126
Acid Acid - - Base Equilibria Base Equilibria Ji Chang Department of Chemistry and Biochemistry Texas State University
AcidAcid--Base EquilibriaBase Equilibria
Ji Chang
Department of Chemistry and BiochemistryTexas State University
Arrhenius DefinitionsArrhenius Definitions
Arrhenius acids are substances which, when Arrhenius acids are substances which, when dissolved in water, increase the Hdissolved in water, increase the H++
concentration.concentration.
HClHCl(g)(g) →→ HH++(aq)(aq) + Cl+ Cl−−(aq)(aq)
Arrhenius bases are substances which, when Arrhenius bases are substances which, when dissolved in water, increase the OHdissolved in water, increase the OH−−
concentration.concentration.
NaOHNaOH(s)(s) →→ NaNa++(aq)(aq) + OH+ OH−−
(aq)(aq)
HCl is an Arrhenius acid.
NaOH is an Arrhenius base.
HH++ in Waterin Water
OHH
H OHHH+
+
Hydronium Ion
Bronsted-Lowry Definitions
A BronstedA Bronsted--Lowry acid is any substance capable Lowry acid is any substance capable of donating a proton.of donating a proton.
A BronstedA Bronsted--Lowry base is any substance that is Lowry base is any substance that is capable of accepting a proton.capable of accepting a proton.
AcidHCl → H+ + Cl-
HAc → H+ + Ac-
NH4+ → H+ + NH3
HCO3- → H+ + CO3
2-
HPO42- → H+ + PO4
3-
Fe(H2O)63+ → H+ + Fe(H2O)5(OH)2+
Base Ac- + H+ → HAc
NH3 + H+ → NH4+
PO43- + H+ → HPO4
2-
ExampleExample
NHNH22−−(aq)(aq) + H+ H22OO(l)(l) →→ NHNH3(aq)3(aq) + OH+ OH−−
(aq)(aq)
HCHC22OO44−−(aq)(aq) + CO+ CO33
22--(aq)(aq) ⇄⇄ CC22OO44
22--(aq)(aq) + HCO+ HCO33
−−(aq)(aq)
NHNH44++
(aq)(aq) + CN+ CN−−(aq)(aq) ⇄⇄ NHNH3(aq)3(aq) + HCN+ HCN(aq)(aq)
BL base
BL base
BL base
BL acid
BL acid
BL acid
Conjugate AcidConjugate Acid--Base Base PairsPairs
NHNH44++
(aq)(aq) + CN+ CN−−(aq)(aq) ⇄⇄ NHNH3(aq)3(aq) + HCN+ HCN(aq)(aq)BL
baseBL
acidBL
baseBL
acid
• NH4+ and NH3 are called a conjugate acid-base
pair, as are CN − and HCN.• A conjugate acid-base pair is an acid and a
base that differ only by a proton.
Example
Determine the conjugate base of:
HClO4
H2SO4
HClHCO3
−
ClO4−
HSO4−
Cl−
CO32-
Subtract H+
Example
Determine the conjugate acid of:
CN−
H2OHCO3
−
SO42-
HCNH3O+
H2CO3
HSO4−
Add H+
AmphiproticSubstances that can act either as an Substances that can act either as an acid or as a base are called acid or as a base are called amphiprotic.amphiprotic.
HCOHCO33−− + H+ H22O O ⇄⇄ HH33OO++ + CO+ CO33
22--
HCOHCO33−− + H+ H22O O ⇄⇄ HH22COCO33 + OH+ OH−−
HCO3− is an amphiprotic substance.
水溶液中的质子传递反应：
酸1 + 碱2 酸2 + 碱1
HAc + H2O H3O+ + Ac-
NH4+ + H2O H3O+ + NH3
H2O + NH3 NH4+ + OH-
H2O + Ac- HAc + OH-
H3O+ + OH- H2O + H2O
H3O+ + Ac- HAc + H2O
NH4+ + OH- H2O + NH3
酸的电离反应
酸碱中和反应
碱的电离反应
非水溶剂中的质子传递：
HI + CH3OH I- + CH3OH2+
酸1 碱2 碱1 酸2
HBr + HAc Br- + H2Ac+
酸1 碱2 碱1 酸2
酸碱电子理论（Lewis酸碱理论, 1923）能够接受电子对的物质是酸，例：AlCl3、BF3、
Cu2+、Fe3+。能够给出电子对的物质是碱，例：
NH3、H2O。扩大了酸的概念。
酸碱反应的实质：碱性物质提供电子对与酸性物质
形成配位共价键
AutoionizationAutoionization is the transfer of a proton from
one molecule to another of the same substance.
2 H2O ⇄ H3O+ + OH−
H2O ⇄ H+ + OH−
H3O+ is equivalent
to H+
2 NH3 ⇄ NH4+ + NH2
−
Equilibrium Constant, Kw
2H2O ⇄ H3O+ + OH−
Kw = [ H3O+ ][ OH− ]
• Kw is called the ion-product constant (or the dissociation constant) for water.
• The value of Kw is 1.0 x 10-14 at 25 °C.
• The low value indicates that the reactants are favored.
Aqueous Solutions
In NEUTRAL solutions, [ H3O+ ] = [ OH− ]In ACIDIC solutions, [ H3O+ ] > [ OH− ]In BASIC solutions, [ H3O+ ] < [ OH− ]
What is the [ H3O+ ] in pure water at 25 °C?Kw = [ H3O+ ][ OH− ] = 1.0 x 10-14
[ H3O+ ] [ H3O+ ]= 1.0 x 10-14
[ H3O+ ] = 1.0 x 10-7
在水溶液中，Kw = [H3O+][OH-] = 1.0 × 10-14
[H3O+] [OH-]
[OH-] [H3O+]
酸性溶液中， [OH-] ≠ 0
碱性溶液中， [H3O+] ≠ 0
The pH Scale
The concentration of the hydrogen ion, expressed as [ H3O+ ], is typically expressed in terms of pH.
pH = − log10 [ H3O+ ]
pH interval: 0-14pH changes by 1 for every power of 10 change in [ H3O+ ]
pH Scale
0 7 14
acid base
At 25 °C, a pH of 7 is neutral.
