숙제 : 7 장 문제 : 답이 있는 30 번 까지 짝수 번 문제

Systematic Systematic Treatment of Treatment of EquilibriumEquilibriumIn 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 : [PO43-] = 0.01, then the negative

charge in

3[PO43-] = 3(0.01) = 0.03 M

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[HPO4

2-]

+ 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)

= 0.0550

Charge BalanceCharge Balance

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 BalanceMass 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 0.100 M solution of acetic acid.

The equilibria are:HOAc H+ + OAc-

H2O H+ + HO-

Equilibrium concentration of acetic acid, CHOAc = sum of the equilibrium concentrations of all its species

= [HOAc] + [OAc-]

= 0.100 M

Equilibrium concentration of H+, CH = [OAc-] +

[HO-]

+

Example :

Write the equations of mass balance for a 1.00 x 10-5 M [Ag(NH3)2]Cl solution

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

-

+

CNH = [NH4+] + [NH3] + [Ag(NH3)+] +

+ 2 [Ag(NH3)2+]

= 2.00 x 10-5 M

[HO-] = [NH4+] + [H+]

3

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 Systematic Approach to Equilibrium to Equilibrium CalculationsCalculations

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 pH계산Step 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-] = 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

[A][B]

[AB]

위 농도는 적정한가 ?

공통이온효과 무시경우 =

10-7 + 10-8 = 1.1 x 10-7

10-7 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 10-7 + 0.95 x 10-7

=1.05 x 10-7

From (1) and (3)

= 3.0 x 10-6

= 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]

[A][B]0.10

Dependence of Dependence of Solubility on pHSolubility 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 3:

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

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

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 SolubleAcid Rain : SOx, NOx ==>

Sulfuric, Nitric Acid in the air

r결과 : 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-]