Acid-Base Equilibria

Abstract

This chapter presents acid-base equilibria and simple methods to calculate the pH of solutions. The inorganic carbon system in water is introduced, including the concept of alkalinity, and methods to solve carbon dioxide equilibria in water are discussed. Earth system science relevant examples such as rainwater, surface waters in equilibrium with calcium carbonate minerals, soda lakes and the physical chemistry of karst systems are presented.

1 Introduction

Although protons are produced and consumed in many chemical processes, most natural waters fall within a narrow range of pH values of 6–9 because of interactions with minerals and transfer of protons (i.e., hydrogen ions) among acids and bases in solution. Most proton transfer reactions in solution are very fast (i.e., reach an equilibrium in milliseconds) and a (homogenous) equilibrium description is thus appropriate. A proton does not exist as such in water because it reacts fast and strongly with water to form a hydrated proton (hydronium), H₃O⁺, which in turn associates with hydrogen bonds to additional water molecules. It is macroscopically not possible to distinguish between the various hydrated proton species and we will therefore use H⁺ and H₃O⁺ interchangeably and refer to H⁺ as the proton or hydrogen ion in solution.

An acid is a substance that dissociates in water releasing protons into solution, i.e., it is a proton donor, while a base accepts protons. Chemical species whose formulas differ only by one hydrogen ion are called conjugate acid-base pairs. An acid HA dissolved in water dissociates into a proton and a base A⁻:

$$ {\text{HA}} \left( {aq} \right) + {\text{H}}_{2} {\text{O}} \left( l \right) \leftrightarrow {\text{H}}_{3} {\text{O}}^{ + } \left( {aq} \right) + {\text{A}}^{ - } \left( {aq} \right) $$

(6.1)

for which we can write the general equilibrium constant K

$$ K = \frac{{\left[ {{\text{H}}_{3} {\text{O}}^{ + } } \right]\left[ {{\text{A}}^{ - } } \right]}}{{\left[ {{\text{HA}}} \right] \left[ {{\text{H}}_{2} {\text{O}} } \right]}} $$

Noting that the liquid medium has a (activity) value of 1, we can then define the acid dissociation or acidity constant

$$ K_{a} = \frac{{\left[ {{\text{H}}^{ + } } \right]\left[ {{\text{A}}^{ - } } \right]}}{{\left[ {{\text{HA}}} \right] }} $$

(6.2)

A similar treatment can be followed for base B added to water

$$ {\text{B}}\left( {aq} \right) + {\text{H}}_{2} {\text{O}} \left( l \right) \leftrightarrow {\text{OH}}^{ - } \left( {aq} \right) + {\text{BH}}^{ + } \left( {aq} \right) $$

(6.3)

with the base dissociation or basicity constant

$$ K_{b} = \frac{{\left[ {{\text{OH}}^{ - } } \right]\left[ {{\text{BH}}^{ + } } \right]}}{{\left[ {\text{B}} \right] }}, $$

(6.4)

but we will elaborate it from the acid side of the story.

The extent of dissociation, i.e., the proton release by the acid HA and its transfer to water acting as the base in Reaction 6.1, depends on the strength of the acid, i.e., the value of K_a. Strong acids nearly fully dissociate into H⁺ and A⁻ and almost no HA remains in solution. Weak acids dissociated partially and after equilibration HA, H⁺ and A⁻ are found in solution. Acids stronger than H₃O⁺ are called strong acids and their conjugated bases are very weak. Weak acids are less strong than H₃O⁺ but stronger than H₂O (Table 6.1) and their conjugated bases are also weak.

Table 6.1 Acid and base dissociation constants of conjugated acid-base pairs

Reactions 6.1 and 6.3 show that water can act as a proton donor as well as proton acceptor. The dissociation or self-ionization of water,

$$ 2{\text{H}}_{2} {\text{O}} \left( l \right) \leftrightarrow {\text{H}}_{3} {\text{O}}^{ + } \left( {aq} \right) + {\text{OH}}^{ - } \left( {aq} \right) $$

(6.5)

leads to the ion-product constant for water, i.e., the equilibrium constant for self-ionisation of water,

$$ K_{w} = \left[ {{\text{H}}_{3} {\text{O}}^{ + } } \right] \left[ {{\text{OH}}^{ - } } \right] = 10^{ - 14} $$

(6.6)

Some acids contain multiple protons, and these are called polyprotic acids (e.g. carbonic acid, phosphoric acid). These polyprotic acids dissociate stepwise, and each dissociation step is characterized by its own acid dissociation constant. In solutions containing multiple acids and bases, protons are transferred from the stronger acid to the stronger base to yield the weaker acid and base.

