숙제 : 7장 문제 :답이 있는 30번 까지 짝수
번 문제
Systematic
Treatment of
In this chapter we Equilibrium
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
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)
= 0.0550
Mass 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 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
3
= [NH4+] + [NH3] + [Ag(NH3)+] +
+ 2 [Ag(NH3)2+]
= 2.00 x 10-5 M
[HO-]
= [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
Example : 간단한 예 : 10-8 M HCl의 pH계산
Step 1:
HCl
---> H+ + Cl-
H2O ---> H+ + OHStep 2: [H+] = [Cl-] + [OH-]
Step 3: [H+] = [Cl-] + [OH-] : same
[Cl-] = 10-8M
Step 4: [H+][OH-] = Kw = 1.0 x 10-14
[A][B]
Step 5: 반응식
수=3
[AB]
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
From (1) and (3)
[A][B]
[AB]
= 3.0 x 10-6
[A][B]
0.10
= 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)
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
[HF][OH-]
[F-]
= 1.5 x 10-11
Step 1: write the balanced chemical
equations
CaF2(s)
Ca2+ + 2FF- + 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
Kb =
[HF][OH-]
= 1.5 x 10-11
-]
[F
+
Kw = [H ] [HO-] = 1.0 x 10-14
(1)
(2)
(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
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)
[HF]
[F-]
=
Kb
[HO-] =
1.5 x 10-11
1.0 x 10-11
= 1.5
[HF]
From (4)
= 1.5[F-]
(6)
[F-] + [HF] = 2[Ca2+]
[F-] + 1.5[F-] = 2[Ca2+]
[F-] = 0.80[Ca2+]
(7)
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 Soluble
Acid Rain : SOx, NOx ==>
Sulfuric, Nitric Acid in the air
r 과 : 대리석 녹임 : Fig 9-3
결
토양 녹임 : Fig 9-4
Solubility of HgS
Step 1: HgS = Hg2+ + S2S2- + H2O = HS- + OH-
HS- + H2O = H2S + OHH2O = H+ + OHIf 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-]