Chemical Equilibrium
Chemical Equilibrium Equilibrium constant • for the reaction: aA + bB + … ⇋ cC + dD + … [C ] c [D ]d ... • equilibrium constant K = a b [ A] [B ] ... ◦ [] = concentration relative to standard state ▪ molarity (M): for dilute solute concentration ▪ in bar: for a gas (i.e., [ ]= P) ▪ 1: pure liquid or solid as a second phase ▪ ≈ 1: solvent (H2O) • equilibrium constant of the reverse reaction would be 1/K. • isf n reactions are added, the new K is the product of the individual equilibrium constants for each of the n reactions What is the standard state for solutes? a) 1.00 b) 1 M c) 100% What is the standard state for gases? a) 1 bar b) 1.00 c) 100% What is the standard state for solids and liquids? a) 1.00 b) pure solid or liquid c) 100% A reaction is thermodynamically favored if: a) K1 Thermodynamics • K = e-ΔGo/RT = e-ΔHo/RT eΔSo/R ◦ because ΔG = ΔH - TΔS ▪ ΔG = Gibbs free energy ▪ H = enthalpy change (+ = endothermic, - = exothermic) ▪ S = entropy change ( + = more disorder, - = less disorder) • A reaction is spontaneous if ΔGo 1 (thermodynamically). • The size of K does not say anything about the kinetics (i.e. rate of reaction)
Le Châtelier’s principle • when a system at equilibrium is disturbed by a change, the system will proceed back to equilibrium in the direction that will partially offset the change • compare the reaction quotient Q to K [C ]c [D ]d ... • Q= [ A]a [B ]b ... ◦ where [] = existing concentration • if Q > K, excess product, reaction goes left • if Q < K, excess reactant, reaction goes right • heat = product for exothermic reaction • heat = reactant for endothermic reaction Solubility product • Ag2CO3 (s) ⇋ 2 Ag+ (aq) + CO32-(aq) • solubility product constant: ◦ Ksp= [CO32-][Ag+]2 ◦ true only if solid is present (independent of amount of solid) • common ion effect = application of Le Châtelier’s principle ◦ if a solid is added to a solution containing one of its constituent ion, less will dissolve The mathematical equation which represents the solubility product when the insoluble compound Mn2S3 is dissolved in water is a) [Mn3+][S2-] = Ksp. b) [Mn3+]3[S2-]2 = Ksp. c) [Mn3+]2[S2-]3 = Ksp. What is the solubility of Ag2CO3 in water? • Ksp = 8.1 x 10-12 • each mol of Ag2CO3 dissolving generates 2 mols Ag+ and one mol of CO32→ solubility = ½ [Ag+] = [CO32-] • Ksp = [CO32-][Ag+]2 = ½ [Ag+] [Ag+]2 = 8.1 10-12 → [Ag+] = 2.5x10-4 M • solubility = ½ (2.5x10-4 M) = 1.3x10-4 M What is the solubility of Ag2CO3 in 0.0200M AgNO3? • molar solubility ≠ ½ [Ag+] • BUT for each mol dissolving → 1 mol of CO32- formed → molar solubility = [CO32-] • Ksp =[CO32-](0.0200 M+2[CO32-])2= 8.1x10-12 ≈ [CO22-](0.0200 M)2 • molar solubility= [CO32-] = 2.0 x 10-8 M ◦ much smaller than when no AgNO3 is present
Which of these statements concerning the solubility is correct? a) A salt is less soluble if one of its ions is already present in solution. b) A salt is no more soluble if one of its ions is already present in solution. c) A salt is more soluble if one of its ions is already present in solution. A beaker contains 250.0 mL of 0.150 molar silver ion (Ag+). To this beaker is added 250.0 mL of 0.300 M bromide ion (Br-). What is the concentration of Ag+ in the final solution? Ksp for AgBr = 5.0 x10-13. a) 7.1 x 10-7 M b) 6.7 x 10-12 M c) 3.3 x 10-12 M solution: 250.0ml x 0.150 M Ag = 0.0375 mol Ag 250.0 ml x 0.300M Br = 0.0750 mol Br final volume is 500 mL Q = [Ag][Br] = (0.0375mol/0.5L)(0.0750/0.5) = 0.01125 > Ksp so will have precipitation at eqm, [Br] = 0.0375 mol (left) / 0.5000L = 0.0750 [Ag+] = Ksp / [Br] = (5.0x10-13) / 0.0750 - 6.7x10-12M = solubility (have an excess of bromine Factors affecting solubility • Temperature ◦ in general, solubility increases with temperature (exception: exothermic rxns) ◦ precipitations carried out at low temperature (if precipitate forms too slowly, an increase in temperature may facilitate its formation). • Solvent ◦ addition of a miscible solvent (MeOH, EtOH) to aqueous solution to decrease solubility. • Composition of the solution ◦ common ion effect ◦ ex.: Solubility of AgCl depends on the concentration of Ag+ and Cl- since [Ag+] [Cl-]=Ksp Remember • each ion has only one concentration in a solution • this concentration must confirm to all equilibrium constants • every solution is electrically neutral
