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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

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Diprotic Acids and Equivalence Points

  • 1. Diprotic Acids Tableaux & Equivalence Points aqion.de updated 2017-04-15
  • 2. monoprotic acid: HA diprotic acid: H2A triprotic acid: H3A our focus most prominent representative: carbonic acid (with A-2 =CO3 -2)
  • 3. pure H2O OH- H+ H2O + H2A H2A HA- A-2 OH- H+ acid H2A 5species 2 components
  • 4. Acid H2A (Concepts & Notation) Part 1
  • 5. Two Dissociation Steps H2A ⇄ H+ + HA- ⇄ 2H+ + A-2 Five Species (NS = 5) H+, OH-, H2A, HA-, A-2
  • 6. (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
  • 7. (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
  • 8. 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
  • 9. 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
  • 10. 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
  • 11. 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
  • 12. 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)
  • 13. 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)
  • 15. NS = NC + NR number of species number of equilibrium reactions number of components (master species)
  • 16. 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 +  ...
  • 17. 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
  • 18. 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.
  • 19. 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
  • 20. Generalization BnH2-n A B+ = Na+, K+, NH4 + n = 0 acid n = 1 ampholyte n = 2 base H2A BHA B2A
  • 21. 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)
  • 22. 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
  • 23. 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
  • 24. 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
  • 25. 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
  • 26. 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
  • 27. 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
  • 28. 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
  • 29. 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                 
  • 30. 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)
  • 31. Equivalence Points (of Carbonic Acid & Co) Part 3
  • 32. 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=
  • 33. 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
  • 34. 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
  • 35. 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
  • 36. 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
  • 37. 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                 
  • 38. 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
  • 39. 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
  • 40. 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)
  • 41. 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: