Version 2012 Updated on 0423 Copyright © All rights reserved
Dong-Sun Lee, Prof., Ph.D. Chemistry, Seoul Women’s University
Chapter 13
Titration Curves for Complex
Acid/Base Systems
The human body contains a
complicated system of buffers
within cells and within bodily
fluids, such as human blood.
A scanning electromicrograph of red blood cells traveling through an artery. The
pH of human blood is controlled to be within the range of 7.35 to 7.45 primarily
by the carbonic acid / bicarbonate buffer system.
CO2(g) + H2O(l)  H2CO3(aq)
H2CO3 + H2O  H3O+ + HCO3
Mixtures of Strong and Weak Acids or Strong and Weak Bases
The dissociation constant for the weak acid or base is somewhat less than about 1.00 ×10–4.
Ex. pH of 0.1200 M HCl and 0.0800 M weak acid HA (Ka= 1.00 ×10–4)
[H3O+] = CHCl + [A–] = 0.1200 + [A–]
If [A–] << 0.1200 M, [H3O+]  0.1200 M  pH = 0.92
Check assumption:
Ka= [H3O+] [A–] / [HA] = 1.00 ×10–4
[A–] / [HA] = Ka / [H3O+] = 1.00 ×10–4 / 0.1200 = 8.33 ×10–4
[HA] = [A–] / 8.33 ×10–4
CHA = [HA] + [A–] = 0.0800 M
= {[A–] / 8.33 ×10–4 } + [A–]
 (1.20×103 ) [A–] = 0.0800 M
 [A–] = 6.7 ×10–5 M << 0.1200 M
Curves for the titration of strong acid / weak acid mixture with 0.1000 M
NaOH. Each titration is on 25.00 mLof a solution that is 0.1200 M in HCl
and 0.0800 M in HA.
Polyfunctional Acids and Bases
The phosphoric acid system
H3PO4 + H2O
H2PO4– + H2O
HPO42 – + H2O
⇌ H2PO4– + H3O+
⇌ HPO42 – + H3O+
⇌ PO43 – + H3O+
Ka1 = [H2PO4–][H3O+] / [H3PO4] = 7.11 ×10–3
Ka2 = [HPO42 –][H3O+] / [H2PO4–] = 6.32×10–8
Ka3 = [PO43 –][H3O+] / [HPO42 –] = 4.5 ×10–13
Ka1 > Ka2 > Ka3
Ka1 Ka2 = [HPO42 –][H3O+] / [H3PO4] = 7.11 ×10–3 × 6.32×10–8 = 4.49×10–10
Ka1 Ka2 Ka3 = [PO43 –][H3O+] / [H3PO4] = 7.11 ×10–3 × 6.32×10–8 × 4.5 ×10–13 =
2.0×10–22
The carbon dioxide carbonic acid system
CO2(aq) + H2O ⇌ H2CO3
Khyd = [H2CO3] / [CO2(aq)] = 2.8 ×10–3
H2CO3 + H2O ⇌ HCO3– + H3O+
K1 = [HCO3–] [H3O+] / [H2CO3] = 1.5×10–4
HCO3– + H2O ⇌ CO32 – + H3O+
K2 = [CO32 – ][H3O+] / [HCO3–] = 4.69 ×10–11
CO2(aq) + 2H2O
HCO3– + H2O ⇌
HCO3– + H3O+ Ka1 = [HCO3–] [H3O+] / [CO2(aq)]
= 2.8 ×10–3 × 1.5×10–4
= 4.2 ×10–7
CO32 – + H3O+
Ka2 = [CO32 – ][H3O+] / [HCO3–] = 4.69 ×10–11
⇌
Buffer Solutions Involving Polyprotic Acids

