Overview Activity Chem Equilibrium Acids Carbonate System Mineral Phases

Diprotic Acids

The math description of diprotic acids (including carbonic acid as a case of exceptional interest) leads to simple analytical equations. The derivation will give us a better understanding of what happens “inside” hydrochemical models and software.¹

Basic Set of Mathematical Equations

When a diprotic acid H₂A (the solute) is added to pure water (the solvent), the equilibrium state of the solution is characterized by five dissolved species: H⁺, OH^-, H₂A, HA^-, and A^-2.

Thus, five equations are required for its math description:²

(1a)	K1	=  {H+} {HA-} / {H2A}	(1st diss. step)
(1b)	K2	=  {H+} {A-2} / {HA-}	(2nd diss. step)
(1c)	Kw	=  {H+} {OH-}	(self-ionization of water)
(1d)	CT	=  [H2A] + [HA-] + [A-2]	(mass balance)
(1e)	0	=  [H+] – [HA-] – 2 [A-2] – [OH-]	(charge balance)

The first three equations are mass-action laws; the last two equations represent the mass balance and the charge balance. While the mass-action laws are based on activities (here denoted by curly braces), the mass-balance and charge-balance equations rely on molar concentrations (denoted by square brackets).

An exact solution in closed form (i.e. an analytical formula) is only obtainable if the activities in the first three equations are replaced by molar concentrations.³ This is valid either in (very) dilute systems or by switching to conditional equilibrium constants ^cK. In the following we assume that this has been done (without explicitly introducing the notation ^cK).

Ionization Fractions

Let’s start with the acid-species distribution as a function of pH. To study this behavior, a subset of the above equation system is sufficient, consisting of three equations only: (1a), (1b), and (1d). From the first two equations one gets (with the abbreviation x = {H⁺} = 10^-pH):

(2)	[H2A] = (x/K1) [HA-]     and     [A-2] = (K2/x) [HA-]

Entering it into 1d yields

(3)	CT  =  (x/K1 + 1 + K2/x) [HA-]

This allows us to write the following simple formulas for the three dissolved species:

(4)

[H2A] = CT a0

[HA-] = CT a1

[A-2] = CT a2

with the three ionization fractions:

(5a)		a0   =   [ 1 + K1/x + K1K2/x2 ]-1
(5b)		a1   =   [ x/K1 + 1 + K2/x ]-1	=	(K1/x) a0
(5c)		a2   =   [ x2/(K1K2) + x/K2 + 1 ]-1	=	(K1K2/x2) a0

It’s easy to check that all three coefficients add up to 1:

(6)

a0 + a1 + a2 = 1

(mass balance)

Because of their elegance and simplicity, diagrams of ionization fractions (also known as Bjerrum plots) appear in almost every textbook on hydrochemistry. Below is an example for the carbonic acid system (with pK₁ = 6.35, pK₂ = 10.33):

The three small circles in the diagram represent equivalence points.

The concentrations of the three acid species in 4 can also be combined in one formula:

(7)

[H2-j A-j]  =  CT aj(x)

for  j = 0, 1, 2

This formula, together with 5, predicts the pH dependence of the three acid species. Aside from the normalization constant C_T, the concentration curves correspond to the ionization curves in the above diagram.

Be careful though, C_T is not a constant, as 7 would suggest; C_T depends on pH — as shown below — and can therefore not be regarded as an independent parameter. This misunderstanding comes from the fact that we have so far ignored both the charge balance and the self-ionization of the water, i.e. 1e and 1c.