For notational simplicity, we will drop the phase descriptions aq for solutes and l for liquid water, but the s for solids is retained. Acid and base dissociation constants are often presented as pK values (− log K), and $\left[ {{\text{H}}^{ + } } \right]$ and $\left[ {{\text{OH}}^{ - } } \right]$ concentrations are traditionally presented as pH and pOH values (− log $\left[ {{\text{H}}^{ + } } \right]$ and − log $\left[ {{\text{OH}}^{ - } } \right]).$ The pH scale is logarithmic implying that a 0.3 increase corresponds to a halving of concentrations (− log(½) = 0.3) and that the absolute change depends on the value: from pH 5 to 5.3 corresponds to a 5 μM proton concentration decrease from 10 to 5 μM, while a pH change from 6 to 6.3 relates to 0.5 μM change in $\left[ {{\text{H}}^{ + } } \right]$ from 1 to 0.5 μM.

pH calculations: When calculating the pH of a solution in which an acid is dissolving, the first step is to evaluate whether the acid is strong (pK_a < 0) or weak.

A strong acid will fully dissociate, i.e., all the acid HA added to the solution will be converted into H⁺ and A⁻ and the eventual pH = − log [HA]. A 0.1 M solution with the strong acid HCl will thus have a pH of 1.

The pH of a solution of a weak acid HA (with concentration C_a) dissolving in pure water can be calculated with a (R)ICE table.

(Reaction)	$\left[ {{\text{HA}}} \right]$	$\left[ {{\text{H}}^{ + } } \right]$	$\left[ {{\text{A}}^{ - } } \right]$
Initial	Ca	≈ 0	0
Change	− x	x	x
Equilibrium	Ca − x	x	x

The first row is for the reaction: ${\text{HA}} \leftrightarrow {\text{H}}^{ + } + {\text{A}}^{ - }$

The second row specifies the initial conditions: the concentration of the acid added (C_a) that is initially undissociated so that $\left[ {{\text{A}}^{ - } } \right]$ is zero and $\left[ {{\text{H}}^{ + } } \right]$ is close to zero (10⁻⁷ for pure water).

The third row gives the change due to reaction; x amount of [HA] is consumed and x $\left[ {{\text{H}}^{ + } } \right]$ and x $\left[ {{\text{A}}^{ - } } \right]$ are produced.

The fourth row, equilibration, is simply the sum of rows two and three.

Next, the acid dissociation constant (K_a) and the concentrations in the equilibrium line in the (R)ICE table are combined in Eq. 6.2.

$$ K_{a} = \frac{{\left[ {{\text{H}}^{ + } } \right]\left[ {{\text{A}}^{ - } } \right]}}{{\left[ {{\text{HA}}} \right]}} = \frac{x \cdot x}{{C_{a} - x }} $$

which can be rewritten as a quadratic equation:

$$ x^{2} + K_{a} x - K_{a} C_{a} = 0 $$

which can be solved for x to obtain the proton concentration, and thus the pH = − log(x). If C_a > 1000 times K_a, then x is very small relative to C_a, and C_a − x ≈ C_a. The proton concentration can then be calculated from $\left[ {{\text{H}}^{ + } } \right] = \sqrt {{\text{K}}_{a} C_{a} }$.

As an example, calculate the pH of 0.1 M Acetic acid solution with a pK_a value of 4.76. The corresponding (R)ICE table reads

(Reaction)	$\left[ {{\text{HAc}}} \right]$	$\left[ {{\text{H}}^{ + } } \right]$	$\left[ {{\text{Ac}}^{ - } } \right]$
Initial	0.1	≈ 0	0
Change	− x	X	x
Equilibrium	0.1 − x	X	x

and the resulting pH = 2.88.

Common ions, i.e., substances that are both in the acid/base system as well as in the solution, shift the equilibrium and thus the pH of the solution. Consider the above example, but now with 0.05 M Na-acetate as well in the solution. Na-acetate has a very high dissociation constant and is fully dissociated, i.e., it is present in the form of acetate $\left[ {{\text{Ac}}^{ - } } \right]$ and sodium ion $\left[ {{\text{Na}}^{ + } } \right]$. Sodium ions are not involved in proton transfer reaction and can be ignored. The (R)ICE table must be modified for the initial conditions to accommodate for the presence of acetate ions.

(Reaction)	$\left[ {{\text{HAc}}} \right]$	$\left[ {{\text{H}}^{ + } } \right]$	$\left[ {{\text{Ac}}^{ - } } \right]$
Initial	0.1	≈ 0	0.05
Change	− x	x	x
Equilibrium	0.1 − x	x	0.05 + x

The corresponding equilibrium

$$ K_{a} = \frac{{\left[ {{\text{H}}^{ + } } \right]\left[ {{\text{Ac}}^{ - } } \right]}}{{\left[ {{\text{HAc}}} \right]}} = \frac{{x \cdot \left( {x + 0.05} \right)}}{0.1 - x } $$

and quadratic equation

$$ x^{2} + (0.05 + K_{a} )x - 0.1\,K_{a} = 0 $$

results in a pH of 4.46. The addition of the common ion acetate shifts the equilibrium to the left and the proton concentration declines (and pH increases).