19
Derivation
Kw = [ H3O+ ][ OH− ] = 1.0 x 10-14
log10 ([ H3O+ ][ OH− ]) = log10 (1.0 x 10-14)log10 [ H3O+ ] + log10 [ OH− ] = −14
− pH + (− pOH) = −14pH + pOH = 14
]OH[K
3
w+
强酸 HX（如：HCl）的水溶液
[H3O+] = CHX + x [OH-] = x
[H3O+] = CHX + x = CHX + [OH−] = CHX +
[H3O+]2 − CHX[H3O+] − Kw = 0
[H3O+]= )4KC(C21
w2HXHX ++
假如 CHX > 1.0 × 10-5 mol/L，水自解离产生的[H3O+]可以忽略，即[H3O+] = CHX
假如 1.0 × 10-9 mol/L < CHX < 1.0 × 10-5 mol/L，水自
解离产生的[H3O+]不可以忽略，即[H3O+] = CHX + x
[H3O+] =25oC, Kw = 1.0 × 10-14
假如CHX < 1.0 × 10-9 mol/L，强酸电离产生的[H3O+]可以忽略，即[H3O+]≈ 1.0 × 10-7 mol/L
)4KC(C21
w2HXHX ++
强碱MOH（如：NaOH）的水溶液
[OH-] = CMOH + y [H3O+] = y
[OH-] = CMOH + y = CMOH + [H3O+] = CMOH +
[OH-]2 − CMOH[OH-] − Kw = 0
[OH-]=
wK[OH ]−
2MOH MOH w
1 (C C 4K )2
+ +
强碱MOH（如：NaOH）的水溶液
假如 CMOH > 1.0 × 10-5 mol/L，水自解离产生的[OH-]可以忽略，即[OH-] = CMOH
假如 1.0 × 10-9 mol/L < CMOH < 1.0 × 10-5 mol/L，水自解
离产生的[OH-]不可以忽略，即[OH-] = CMOH + y
[OH-] =25oC, Kw = 1.0 × 10-14
假如CMOH < 1.0 × 10-9 mol/L，强碱电离产生的[OH-]可以忽略，即[OH-] ≈ 1.0 × 10-7 mol/L
)4KC(C21
w2MOHMOH ++
Weak Acids
The strength of an acid is ranked by the value of the acid dissociation constant, abbreviated Ka.
For a general weak acid, HA,
HA(aq) +H2O H3O+(aq) + A−
(aq)
Ka = [ HA ]
[ H3O+ ] [ A− ]
Example
HF Ka = 6.8 x 10-4
HC2H3O2 Ka = 1.8 x 10-5
HNO2 Ka = 4.5 x 10-4
Rank the acids from highest acidity to lowest acidity.
HF > HNO2 > HC2H3O2
Weak Bases
NH3(aq) + H2O(l) ⇄ NH4+
(aq) + OH−(aq)
Kb =[ NH4
+ ] [ OH− ]
[ NH3 ]
Kb is called the base-dissociation constant.
HC2H3O2 + H2O ⇄ H3O+ + C2H3O2−
C2H3O2− + H2O ⇄ HC2H3O2 + OH−
Ka = [ H3O+ ] [ C2H3O2
− ][ HC2H3O2]
Kb =[ HC2H3O2 ] [ OH− ]
[ C2H3O2− ]
Ka and Kb for Conjugate Acid/Base
[ H3O+ ] [ C2H3O2− ]
[ HC2H3O2][ HC2H3O2 ] [ OH− ]
[ C2H3O2− ]
Ka Kb =
Ka Kb = [ H3O+ ] [ OH− ]
Ka Kb = Kw
The stronger an acid, the weaker its conjugate base.The weaker an acid, the stronger its conjugate base.
Strength of Acids and Bases
水的 “拉平效应”
HX + H2O → H3O+ + X-
非水溶剂的 “区分效应”
HCl + HAc H2Ac+ + Cl-
酸强度的递减顺序：
HClO4 > HI > HBr > HCl > HNO3
Calculating [H3O+] for Weak Monoprotic Acids/Bases
HA水溶液中存在的平衡：
(1) HA(aq) + H2O(l) H3O+(aq) + A-(aq)Ka = [H3O+][A-] / [HA]
(2) H2O(l) + H2O(l) H3O+(aq) + OH-(aq)Kw = [H3O+][OH-]
mass balance： CHA = [HA] + [A-] （3）charge balance： [H3O+] = [OH-] + [A-] （4）
由(1)得，[HA] = [H3O+][A-] / Ka 代入(3)后得：
[A-] = KaCHA / (Ka + [H3O+]) 代入（4）得：
]O[HKCK
]OH[K]OH[
3a
HAa
3
w3 ++
+
++=
------- 计算[H3O+]的精确表达式
2CK4KK-
]OH[ HAa2aa
3
++=+
弱酸溶液中，水自电离产生的[H3O+]可以忽略！
计算[H3O+]的近似式（CHA < 400Ka）
（忽略水自电离产生的[H3O+]）
HA(aq) + H2O(l) H3O+(aq) + A-(aq)起始 CHA 0 0转化 x x x平衡 CHA − x x x
Ka = x2 / (CHA − x)
推导：
2a a a HA
3
-K K 4K C[H O ]
2x + + += =
x2 + Kax − KaCHA = 0
计算[H3O+]的最简式（CHA ≥ 400Ka）
HAa3 CK]OH[ =+
（忽略水自电离产生的[H3O+]，且[HA] ≈ CHA）
2 2a a a HA a a a a
3
a
-K K 4K C -K K 4K 400K[H O ]
2 219.51K
+ + + + + ×= =
=
用最简式计算[H3O+]时，为什么须满足CHA≥ 400Ka？
3 a HA a a a[H O ] K C K 400K 20K+ = = × =
当CHA = 400Ka时，
按最简式计算的相对误差：
(20Ka – 19.51Ka) / (19.51Ka) = 2.5%
CHA ≥ 400Ka时，按最简式计算的相对误差≤ 2.5%!
