diprotic acids and equivalence points
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Diprotic AcidsTableaux & Equivalence Points
aqion.de
updated 2017-04-15
monoprotic acid: HAdiprotic acid: H2Atriprotic acid: H3A
our focusmost prominent representative: carbonic acid (with A-2
= CO3-2)
pure H2O
OH-
H+
H2O + H2A
H2A
HA-
A-2
OH-
H+
acid H2A5 species
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
mass balance
[H2 A] + [HA
-] + [A-2] = C
T
[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-}
Equilibrium Reactions
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 (I0) or apparent equilibrium constants.
Ionization Fractions (a0, a1, a2)exact relation between pH and CT
Ionization Fractionsa0 = [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 – CT K1 – Kw} x2
– K1 {2CTK2 + Kw} x – K1K2Kw
x/K21x/K1K/x
xKxC
2
21wT
total amount CT (of acid H2A) for a given pH
measured quantities
x/K21x/K1K/x
xKxC
2
21wT
constraint: CT 0
CT (x – Kw/x)
x2 Kw x Kw1/2 x 10-7 pH 7
(pure water: CT = 0 pH = 7)
Remark: Don’t confuse CT with [H2A]
CT > [H2A]
total amount of acid H2Athat 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 equilibriumreactions
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 matrixof stoichiometric
coefficients NC components N
S spe
cies
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 N
S spe
cies
s2 = 21 C1 + 22 C2 + ...
equilibrium constants
mass-action law
Example: pure H2A solution
choice is arbitrary(H+ should be included)
H+
H2ANC = 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
charge balance = proton balance
Generalization
BnH2-n A
B+ = Na+, K+, NH4+
n = 0 acid n = 1 ampholyte n = 2 base
H2ABHAB2A
A clever choice of componentssimplifies the calculations:
n C1 C2
0 H2A H+ H2A
1 BHA H+ HA-
2 B2A H+ A-2
also calledProton Reference Level (PRL)
H+ H2A B+
H+ 1
OH- -1 -Kw
H2A 1
HA- -1 1 -K1
A-2 -2 1 -K1K2
B+ 0H2A 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+ 2B2A 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+ nBnH2-n A 0 1 n
General Case: pure BnH2-n A 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-n A solution
(proton balance)
Set of 5 equations
only for n = 0: proton balance = charge balance
The general case (BnH2-n A) differs from thepure 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 + n CT } x3 + {K1K2 + (n-1) CT K1 – 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
12wT K/x)n0()n1(x/K)n2(
K/x1x/KxKxC
total amount of BnH2-n A as a function of pH = – log x
1
21
2wT n
x/KK/x1x/K21
xKxC
Ionization Fractionsa0 = [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
amountof 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 from1
21
2wT n
x/KK/x1x/K21
xKxC
3 numerical model (e.g. PhreeqC)(incl. activity corrections)
exactapproxim
ation
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
C T [M
]
[H+] = [HA-][H2A] = [A-2]
[HA-] = [OH-]
Example: Carbonic Acid
This is an approximationand fails for CT 10-7 M.
1
outside range of applicability
self-ionization of H2O is ignored
pH
C T [M
]
H2CO3 NaHCO3
Na2CO3
n = 0n = 1
n = 2
)pH(fnx/KK/x1
x/K21xKxC n
1
21
2wT
Equivalence Points (EP) based on2
n = 0 H2CO3
n = 1 NaHCO3
n = 2 Na2CO3
versusEP based on ionization fractions(approximate)
Equivalence Points (exact)
1
pH
C T [M
]
H2CO3 NaHCO3
Na2CO3
n = 0n = 1
n = 2
2
dashed lines
)pH(fnx/KK/x1
x/K21xKxC n
1
21
2wT
Equivalence Points (EP)
pH
C T [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
C T [M
]
H2CO3 NaHCO3
Na2CO3
activity corrections NaHCO3 and NaCO3
- species 3
lines: xlogpHwithnx/KK/x1
x/K21xKxC
1
21
2wT
2
dots: PhreeqC calculations with:
3 EP based on Numerical Model
Summary of Assumptions done in previous EP Calculations
approach self-ionizationof 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)
Refwww.aqion.de/file/acid-base-systems.pdf
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