Buffer solutions: Solutions containing weak acids with their conjugated bases are buffer solutions because they resist drastic changes in pH. Quantifying the buffering capacity of natural waters containing multiple acid-base systems can be quite cumbersome. Therefore, we restrict ourselves to the Henderson-Hasselbalch equation for a single acid-base system to illustrate the principle.

Starting with re-arranging Eq. 6.2

$$ {\text{K}}_{a} = \frac{{\left[ {{\text{H}}^{ + } } \right]\left[ {{\text{A}}^{ - } } \right]}}{{\left[ {{\text{HA}}} \right]}} $$

to isolate the proton concentration on the left-hand side

$$ \frac{1}{{\left[ {{\text{H}}^{ + } } \right]}} = \frac{{\left[ {A^{ - } } \right]}}{{{\text{K}}_{a} \left[ {{\text{HA}}} \right] }} = \frac{1}{{{\text{K}}_{a} }}\frac{{\left[ {{\text{A}}^{ - } } \right]}}{{\left[ {{\text{HA}}} \right] }} $$

followed by taking the logarithms

$$ {\text{pH}} = - \log K_{a} + \log \frac{{\left[ {{\text{A}}^{ - } } \right]}}{{\left[ {{\text{HA}}} \right] }} = pK_{a} + \log \frac{{\left[ {{\text{A}}^{ - } } \right]}}{{\left[ {{\text{HA}}} \right] }} $$

(6.7)

we have derived the Henderson-Hasselbalch equation.

Dissolution of carbon dioxide in water results in the formation of carbonic acid and this provides most of the buffering of natural waters. In the next section we will elaborate the carbon dioxide-water system in quite some detail, here we illustrate the buffering provided by the carbonic acid-bicarbonate system:

$$ {\text{H}}_{2} {\text{CO}}_{3} \leftrightarrow {\text{H}}^{ + } + {\text{HCO}}_{3}^{ - } $$

Consider a system with initial H₂CO₃ and HCO₃⁻ concentrations of 1 mM each and a pH of 6.35, i.e., the pK_a value. Using the Henderson-Hasselbalch equation, we obtain

$$ {\text{pH}} = pK_{a} + \log \frac{{\left[ {{\text{HCO}}_{3}^{ - } } \right]}}{{\left[ {{\text{H}}_{2} {\text{CO}}_{3} } \right] }} = 6.35 + \log \frac{{\left[ {10^{ - 3} } \right]}}{{\left[ {10^{ - 3} } \right] }} = 6.35 $$

Addition of 10 $\mu$ mol strong acid to 1 liter would lower HCO₃⁻ and increase H₂CO₃ concentrations, but hardly impact the pH

$$ {\text{pH}} = 6.35 + \log \frac{{\left[ {10^{ - 3} - 10^{ - 5} } \right]}}{{\left[ {10^{ - 3} + 10^{ - 5} } \right] }} = 6.34 $$

The ΔpH is 0.01 at the initial pH of 6.35, but it would be ≈ 0.03 at a pH of 5.35 and ≈ 0.31 at a pH of 4.35. Accordingly, a buffer is optimal close to its pK_a value, but still limits proton concentration changes to within a factor of two at pH values two units higher or lower than the pK_a. Carbonic acid is a diprotic acid, i.e. a polyprotic acid with two protons, with pK_a values of 6.35 and 10.33. It thus has the potential to buffer natural waters over a wide range of pH values.

2 Carbon Dioxide Equilibria

Carbon dioxide is a gas that exchanges between water and air. When dissolved in water, it reacts with water to form carbonic acid, which in turn dissociates into bicarbonate and carbonate ions. This carbon-dioxide-water system provides much of the buffering capacity for natural waters, governs the pH of most natural waters, is pivotal to precipitation and dissolution of carbonate minerals, and governs the uptake of (anthropogenic) carbon by the ocean.

The carbon-dioxide-water system is described by a few equilibria. Carbon dioxide gas dissolves in water and will then react with water to form carbonic acid (H₂CO₃) according to

$$ {\text{CO}}_{2 } \left( {aq} \right) + {\text{H}}_{2} {\text{O}} \leftrightarrow {\text{H}}_{2} {\text{CO}}_{3 } \left( {aq} \right) $$

(6.8)

This equilibrium lies rather far to the left, hence most of the dissolved carbon dioxide remains in the form of ${\text{CO}}_{2 } \left( {aq} \right)$, and we can analytically not distinguish between ${\text{H}}_{2} {\text{CO}}_{3 } \left( {aq} \right)$ and ${\text{CO}}_{2 } \left( {aq} \right)$; these two pools are therefore lumped together and termed carbonic acid (${\text{H}}_{2} {\text{CO}}_{3}$) in this course. Other treatments of carbon-dioxide-water system sometimes use the terms ${\text{H}}_{2} {\text{CO}}_{3}^{*}$ or ${\text{CO}}_{2}^{*}$ for the lumped pool. Accordingly, we can re-write Eq. 6.8 to directly link atmospheric carbon dioxide and carbonic acid as follows:

$$ {\text{CO}}_{2 \left( g \right)} + {\text{H}}_{2} {\text{O}} \leftrightarrow {\text{H}}_{2} {\text{CO}}_{3} $$