2CK4KK-
]OH[ Bb2bb ++
=−
弱碱（B）溶液中，水自电离产生的[OH-]可以忽略！
计算[OH-]的近似式（CB < 400Kb）
（忽略水自电离产生的[OH-]）
B(aq) + H2O(l) OH-(aq) + BH+(aq)起始 CB 0 0转化 y y y平衡 CB − y y y
Kb = y2 / (CB − y)
推导：
2b b b B-K K 4K C
[OH ]2
y − + += =
y2 + Kby − KbCB = 0
计算[OH-]的最简式（CB ≥ 400Kb）
BbCK]OH[ =−
（忽略水自电离产生的[OH-]，且[B] ≈ CB）
Percent Ionization
Defined as:
Amount dissociated (M)
Initial concentration (M)x 100% Ioniz =
例1: 298 K时，HAc Ka = 1.8 × 10-5, 计算 0.1 mol/LHAc溶液中的 [H3O+]、[Ac-]、[HAc] 和 [OH-]以及HAc的解离度？
解: CHAc = 0.1 mol/LCHAc ≥ 400Ka
5 33 a HAc[H O ] K C 1.8 10 0.1 1.34 10 (mol/L)+ − −= = × × = ×
HAc(aq) + H2O(l) H3O+(aq) + Ac-(aq)平衡 CHAc − x x x
[Ac-] = [H3O+] = 1.34 × 10-3 (mol/L)
[HAc] = CHAc − [H3O+] = 0.1 − 1.34 × 10-3
= 0.0987 (mol/L)
[OH-] = Kw / [H3O+] = 1 × 10-14 / (1.34 × 10-3)= 7.46 × 10-12 (mol/L)
33
HAc
[H O ] 1.34 10α 100% 100% 1.34%C 0.1
+ −×= × = × =
例2： 298 K时 HNO2 Ka = 4.6 × 10-4, 计算 0.05 mol/LHNO2溶液的pH值 ? [HNO2]?
解： CHNO2 < 400Ka
2a a a
3
4 4 2 4
3
-K K 4K C[H O ]
24.6 10 (4.6 10 ) 4 4.6 10 0.05
24.57 10 (mol/L)
+
− − −
−
+ +=
− × + × + × × ×=
= ×
pH = -lg[H3O+] = 2.34
HNO2(aq) + H2O(l) H3O+(aq) + NO2-(aq)
平衡 CHNO2 − x x x
[HNO2] = CHNO2 − [H3O+] = 0.05 − 4.57 × 10-3
= 0.0454 (mol/L)
例3： 298 K时 HAc Ka = 1.8 × 10-5, 计算 0.1 mol/L NaAc 溶液的pH值和[Ac-] 、[HAc] 、α?
Ac-(aq) + H2O(l) OH-(aq) + HAc(aq) 平衡 CAc- − x x x
NaAc → Na+ + Ac- CAc- = 0.1 mol/L
10 6b Ac-[OH ] K C 5.56 10 0.1 7.46 10 (mol/L)− − −= = × × = ×
CAc- ≥ 400 Kb
Kb = Kw / Ka = 1×10-14 / (1.8×10-5) = 5.56 × 10-10
pOH = -lg[OH-] = 5.13pH = 14 − pOH = 8.87
或：[H3O+] = Kw / [OH-] = 1×10-14 / (7.46×10-6)= 1.34×10-9 (mol/L)
pH = -lg[H3O+] = 8.87
[HAc] = [OH-] = 7.46×10-6 (mol/L)
[Ac-] = CAc- − [OH-] = 0.1 − 7.46×10-6 = 0.1 (mol/L)
6
Ac-
[OH ] 7.46 10α 100% 100% 0.0076%C 0.1
− −×= × = × =
作业：P313：1，2，3，4 --- a, b1: (d) ClO2
-改为(e) ClO2-；(e) NH4
+改为(f) NH4+
3: (d) NH2CH2COOH改为(e) NH2CH2COOHP318：4P320: 1, 31：NH3，Kb = 1.8 × 10-5
3：HClO2, Ka = 1.1×10-2; HNO2, Ka = 4.6×10-4
P330 − 331：2，5，6，102: HC6H7O6的摩尔质量为176.12 g/mol10: 一元弱碱, CH3COOH的Ka = 1.76×10-5
对应书上的内容：
P311 − 317上 P318 − 320
Example
Predict whether the following salt solutions are acidic, basic, or neutral
NaC2H3O2
NH4BrNaClNH4CN
Polyprotic Acids
二元(diprotic)弱酸的电离平衡： H2A
H2A(aq) + H2O(l) H3O+(aq) + HA-(aq)
Ka1 = [H3O+][HA-]/[H2A] H2A的一级解离常数
HA-(aq) + H2O(l) H3O+(aq) + A2-(aq)
Ka2 = [H3O+][A2-]/[HA-] H2A的二级解离常数
H2O(l) + H2O(l) H3O+(aq) + OH-(aq)
Kw = [H3O+][OH-]
a2a13a12
3
3H2Aa13 KK]O[HK]O[H
]O[HCK]OH[++
= ++
++
物料平衡： CH2A = [H2A] + [HA-] + [A2-]
电荷平衡： [H3O+] = [HA-] + 2[A2-] + [OH-]
]OH[K
KK]O[HK]OH[CK2K
3
w
a2a13a12
3
H2Aa2a1+++ +
+++
等式右边三项的物理意义：
第1项：一级电离的贡献
第2项：二级电离的贡献
第3项：水自电离的贡献
一般Ka1 >> Ka2，差3 ∼ 4个数量级
例：H2CO3，Ka1 = 4.3 ×10-7，Ka2 = 4.8 ×10-11
二元弱酸作为一元弱酸来处理（忽略二级电离）！
2CK4KK-
]OH[ H2Aa12a1a1
3
++=+
H2Aa13 CK]OH[ =+
CH2A < 400Ka1，近似式：
CH2A ≥ 400Ka1，最简式：
例3: 计算 0.1 mol/L 的 H2C2O4溶液中各离子的浓度?已知: Ka1 = 5.9 × 10-2 Ka2 = 6.4 × 10-5
解： CH2C2O4 < 400Ka1
2a1 a1 a1
3
2 2 2 2
-K K 4K C[H O ]
25.9 10 (5.9 10 ) 4 5.9 10 0.1
20.0528 (mol/L)
+
− − −
+ +=
− × + × + × × ×=
=
H2C2O4(aq)+H2O(l) H3O+(aq)+HC2O4-(aq) (1)
平衡 C − x x x
[HC2O4-] = [H3O+] = 0.0528 (mol/L)
[H2C2O4] = c − [H3O+] = 0.1 − 0.0528 = 0.0472 (mol/L)
[OH-] = Kw / [H3O+] = 1.89×10-13 (mol/L)
HC2O4-(aq)+H2O(l) H3O+(aq)+C2O4
2-(aq) (2)
Ka2 = [H3O+]×[C2O42-] / [HC2O4
-] = [C2O42-]
[C2O42-] = Ka2 = 6.4×10-5 (mol/L)
纯二元弱酸（H2A）水溶液中的 [A2-] = Ka2 !
