial pressure 0% CO 2 1% CO 2 10% CO 2

70. 1-70 Comparison of measured and calculated uranyl organic colliod

71. 1-71 Energy terms Constants can be used to evaluate energetic of reaction §From Nernst equation à∆G=-RTlnK §∆G=∆H-T∆S à-RTlnK = ∆H-T∆S àRlnK= - ∆H/T + ∆S *Plot RlnK vs 1/T

72. 1-72

73. 1-73 Solubility Products Equilibrium involving a solid phase §AgCl(s) Ag + + Cl - §AgCl concentration is constant àSolid activity and concentration is treated as constant àBy convention, reaction goes from solid to ionic phase in solution §Can use K sp for calculating concentrations

74. 1-74 Solubility calculations AgCl(s) at equilibrium with water at 25°C gives 1E-5 M silver ion in solution. What is the K sp ?? §AgCl(s) Ag + + Cl - : [Ag + ] = [Cl - ] §K sp = 1E-5 2 = 1E-10 What is the [Mg 2+ ] from Mg(OH) 2 at pH 10? § K sp = 1.2E-11= [Mg 2+ ] [OH] 2 §[OH] = 10 -(14-10)

75. 1-75 Solubility Calculations 50 mL of 0.02 M Ag + added to 50 mL 0.1 M HCl, what is the silver ion concentration in solution?? §So assume all Ag + forms AgCl §0.001 moles Ag +, 0.005 moles Cl - §[Cl - ] = 0.004 moles/0.1 L =0.04 M §K sp = 1E-10 §[Ag + ] =1E-10/0.04 =2.5E-9 M

76. 1-76 Solubility Calculations What is the maximum pH of a 0.1 M Mg 2+ solution in which Mg(OH) 2 will not precipitate?? §K sp = 1.2E-11= [Mg 2+ ][OH - ] 2 §[OH - ]=(1.2E-11/0.1) 0.5 =1.09E-5 §pOH = 4.96 §pH = 9.04 How can this be used to separate metal ions? §Separate Mg 2+ from Fe 2+, 0.1 M each metal ion. Want minimum 500:1 Mg 2+ :Fe 2+ §K sp (Fe(OH) 2 )=1.6E-14

77. 1-77 Limitations of K sp Solid phase formation limited by concentration §below ≈1E-5/mL no visible precipitate forms àcolloids formation of supersaturated solutions §slow kinetics Competitive reactions may lower free ion concentration Large excess of ligand may form soluble species §AgCl(s) + Cl - AgCl 2 - (aq) K sp really best for slightly soluble salts