Exact Analytical Solution

The problem mentioned above is solved by incorporating two constraints: charge balance and the self-ionization of water. Setting 1d into 1e and using 7, together with the shorthand y_j = [H_2-jA^-j], yields:

		0	=   x – y1 – 2 y2 – Kw/x
			=   x – Kw/x – (y1 + 2 y2)
			=   x – Kw/x – CT (a1 + 2 a2)

This provides the exact relationship between the total amount of acid C_T and the pH value (= –lg x):

(8)

\(C_T(x) \ =\ \dfrac{x-K_w/x}{a_1 + 2a_2} \ =\ \left(x-\dfrac{K_w}{x}\right) \ \dfrac{K_2/x + 1 + x/K_1} {1 + 2K_2/x}\)

In fact, this one-liner encapsulates the entire information contained in the set of five nonlinear equations, i.e. 1a to (1e).

Based on 8 we are in a position to replace the approximate formula in 7 by an exact formula valid for all three acid species:

(9)

[H2-j A-j]  =  \(\left( \dfrac{x-K_w/x}{a_1 + 2a_2} \right) \ a_j\)

for   j = 0, 1, 2

[Example: The equations above were applied for the description of the closed and the open CO₂ system.]

Inverse Task. Given the pH (or x), 8 calculates C_T. The inverse task to calculate the pH (or x) for a given C_T, however, is intricate, because an explicit function, such as pH = f(C_T), does not exist. The only thing we can offer is an implicit function in form of a polynomial of degree 4 in x, which is a quartic equation:

(10)

x4 + K1 x3 + (K1K2 – CT K1 – Kw) x2 – K1 (2CT K2 + Kw) x – K1K2Kw  =  0

To recap: There is 8, there is 10, and there is the set of five equations defined in (1). All three entities are equivalent; they represent one and the same thing: the complete math description of a diprotic acid. Surely, calculating C_T for a given x (or pH) by 8 is much easier than to solve a 4^th order equation to get x (or pH) for a given value of C_T.

Diprotic Acids including Ampholytes and Conjugate Bases

Any diprotic acid is tight-knit with its conjugate base(s), H₂A ⇔ BHA ⇔ B₂A, where B refers to the cation of a monoacidic base (B⁺ = Na⁺, K⁺, or NH₄⁺). For example: H₂CO₃, NaHCO₃, and Na₂CO₃ represents such an acid-ampholyte-base triple.

Let’s denote the stoichiometric coefficient of B⁺ by n, then we get the compact notation:

(11)	BnH2-n A	(or  H2-n A-n)	with	n = 0	for acid	(H2A)
				n = 1	for ampholyte	(BHA)
				n = 2	for base	(B2A)

The set of equations to describe this system is

(12a)	K1	=  {H+} {HA-} / {H2A}	(1st diss. step)
(12b)	K2	=  {H+} {A-2} / {HA-}	(2nd diss. step)
(12c)	Kw	=  {H+} {OH-}	(self-ionization)
(12d)	CT	=  [H2A] + [HA-] + [A-2]	(mass balance)
(12e)	0	=  [H+] + n [H2A] + (n-1) [HA-] + (n-2) [A-2] – [OH-]	(proton balance)

It differs from the set of equations (1) only by a single equation, namely the last line, where “charge balance” is replaced by the more general concept of proton balance.⁴

Remarkably enough, the last equation (12e) is the sole equation that explicitly depends on n. The other four equations are independent on the type of reactant we add to water (acid, ampholyte, or base). In particular, the ionization fractions derived in 5a to (5c) for the diprotic acid, H₂A, are independent of n; they remain the same in our extended approach.

The set of equations (12) represents the core for the math description of buffer systems.

Exact Relationship between pH and C_T

The entire set of equations defined in 12a to (12e) can be condensed into a single formula, much like it was done in 8 above:

(13)	CT(n,x)  =  \(\dfrac{x-K_w/x}{a_1 + 2a_2 - n} \, =\, \left(x-\dfrac{K_w}{x}\right)\, \left(\dfrac{1+2K_2/x} {x/K_1 + 1 + K_2/x} - n\right)^{-1}\)

For n=0, it falls back to 8. Based on 13 we get — in place of 9 — the generalized formula for the three acid species:

(14)