三元(triprotic)弱酸的电离平衡： H3A
H3A(aq) + H2O(l) H3O+(aq) + H2A-(aq)
Ka1 = [H3O+][H2A-]/[H3A] H3A的一级解离常数
H2A-(aq) + H2O(l) H3O+(aq) + HA2-(aq)
Ka2 = [H3O+][HA2-]/[H2A-] H3A的二级解离常数
HA2-(aq) + H2O(l) H3O+(aq) + A3-(aq)
Ka3 = [H3O+][A3-]/[HA2-] H3A的三级解离常数
二元弱碱的电离平衡： Na2S Na2S → 2Na+ + S2-
S2-(aq) + H2O(l) OH-(aq) + HS-(aq)Kb1 = [OH-][HS-]/[S2-] S2-的一级解离常数
Kb1 = Kw / Ka2
HS-(aq) + H2O(l) OH-(aq) + H2S(aq) Kb2 = [OH-][H2S]/[HS-] S2-的二级解离常数
Kb2 = Kw / Ka1
Ka1, Ka2分别为H2S的一级解离常数和二级解离常数！
一般Kb1 >> Kb2
二元弱碱作为一元弱碱来处理（忽略二级电离）！
2b1 b1 b1 S2-K K 4K C
[OH ]2
−− + +=
b1 S2[OH ] K C−−=
CS2- < 400Kb1，近似式：
CS2- ≥ 400Kb1，最简式：
例4：算 0.1 mol/L Na2S 溶液中[OH-]、[S2-]、[HS-]、[H3O+]和[H2S]的浓度? 已知: H2S 的 Ka1 = 9.1 × 10-8 Ka2 = 1.1 × 10-12
S2-(aq) + H2O(l) OH-(aq) + HS-(aq) 平衡 C − x x x
Kb1 = KW / Ka2 = 9.1×10-3
Na2S → 2Na+ + S2- C = 0.1 < 400Kb12
b1 b1 b1
3 3 2 3
-K K 4K C[OH ]
29.1 10 (9.1 10 ) 4 9.1 10 0.1
20.026 (mol/L)
−
− − −
+ +=
− × + × + × × ×=
=
[HS-] = [OH-] = 0.026 (mol/L) [S2-] = C − [OH-] = 0.1 − 0.026 = 0.074 (mol/L)
HS-(aq) + H2O(l) OH-(aq) + H2S(aq) Kb2 = Kw / Ka1 = 1.1×10-7
Kb2 = [OH-][H2S] / [HS-] = [H2S]所以，[H2S] = 1.1×10-7 (mol/L)[H3O+] = Kw / [OH-] = 1×10-14 / 0.026
= 3.8×10-13 (mol/L)
三元弱碱的电离平衡：Na3PO4 → 3Na+ + PO43-
PO43-(aq) + H2O(l) OH-(aq) + HPO4
2-(aq)
Kb1 = [OH-][HPO42-]/[PO4
3-] PO43-的一级解离常数
HPO42-(aq) + H2O(l) OH-(aq) + H2PO4
-(aq)
Kb2 = [OH-][H2PO4-]/[HPO4
2-] PO43-的二级解离常数
H2PO4-(aq) + H2O(l) OH-(aq) + H3PO4(aq)
Kb3 = [OH-][H3PO4]/[H2PO4-] PO4
3-的三级解离常数
Kb1 = Kw / Ka3 Kb2 = Kw / Ka2 Kb3 = Kw / Ka1
Ka1, Ka2, Ka3分别为H3PO4的一级, 二级, 三级解离常数！
酸式盐的电离平衡（酸碱两性物种, amphiprotic）
H2A是二元弱酸，NaHA Na+ + HA−
H2A的一级解离常数和二级常数分别为Ka1和Ka2！
HA-(aq) + H2O(l) H3O+(aq) + A2-(aq) Ka = Ka2
HA-(aq) +H2O(l) H2A(aq) + OH-(aq) Kb = Kw / Ka1
Ka2 > Kw / Ka1 Ka1 Ka2 > Kw 酸性
Ka2 < Kw / Ka1 Ka1 Ka2 < Kw 碱性
NaH2PO4水溶液：NaH2PO4 Na+ + H2PO4-
H2PO4- + H2O H3O+ + HPO4
2- Ka = Ka2 = 6.2 × 10-8
H2PO4- + H2O H3PO4 + OH- Kb = Kw / Ka1 = 1.3×10-12
Ka2 > Kw / Ka1 酸性
Na2HPO4水溶液： Na2HPO4 2Na+ + HPO42-
HPO42- + H2O H3O+ + PO4
3- Ka = Ka3 = 1 × 10-13
HPO42- + H2O H2PO4
- + OH- Kb = Kw/Ka2 = 1.6 × 10-7
Ka3 < Kw/Ka2 碱性
H2CO3 的 Ka1 = 4.3 × 10-7，Ka2 = 5.6 × 10-11
NaHCO3水溶液: Ka2 < Kw / Ka1 碱性
Ka1, Ka2, Ka3分别为H3PO4的一级, 二级, 三级解离常数！
[H3O+]计算公式的推导：
H2A是二元弱酸（Ka1，Ka2）， NaHA的浓度CNaHA Na+ + HA− HA-的初始浓度为CHA- + H2O H3O+ + A2- Ka = Ka2
HA- + H2O H2A + OH- Kb = Kw/Ka1
H2O + H2O H3O+ + OH- Kw = [H3O+][OH-]物料平衡： C = [HA-] + [H2A] + [A2-]电荷平衡： [H3O+] + [Na+] = [HA-] + 2[A2-] + [OH-][Na+] = C = [HA-] + [H2A] + [A2-]代入电荷平衡式得：
[H3O +] + [ H2A ] = [A2-] + [OH-] (*)
由Ka = Ka2 = [H3O+][A2-] / [HA-] 得：
[A2-] = Ka2[HA-] / [H3O+]
由Kb = Kw / Ka1 = [OH-][H2A] / [HA-] = Kw[H2A] / ([H3O+][HA-])得：
