Diprotic Acids
Tableaux & Equivalence Points
aqion.de
updated 2017-04-15

monoprotic acid: HA
diprotic acid: H2A
triprotic acid: H3A
our focus
most prominent representative:
carbonic acid (with A-2 =CO3
-2)

pure H2O
OH-
H+
H2O + H2A
H2A
HA-
A-2
OH-
H+
acid H2A
5species
2 components

Acid H2A
(Concepts & Notation)
Part 1

Two Dissociation Steps
H2A ⇄ H+ + HA- ⇄ 2H+ + A-2
Five Species (NS = 5)
H+, OH-, H2A, HA-, A-2

(1) Species are interrelated by
conservation rules:
H+ OH- H2A HA- A-2
charge balance
massbalance
[H2A]+[HA-]+[A-2]=CT
[H+] – [OH-] – [HA-] – 2[A-2] = 0
total amount of acid

(2) Species are interrelated by
Mass-Action Laws:
H2A = H+ + HA- HA- = H+ + A-2
H2O = H+ + OH- Kw = {H+
}{OH-
}
K1 = {H+
}{HA-
}/{H2A}
H+ OH- H2A HA- A-2
K2 = {H+
}{A-2
}/{HA-
}
EquilibriumReactions
self-ionization of water

H+ OH- H2A HA- A-2
Kw = {H+} {OH-} = 10-14
K1 = {H+} {HA-} / {H2A} (1st diss. step)
K2 = {H+} {A-2} / {HA-} (2nd diss. step)
CT = [H2A] + [HA-] + [A-2] (mass balance)
0 = [H+] – [HA-] – 2[A-2] – [OH-] (charge balance)
5 species (unknowns)  5 equations
{..} = activities, [..] = molar concentrations

Replace: Activities {..}  Concentrations [..]
Kw = [H+] [OH-]
K1 = [H+] [HA-] / [H2A]
K2 = [H+] [A-2] / [HA-]
CT = [H2A] + [HA-] + [A-2]
0 = [H+] – [HA-] – 2[A-2] – [OH-]
This is valid for small ionic strengths (I0) or apparent equilibrium constants.
Ionization Fractions (a0, a1, a2)
exact relation between pH and CT

Ionization Fractions
a0 = [1 + K1/x + K1K2/x2]-1 [H2A] = CT a0
a1 = [x/K1 + 1 + K2/x]-1 [HA-] = CT a1
a2 = [x2/(K1K2) + x/K2 + 1]-1 [A-2] = CT a2
x = [H+] = 10-pH
K1 = 10-6.35
K2 = 10-10.33
carbonic acid
mass balance
a0 + a1 + a2 = 1
pK1 pK2

Exact Relations between pH and CT
4th order equation in x = [H+] = 10-pH
0 = x4 + K1x3 + {K1K2– CTK1– Kw} x2
– K1 {2CTK2 + Kw} x – K1K2Kw
x/K21
x/K1K/x
x
K
xC
2
21w
T









total amount CT (of acid H2A) for a given pH
measured quantities

x/K21
x/K1K/x
x
K
xC
2
21w
T









constraint: CT  0
CT  (x – Kw/x)
x2  Kw x  Kw
1/2 x  10-7 pH  7
(pure water: CT = 0  pH = 7)

Remark:
Don’t confuse CT with [H2A]
CT > [H2A]
total amount of acid H2A
that enters the solution:
CT = [H2A] + [HA-] + [A-2]
amount of
aqueous species H2A(aq)
(dissolved,
but non-dissociated)(measured quantity)

Generalization
(Acids, Ampholytes, Bases)
Part 2

NS = NC + NR
number of
species
number of
equilibrium
reactions
number of
components
(master species)

C1 C2 ...
s1 11 12
s2 21 22
s3 31 32
...
Tableaux Method
TOT C1 = 11 s1 + 21 s2 + 31 s3 + ...
NSNC matrix
of stoichiometric
coefficientsNC components
NSspecies
s2 = 21 C1 + 22 C2 + ...