(6.9)

with the equilibrium expression based on Henry’s law (Eq. 5.13):

$$ \left[ {{\text{H}}_{2} {\text{CO}}_{3} } \right] = K_{{\text{H}}} \cdot P_{{{\text{CO}}_{2} }} $$

(6.10)

where $P_{{{\text{CO}}_{2} }}$ is the partial pressure of carbon dioxide in equilibrium with water and $K_{{\text{H}}}$ is Henry’s Law constant (mol kg⁻¹ atm⁻¹) with a pK_H value of 1.47 at 25 °C in pure water. Carbonic acid is a weak acid that partly dissociates into a proton and bicarbonate (${\text{HCO}}_{3}^{ - } )$, the latter being another weak acid that dissociates into another proton and carbonate (${\text{CO}}_{3}^{2 - } )$. The relevant reactions are

$$ {\text{H}}_{2} {\text{CO}}_{3} \leftrightarrow {\text{H}}^{ + } + {\text{HCO}}_{3}^{ - } $$

(6.11)

and

$$ {\text{HCO}}_{3}^{ - } \leftrightarrow {\text{H}}^{ + } + {\text{CO}}_{3}^{2 - } $$

(6.12)

with the corresponding equilibria and equilibrium constants (in freshwater and at 25 °C)

$$ {\text{K}}_{1} = \frac{{\left[ {{\text{H}}^{ + } } \right]\left[ {{\text{HCO}}_{3}^{ - } } \right]}}{{\left[ {{\text{H}}_{2} {\text{CO}}_{3} } \right] }} = 10^{ - 6.35} $$

(6.13)

$$ \text{K}_{2} = \frac{{\left[ {\text{H}^{ + } } \right]\left[ {\text{CO}_{3}^{2 - } } \right]}}{{\left[ {\text{HCO}_{3}^{ - } } \right] }} = 10^{ - 10.33} $$

(6.14)

Accordingly, the carbon-dioxide-water system is characterized by five aqueous species: $\left[ {{\text{H}}^{ + } } \right]$, $\left[ {{\text{OH}}^{ - } } \right]$, $\left[ {{\text{H}}_{2} {\text{CO}}_{3} } \right]$, $\left[ {{\text{HCO}}_{3}^{ - } } \right]$ and $\left[ {{\text{CO}}_{3}^{2 - } } \right]$ that are linked via three equilibrium relations (the self- ionization of water, Eq. 6.6, and the first and second acid-base equilibria, Eqs. 6.13 and 6.14) with known constants. To calculate the abundance of the five species, we must solve the system and for this we need more information, specifically, two more relations.

Various types of information can be used. These can either be information on the measured or known concentration of the aqueous species (e.g., known pH), or additional relations. For a system open to exchange with the air, Henrys’ law (Eq. 6.10) provides a link of carbonic acid (${\text{H}}_{2} {\text{CO}}_{3}$) to the partial pressure of carbon dioxide in the air ($P_{{{\text{CO}}_{2} }} )$. Solubility equilibria involving carbonate minerals can constrain the carbonate ion concentration. Natural waters are uncharged, and we can thus use the charge balance equation as an additional constraint: the positive charge of protons should be balanced by the negative charge of hydroxide, bicarbonate and carbonate ions.

$$ \left[ {{\text{H}}^{ + } } \right] = \left[ {{\text{OH}}^{ - } } \right] + \left[ {{\text{HCO}}_{3}^{ - } } \right] + 2 \left[ {{\text{CO}}_{3}^{2 - } } \right] $$

(6.15)

Note that the carbonate ion is counted twice in a charge balance because of its double charge. Alternatively, one can define a proton balance equation, a mass balance for protons

$$ \left[ {{\text{H}}^{ + } } \right] = \left[ {{\text{H}}^{ + } } \right]_{{{\text{H}}_{2} {\text{O}}}} + \left[ {{\text{H}}^{ + } } \right]_{{{\text{H}}_{2} {\text{CO}}_{3} }} $$

(6.16a)

or its equivalent

$$ \left[ {{\text{H}}^{ + } } \right] = \left[ {{\text{OH}}^{ - } } \right] + \left[ {{\text{HCO}}_{3}^{ - } } \right] + 2 \left[ {{\text{CO}}_{3}^{2 - } } \right], $$

(6.16b)

This proton conservation equation balances excess protons on the left-hand side with the recipe on the right-hand side and is in this case identical to the charge balance (Eq. 6.15). However, the proton conservation and charge balance equations can be different in more complex solutions such as seawater.