[H2A] = [H3O+][HA-] / Ka1
或直接由Ka1的定义可得：[H2A] = [H3O+][HA-] / Ka1
将[A2-]的表达式，[H2A] 的表达式以及[OH-] = Kw / [H3O+]代入等式(*), 可得：
][[
−
−+
++
=HA
HA
1
w213 Ka
)K ](KaKa]O[H
CHA- ≈ [HA-] (CHA-不是太小)
精确表达式：
C++
=+
1
w213 Ka
)K C(KaKa]O[H ⎯近似式
[H3O+] + [H3O+][HA-] / Ka1 = Ka2[HA-] / [H3O+] + Kw / [H3O+]
213 KaKa]O[H =+
当Ka2C ≥ 20Kw，C ≥ 20Ka1时，
近一步近似：
⎯最简式
例5：计算0.10 mol/L NaHCO3溶液中各组分的浓度？
已知：H2CO3 的 Ka1 = 4.3 × 10-7 、 Ka2 = 5.6 × 10-11
NaHCO3 → Na+ + HCO3-
[Na+] = 0.10 mol/L, HCO3-的初始浓度为0.10 mol/L
9a1 a2K K 4.9 10 (mol/L)+ −== = ×3[H O ]
(1) HCO3- + H2O H3O+ + CO3
2- Ka = Ka2
(2) HCO3- + H2O OH- + H2CO3 Kb = Kw / Ka1
物料平衡：C = [HCO3-] + [CO3
2-] + [H2CO3] (3)
由(1)得：Ka2 = [CO32-][H3O+] / [HCO3
-]
于是， [CO32-] = Ka2[HCO3
-] / [H3O+] 由(2)得：Kw / Ka1 = [H2CO3][OH-] / [HCO3
-]= [H2CO3]Kw / ([H3O+][HCO3
-])于是， [H2CO3] = [H3O+][HCO3
-] / Ka1或者直接由Ka1的定义得[H2CO3] = [H3O+][HCO3
-] / Ka1
将[CO32-], [H2CO3]的表达式代入(3)得：
a2 3
3 a1
C 0.098 (mol/L)K [H O ]1
[H O ] K
−+
+
= =+ +
3[HCO ]
[CO32-] = Ka2[HCO3
-] / [H3O+] = 1.1×10-3 (mol/L)
[H2CO3] = [H3O+][HCO3-] / Ka1 = 1.1×10-3 (mol/L)
[OH-] = Kw / [H3O+] = 2.0×10-6 (mol/L)
问题：NaH2PO4水溶液的浓度C，[H3O+] = ?NaH2PO4 Na+ + H2PO4
-
H2PO4- + H2O H3O+ + HPO4
2- Ka = Ka2H2PO4
- + H2O H3PO4 + OH- Kb = Kw/Ka1
H3PO4 Ka1 Ka2 Ka3
C++
=+
1
w213 Ka
)K C(KaKa]O[H
当Ka2C ≥ 20Kw，C ≥ 20Ka1时，
213 KaKa]O[H =+ ⎯最简式
⎯近似式
问题：Na2HPO4水溶液的浓度C，[H3O+] = ?Na2HPO4 2Na+ + HPO4
2-
HPO42- + H2O H3O+ + PO4
3- Ka = Ka3HPO4
2- + H2O H2PO4- + OH- Kb = Kw/Ka2
C++
=+
2
w323 Ka
)K C(KaKa]O[H ⎯近似式
当Ka3C ≥ 20Kw，C ≥ 20Ka2时，
+ =3 2 3[H O ] Ka Ka ⎯最简式
H3PO4 Ka1 Ka2 Ka3
作业：
P346：1，3，
3：H2CO3，Ka1 = 4.3 × 10-7，Ka2 = 4.8 × 10-11
The Common-Ion Effect
NaC2H3O2 (aq) → Na+ (aq) + C2H3O2− (aq)
HC2H3O2 (aq) ⇄ H+ (aq) + C2H3O2− (aq)
Since the acetate ion is present in both reactions, it is called a common ion.
HC2H3O2 (aq) ⇄ H+ (aq) + C2H3O2− (aq)
If we increase the concentration of the acetate ion, which direction will the equilibrium shift?
What effect would the presence of a common ion have?
Decrease in [H+]Increase in pHDecrease in the percent ionization
The Common-Ion Effect
The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte.
The Common-Ion Effect
在一个 HA~ A- 的体系中，总是满足这样的平衡关
系：
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
Ka / [H3O+] = [A-] / [HA]
调节 [H3O+] ， 调节 [A-]/[HA]比值
调节 [A-]/[HA]比值，调节[H3O+]
Buffered SolutionsBuffers are the solutions that can resist a
change in pH upon addition of small amounts of strong acid or base.
There must be an acidic species to neutralize OH− ions and there must be a basic species to neutralize H+ ionsThe acidic and basic species must not consume each other
Buffers must contain a weak conjugate acid-base pair!