[H2-j A-j]  =  \(\left( \dfrac{x-K_w/x}{a_1 + 2a_2 - n} \right)\ a_j\)

for   j = 0, 1, 2

Inverse Task. The conversion of C_T(n,x) into its inverse form x(n,C_T) leads again to a polynomial of degree 4 in x (quartic equation):

(15)

x4 + {K1 + nCT} x3 + {K1K2 + (n–1)CT K1 – Kw} x2 + K1 {(n–2)CT K2 – Kw} x – K1K2Kw  =  0

Each formula, whether 13 or 15, mimics three equations in compact form: one for an acid (n=0) — already presented in 8 and (10), one for an ampholyte (n=1), and one for a base (n=2).

Plots. The diagram below displays C_T as a function of pH. The solid lines represent 13 for n = 0, 1, and 2. The dots are exact results calculated with aqion (or PhreeqC), where activity corrections are considered. [Note: Activity corrections are especially relevant for high concentrations, i.e. high ionic strengths.]

The prefactor in the parenthesis of 13, x – K_w/x, becomes zero at pH = 7.0, i.e. at x = 10^-7. This must be the case, because C_T=0 means “pure water” (where all the curves come together).

Proton Balance Equation (Proton Condition)

The proton balance was used in 12e. It is a balance between the species that have excess protons versus those that are deficient in protons relative to a defined proton reference level (PRL).

Example 1. The simplest case is pure water with its three species H⁺, OH^-, and H₂O. Choosing H₂O as the reference level, the species H⁺ (or H₃O⁺) is enriched in 1 proton (excess proton), while OH^- is depleted in 1 proton (deficient proton). The proton balance equation becomes:⁵

	PRL	excess protons	=	deficient protons
(16)	H2O	[H+]	=	[OH-]

Because water is ever-present in a acid-base system, H⁺ and OH^- always enter the proton balance, one on the left- and the other on the right-hand side of the equation.

Example 2. The carbonic acid system has three distinct reference levels:⁶

	PRL	excess protons	=	deficient protons
(17a)	H2CO3	[H+]	=	[HCO3-] + 2 [CO3-2] + [OH-]
(17b)	HCO3-	[H+] + [H2CO3]	=	[CO3-2] + [OH-]
(17c)	CO3-2	[H+] + 2 [H2CO3] + [HCO3-]	=	[OH-]

How do you obtain these equations?

First, the two species H⁺ and OH^- that appear in each equation trace back from the H₂O-reference level in 16.⁷ They have their permanent place on opposite sides in any proton balance. Thus, all we have to do is to add the carbonic-acid species (H₂CO₃, HCO₃^-, CO₃^-2) to the correct side of the equation.

In 17a, H₂CO₃ is the reference level. There are no carbonate species that have more protons than H₂CO₃, hence, there is nothing to add to the left-hand side. Conversely, HCO₃^- is deficient by 1 proton and CO₃^-2 by 2 protons; therefore, both species enter the right-hand side.⁸

In 17b, HCO₃^- is the reference level. From this perspective, H₂CO₃ has 1 excess proton (species enters the left-hand side), while CO₃^-2 is deficient by 1 proton (species enters the right-hand side).

In 17c, CO₃^-2 is the reference level. Now, H₂CO₃ has 2 excess protons and HCO₃^- has 1 excess proton (both species enter the left-hand side); but there are no species that have less protons than CO₃^-2 (i.e. no carbonate species enters the right-hand side).

General Case. Given the proton-reference level by H_2-nA^-n, the proton balance equation becomes (for n = 0, 1, 2):

	PRL	0	=	excess protons – deficient protons
(18)	H2-n A-n	0	=	[H+] + n [H2A] + (n-1) [HA-] + (n-2) [A-2] – [OH-]

This one-liner comprises all three equations of Example 2. Equation (18) was adopted in 12e above.

The proton reference level (PRL) is closely related to the concept of alkalinity and equivalence points (often both terms are used as synonyms).

Remarks & Footnotes