Transcription of Equilibrium constant Chemical equil .: solubility product ...
1 Chem215 product /P 1 Chemical equil .: solubility product (Ch 6-1 to 6-4) Equilibrium constant calculation of free ion concentration in a saturatedsolution. common ion effectSystematic treatment of Equilibrium (Ch 9) Mass balance Charge balance Dependence of solubility on pHChem215 product /P 2 Equilibrium constantFor the reversible reaction,aA + bB cC + dDbeqaeqdeqceqBADCK tconsmequilibriuwhere][][][][,tan=A convention defines K to be a dimensionless examples of Chemical equilibria: solubility , acid-base, complexWhen the forward reaction predominates,KBADCQ badc>=][][][][When the backward reaction predominates,KBADCQ badc<=][][][][Chem215 product /P 3 free energy change and Equilibrium constantwhere G0 is the change in Gibbs free energy at standard =pressureandetemperaturtconsatSTHG andtan000 = Note than K > 1 ( the forward reaction predominates) if G0 <0 ( products are thermodynamically more stable than thereactants)Chem215 product /P 4 solubility ProductHg2Cl2 (s) Hg22+ + 2 Cl-222]][[ +=ClHgKspInitial conc.
2 (M)00 Final conc. (M)x2xwhere x equals solubility218222)2( ]][[xxClHgKsp = = +Since some dissolved Hg2Cl2 may not dissociate into free ion, wesay that its solubility is at least % 10-7 a solution saturated with CaSO4 (s), there is 9 mM [Ca2+], but thetotal Ca in solution (or solubility ) is 15-19 if there is little ion-pair =Chem215 product /P 5 Common ion effectInitial conc. (M) conc. (M)x2x + (s) Hg22+ + 2 Cl-218222) ( ]][[+ = = +xxClHgKspLe Ch telier s Principle states that the direction the system re-establishequilibrium after a change is in such a way that the change is partially predicts that, in the presence of excess Cl-, [Hg22+] < % 10-7 2x ^ , we can make the approximation as 2x + ( )2 = % 10-18x = % 10-15 M This justifies the above product /P 6 Dependence of solubility on pHSolubility of CaF2 CaF2(s) Ca2+ + 2F-Ksp = [Ca2+][F-]2 (9-17)= x 10-11F- + H2O HF + OH- 10 )189(][]][[11- = = FOHHFKb H2O H+ + OH-Kw = [H+][OH-] (9-19)
3 = x 10-14 Chem215 product /P 7 Systematic treatment of equilibriumCharge balanceElectroneutrality of the solution)159(][][][2][2 +=+ ++FOHCaHMass balanceConservation of atoms)169(][2][][2 =++ CaHFFWe can solve the 5 unknowns, [Ca2+], [F-], [HF], [H+]and [OH-] using the above 5 product /P 8 Systematic treatment of equilibriumCombine 9-16 and 9-18, we haveCombine (B) and (9-17), we haveWe can solve the 5 unknowns, [Ca2+], [F-], [HF], [H+] and [OH-]using the above 5 equations.][2][][][2+ =+CaOHFKFb)(][1][2][2 BOHKCaFb + +=222][1][2][ += ++OHKCaCaKbsp)(][14][3122 COHKKC absp += +Chem215 product /P 9 Dependence of solubility on pHFig. 9-2: pH dependence of [Ca2+],[F-], [HF] in a saturated solution ofCaF2 Chem215 product /P 10 Summary: solubility product Equilibrium To write an expression for an Equilibrium constant .
4 To calculate free ion concentration in a saturated solution. (DoExercise 6-20a) To describe the common ion effect. To calculate free ion concentration in a saturated solution in thepresence of a common ion. (Do Exercise 6-20b) To do calculations related to the dependence of solubility on pH(Do Exercise 9-17 & 9-22)