Composition of Buffers
Buffer solutions are composed of either
Weak acids and the salt of the conjugate base
HC2H3O2 and NaC2H3O2
Weak bases and the salt of the conjugate acid
NH3 and NH4Cl
HA − NaA HA：Ca；NaA：Cb
NaA Na+ + A-
HA + H2O H3O+ + A-
A- + H2O OH- + HA
H2O + H2O H3O+ + OH- Kw = [H3O+][OH-]
物料平衡：Ca + Cb = [HA] + [A-]电荷平衡：[A-] + [OH-] = [H3O+] + [Na+][Na+] = Cb, 由电荷平衡得：
[A-] = Cb + [H3O+] − [OH-]
[HA]]][AO[HK 3
a
−+
=
][A][HA][OHKb −
−
=
于是， Cb = [A-] − [H3O+] + [OH-], 代入物料平衡
得：
[HA] = Ca − [H3O+] + [OH-]由Ka = [H3O+][A-] / [HA]得：
a 33 a a
b 3
C [H O ] [OH ][HA][H O ] K =K[A ] C [H O ] [OH ]
+ −+
− + −
− +=
+ −
b
aa3 C
CK]O[H =+近似得：
b
aa3 C
CK]O[H =+
-24
-42
HPO
POHa23 C
CK]O[H =+
3
4
NH
NHa3 C
CK]O[H
+
=+
-3
32
HCO
COHa13 C
CK]O[H =+
H2PO4- − HPO4
2-
(NaH2PO4 − Na2HPO4)
H2CO3 − HCO3-
(H2CO3 − NaHCO3)
NH4+ − NH3
(NH4Cl − NH3)
Buffers
Buffers
关于缓冲溶液：
1 由共轭弱酸弱碱对（缓冲对）组成
例如: NH4+ / NH3 HAc / Ac- H3PO4 / H2PO4
-
H2PO4- / HPO4
2- HCO3- / CO3
2- ……
2 缓冲能力(buffer capacity)
与缓冲对浓度和 C共轭酸 /C共轭碱比值有关
a. 缓冲对浓度（C共轭酸、C共轭碱）越大，缓冲能
力越强 (由于浓度过高对化学反应可能造成不
利影响，一般选择0.1 ∼ 1 mol/L)
b. C共轭酸 /C共轭碱比值（缓冲比）越接近于 1 ，缓冲
能力越强。共轭酸碱对的浓度比必须控制在 0.1到10的范围内，超出此范围，少量外来酸或碱的加
入，会导致溶液pH值的显著变化，即不具有缓冲
能力。
pH Range for Buffer
0.1 ≤ Ca/Cb ≤ 100.1Ka ≤ [H3O+] ≤ 10Ka
lgKa − 1 ≤ lg[H3O+] ≤ lgKa + 1-lgKa − 1 ≤ -lg[H3O+] ≤ -lgKa + 1
pH range: pKa − 1 ≤ pH ≤ pKa + 1
b
aa3 C
CK]O[H =+
例2：已知： HAc ，Ka = 1.8 × 10-5
H2CO3 的 Ka1 = 4.3 × 10-7 、Ka2 = 5.6 × 10-11
H3PO4的 Ka1 = 7.5 × 10-3 、Ka2 = 6.2 × 10-8 、
Ka3 = 2.0 × 10-13
要配制pH = 7.5 的缓冲溶液，选择什么缓冲对最好？
此时缓冲比是多少？
解：pH与pKa越接近越好！
选择缓冲对：H2PO4- − HPO4
2- (NaH2PO4 − Na2HPO4)
a3 a2
b
C[H O ] KC
+ =7.5
a 38
b a2
C [H O ] 10 0.51C K 6.2 10
+ −
−= = =×
ExampleConsider a buffer made by adding 0.140 mol cyanic acid, HCNO, 0.110 mol potassium cyanate, KCNO, and sufficient water to produce 1.00 liters of solution. Calculate the pH of the buffer:
a) Before any acid or base is addedb) After the addition of 0.015 mol of HNO3c) After the addition of 0.015 mol KOH.
In parts (b) and (c), assume the volume does not change. The Ka for HCNO is 3.5 x 10-4.
作业：
P351 − 352：1，2(CH3COOH, Ka = 1.8×10-5)
4 (HCOOH, Ka = 1.77×10-4)
对应书上的内容：
P347 − 349上
P349下 − 351
88
Acid-Base TitrationsThe solution being added is called the
titrant.
The indicator changes color when the
endpoint is reached.
强酸---强碱； 强酸滴定弱碱； 强碱滴定弱酸
酸碱滴定
----- 定量分析中最常用的一种分析方法
HCl — NaOH, NaOH — HAc, HCl — CH3NH2
C1 V1 = C2 V2 （化学计量点: 刚好完全反应）
酸碱指示剂(indicator)作用原理
HIn(aq) + H2O(l) H3O+(aq) + In-(aq)
酸式 碱式
Ka = [H3O+][In-] / [HIn]
[In-] / [HIn] > 10 显碱色
< 0.1 显酸色
0.1 ≤ [In-] / [HIn] ≤ 10 观察到颜色变化
Ka / [H3O+] = [In-] / [HIn]
指示剂变色范围： pKa − 1 ≤ pH ≤ pKa + 1
(pH transition range) pKa = -lgKa
Acid – Base Indicators
甲基橙 3.1 (红) ---- 4.4 (黄)
甲基红 4.4 (红) ---- 6.2 (黄)
酚 酞 8.0 (无) ---- 9.8 (紫红)
Strong Acid – Strong Base
HCl + NaOH → NaCl + H2O
Reaction proceeds to completionFormed a salt in the processThe pH at the equivalence point will be equal to 7
(i) 强碱滴定强酸
用0.1000 mol/L NaOH 滴定20.00 ml 0.1000 mol/L HCl
滴定前（ VNaOH = 0 ）：
[H3O+] = 0.1 mol/L pH = 1.00
加入NaOH(aq)至化学计量点之前(V < 20.00 ml)
NaOH3
NaOH
0.1000 20.00 0.1000 V[H O ]20.00 V
+ × − ×=
+
VNaOH = 10.00 ml，[H3O+] = 1/30 mol·L-1，pH = 1.48VNaOH = 19.98 ml（化学计量点之前半点(0.02 ml）)[H3O+] = 5.00×10-5 mol·L-1， pH = 4.30
加入NaOH(aq)至化学计量点(刚好完全反应, 需加
入 20.00 ml NaOH )：pH = 7.00
加入NaOH(aq)至化学计量点之后(V > 20.00 ml)：
NaOH
NaOH
V20.0000.201000.0V0.1000]OH[
+×−×
=−
VNaOH = 20.02 ml（化学计量点之后半点(0.02 ml)）[OH-] = 5.00×10-5 mol·L-1， pOH = -lg[OH-] = 4.30，pH = 14 − pOH = 9.70VNaOH = 40.00 ml，[OH-] = 1/30 mol·L-1，pOH = -lg[OH-] = 1.48，pH = 14 − pOH = 12.52
