1. 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-

2. 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

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

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

5. 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

6. 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

7. Example : Write the equation of mass balance for a 0
Example : Write the equation of mass balance for a 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-]
= M
Equilibrium conc of H+, CH = [OAc-] + [HO-] +

8. 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) Cl-
Ag(NH3) Ag(NH3) NH3
Ag(NH3) Ag NH3
NH3 + H2O NH HO-
H2O H HO- -

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

10. [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) Ag(NH3)+ + NH3
Ag(NH3) Ag+ + NH3
NH3 + H2O NH4+ + HO-
H2O H+ + HO-
[Ag+] + [Ag(NH3)+] + [Ag(NH3)2+] + [NH4+]+ [H+]
= [HO-] + [Cl-]

11. 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

12. 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+] = [OH-]
[OH-] = 1.0 x /[H+]
답: [H+] = 1.05 x 10-7 (, x 10-7)
Disregard negative number

13. 위 농도는 적정한가?
공통이온효과 무시경우 = = 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 x 10-7
=1.05 x 10-7

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

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

16. Step 4: write the mass balance equation CF =[F-] + [HF] = 2[Ca2+] (4)
Ksp = [Ca2+][F-]2 = 3.9 x (1)
Kb = = 1.5 x (2)
Kw = [H+] [HO-] = 1.0 x (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-]

17. 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

18. 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

19. 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

20. 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

21. 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

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