숙제 : 7 장 문제 : 답이 있는 30 번 까지 짝수 번 문제

Systematic Treatment of Equilibrium In this chapter we will learn: -some of the tools to deal with all types of chemical equilibria Systematic procedure : -balanced chemical equation -charge balance equation -mass balance equation Charge Balance : -principle electroneutrality in a reaction -sum of positive charge = sum of negative charge -the coefficient in front of each species always equals the magnitude of the charge on the ion eg If : [PO 4 3- ] = 0.01, then the negative charge in 3[PO 4 3- ] = 3(0.01) = 0.03 M

Example : Suppose a solution contains ionic species at the following concentrations: [H + ] = 5.1 x M [H 2 PO 4 - ] = 1.3 x M [K + ] = M [HPO 4 2- ] = M [OH - ] = M [PO 4 3- ] = M The charge balance is: [H + ] +[K + ] = [OH - ] + [H 2 PO 4 - ] + 2[HPO 4 2- ] + 3[PO 4 3- ] Are the charges balanced ? Considering the positive charges : ( 5.1 x ) = Considering the negative charges : (1.3 x ) + 2(0.0220) + 3(0.0030) = Charge Balance

In general, the charge balance for a solution is : n 1 [C 1 ] + n 2 [C 2 ] + n 3 [C 3 ] +… = m 1 [A 1 ] + m 2 [A 2 ] + m 3 [A 1 ] +….. 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 M solution of acetic acid. The equilibria are: HOAc H + + OAc - H 2 OH + + HO - Equilibrium concentration of acetic acid, C HOAc = sum of the equilibrium concentrations of all its species = [HOAc] + [OAc - ] = M Equilibrium concentration of H +, C H = [OAc - ] + [HO - ] + Example : Write the equations of mass balance for a 1.00 x M [Ag(NH 3 ) 2 ]Cl solution

The equilibria are : [Ag(NH 3 ) 2 ]Cl Ag(NH 3 ) Cl - Ag(NH 3 ) 2 + Ag(NH 3 ) + + NH 3 Ag(NH 3 ) + Ag + + NH 3 NH 3 + H 2 O NH HO - H 2 O H + + HO - C Cl = 1.00 x M C Ag = [Ag + ] + [Ag(NH 3 ) + ] + [Ag(NH 3 ) 2 + ] = 1.00 x M - + C NH = [NH 4 + ] + [NH 3 ] + [Ag(NH 3 ) + ] [Ag(NH 3 ) 2 + ] = 2.00 x M [HO - ] = [NH 4 + ] + [H + ] 3

What is the charge balance equation for : [Ag(NH 3 ) 2 ]Cl Ag(NH 3 ) Cl - Ag(NH 3 ) 2 + Ag(NH 3 ) + + NH 3 Ag(NH 3 ) + Ag + + NH 3 NH 3 + H 2 O NH HO - H 2 O H + + HO - [ Ag + ] + [ Ag(NH 3 ) + ] + [ Ag(NH 3 ) 2 + ] + [ NH 4 + ] + [ 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

Example : 간단한 예 : 10-8 M HCl 의 pH 계산 Step 1: HCl ---> H+ + Cl- H2O ---> H+ + OH- Step 2: [H+] = [Cl-] + [OH-] Step 3: [H+] = [Cl-] + [OH-] : same [Cl-] = 10-8M Step 4: [H+][OH-] = K w = 1.0 x Step 5: 반응식 수 = 3 미지수 수 = 3 Step 6: [H+] = [OH-] [OH-] = 1.0 x /[H+] 답 : [H+] = 1.05 x 10-7 (, x 10-7) Disregard negative number [A][B] [AB]

위 농도는 적정한가 ? 공통이온효과 무시경우 = = 1.1 x 과 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 x 10-7 =1.05 x 10-7

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

Dependence of Solubility on pH Find the solubility of CaF 2 in water at pH3, given: K sp = [Ca 2+ ][F - ] 2 = 3.9 x and = 1.5 x [HF][OH - ] [F-][F-] Step 1: write the balanced chemical equations CaF 2 (s) Ca F - F - + H 2 OHF + OH - H 2 O H + + HO - If there are other equations? Step 2: To find [Ca 2+ ], [F - ], [OH - ], [H + ] and [HF]

Step 3: K sp = [Ca 2+ ][F - ] 2 = 3.9 x (1) K b = = 1.5 x (2) K w = [H + ] [HO - ] = 1.0 x (3) Step 4 : write the mass balance equation C F =[F - ] + [HF] = 2[Ca 2+ ](4) Step 5: write the charge balance equation [H + ] + 2[Ca 2+ ] = [F - ] + [HO - ] (5) Step 6 : count the equations and chemical species There are five equations and and five chemical species [HF][OH - ] [F-][F-]

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

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

pH and Tooth Decay 치아각질 (enamel): Hydroxyapatite: ---> Acid Soluble ( No Soda!!) Ca 10 (PO 4 ) 6 (OH) H + = 10Ca H2PO4- + 2H2O What Acid? Lactic!!! OH CH 3 CHCO 2 H

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 = Hg 2+ + S 2- S 2- + H2O = HS- + OH- HS- + H2O = H2S + OH- H2O = H+ + OH- If other reactions? Step 2: [Hg 2+] + [H+] = 2[S 2- ] + [HS-] + [OH-] Step 3: [Hg 2+ ] = [S 2- ] + [HS-] + [H2S] Step 4: Ksp = Kb1 = Kb2 = Kw = Step 5: 6 eqn, 6 unknown

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