滴定曲线；突跃范围(化学计量点之前半点的pH至化学
计量点之后半点的pH)：4.30 ∼ 9.70指示剂的选择原则：指示剂的变色范围全部或大部分
落在突跃范围内。为什么？
Selection of Indicators
选指示剂：
甲基红(4.4 − 6.2) 红色变为黄色
酚 酞(8.0 − 9.8) 无色变为紫红色
滴定终点，滴定误差
(ii) 用0.1000 mol/L HCl 滴定20.00 ml 0.1000 mol/L NaOH突跃范围(化学计量点之前半点的pH至化学计量点之后
半点的pH)： 9.70 ∼ 4.30选指示剂：
甲基红 黄色变为红色
酚 酞 紫红色变为无色
Weak Acid – Strong Base
HC2H3O2 + NaOH → NaC2H3O2 + H2O
Reaction proceeds to completionFormed a salt in the process
C2H3O2− + H2O HC2H3O2 + OH−
The pH at the equivalence point will be greater than 7
(iii) 强碱滴定弱酸（弱酸的Ka不能太小）
用0.1000 mol/L NaOH 滴定20.00 ml 0.1000 mol/L HAcHAc，Ka = 1.8 × 10-5
滴定前（VNaOH = 0）：一元弱酸，C ≥ 400Ka
pH = 2.87加入NaOH(aq)至化学计量点之前 (V < 20.00 ml) ：
HAc + NaAc, 共轭弱酸弱碱水溶液
HAc3 a a
NaAc
a
20.00 0.1000 0.1000VC 20.00 V[H O ] K K 0.1000VC
20.00 V20.00 0.1000 0.1000V=K
0.1000V
+
× −+= × = ×
+× −
×
5 33 a[H O ] K C 1.8 10 0.1 1.34 10 (mol/L)+ − −= = × × = ×
VNaOH = 10.00 ml，[H3O+] = 1.8×10-5 mol/L，pH = 4.74VNaOH = 19.98 ml（化学计量点之前半点(0.02 ml)）[H3O+] = 1.8×10-8 mol/L，pH = 7.74
加入NaOH(aq)至化学计量点(刚好完全反应, 需加入 20.00 ml NaOH)： NaAc, 一元弱碱Kb = Kw /Ka = 1×10-14 / (1.8×10-5) = 5.56×10-10
CNaAc = 0.05000 mol/L ≥ 400Kb
10 6b[OH ] K C 5.56 10 0.05 5.27 10 (mol/L)− − −= = × × = ×
pOH = -lg[OH-] = 5.28， pH = 14 − pOH = 8.72
加入NaOH(aq)至化学计量点之后(V > 20.00 ml)：NaAc + NaOH
由于Ac-是很弱的碱，溶液中的 [OH-]是由过量的NaOH提供的。
0.1000V 20.00 0.1000[OH ]20.00 V
− − ×=
+
VNaOH = 20.02 ml（化学计量点之后半点(0.02 ml)）[OH-] = 5.00×10-5 mol/L, pOH = 4.30，pH = 9.70VNaOH = 40.00 ml[OH-] = 1/30 mol·L-1, pOH = 1.48，pH = 12.52
突跃范围(化学计量点之前半点的pH至化学计量点之后
半点的pH)：7.74 ∼ 9.70
选指示剂：
酚 酞 无色变为紫红色
Titration Curve
Effect of Ka
With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle.
106
Strong Acid – Weak Base
HCl + NH3 → NH4Cl
Reaction proceeds to completionFormed a salt in the process
NH4+ + H2O → NH3 + H3O+
The pH at the equivalence point will be less than 7
(iiii) 强酸滴定弱碱（弱碱的Kb不能太小）
用0.1000 mol/L HCl滴定20.00 ml 0.1000 mol/L CH3NH2
CH3NH2，Kb = 4.17×10-4
滴定前（VHCl = 0）：一元弱碱
C < 400Kb2
b b b
4 4 2 43
-K K 4K C[OH ]
24.17 10 (4.17 10 ) 4 4.17 10 0.1000
6.25 10 (mol/L)2
−
− − −−
+ +=
− × + × + × × ×= = ×
pOH = -lg[OH-] = 2.20，pH = 14 − pOH = 11.80
加入HCl(aq)至化学计量点之前(V < 20.00 ml)：CH3NH3Cl + CH3NH2, 共轭弱酸弱碱水溶液
3 3
3 2
CH NH Cl3 a a
CH NH
a
0.1000VC 20.00 V[H O ] K K 20.00 0.1000 0.1000VC
20.00 V0.1000VK
20.00 0.1000 0.1000V
+ += × = ×× −
+
= ×× −
Ka = Kw / Kb = 1×10-14 / (4.17×10-4) = 2.4×10-11
VHCl = 10.00 ml，[H3O+] = 2.4×10-11 mol/L，pH = 10.62VHCl = 19.98 ml（化学计量点之前半点(0.02 ml)）[H3O+] = 2.4×10-8 mol/L，pH = 7.62
加入HCl(aq)至化学计量点(刚好完全反应, 需加入
20.00 ml HCl)： CH3NH3Cl, 一元弱酸
C = 0.05000 mol/L ≥ 400Ka
pH = 5.96
加入HCl(aq)至化学计量点之后(V > 20.00 ml)：HCl + CH3NH3Cl
由于CH3NH3Cl是很弱的酸，溶液中的[H3O+]是由过量
的HCl提供的。
11 63 a[H O ] K C 2.40 10 0.05000 1.095 10 (mol/L)+ − −= = × × = ×
+3
0.1000V 20.00 0.1000[H O ]20.00 V− ×
=+
VHCl = 20.02 ml（化学计量点之后半点(0.02 ml)）[H3O+] = 5.00 × 10-5 mol/L，pH = 4.30VHCl = 40.00 ml [H3O+] = 1/30 mol·L-1，pH = 1.48
突跃范围(化学计量点之前半点的pH至化学计量点之后
半点的pH)：7.62 ∼ 4.30
选指示剂：
甲基红 黄色变为红色
Titration Curve
多元弱酸的中和: H3PO4中加NaOH 逐步反应！
H3PO4 + NaOH = NaH2PO4 + H2ONaH2PO4 + NaOH = Na2HPO4 + H2ONa2HPO4 + NaOH = Na3PO4 + H2O已知：H3PO4, Ka1, Ka2, Ka3
NaOH与H3PO4的摩尔数之比为nn < 1, H3PO4/NaH2PO4 (共轭弱酸弱碱对)
n = 1, NaH2PO4 (酸式盐)
H3PO4a1
NaH2PO4
CKC
+ = ×3[H O ]
+ =3 1 2[H O ] Ka Ka
1 < n < 2, NaH2PO4/Na2HPO4 (共轭弱酸弱碱对)
n = 2, Na2HPO4 (酸式盐)
2 < n < 3, Na2HPO4/Na3PO4 (共轭弱酸弱碱对)
n = 3, Na3PO4 (三元弱碱)
Na2HPO4a3
Na3PO4
CKC
+ = ×3[H O ]
NaH2PO4a2
Na2HPO4
CKC
+ = ×3[H O ]
+ =3 2 3[H O ] Ka Ka
Polyprotic Acid Titration Curve
In the titration curve of a polyprotic acid, there will be more than one equivalence point.