Ex. [H3O+] of mixture of 2.00 M phosphoric acid + 1.50 M potassium
dihydrogen phosphate
H3PO4 + H2O ⇌ H2PO4– + H3O+
Ka1 = [H2PO4–] [H3O+] / [H3PO4] = 7.11 ×10–3
[H3O+] = Ka1 × [H3PO4] / [H2PO4–]
= 7.11 ×10–3 × (2.00 / 1.50)
= 9.48 ×10–3 M
Diprotic Buffers
Henderson-Hasselbalch Equation
pH = pKa1 + log [HA ]
[H2A]
2pH = pKa2 + log [A ]
[HA-]
Polyprotic Acids and Bases
H3 L
H2LHL2-
⇌ H2L- + H+
⇌ HL2- + H+
⇌ L3- + H+
Ka1 = K1
Ka2 = K2
Ka3 = K3
L3- + H2O ⇌ HL2- + OHHL2- + H2O ⇌ H2L- + OHH2L- + H2O ⇌ H3L + OH-
Kb1 = Kw / Ka3
Kb2 = Kw / Ka2
Kb3 = Kw / Ka1
Remember !
Ka1 ˙ Kb3 = Kw
Ka2 ˙ Kb2 = Kw
Ka3˙ Kb1 = Kw
Diprotic Acids and Bases
+
CH COO–
R
H3 N
NH 2 CH C O O H
R
Zwitterion
H2L+
Leucine HL
L
Diprotic acid
H2L+ ⇌ HL + H+
HL ⇌ L– + H+
Ka1 = K1= 4.69×10–3
Ka2 = K2= 1.79×10–10
Diprotic base
L– + H2O ⇌ HL + OH–
HL + H2O ⇌ H2L+ + OH–
Remember !
Ka1 = Kw / K2
Kb2 = Kw / K1
Ka1 Kb2 = Kw
Ka2 Kb1 = Kw
To calculate the pH and composition of individual solution of 0.050 0M H2L+, HL and L–.
The Acid Form, H2L
N+H3 CH COO -
N+H3 CH COOH
H2L+
0.0500- x
Ka1 =
HL
x
[HL][H+]
[H2L+]
=
x 2
0.05 - x
[HL] = [H+]= 1.31×10-2 M
= 4.69 × 10-3
x
∴ x = 1.31 × 10-2 M
⇒ pH=1.88
[H2L+] = 0.05 - x = 3.69×10-2 M
▶
The concentration of L– ?
Ka2 = K2 =
[
L-]
=
[ L-][H+]
[HL]
1.79×10-10 [HL]
[H+]
+ H+
= 1.79×10-10
= 1.79×10-10 M = K2
The Basic Form, LL- + H2O ⇌ HL + OH-
Kb1 = Kw / K2 = 5.59×10-5
HL + H2O ⇌ H2L+ + OH-
Kb2 = Kw / K1 = 2.13×10-12
+
+
H2O
Kb1 =
[HL] =
+
[HL][OH–]
[L–]
1.64×10-3 M
H2O
HL
x
x 2 -0.05
=
= 5.59×10-5
2
x
Kw
[H+] =
= 6.08×10-12
x
[L-] = 0.05 - x = 4.84×10-2
▶ The concentration of
+
2.13×10-12[HL]
[OH-]
[H2L+][OH-]
[HL]
= 2.13×10-12 = Kb2
OHx
x = 1.64×10-3 M
⇒ pH=11.22
H2L+ ?
Kb2 = Kw / K1 = 2.13×10-12 =
[ H2L+] =
-
N+H3 CH COOH
NH 2 CH COO -
L0.05-x
OH
([HL]=[OH-])
The Intermediate Form, HL
HL ⇌ L- + H+
HL + H2O ⇌ H2L+ + OH-
Ka2 ≡ K2= 1.79×10-10
Kb2 ≡ Kw / K1 = 2.13×10-12
[H+] + [H2L+] = [L–] +[OH–]
[H2L+]–[L–] +[H+] –[OH–] = 0
[HL][H+]
K1
–
[HL]K2
[H+]
[HL][H+]2
K1
–
[HL]K2
[HL]
K1
[H+]2 (
[H+]2
=
+ [H+]
+ [H+]2
–
–
KW
[H+]
=0
KW
=0
+ 1 ) = K2 [HL] + Kw
K2 [HL] + Kw
[HL]
(
+1)
K1
([HL] = F)
[H+] =
K1K2 [HL] + K1 Kw
( K1 + [HL] )
[H+] ≈
K1K2 F + K1 Kw
( K1 + F )
▶ The pH of 0.0050 0 M Leucine?
(4.69 × 10-3)(1.79×10-10 )(0.005 0 M) + (4.69 × 10-3)(1.0 ×10-14)
[H+] =
4.69 × 10-3 + 0.005 0M
= 8.76 × 10-7 M
[
H2L+]
⇒ pH=6.06
[HL][H+]
= 9.34 x 10-6 M
K1
[HL]K2
= 1.02 x 10-5 M
+
[H ]
=
[L-] =
Simplified Calculation for the Intermedate Form
K2F > Kw
K1K2F + K1Kw ≒ K1 K2F
F > K1
K1+F + ≒ F
[H+]
=
K1K2 F + K1 Kw
( K1 + F )
pH =
1
2
( pK1 + pK2 )
=
K1K2 F
F
= K1K2