Unfortunately, not all species of the system can be measured directly, but there is also no need because they are interlinked via equilibrium, mass balance and charge conservation equations. The four parameters most often measured are: the pH, the partial pressure of $(P_{{{\text{CO}}_{2} }} )$, the dissolved inorganic carbon content and the titration alkalinity. The total dissolved inorganic carbon content, abbreviated as C_T or DIC (dissolved inorganic carbon) is defined as:

$$ DIC = C_{T} = \left[ {{\text{H}}_{2} {\text{CO}}_{3} } \right] + \left[ {{\text{HCO}}_{3}^{ - } } \right] + \left[ {{\text{CO}}_{3}^{2 - } } \right]. $$

(6.17)

The titration alkalinity is the excess of proton acceptors over donors of a solution and is normally derived from an acidimetric titration. Alkalinity (TA) for our simple system is defined as:

$$ TA = \left[ {{\text{OH}}^{ - } } \right] + \left[ {{\text{HCO}}_{3}^{ - } } \right] + 2 \left[ {{\text{CO}}_{3}^{2 - } } \right] - \left[ {{\text{H}}^{ + } } \right] $$

(6.18)

This alkalinity definition is rooted in the charge and proton balances presented earlier (Eqs. 6.15 and 6.16). Carbonate alkalinity ($CA =$ $\left[ {{\text{HCO}}_{3}^{ - } } \right] + 2$ $\left[ {{\text{CO}}_{3}^{2 - } } \right]$) predominates alkalinity of natural waters because of the presence of carbon dioxide in the atmosphere.

3 Solving Carbon Dioxide Equilibria

The previous section has provided the tools to solve carbon dioxide equilibria. However, solving carbon dioxide equilibria in natural waters, particularly in seawater, can be challenging because of the presence of multiple acid-base systems and other species interacting with the carbonate system. Fortunately, there are multiple computer programs available to assist us. Below we present a few simple, Earth-sciences relevant, cases that are instructive and can be done without such computer programs. First, it is important to identify whether the system is open (to an atmosphere with known composition) or closed (when interactions with the atmosphere are negligible). The open system approach applies to surface waters of streams, rivers, lakes, reservoirs, estuaries, and the surface ocean, while a closed approach is more appropriate for groundwater and subsurface waters in lakes and in the ocean interior that are isolated from the atmosphere. Second, calculations for problems with known pH are usually simpler than those without a priori knowledge on pH, and we therefore start with the former.

Case 1 The pH is known

This type of problems is most conveniently solved using the ionization fraction approach. The derivation of the ionization fraction starts with rewriting the first carbonic acid equilibrium (Eq. 6.13) to isolate bicarbonate on the left-hand side:

$$ {\text{K}}_{1} = \frac{{\left[ {{\text{H}}^{ + } } \right]\left[ {{\text{HCO}}_{3}^{ - } } \right]}}{{\left[ {{\text{H}}_{2} {\text{CO}}_{3} } \right]}} \Rightarrow \left[ {{\text{HCO}}_{3}^{ - } } \right] = \frac{{{\text{K}}_{1} \left[ {{\text{H}}_{2} {\text{CO}}_{3} } \right] }}{{\left[ {{\text{H}}^{ + } } \right]}} $$

(6.19)

Next, we substitute this equation in the second carbonic acid equilibrium (Eq. 6.14)

$$ {\text{K}}_{2} = \frac{{\left[ {{\text{H}}^{ + } } \right]\left[ {{\text{CO}}_{3}^{2 - } } \right]}}{{\left[ {{\text{HCO}}_{3}^{ - } } \right] }} \Rightarrow \left[ {{\text{CO}}_{3}^{2 - } } \right] = \frac{{{\text{K}}_{2} \left[ {{\text{HCO}}_{3}^{ - } } \right] }}{{\left[ {{\text{H}}^{ + } } \right]}} $$

(6.20)

and we obtain after re-arrangement an expression with the carbonate ion isolated on the left-hand side:

$$ \left[ {{\text{CO}}_{3}^{2 - } } \right] = \frac{{{\text{K}}_{1} {\text{K}}_{2} \left[ {{\text{H}}_{2} {\text{CO}}_{3} } \right] }}{{\left[ {{\text{H}}^{ + } } \right]^{2} }} $$

(6.21)

Combining these equations with the definition of the dissolved inorganic carbon content (Eq. 6.17)

$$ \begin{aligned} DIC & = \left[ {{\text{H}}_{2} {\text{CO}}_{3} } \right] + \left[ {{\text{HCO}}_{3}^{ - } } \right] + \left[ {{\text{CO}}_{3}^{2 - } } \right] \\ & = \left[ {{\text{H}}_{2} {\text{CO}}_{3} } \right] + \frac{{K_{1} \left[ {{\text{H}}_{2} {\text{CO}}_{3} } \right] }}{{\left[ {{\text{H}}^{ + } } \right]}} + \frac{{K_{1} K_{2} \left[ {{\text{H}}_{2} {\text{CO}}_{3} } \right] }}{{\left[ {{\text{H}}^{ + } } \right]^{2} }} \\ & = \left[ {{\text{H}}_{2} {\text{CO}}_{3} } \right]\left( {1 + \frac{{K_{1} }}{{\left[ {{\text{H}}^{ + } } \right]}} + \frac{{K_{1} K_{2} }}{{\left[ {{\text{H}}^{ + } } \right]^{2} }}} \right) \\ \end{aligned} $$