Other factors affecting solubility • Other ion effect ◦ Ksp is defined as a product of activities ▪ the presence of other ions will change μ and the activity coefficients → change in concentration • pH effect ◦ if solid = salt from weak acid: solubility increases as pH decreases ▪ ex.: CaCO3 (s) + H+ ⇋ Ca2+ + HCO3▪ Mg(OH)2(s) + 2H+ ⇋ Mg2+ + 2H2O lewis base ▪ NiS(s) + H+ ⇋ Ni2+ + HSs adduct ◦ Complex formation lewis acid ▪ increases solubility • ex.: AgI(s) + 2CN- → Ag(CN)2- + I• AgCl(s) + 2NH3 → Ag(NH3)2+ + ClWhat is the definition of pH? a) -log [H+] Strictly, answer c) b) -log [H3O+] For most purposes d is good c) -log aH3O+ d) a or b Which of these compounds should increase in solubility in acid solution? a) KCl: no basic propertie b/c Cl is the conjugate base of a strong acid b) CaF2: F is the conjugate base of a weak acid → basic properties c) Ba(OH)2: is a strong base d) NaClO4: perchlorate is the conjugate base of a strong acid e) two of the above Example: dissolution of precipitate by complex formation • AgCl(s) + 2NH3 ⇋ Ag(NH3)2+ + Cl• Ksp=[Ag+] [Cl-] = 10-9.7 • Kf=[Ag(NH3)2+]/{[Ag+][NH3]2} = 107.2 • which concentration of NH3 should be used to dissolve AgCl so as to obtain a 1.0 x 10-2 M solution at the end of dissolution? Solution • [Cl-]= 10-2 M at end of the dissolution • [Cl-]=cAg=[Ag+] + [Ag(NH3)2+] ≈ [Ag(NH3)2+] • From Ksp,[Ag+]=10-9.7/[Cl-]=10-9.7/10-2=10-7.7 M • From Kf=[Ag(NH3)2+]/{[Ag+][NH3 ]2} = 10-2/{10-7.7 [NH3]2}= 107.2 • → [NH3] = 0.18 M • AgCl would not precipitate if ≤ 10-2 M Cl- was added to a solution of 10-2 M Ag+ containing 0.20 M of NH3
Precipitation of hydroxides • most metallic hydroxides are not very soluble ◦ cations separation can be carried out based on the difference in solubility of the corresponding hydroxides • example: Mg2+ + 2OH- ⇋ Mg(OH)2(s) ◦ Ksp = [Mg2+][OH-]2 = 10-10.9 ◦ if OH- is added to 0.010 M Mg2+, Mg(OH)2 will start to precipitate at: ▪ [OH-]2=10-10.9/0.010=10-8.9 → [OH-] = 10-4.45 or pH = 9.55 ▪ if addition of OH- continues, Mg(OH)2 precipitates Separation by precipitation • precipitation is considered complete when 0.01% of initial [Mg2+] is left ◦ ie. when [Mg2+]= 10-5 M (pH = 11.05) • precipitation in the range:9.55
Le Châtelier’s principle • when a system at equilibrium is disturbed by a change, the system will proceed back to equilibrium in the direction that will partially offset the change • compare the reaction quotient Q to K [C ]c [D ]d ... • Q= [ A]a [B ]b ... ◦ where [] = existing concentration • if Q > K, excess product, reaction goes left • if Q < K, excess reactant, reaction goes right • heat = product for exothermic reaction • heat = reactant for endothermic reaction Solubility product • Ag2CO3 (s) ⇋ 2 Ag+ (aq) + CO32-(aq) • solubility product constant: ◦ Ksp= [CO32-][Ag+]2 ◦ true only if solid is present (independent of amount of solid) • common ion effect = application of Le Châtelier’s principle ◦ if a solid is added to a solution containing one of its constituent ion, less will dissolve The mathematical equation which represents the solubility product when the insoluble compound Mn2S3 is dissolved in water is a) [Mn3+][S2-] = Ksp. b) [Mn3+]3[S2-]2 = Ksp. c) [Mn3+]2[S2-]3 = Ksp. What is the solubility of Ag2CO3 in water? • Ksp = 8.1 x 10-12 • each mol of Ag2CO3 dissolving generates 2 mols Ag+ and one mol of CO32→ solubility = ½ [Ag+] = [CO32-] • Ksp = [CO32-][Ag+]2 = ½ [Ag+] [Ag+]2 = 8.1 10-12 → [Ag+] = 2.5x10-4 M • solubility = ½ (2.5x10-4 M) = 1.3x10-4 M What is the solubility of Ag2CO3 in 0.0200M AgNO3? • molar solubility ≠ ½ [Ag+] • BUT for each mol dissolving → 1 mol of CO32- formed → molar solubility = [CO32-] • Ksp =[CO32-](0.0200 M+2[CO32-])2= 8.1x10-12 ≈ [CO22-](0.0200 M)2 • molar solubility= [CO32-] = 2.0 x 10-8 M ◦ much smaller than when no AgNO3 is present