Isoelectric and isoionic pH
isoelectric pH : The pH value of the dispersion medium of a colloidal suspension
at which the colloidal particles do not move in an electric field.
At this point, the average charge of the polyprotic acid is zero.
Isoionic point : The pH obtained when the pure, neutral polyprotic acid HA(the
neutral zwitter ion) is dissolved in water.
Ex. 0.10 M Alanine
+
+
H3NCH(CH) 3COOH  H3NCH(CH) 3COO–  H2NCH(CH) 3COO–
Isoionic pH : [H+] =  (Ka1Ka2 F + Ka1Kw) / (Ka1+F)
Isoelectric pH : [H+] =  Ka1Ka2
The isoelectric pH is not concentration dependent.
 pH = 6.12
 pH = 6.11
Acid-base titration of the enzyme ribonuclease. The isoionic point is the
pH of the pure protein with no ions present except H+ and OH–. The
isoelectric point is the pH at which the average charge on the protein is 0.
Isoelectric focusing of a mixture of proteins: (1) soybean trypsin inhibitor,
(2) -lactoglobulin A, (3) -lactoglobulin B, (4) ovotransferrin, (5) horse
myoglobin, (6) whale myoglobin, (7) cytochrome C.
Principles of electrophoresis
Electrophoresis is a separation method based on the fact that charged particles migrate
through solution under the influence of an electric field. Their rate of migration depends on
the strength of the electric field, their net charge, the size and shape of the molecules, the
viscosity and temperature of the medium in which the molecules are moving, and on the
ionic strength of the buffer used. If the ion were placed in a vacuum, and if an electric field
were imposed, the ion would respond by being accelerated toward the electrode of opposite
charge.
Upper buffer chamber
F = qE

F= accelerating force
q= the charge on the
spherical ion
E= electrical field strength
sample
Power
supply
Gel tube
(V/cm)
F=mA
+
m= mass of spherical ion
A= acceleration
qE= mA
A = qE/m
Lower buffer chamber
A column gel for polyacrylamide electrophoresis