Introducing the ionization fraction and $\alpha$

$$ \alpha = \left( {1 + \frac{{K_{1} }}{{\left[ {{\text{H}}^{ + } } \right]}} + \frac{{K_{1} K_{2} }}{{\left[ {{\text{H}}^{ + } } \right]^{2} }}} \right) $$

(6.22)

we can express carbonic acid, bicarbonate and carbonate concentrations as a function of DIC, $K_{1}$, $K_{2}$ and $\left[ {H^{ + } } \right]$:

$$ \left[ {{\text{H}}_{2} {\text{CO}}_{3} } \right] = \frac{DIC}{\alpha } $$

(6.23a)

$$ \left[ {{\text{HCO}}_{3}^{ - } } \right] { } = \frac{{DIC \cdot K_{1} }}{{\alpha \cdot \left[ {{\text{H}}^{ + } } \right]}} $$

(6.23b)

$$ \left[ {{\text{CO}}_{3}^{2 - } } \right] = \frac{{DIC \cdot K_{1} \cdot K_{2} }}{{\alpha \cdot \left[ {{\text{H}}^{ + } } \right]^{2} }} $$

(6.23c)

This ionization approach can not only be used for closed systems with known DIC, but also for open systems (without a priori information on DIC, but with known pH). In an open system, the carbonic acid concentration is set by Henry’s Law (Eq. 6.10), and via Eq. 6.23a one can then estimate DIC first and subsequently the bicarbonate and carbonate ion concentrations.

Figure 6.1 shows the distribution of species contributing to DIC as a function of pH, a Bjerrum plot, calculated using the ionization fraction approach. Carbonic acid is the dominant species at pH values below the pK₁, bicarbonate dominates between the pK₁ and pK₂ values and the carbonate ion dominates at pH values above the pK₂. Moreover, the pH dependent distributions of carbonic acid, bicarbonate and carbonate differ between freshwater and seawater because the many ion interactions, i.e., non-ideal behavior, in seawater shift equilibria to lower pH values.

Fig. 6.1

Distribution of carbonic acid, bicarbonate and carbonate as a function of pH in ocean and freshwater for a total DIC of 2000 ${\upmu}\text{M}$ and a temperature of 15 °C

Case 2 Rainwater pH: an open system

Carbon dioxide in rainwater is in equilibrium with the atmosphere, and we have five unknown aqueous species: $\left[ {{\text{H}}^{ + } } \right]$, $\left[ {{\text{OH}}^{ - } } \right]$, $\left[ {{\text{H}}_{2} {\text{CO}}_{3} } \right]$, $\left[ {{\text{HCO}}_{3}^{ - } } \right]$ and $\left[ {{\text{CO}}_{3}^{2 - } } \right].$ We can start the calculation using Eq. 6.10 from the known atmospheric partial pressure of carbon dioxide (420 μ${\text{atm}})$ and Henry’s Law constant (3.4 × 10⁻² mol kg⁻¹ atm⁻¹) to arrive at a carbonic acid concentration of 1.42 × 10⁻⁵ M (= 10^−4.85). Next, we rearrange the first equilibrium relation:

$$ \begin{aligned} K_{1} & = \frac{{\left[ {{\text{H}}^{ + } } \right]\left[ {{\text{HCO}}_{3}^{ - } } \right]}}{{\left[ {{\text{H}}_{2} {\text{CO}}_{3} } \right]}} \Rightarrow \left[ {{\text{HCO}}_{3}^{ - } } \right] \cdot \left[ {{\text{H}}^{ + } } \right] = K_{1} \cdot \left[ {{\text{H}}_{2} {\text{CO}}_{3} } \right] \\ & = 10^{{ - 6.35}} \times 10^{{ - 4.85}} = 10^{{ - 11.20}} \\ \end{aligned} $$

and see that the product of $\left[ {{\text{HCO}}_{3}^{ - } } \right] \cdot \left[ {{\text{H}}^{ + } } \right] = 10^{ - 11.20}$. Next, we consider the charge balance (Eq. 6.15),

$$ \left[ {{\text{H}}^{ + } } \right] = \left[ {{\text{OH}}^{ - } } \right] + \left[ {{\text{HCO}}_{3}^{ - } } \right] + 2 \left[ {{\text{CO}}_{3}^{2 - } } \right] $$

and realize that $\left[ {{\text{HCO}}_{3}^{ - } } \right] \gg$ $\left[ {{\text{CO}}_{3}^{2 - } } \right]$ and $\left[ {{\text{H}}^{ + } } \right] \gg$ $\left[ {{\text{OH}}^{ - } } \right],$ because after the addition of carbon dioxide pH should be lower than 7. The charge balance can thus be simplified to

$$ \left[ {{\text{H}}^{ + } } \right] \approx \left[ {{\text{HCO}}_{3}^{ - } } \right] \approx 10^{ - 5.6} \Rightarrow {\text{pH}} = 5.6 $$

Pure rainwater (not impacted by atmospheric dust or pollution) is thus slightly acidic. Increasing atmospheric carbon dioxide levels to 1000 or 5000 μatm would lower rainwater pH values to 5.4 and 5.06, respectively.