C1 C2 ...
s1 11 12 K1
s2 21 22 K2
s3 31 32 K3
...
Tableaux Method
{C1}21 {C2}21 / {s2} = K2
NC components
NSspecies
s2 = 21 C1 + 22 C2 + ...
equilibrium
constants
mass-action law

Example: pure H2A solution
choice is arbitrary
(H+ should be included)
H+
H2A
NC = 2
A set of NC components
defines a basis in the
„vector space“ of chemical species.

H+ H2A
H+ 1
OH- -1 -Kw
H2A 1
HA- -1 1 -K1
A-2 -2 1 -K1K2
H2A 0 1
pure H2A solution (NS = 5, NC = 2)
TOT H = [H+] – [HA-] – 2[A-2] – [OH-] = 0
chargebalance=protonbalance

Generalization
BnH2-n A
B+ = Na+, K+, NH4
+
n = 0 acid
n = 1 ampholyte
n = 2 base
H2A
BHA
B2A

A clever choice of components
simplifies the calculations:
n C1 C2
0 H2A H+ H2A
1 BHA H+ HA-
2 B2A H+ A-2
also called
Proton Reference Level (PRL)

H+ H2A B+
H+ 1
OH- -1 -Kw
H2A 1
HA- -1 1 -K1
A-2 -2 1 -K1K2
B+ 0
H2A 0 1 0
pure H2A solution
TOT H = [H+] – [HA-] – 2[A-2] – [OH-] = 0
n=0

H+ HA- B+
H+ 1
OH- -1 -Kw
H2A 1 1 K1
HA- 1
A-2 -1 1 -K2
B+ 1
BHA 0 1 1
pure BHA (or HA-) solution
TOT H = [H+] + [H2A] – [A-2] – [OH-] = 0
n=1

H+ A-2 B+
H+ 1
OH- -1 -Kw
H2A 2 1 K1K2
HA- 1 1 K2
A-2 1
B+ 2
B2A 0 1 2
pure B2A (or A-2) solution
TOT H = [H+] + 2[H2A] + [HA-] – [OH-] = 0
n=2

H+ H2-nA-n B+ n=0 n=1 n=2
H+ 1
OH- -1 -Kw -Kw -Kw
H2A n 1 1 K1 K1K2
HA- n-1 1 -K1 1 K2
A-2 n-2 1 -K1K2 -K2 1
B+ n
BnH2-nA 0 1 n
General Case: pure BnH2-nA solution
TOT H = [H+] – [OH-] + n[H2A]
+ (n-1)[HA-] + (n-2)[A-2] = 0

Kw = {H+} {OH-} = 10-14
K1 = {H+} {HA-} / {H2A} (1st diss. step)
K2 = {H+} {A-2} / {HA-} (2nd diss. step)
CT = [H2A] + [HA-] + [A-2] (mass balance)
0 = [H+] + n[H2A] + (n-1)[HA-] + (n-2)[A-2] – [OH-]
General Case: pure BnH2-nA solution
(proton balance)
Set of 5 equations

only for n=0:
proton balance = charge balance
The general case (BnH2-nA) differs from the
pure acid case (H2A) by the proton balance only
TOT H = [H+] + n[H2A] + (n-1)[HA-] + (n-2)[A-2] – [OH-] = 0
[H+] – [HA-] – 2[A-2] – [OH-] = 0
Proton Balance

Set of 5 equations
replace activities {..} by concentrations [..]
(or K by conditional equilibrium constants cK)
closed-form 4th order equation in x=[H+]
0 = x4 + {K1+ nCT} x3
+ {K1K2+ (n-1)CTK1– Kw} x2
+ K1 {(n-2)CTK2– Kw} x – K1K2Kw
exact relation between CT and pH=-log x

C
C
convert 4th-order equation to CT
12
12w
T
K/x)n0()n1(x/K)n2(
K/x1x/K
x
K
xC









total amount of BnH2-nA
as a function of pH = – log x
1
21
2w
T n
x/KK/x1
x/K21
x
K
xC


