2
1
albumin
albumin
1
2 

Electrophoresis and densitometry for serum A/G ratio data (44.9: 54.2= 0.83:1)
of young Dong-Sun Lee in 1972. Normal adult range of A/G ratio is 0.8~2.0.
Reactions of weak acids with weak bases
HA  A– + H+
Ka = [A–][H+] / [HA]
B + H+  BH+
K = 1/ Ka(BH+) = Kb/ Kw
HA + B  A– + BH+
K = KaKb/ Kw
1) K >>1 , the reaction will go completion : like the weak + strong
Ex. ClCH2COOH + CH3NH2  ClCH2COO– + CH3NH3+
0.0500 M 100.0 mL 0.0600 M 60.0 mL
5.00
3.60
-
1.40
-
3.60
3.60
K = KaKb/ Kw = 6.0 ×107
pH = pKa + log( [A–] / [HA]) = 2.86 + log (3.60/1.40) = 3.27
2) K is not large : the reaction goes to completion
Ex. NH4 + + (CH3)3N  NH3 + (CH3)3NH+
0.0100M
0.0160 M
0.0100 – x 0.0160 – x
-
-
x
x
K = KaKb/ Kw = 3.60 = x2/(0.0100 – x)(0.0160 – x)
x = 0.00788
[NH4 +] = 0.00212 M [(CH3)3N] = 0.00812 M
[NH3] = 0.00788 M [(CH3)3NH+] = 0.00788 M
Aha ! Two buffers !
pH = pKa + log( [NH3] / [NH4 +])
= 9.244 + log (0.00788/0.00212) = 9.81
pH = pKa + log([(CH3)3N] / [(CH3)3NH+])
= 9.800 + log (0.00812/0.00788) = 9.81
3) Equimolar mixture of HA and B
Ex. 0.10M ammonium bicarbonate
NH4 HCO3  NH4+ + HCO3–
NH4 +  NH3 + H+
HCO3– + H2O  CO32– + OH–
mass balance: [NH4+] + [H+] = [HCO3–] + [OH–]
K1 > K1’
K1’ = [NH3] [H+] / [NH4+] = 5.70×10–10
K1 = [HCO3–] [H+] / [CO32–] = 4.45 ×10–7
K2 = [CO32–] [H+] / [HCO3–] = 4.69 ×10–11
[H+] =  (K1 K1’ F + K1 Kw ) / (K1 + F )
= {(4.45×10–7)(5.70×10–10)(0.10) + (4.45×10–7)(Kw )} / (4.45×10–7+0.10)
= 7.80
Curves for the titration of polyprotic acid. A) 0.1000M NaOH solution is used to titrate
25.00 ml of 0.1000 M H3PO4, B) 0.1000 M oxalic acid, and C) 0.1000 M H2SO4
Titration curves for
polyfunctional acids
Titration of 20.00 ml of 0.1000 M H2A with 0.1000 M
NaOH. For H2A, Ka1= 1.00 × 10–3 and Ka2 = 1.00 × 10–
7.
Molecular model of maleic acid,
or (Z)-butenedioic acid and
fumaric acid,
or (E)-butenedioic acid.
Titration curve for 25.00 ml of 0.1000 M maleic acid H2A with 0.1000 M NaOH
Curve for the titration of 5.00 mL of 0.0500 M
Na2CO3 with 0.1000 M HCl
Titraions in diprotic system 
1) Titration reactions :
B + H+  BH+
BH+ + H+  BH22+
2) Equivalence point :
FB N×VB mL = FH+N×Ve mL
2Ve1 = Ve2
3) Titration curve :
A. Before the addition of any titrant
B + H2O  BH+ + OH–
[OH–] = Kb1 F
[H+] = Kw / [OH–]
Ex. Na2CO3(mw=105.99)
0.0500 M =0.1000 N
pH = 11.51
The calculated pH titration curve for the titration
of 5.00 mL of 0.0500 M Na2CO3 with 0.1000 M HCl.
B. Before the 1st equivalence point
(First buffer region) 0<Va< Ve
CO32 – + H+  HCO3–
Major constituents: CO32 – + HCO3–
Ka2 = [H+][CO32 –]/[HCO3–] = Kw /Kb1= 4.7×10-11
pH = pKa2 + log [CO32 –]/[HCO3–]
initial CO32 – amount = F×Ve
– added HCl amount
= used amount CO32 – = F×Va
Remaining CO32 – amount = F(Vi–Va)