Which of these statements concerning the solubility is correct? a) A salt is less soluble if one of its ions is already present in solution. b) A salt is no more soluble if one of its ions is already present in solution. c) A salt is more soluble if one of its ions is already present in solution. A beaker contains 250.0 mL of 0.150 molar silver ion (Ag+). To this beaker is added 250.0 mL of 0.300 M bromide ion (Br-). What is the concentration of Ag+ in the final solution? Ksp for AgBr = 5.0 x10-13. a) 7.1 x 10-7 M b) 6.7 x 10-12 M c) 3.3 x 10-12 M solution: 250.0ml x 0.150 M Ag = 0.0375 mol Ag 250.0 ml x 0.300M Br = 0.0750 mol Br final volume is 500 mL Q = [Ag][Br] = (0.0375mol/0.5L)(0.0750/0.5) = 0.01125 > Ksp so will have precipitation at eqm, [Br] = 0.0375 mol (left) / 0.5000L = 0.0750 [Ag+] = Ksp / [Br] = (5.0x10-13) / 0.0750 - 6.7x10-12M = solubility (have an excess of bromine Factors affecting solubility • Temperature ◦ in general, solubility increases with temperature (exception: exothermic rxns) ◦ precipitations carried out at low temperature (if precipitate forms too slowly, an increase in temperature may facilitate its formation). • Solvent ◦ addition of a miscible solvent (MeOH, EtOH) to aqueous solution to decrease solubility. • Composition of the solution ◦ common ion effect ◦ ex.: Solubility of AgCl depends on the concentration of Ag+ and Cl- since [Ag+] [Cl-]=Ksp Remember • each ion has only one concentration in a solution • this concentration must confirm to all equilibrium constants • every solution is electrically neutral
Other factors affecting solubility • Other ion effect ◦ Ksp is defined as a product of activities ▪ the presence of other ions will change μ and the activity coefficients → change in concentration • pH effect ◦ if solid = salt from weak acid: solubility increases as pH decreases ▪ ex.: CaCO3 (s) + H+ ⇋ Ca2+ + HCO3▪ Mg(OH)2(s) + 2H+ ⇋ Mg2+ + 2H2O lewis base ▪ NiS(s) + H+ ⇋ Ni2+ + HSs adduct ◦ Complex formation lewis acid ▪ increases solubility • ex.: AgI(s) + 2CN- → Ag(CN)2- + I• AgCl(s) + 2NH3 → Ag(NH3)2+ + ClWhat is the definition of pH? a) -log [H+] Strictly, answer c) b) -log [H3O+] For most purposes d is good c) -log aH3O+ d) a or b Which of these compounds should increase in solubility in acid solution? a) KCl: no basic propertie b/c Cl is the conjugate base of a strong acid b) CaF2: F is the conjugate base of a weak acid → basic properties c) Ba(OH)2: is a strong base d) NaClO4: perchlorate is the conjugate base of a strong acid e) two of the above Example: dissolution of precipitate by complex formation • AgCl(s) + 2NH3 ⇋ Ag(NH3)2+ + Cl• Ksp=[Ag+] [Cl-] = 10-9.7 • Kf=[Ag(NH3)2+]/{[Ag+][NH3]2} = 107.2 • which concentration of NH3 should be used to dissolve AgCl so as to obtain a 1.0 x 10-2 M solution at the end of dissolution? Solution • [Cl-]= 10-2 M at end of the dissolution • [Cl-]=cAg=[Ag+] + [Ag(NH3)2+] ≈ [Ag(NH3)2+] • From Ksp,[Ag+]=10-9.7/[Cl-]=10-9.7/10-2=10-7.7 M • From Kf=[Ag(NH3)2+]/{[Ag+][NH3 ]2} = 10-2/{10-7.7 [NH3]2}= 107.2 • → [NH3] = 0.18 M • AgCl would not precipitate if ≤ 10-2 M Cl- was added to a solution of 10-2 M Ag+ containing 0.20 M of NH3
Precipitation of hydroxides • most metallic hydroxides are not very soluble ◦ cations separation can be carried out based on the difference in solubility of the corresponding hydroxides • example: Mg2+ + 2OH- ⇋ Mg(OH)2(s) ◦ Ksp = [Mg2+][OH-]2 = 10-10.9 ◦ if OH- is added to 0.010 M Mg2+, Mg(OH)2 will start to precipitate at: ▪ [OH-]2=10-10.9/0.010=10-8.9 → [OH-] = 10-4.45 or pH = 9.55 ▪ if addition of OH- continues, Mg(OH)2 precipitates Separation by precipitation • precipitation is considered complete when 0.01% of initial [Mg2+] is left ◦ ie. when [Mg2+]= 10-5 M (pH = 11.05) • precipitation in the range:9.55
