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