Ionization Fractions
a0 = [1 + K1/x + K1K2/x2]-1 [H2A] = CT a0
a1 = [x/K1 + 1 + K2/x]-1 [HA-] = CT a1
a2 = [x2/(K1K2) + x/K2 + 1]-1 [A-2] = CT a2
x = [H+] = 10-pH
are independent of n
(same for pure H2A, HBA, B2A solutions)

Equivalence Points
(of Carbonic Acid & Co)
Part 3

Equivalence Points (EP)
n=0 pH of pure H2A solution: [H+] = [HA-]
n=1 pH of pure BHA solution: [H2A] = [A-2]
n=2 pH of pure B2A solution: [HA-] = [OH-]
amount
of acid
amount
of base=

Carbonic Acid & Co
pH of pure H2CO3 solution: [H+] = [HCO3
-]
pH of pure NaHCO3 solution: [H2CO3] = [CO3
-2]
pH of pure Na2CO3 solution: [HCO3
-] = [OH-]
Equivalence
Points
pK1 = -log K1 6.35
pK2 = -log K2 10.33
pKw = -log Kw 14.0

Calculation of
Equivalence Points
1 from Ionization Fractions (a0, a1, a2)
2 from
1
21
2w
T n
x/KK/x1
x/K21
x
K
xC

















3 numerical model (e.g. PhreeqC)
(incl. activity corrections)
exact
approximation

x = [H+] = 10-pH
EP based on Ionization Fractions
[H+] = [HA-]  x = CT a1  CT = x2/K1 + x + K2
[H2A] = [A-2]  a0 = a2  x2 = K1K2 (independent of CT)
[HA-] = [OH-]  CT a1 = Kw/x  CT = (Kw/x2)(x2/K1 + x + K2)
pH = -log x
CT[M]
[H+] = [HA-]
[H2A] = [A-2]
[HA-] = [OH-]
Example: Carbonic Acid
This is an
approximation
and fails
for CT  10-7 M.
1
outside range
of applicability
self-ionization
of H2O is ignored

pH
CT[M]
H2CO3
NaHCO3
Na2CO3
n= 0
n=1
n=2
)pH(fn
x/KK/x1
x/K21
x
K
xC n
1
21
2w
T 
















Equivalence Points (EP) based on2
n = 0 H2CO3
n = 1 NaHCO3
n = 2 Na2CO3

versus
EP based on
ionization fractions
(approximate)
Equivalence Points (exact)
1
pH
CT[M]
H2CO3
NaHCO3
Na2CO3
n=0
n=1
n=2
2
dashed lines
)pH(fn
x/KK/x1
x/K21
x
K
xC n
1
21
2w
T 

















Equivalence
Points (EP)
pH
CT[M]
H2CO3
NaHCO3
Na2CO3
I = 0 seawater
pK1 6.35 6.00
pK2 10.33 9.10
pKw 14.0 13.9
dashed: seawater
conditional (apparent)
equilibrium constants at 25 °C
[Millero 1995]
2

pH
CT[M]
H2CO3 NaHCO3
Na2CO3
 activity corrections
 NaHCO3 and NaCO3
- species
3
lines: xlogpHwithn
x/KK/x1
x/K21
x
K
xC
1
21
2w
T 
















2
dots: PhreeqC calculations with:
3 EP based on
Numerical
Model

Summary of Assumptions
done in previous EP Calculations
approach
self-ionization
of water
activity
corrections
formation of
complexes (e.g.
NaHCO3
-)
no no no
yes no no
yes yes yes
1
2
3
determines behavior
at very low CT
determines behavior
at very high CT
(especially for Na2CO3)

CT = 10-4 M
CT = 10-3 M
[H+] = [HCO3
-] [HCO3
-] = [OH-]
Equivalence
Points (EP)
as intersection points
EP
pH at
CT = 10-4 M
pH at
CT = 10-3 M
H2CO3 5.16 4.68
Na2CO3 9.86 10.52
plots based on PhreeqC or aqion calculations
demonstrated for two total
concentrations CT:

www.aqion.de/site/122 (EN)
www.aqion.de/site/59 (DE)
Ref
www.aqion.de/file/acid-base-systems.pdf