9. Systematic Treatment of Equilibrium Purpose : To solve complex equilibriums Systematic procedure : balanced chemical equation charge balance equation mass balance equation Example : 간단한 예 : 10-8 M HCl의 pH계산 HCl ---> H+ + Cl- H2O ---> H+ + OH-

Charge Balance : 양전하 합 = 음전하 합 principle of electroneutrality in a reaction 양전하 합 = 음전하 합 the coefficient in front of each species always equals the magnitude of the charge on the ion eg If : [PO43-] = 0.01, then the negative charge in 3[PO43-] = 3(0.01) = 0.03 M

Charge Balance Example : Suppose a solution contains ionic species at the following concentrations: [H+] = 5.1 x 10-12 M [H2PO4-] = 1.3 x 10-6 M [K+] = 0.0550 M [HPO42-] = 0.0220 M [OH-] =0.0020 M [PO43-] = 0.0030 M The charge balance is: [H+] +[K+] = [OH-] + [H2PO4-] + 2[HPO42-] + 3[PO43-] Are the charges balanced ?

Considering the positive charges : (5.1 x 10-12) + 0.0550 = 0.0550 Considering the negative charges : 0.0020 + (1.3 x 10-6) + 2(0.0220) + 3(0.0030)

In general, the charge balance for a solution is : n1[C1] + n2[C2] + n3[C3] +… = m1[A1] + m2[A2] + m3[A1] +… where [C] = concentration of a cation n = charge of the cation [A] = concentration of a anion m = charge of the anion

Mass Balance Mass balance : principle is based on the law of mass conservation states that the quantity of all species in a solution containing a particular atom or group of atoms must equal the amount of that atoms or group of atoms the is delivered to the solution

Example : Write the equation of mass balance for a 0 Example : Write the equation of mass balance for a 0.100 M solution of acetic acid. The equilibria are: HOAc H+ + OAc- H2O H+ + HO- Equilibrium conc of acetic acid, CHOAc = sum of the equilibrium concs of all its species = [HOAc] + [OAc-] = 0.100 M Equilibrium conc of H+, CH = [OAc-] + [HO-] +

Example : Write the equations of mass balance for a 1 Example : Write the equations of mass balance for a 1.00 x 10-5 M [Ag(NH3)2]Cl solution [Ag(NH3)2]Cl Ag(NH3)2+ + Cl- Ag(NH3)2+ Ag(NH3)+ + NH3 Ag(NH3)+ Ag+ + NH3 NH3 + H2O NH4+ + HO- H2O H+ + HO- -

- The equilibria are : [Ag(NH3)2]Cl Ag(NH3)2+ + Cl- Ag(NH3)2+ Ag(NH3)+ + NH3 Ag(NH3)+ Ag+ + NH3 NH3 + H2O NH4+ + HO- H2O H+ + HO- CCl = 1.00 x 10-5 M CAg = [Ag+] + [Ag(NH3)+] + [Ag(NH3)2+] = 1.00 x 10-5 M - + CNH3 = [NH4+] + [NH3] + [Ag(NH3)+] + 2 [Ag(NH3)2+] = 2.00 x 10-5 M [HO-] = [NH4+] + [H+]

[Ag+] + [Ag(NH3)+] + [Ag(NH3)2+] + [NH4+]+ [H+] What is the charge balance equation for : [Ag(NH3)2]Cl Ag(NH3)2+ + Cl- Ag(NH3)2+ Ag(NH3)+ + NH3 Ag(NH3)+ Ag+ + NH3 NH3 + H2O NH4+ + HO- H2O H+ + HO- [Ag+] + [Ag(NH3)+] + [Ag(NH3)2+] + [NH4+]+ [H+] = [HO-] + [Cl-]

Systematic Approach to Equilibrium Calculations Step 1: 적절한 반응식 쓰기 Step 2: charge balance equation 쓰기 Step 3: mass balance equation 쓰기 Step 4: step1 반응평형식 쓰기 Step 5: 반응식 수와 미지수 수 비교 Step 6: calculate the answer 반응식 쓰기 : 화학지식 필요 강의에서 주어짐

Step 1: HCl ---> H+ + Cl- H2O ---> H+ + OH- Example : 간단한 예 : 10-8 M HCl의 pH계산 Step 1: HCl ---> H+ + Cl- H2O ---> H+ + OH- Step 2: [H+] = [Cl-] + [OH-] Step 3: [H+] = [Cl-] + [OH-] : same [Cl-] = 10-8 M Step 4: [H+][OH-] = Kw = 1.0 x 10-14 Step 5: 반응식 수 = 3 미지수 수 = 3 Step 6: [H+] = 10-8 + [OH-] [OH-] = 1.0 x 10-14 /[H+] 답: [H+] = 1.05 x 10-7 (, -0.95 x 10-7) Disregard negative number