多元弱碱的中和: Na3PO4中加HCl 逐步反应！
Na3PO4 + HCl = Na2HPO4 + NaClNa2HPO4 + HCl = NaH2PO4 + NaClNaH2PO4 + HCl = H3PO4 + NaCl已知：H3PO4, Ka1, Ka2, Ka3
HCl与Na3PO4的摩尔数之比为nn < 1, Na2HPO4/Na3PO4 (共轭弱酸弱碱对)
n = 1, Na2HPO4 (酸式盐)
Na2HPO4a3
Na3PO4
CKC
+ = ×3[H O ]
2 3Ka Ka+ =3[H O ]
1 < n < 2, NaH2PO4/Na2HPO4 (共轭弱酸弱碱对)
n = 2, NaH2PO4 (酸式盐)
2 < n < 3, H3PO4/NaH2PO4 (共轭弱酸弱碱对)
n = 3, H3PO4 (三元弱酸)
H3PO4a1
NaH2PO4
CKC
+ = ×3[H O ]
NaH2PO4a2
Na2HPO4
CKC
+ = ×3[H O ]
1 2Ka Ka+ =3[H O ]
注意：
H2CO3 + Na2CO3 → 2NaHCO3
H3PO4 + Na2HPO4 → 2NaH2PO4
Na3PO4 + NaH2PO4 → 2Na2HPO4
等摩尔反应时形成酸式盐！不等摩尔反应时形成共轭
弱酸弱碱对水溶液！
例1：已知 H3PO4 Ka1 = 7.5 × 10-3 Ka2 = 6.2 × 10-8
Ka3 = 2.0 × 10-13
（1）50 mL 2.4 mol/L 的H3PO4中加入60 mL 4.0 mol/L的NaOH
（2）50 mL0.2 mol/L的NaH2PO4中加入50 mL 0.1 mol/L的NaOH
（3）50 mL 3.0 mol/L的Na3PO4中加入50 mL 6.0 mol/L的HCl
求以上三种溶液的pH值？
（1）H3PO4 中加 NaOH， 逐步反应！
H3PO4 + NaOH = NaH2PO4 + H2ONaH2PO4 + NaOH = Na2HPO4 + H2ONa2HPO4 + NaOH = Na3PO4 + H2O
NaOH与H3PO4的摩尔数之比为2，生成Na2HPO4酸式盐
C = 50×2.4 / (50+60) = 1.091 mol/L
Na2HPO4 2Na+ + HPO42-
HPO42- + H2O H3O+ + PO4
3- Ka = Ka3HPO4
2- + H2O H2PO4- + OH- Kb = Kw/Ka2
8 13 102 3Ka Ka 6.2 10 2.0 10 1.11 10 (mol/L)+ − − −= = × × × = ×3[H O ]
pH = -lg[H3O+] = 9.95
（2）NaH2PO4 中加 NaOH， 逐步反应！
NaH2PO4 + NaOH = Na2HPO4 + H2ONa2HPO4 + NaOH = Na3PO4 + H2O
NaOH与NaH2PO4的摩尔数之比小于1，为共轭弱酸弱
碱对： NaH2PO4/Na2HPO4
8NaH2PO4a2
Na2HPO4
8
50 0.2 50 0.1C 50 50K 6.2 10 50 0.1C
50 506.2 10 (mol/L)
+ −
−
× − ×+= × = × ××+
= ×
3[H O ]
pH = -lg[H3O+] = 7.21
（3）Na3PO4 中加HCl， 逐步反应！
Na3PO4 + HCl = Na2HPO4 + NaClNa2HPO4 + HCl = NaH2PO4 + NaClNaH2PO4 + HCl = H3PO4 + NaCl
HCl与Na3PO4的摩尔数之比为2，生成NaH2PO4酸式盐
C = 50×3.0 / (50+50) = 1.5 mol/L
NaH2PO4 Na+ + H2PO4-
H2PO4- + H2O H3O+ + HPO4
2- Ka = Ka2H2PO4
- + H2O H3PO4 + OH- Kb = Kw/Ka1
3 8 57.5 10 6.2 10 2.16 10 (mol/L)+ − − −= = × × × = ×3 a1 a2[H O ] K K
pH = -lg[H3O+] = 4.67
例2：将一未知一元弱酸溶于未知量的水中，并用一未
知浓度的一元强碱去滴定。已知当用去 3.05 ml 强碱
时，溶液pH = 4.00；当用去 12.91 ml 强碱时，溶液 pH = 5.00。问该弱酸的 Ka为多少？
解： HA + NaOH → H2O + NaA化学计量点（刚好完全反应）时，溶液的pH > 7当用去3.05 ml和12.91 ml强碱时，溶液的pH分别为 4.00和5.00，说明加入强碱的量在化学计量点之前，此时生
成的是共轭弱酸弱碱对水溶液。假定HA：Ca (mol/L), Va (ml)；NaOH: Cb (mol/L)
3 a
C[H O ] K
C+ = 共轭弱酸
共轭弱碱
a a b
4.00 a a a ba a
b b
a
C V 3.05CV +3.05 C V 3.05C10 K K (1)3.05C 3.05CV +3.05
−
−−
= =
a a b
5.00 a a a ba a
b b
a
C V 12.91CV +12.91 C V 12.91C10 K K (2)12.91C 12.91CV +12.91
−
−−
= =
4.00 b ba
b
20.15C 3.05C10 K3.05C
− −=
(1) / (2), 得CaVa = 20.15Cb, 代入(1)得：
所以，Ka = 1.78 × 10-5
对应书上的内容：
P. 347 − 350
作业：
P331：13，14 (滴定曲线不需要作)13：0.00, 20.00, 39.98, 40.00, 40.02, 50.00 mL14: 0.00, 25.00, 49.98, 50.00, 50.02, 60.00 mL上述两题的指示剂从甲基橙，甲基红和酚酞
中选择，并指出指示剂的变色情况。