[CO32 –] = {(F×Ve) – (F×Va)}/(Vi +Va)
[HCO3–] = (F×Va) / (Vi +Va)
The calculated pH titration curve for the titration
of 5.00 mL of 0.0500 M Na2CO3 with 0.1000 M HCl.
이 완충영역 구간에 대한 Henderson－Hasselbalch식 log항의 [CO32–]/[HCO3–] 비를
계산하려면 적정반응의 진행정도에 관한 분율비를 구하면 된다.
0< Va <Ve1 구간에서 분율비는 다음과 같다.
[CO32–]/[HCO3–] = (Ve1 － Va) / Va
여기서 pKa2 = 10.329이므로 만약 Ve1 = 2.50 mL이고 Va = 0.50 mL라고 가정하고 대
입해서 구하면 pH = 10.93이다. 같은 방식으로 Va가 1.00, 1.25, 1.50, 1.75, 2.00, 2.25
mL일 때의 pH를 각각 10.51, 10.329, 10.15, 9,96, 9.73, 9.37로 계산할 수 있다.
특히 Ve1/2에 해당하는 Va = 1.25 mL인 점에서는 [CO32–]/[HCO3–] = 1이므로
pH = pKa2 이다.
C. First equivalence point Va=Ve1
Major constituents: HCO3–
[H+]= (Ka1 Ka2 F+ Ka1Kw ) /(Kb1 + F)
 (Ka1 Ka2
pH = (1/2)(pKa1 +pKa2 )
ex. 0.1000 M Na2HCO3
[H+]=  Ka1 Ka2
= (4.4×10–7× 4.7×10–7)
= 4.5×10–9 M
pH = 8.34
The calculated pH titration curve for the titration
of 5.00 mL of 0.0500 M Na2CO3 with 0.1000 M HCl.
D. Between the 1st and 2nd
equivalence points
(2nd buffer region) Ve1<Va<Ve2
HCO3 – + H+  H2CO3
Major constituents: HCO3– + H2CO3
Ka1 = [H+][HCO3 –]/[H2CO3]
= Kw /Kb2= 4.4×10-7
pH = pKa1 + log [HCO3–] / [H2CO3]
added HCl amount = F×Va
– initial CO32 – amount
= used HCl amount = F×Ve1
Remaining HCl amount = F(Va–Ve1)
Remaining HCl reacts with HCO3 – to form
H2CO3
[HCO3
–]
= {(1– F(Va–Ve1)}/(Vi +Va)
[H2CO3] = F(Va–Ve1) / (Vi +Va)
The calculated pH titration curve for the titration
of 5.00 mL of 0.0500 M Na2CO3 with 0.1000 M HCl.
제1당량점을 지나 제2당량점 이전까지 더 많은 HCl이 첨가되면 제1당량점에서 생성된
NaHCO3의 일부가 H2CO3로 변화된다. 따라서 NaHCO3와 H2CO3로 구성된 새로운 완충
용액이 만들어진다.
이 완충용액의 pH는 Kb2 또는 Ka1을 이용한 Henderson-Hasselbalch식으로 구할 수 있다.
Ka1 = [H+][HCO3 –]/[H2CO3] = Kw /Kb2= 4.4×10-7
pH = pKa1 + log [HCO3–] / [H2CO3]
이 Henderson－Hasselbalch식 log항의 비는 다음과 같이 계산된다.
제1당량점 이후 추가된 HCl이 HCO3–과 반응하여 H2CO3를 생성한다.
Ve1 < Va < Ve2 구간에서 [HCO3–] 분율은 (Ve2−Va)이고 [H2CO3] 분율은 (Va−Ve1)이다.
pH = pKa1 + log[HCO3–] / [H2CO3] = pKa1 + log (Ve2−Va)/ (Va−Ve1)
여기서 pKa1 = 6.352이다. 만약 Ve1 = 2.50 mL, Ve2 = 5.00 mL, Va = 2.75 mL라 가정하고 대
입하면 7.31이 얻어진다. 같은 방식으로 Va가 3.00, 3.25, 3.50, 3.75, 4.00, 4.25, 4.50, 4.75
mL일 때 pH를 각각 6.95, 6.72, 6.53, 6.352, 6.18, 5.98, 5.75, 5.37로 계산할 수 있다.
특히 (Ve1+Ve2)/2에 해당하는 Va = 3.75 mL인 점에서는
[HCO3–]/[H2CO3] = 1이므로 pH = pKa1 이다.
E. Second equivalence point Va=Ve2
Major constituents: H2CO3
H2CO3 = HCO3 – + H+
[H+] = Ka1F’
F’= ( F×Vi ) / (Vi +Va)
The calculated pH titration curve for the titration
of 5.00 mL of 0.0500 M Na2CO3 with 0.1000 M HCl.
F. After 2nd equivalence point
Va>Ve2
added HCl amount = F×Va
used HCl amount = F×Ve
Excess HCl amount = F×(Va –Ve)
[H+] = FHCl ×(Va– Ve) / (Vi +Va)
The calculated pH titration curve for the titration
of 5.00 mL of 0.0500 M Na2CO3 with 0.1000 M HCl.
Calculation of concentration (Finding of HCl concentration)
 