Humans have released large quantities of SO₂ to the atmosphere due to burning of sulfur bearing fossil fuels for industrial activities and transport. In the air some of the SO₂ gas is oxidized to SO₃ and both these gases have dissolved in rain. This had led to the production of sulfurous (H₂SO₃) and sulfuric acid (H₂SO₄). As a result, pH in rain declined to values down to 3, i.e., acid rain, in some industrial areas. Such low pH values have consequences for mineral equilibria, availability of essential nutrients, and organisms’ physiology. In the 1980/90s environmental measures such as sulfur removal from exhausts have been implemented resulting in a gradual return to more natural pH values, but some of the impacted ecosystems (e.g., forests) are still recovering from that perturbation.

Case 3 The pH of a solution in equilibrium with calcium carbonate and the atmosphere

This case would correspond to a pool or shallow lake in a carbonate rock setting, or a coastal ocean with carbonate minerals at the seafloor. It also applies to travertine formation in rivers or stalagmite and stalactite formation in caves (see Box 4). The system is characterized by six unknown aqueous species: $\left[ {{\text{H}}^{ + } } \right]$, $\left[ {{\text{OH}}^{ - } } \right]$, $\left[ {{\text{H}}_{2} {\text{CO}}_{3} } \right]$, $\left[ {{\text{HCO}}_{3}^{ - } } \right]$ and $\left[ {{\text{CO}}_{3}^{2 - } } \right]$ as in the previous case of rainwater, plus $\left[ {{\text{Ca}}^{2 + } } \right]$ from the equilibrium with carbonate minerals. Consequently, we need six equations to solve the problem. The first four are the self-ionization of water (Eq. 6.6), the first and second acid-base equilibria (Eqs. 6.13 and 6.14) and Henry’s Law combined with the $P_{{{\text{CO}}_{2} }}$ (Eq. 6.10). Since the water bodies are in contact and equilibrium with calcium carbonate minerals, the solubility product (Eq. 5.16) provides the fifth constraint:

$$ K_{sp} = \left[ {{\text{Ca}}^{2 + } } \right]\left[ {{\text{CO}}_{3}^{2 - } } \right] $$

(6.24)

The charge balance provides the sixth, final constraint.

$$ \left[ {{\text{H}}^{ + } } \right] + 2\left[ {{\text{Ca}}^{2 + } } \right] = \left[ {{\text{OH}}^{ - } } \right] + \left[ {{\text{HCO}}_{3}^{ - } } \right] + 2 \left[ {{\text{CO}}_{3}^{2 - } } \right] $$

(6.25)

Notice that $\left[ {{\text{Ca}}^{2 + } } \right]$ and $\left[ {{\text{CO}}_{3}^{2 - } } \right]$ are counted twice because of their double charge.

With six unknown and six equations, the problem is well posed, but the mathematical solution is somewhat cumbersome. We will first present the full solution and then show that by making use of chemical insight, we can simplify the solution.

Full solution: Since it is an open system in equilibrium with the atmosphere, we can start with Eq. 6.10:

$$ \left[ {{\text{H}}_{2} {\text{CO}}_{3} } \right] = K_{H} \cdot P_{{{\text{CO}}_{2} }} $$

and insert it in Eqs. 6.19 and 6.21 to obtain expressions for bicarbonate and carbonate ions:

$$ \left[ {{\text{HCO}}_{3}^{ - } } \right] = \frac{{K_{1} \left[ {{\text{H}}_{2} {\text{CO}}_{3} } \right] }}{{\left[ {{\text{H}}^{ + } } \right]}} = \frac{{K_{1} K_{H} \cdot P_{{{\text{CO}}_{2} }} }}{{\left[ {{\text{H}}^{ + } } \right]}} $$

(6.26)

$$ \left[ {CO_{3}^{2 - } } \right] = \frac{{K_{1} K_{2} \left[ {H_{2} CO_{3} } \right] }}{{\left[ {H^{ + } } \right]^{2} }} = { }\frac{{K_{1} K_{2} K_{H} \cdot P_{{CO_{2} }} }}{{\left[ {H^{ + } } \right]^{2} }} $$

(6.27)

Using Eqs. 6.24, 6.6, 6.26, and 6.27, we can rewrite the charge balance Eq. (6.25):

$$ \left[ {{\text{H}}^{ + } } \right] + 2\frac{{K_{sp} }}{{\frac{{K_{1} K_{2} K_{H} \cdot P_{{{\text{CO}}_{2} }} }}{{\left[ {{\text{H}}^{ + } } \right]^{2} }}}} = \frac{{K_{w} { }}}{{\left[ {{\text{H}}^{ + } } \right]}} + \frac{{K_{1} K_{H} \cdot P_{{{\text{CO}}_{2} }} }}{{\left[ {{\text{H}}^{ + } } \right]}} + 2\frac{{K_{1} K_{2} K_{H} \cdot P_{{{\text{CO}}_{2} }} }}{{\left[ {{\text{H}}^{ + } } \right]^{2} }} $$