위 농도는 적정한가? 공통이온효과 무시경우 = 10-7 + 10-8 = 1.1 x 10-7 10-7과 1.1 x 10-7 사이임 다른 이온농도 계산 [OH-] = 0.95 x 10-7 ***최종확인 1. 전하균형: [H+] = [Cl-] + [OH-] 양이온 : 1.05 x 10-7 음이온 : 0.1 x 10-7 + 0.95 x 10-7 =1.05 x 10-7

Step 9: check the validity of the assumption From (1) and (3) = 3.0 x 10-6 [A] = 3.0 x 10-7 = 5.5 x 10-4 M [B] = 5.5 x 10-4 M Step 9: check the validity of the assumption [AB] = CAB - [A] = 0.10 – [A] = 0.10 – 5.5 x 10-4 = 0.10 (within significant figures) [A][B] [AB] 0.10

Dependence of Solubility on pH Find the solubility of CaF2 in water at pH3, given: Ksp = [Ca2+][F-]2 = 3.9 x 10-11 and = 1.5 x 10-11 [HF][OH-] [F-] Step 1: write the balanced chemical equations CaF2(s) Ca2+ + 2F- F- + H2O HF + OH- H2O H+ + HO- If there are other equations? Step 2: To find [Ca2+], [F-], [OH-], [H+] and [HF]

Step 4: write the mass balance equation CF =[F-] + [HF] = 2[Ca2+] (4) Ksp = [Ca2+][F-]2 = 3.9 x 10-11 (1) Kb = = 1.5 x 10-11 (2) Kw = [H+] [HO-] = 1.0 x 10-14 (3) Step 4: write the mass balance equation CF =[F-] + [HF] = 2[Ca2+] (4) Step 5: write the charge balance equation [H+] + 2[Ca2+] = [F-] + [HO-] (5) Step 6: count the equations and chemical species There are five equations and and five chemical species [HF][OH-] [F-]

Step 7: make suitable approximations to simplify the mathematics Let the pH of the solution be 3.00 [H+] = 1.0 x 10-3 M From (3) [HO-] = 1.0 x 10-14/ 1.0 x 10-3 = 1.0 x 10-11 From (2) = = = 1.5 [HF] = 1.5[F-] (6) From (4) [F-] + [HF] = 2[Ca2+] [F-] + 1.5[F-] = 2[Ca2+] [F-] = 0.80[Ca2+] (7) [HF] [F-] Kb [HO-] 1.5 x 10-11 1.0 x 10-11

At high pH, there is very little HF, so [F-]  2[Ca2+] From (1) Ksp = [Ca2+][F-]2 = 3.9 x 10-11 [Ca2+](0.80 [Ca2+])2 = 3.9 x 10-11 [Ca2+] = 3.9 x 10-4 M From (7) [F-] = 0.80[Ca2+] = 3.1 x 10-4 M From (6) [HF] = 1.5[F-] = 4.7 x 10-4 M At high pH, there is very little HF, so [F-]  2[Ca2+] At low pH, there is very little F-, so [HF]  2[Ca2+] ---> See Fig 9-2

pH and Tooth Decay Ca10(PO4)6(OH)2 + 14H+ = 10Ca2+ + 6H2PO4- + 2H2O OH 치아각질(enamel): Hydroxyapatite: ---> Acid Soluble (No Soda!!) Ca10(PO4)6(OH)2 + 14H+ = 10Ca2+ + 6H2PO4- + 2H2O What Acid? Lactic!!! OH CH3CHCO2H

Acid Rain Salts of Basic Anion : F-, OH-, S2-, CO32- PO43- ===> Acid Soluble Acid Rain : SOx, NOx ==> Sulfuric, Nitric Acid in the air r 결과 : 대리석 녹임 : Fig 9-3 토양 녹임 : Fig 9-4

Solubility of HgS Step 1: HgS = Hg2+ + S2- S2- + H2O = HS- + OH- HS- + H2O = H2S + OH- H2O = H+ + OH- If other reactions? Step 2: [Hg2+] + [H+] = 2[S2- ] + [HS-] + [OH-] Step 3: [Hg2+] = [S2- ] + [HS-] + [H2S] Step 4: Ksp = Kb1 = Kb2 = Kw = Step 5: 6 eqn, 6 unknown

At <pH = 6, [Hg2+] = [H2S] At >pH=8, [Hg2+] = [HS-]