Na2CO3 (mw 105.99)  2 HCl (mw 36.361)  2 Eq wt.
105.99 g
105.99 g / 2
10.599 mg / 2
 2 × 1 M × 1000 mL
 1 M × 1000 mL
 0.1 M × 1 mL
a g  x M × Ve mL
 x M = ( a g ×1 M × 1000 mL) / (52.995 g × Ve mL)
Principal species 
The principal form of weak acid or base in aqueous solution under given
pH conditions.
COOH
COOH
COOH
pKa = 4.20
pH = pKa + log( [A–] / [HA])
pKa1 = 1.25
pKa2 = 4.27
pH = pKa1 + log( [HA–] / [H2A])
pH = pKa2 + log( [A2–] / [HA–])
pH
Major species
pH
pH < pKa
HA
pH < pKa1
H 2A
pH= pKa
HA and A–
pH= pKa1
H2A and HA–
pKa1<pH < pKa2
HA–
pH> pKa2
A2–
pH > pKa
A–
Major species
The predominent molecular form of a triprotic system in the various pH intervals.
Amino acid arginine
H3Arg2+

H2Arg+
pKa1 =1.82

pKa2 =8.99

HArg
Arg2–
pKa3 =12.48
pH = 1/2(pKa1+ pKa2) pH = 1/2(pKa2+ pKa3)
Acidic
Basic
Fractional composition equation
1) Monoprotic systems
Dissociation equilibria :
HA  A– + H+
Charge balance :
[H+] = [A–] + [OH–]
Mass balance :
F = [HA] + [A–]
Dissociation equilibria constants :

[A–] = F – [HA]
Ka = [A–][H+] / [HA]
Ka = [A–][H+] / [HA] = (F – [HA])[H+] / [HA]
[HA] = F [H+] / ([H+]+ Ka )
0 = fraction of species in the form HA = fraction of dissociation
1 = fraction of species in the form A–
0 + 1 = 1
0 = [HA] / F = [H+] / ([H+]+ Ka )
1 = Ka / ([H+]+ Ka )
Ka= 1.00E-05
pH
1
2
3
4
5
6
7
8
9
10
0
1
1.00E+00 1.00E-04
9.99E-01 9.99E-04
9.90E-01 9.90E-03
9.09E-01 9.09E-02
5.00E-01 5.00E-01
9.09E-02 9.09E-01
9.90E-03 9.90E-01
9.99E-04 9.99E-01
1.00E-04 1.00E+00
1.00E-05 1.00E+00
1.0
0.8
Fraction of species
[H+]
1.00E-01
1.00E-02
1.00E-03
1.00E-04
1.00E-05
1.00E-06
1.00E-07
1.00E-08
1.00E-09
1.00E-10
0.6
0.4
0.2
0.0
pH
Fractional composition diagram of a
monoprotic system with pKa=5.00
2) Diprotic system
H2A  HA– + H+
Dissociation equilibria :
HA–  A2– + H+
Dissociation equilibria constants :
Ka1 = [HA–][H+] /[H2A]
 [HA–] = Ka1[H2A] / [H+]
Ka2 = [A2–][H+] / [HA–]

[A2–] = Ka2 [HA–] / [H+]
= Ka1 Ka2[H2A] / [H+]2
Mass balance :
F = [H2A] + [HA–] + [A2–]
= [H2A] + (Ka1[H2A] / [H+]) +(Ka1 Ka2[H2A] / [H+]2)
= [H2A](1+Ka1/ [H+] + Ka1 Ka2 / [H+]2)
0 + 1 + 2 = 1
0 = [H2A] / F = [H+]2 / ([H+]2 + [H+]Ka1 + Ka1 Ka2 )
1 = [HA–] / F= Ka1 [H+] / ([H+]2 + [H+]Ka1 + Ka1 Ka2 )
2 = [A2–] / F= Ka1 Ka2 / ([H+]2 + [H+]Ka1 + Ka1 Ka2 )
Fractional composition diagram for
fumaric acid (trans-butenedioic acid).
Composition of H2M solutions
as a function of pH.
Titration of 25.00 ml of 0.1000M maleic acid with 0.1000M NaOH.
Logarithmic concentration diagram for 0.100M maleic acid.
Summary
Polyfunctional acid base
diprotic acid system
triprotic system
Principal species
Fractional composition equations
Ioselectric point
Isoionic pH
Isoelectric focusing
Electrophoresis
pH of the salt of weak acid + weak base