Overview Activity Chem Equilibrium Acids Carbonate System Mineral Phases

The Closed Carbonate System

In a closed carbonate system there is no CO₂ exchange between the solution and the atmosphere. Thus, in contrast to the open carbonate system, the total amount of dissolved inorganic carbon (DIC) remains constant when pH is varied by addition of acids or bases:

(1)	CT  =  DIC  =  [H2CO3*] + [HCO3-] + [CO3-2]  =  const

The closed CO₂-H₂O system is characterized by five species:¹

(2)	H2CO3*,   HCO3-,   CO3-2,   H+   and   OH-

Thus, at least five equations are required for an appropriate mathematical description.²

The Nonlinear System of Equations

The chemical equilibrium state is completely determined by five equations (with equilibrium constants valid for 25):

(3a)	K1	=   {H+} {HCO3-} / {H2CO3*}	=   10-6.35
(3b)	K2	=   {H+} {CO3-2} / {HCO3-}	=   10-10.33
(3c)	Kw	=   {H+} {OH-}	=   10-14.0
(3d)	CT	=   [H2CO3*] + [HCO3-] + [CO3-2]	(mole balance)
(3e)	0	=   [H+] – [HCO3-] – 2 [CO3-2] – [OH-]	(charge balance)

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

It is just the presence of the ion activities (in the curly braces) that makes the numerical solution so difficult and precludes any pencil-and-paper calculation.³ The situation changes/simplifies for so-called “ideal aqueous solutions” when activities collapse to concentrations. Another way out is to use conditional equilibrium constants ^cK (which is the standard approach in seawater studies).

Conditional Equilibrium Constants (Seawater)

Principally, it is legitimate to replace the activities by concentrations in 3a to (3c), that is, to substitute the curly braces by square brackets. The price we have to pay for this simplification is to switch from thermodynamic equilibrium constants K to conditional or apparent equilibrium constants ^cK:

(4a)	cK1	=   [H+] [HCO3-] / [H2CO3*]
(4b)	cK2	=   [H+] [CO3-2] / [HCO3-]
(4c)	cKw	=   [H+] [OH-]
(4d)	CT	=   [H2CO3*] + [HCO3-] + [CO3-2]	(mole balance)
(4e)	0	=   [H+] – [HCO3-] – 2 [CO3-2] – [OH-]	(charge balance)

The disadvantage of ^cK is that it depends on the actual composition of the water, notably the ionic strength or salinity. For very dilute waters with negligibly small ionic strength, I ≈ 0, the thermodynamic and conditional equilibrium constants are the same: ^cK = K. The larger I the more both values drift apart.

Seawater. Seawater has I ≈ 0.7 M, which is just on the upper bound of the validity range of common activity models. Thus, in oceanography, chemists prefer conditional equilibrium constants as an empirical function of temperature, pressure, and salinity: ^cK = f (T,P,S).

The following example shows the deviations of ^cK from the thermodynamic equilibrium constants K when the salinity is enhanced from 0 (pure water) to 35 g/L (seawater at 25, 1 atm):⁴

		thermodynamic K (pure water, I=0)		conditional cK (seawater)
	pK1	6.35		6.00
	pK2	10.33		9.10
	pKH	1.47		1.53
	pKW	14.0		13.9

[The thermodynamic equilibrium constants in the left column refer to 3a to (3c), and Henry’s constant K_H, whereby pK = –log K.]

Closed-Form Expression & Ionization Fractions

Once all activities in the formulas are expressed by concentrations (which can be done either for a dilute system or for conditional equilibrium constants), we are able to convert the equation system into a single analytical formula.

To do this, insert 4a to (4c) into 4d and 4e, and then merge the last two equations into one. In this way we get rid of all concentrations except [H⁺]. Replacing [H⁺] simply by x, the entire equation system coalesces into one single line:⁵

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

This is an equation of 4^th degree in x. All other quantities are either physico-chemical constants (K₁, K₂, K_W) or C_T (i.e. DIC). Thus, for any given value of DIC the equation predicts x, and via pH = –log x, the exact pH value of the water.

[Besides: 5 is valid for any diprotic acid, H₂A, with two equilibrium constants K₁ and K₂.]

Ionization Fractions. 5 predicts the pH value for a given DIC. Now we ask the question: What are the concentrations of all carbonate species as a function of pH?

For this purpose we insert 4a and (4b) into the mole balance (4d), which yields:

(6)

[H2CO3*] = CT α0

[HCO3-] = CT α1

[CO3-2] = CT α2

based on three ionization fractions (and x = [H⁺] = 10^-pH):

(7a)		α0   =   ( 1 + K1/x + K1K2/x2 )-1	=	(x/K1) α1
(7b)		α1   =   ( x/K1 + 1 + K2/x )-1
(7c)		α2   =   ( x2/(K1K2) + x/K2 + 1 )-1	=	(K2/x) α1

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

(8)	α0 + α1 + α2 = 1

The pH dependence of the ionization fractions are plotted in the second diagram below (for the special case C_T = DIC = 1 mM):

green curve:	CT α0 = α0 in mM
blue curve:	CT α1 = α1 in mM
orange curve:	CT α2 = α2 in mM

As the equation above suggest, the shape of the curves does not alter when C_T changes.

Example: Equilibrium Speciation of 1 mM DIC

The closed H₂O-CO₂ system, which is uniquely determined by the set of equations (3a) to (3e), yields the following speciation for pure water with 1 mM DIC (at 25 °C):

pH		4.68
CO2		0.979	mM	( = H2CO3* )
HCO3-		0.021	mM
CO3-2		4.8·10-8	mM
DIC		1.00	mM	( =  CO2 + HCO3- + CO3-2 )

The program provides three ways to obtain this result. All three calculations start with pure water (button New), but differ afterward as follows

Way 1.  Switch to molar units (checkbox Mol) and enter 1 mM DIC, then button Start and perform charge balance with parameter pH. The carbonate speciation is displayed in table Ions.

Way 2.  Add 1 mM CO2 to pure water by the reaction module (button Reac).

Way 3.  Add 1 mM H2CO3 to pure water by the reaction module (button Reac).

Example: Titration Calculation

Given is a closed carbonate system with DIC = C_T = 10^-3 M. The results of a titration calculation (based on successive addition of HCl and NaOH) are presented in the two diagrams below.

Example: Equivalence Points (EP)

Equivalence points are pH values at which the amount of two molar concentrations coincide.⁶ Let’s consider two equivalence points defined by:

EP   H2CO3:	[H+] = [HCO3-]
EP   Na2CO3:	[HCO3-] = [OH-]

The corresponding pH values depend on the total concentration of C_T as shown below in the first diagram. Thus, what we call an EP is one curve in the pH-C_T-diagram. (The top diagram contains three curves, but at this moment, we ignore the middle curve that belongs to the EP of NaHCO₃.)

In the first diagram, two points on the left and two points on the right EP-curve are selected:

EP   H2CO3:	pH = 5.16	(for CT = 10-4 M)	and	pH = 4.68	(for CT = 10-3 M)
EP   Na2CO3:	pH = 9.86	(for CT = 10-4 M)	and	pH = 10.52	(for CT = 10-3 M)

The selected points refer to the species distribution in the two lower diagrams (one for C_T = 10^-3 M and one for C_T = 10^-3 M). They are just points of intersection, that is, points where two concentrations coincide.

Calculations of equivalence points are performed here and here.

Remarks & References