(6.28)

Multiplying by $\left[ {H^{ + } } \right]^{2}$ and re-arranging so that all terms are on the left-hand side results in

$$ \begin{aligned} \left[ {{\text{H}}^{ + } } \right]^{3} & + 2\frac{{K_{{sp}} \left[ {{\text{H}}^{ + } } \right]^{4} }}{{K_{1} K_{2} K_{H} \cdot P_{{{\text{CO}}_{2} }} }} - K_{w} \left[ {{\text{H}}^{ + } } \right] - K_{1} K_{H} \cdot P_{{{\text{CO}}_{2} }} \left[ {{\text{H}}^{ + } } \right] \\ & - 2K_{1} K_{2} K_{H} \cdot P_{{{\text{CO}}_{2} }} = 0 \\ \end{aligned} $$

(6.29)

This is a fourth-order polynomial in $\left[ {{\text{H}}^{ + } } \right]$ that can be solved by trial-and-error or numerical techniques.

Approximate solution: For a system open to the atmosphere, the final pH is likely close to neutral (7 ± 2), and we can thus simplify the charge balance (6.25) to:

$$ 2\left[ {{\text{Ca}}^{2 + } } \right] = \left[ {{\text{HCO}}_{3}^{ - } } \right] $$

(6.30)

because protons and hydroxide ions are on the order of 10⁻⁵ to 10⁻⁹ M and $\left[ {{\text{HCO}}_{3}^{ - } } \right] >$ $\left[ {{\text{CO}}_{3}^{2 - } } \right]$(see Fig. 6.1). Next, we rewrite the second acid equilibrium (Eq. 6.14) to isolate carbonate on the left-hand side:

$$ K_{2} = \frac{{\left[ {{\text{H}}^{ + } } \right]\left[ {{\text{CO}}_{3}^{2 - } } \right]}}{{\left[ {{\text{HCO}}_{3}^{ - } } \right]}} \Rightarrow \left[ {{\text{CO}}_{3}^{2 - } } \right] = \frac{{K_{2} \left[ {{\text{HCO}}_{3}^{ - } } \right]}}{{\left[ {{\text{H}}^{ + } } \right]}} $$

(6.31)

Next, we substitute Eqs. 6.30 and 6.31 into the solubility Eq. 6.25

$$ K_{sp} = \left[ {{\text{Ca}}^{2 + } } \right]\left[ {{\text{CO}}_{3}^{2 - } } \right] \Rightarrow K_{sp} = 0.5 \left[ {{\text{HCO}}_{3}^{ - } } \right]\frac{{K_{2} \left[ {{\text{HCO}}_{3}^{ - } } \right]}}{{\left[ {{\text{H}}^{ + } } \right]}} = 0.5 \left[ {{\text{HCO}}_{3}^{ - } } \right]^{2} \frac{{K_{2} }}{{\left[ {{\text{H}}^{ + } } \right]}} $$

(6.32)

As above, equation (6.26) is used to express bicarbonate in terms of $P_{{{\text{CO}}_{2} }}$ and $\left[ {{\text{H}}^{ + } } \right]$, and (6.32) is then rewritten as

$$ K_{sp} = 0.5 \left[ {\frac{{K_{1} K_{H} P_{{{\text{CO}}_{2} }} { } }}{{\left[ {{\text{H}}^{ + } } \right]}}} \right]^{2} \frac{{K_{2} }}{{\left[ {{\text{H}}^{ + } } \right]}} $$

(6.33)

Or after some further re-arrangements to isolate the proton concentration on the left-hand side:

$$ \left[ {{\text{H}}^{ + } } \right]^{3} = P_{{{\text{CO}}_{2} }} { }^{2} \left[ {\frac{{K_{1}^{2} K_{2} K_{H}^{2} { } }}{{2K_{sp} }}} \right] $$

(6.34)

Solving this equation for a $P_{{{\text{CO}}_{2} }}$ of 420 $\mu atm$ and using the standard values for pK₁ (6.35), pK₂ (10.33), pK_H (1.47) and the solubility product of calcite (pK = 8.48), yields a pH value of 8.18, consistent with observations for freshwaters in carbonate terrains. Moreover, it is also very similar to that of modern seawater despite seawater being a non-ideal solution.

A more accurate way to calculate the pH for seawater in equilibrium with calcite and the modern atmosphere is to use equilibrium constants specific for seawater at 25 °C: $pK_{1}^{sw}$(5.87), $pK_{2}^{sw}$ (8.76), $pK_{H}^{sw}$ (1.54) and the solubility product of calcite (p $K_{sp}^{sw}$ = 6.36). These seawater specific constants are shifted towards lower pK values, in other words higher K values (see Fig. 6.1). Modern seawater is estimated to have a pH of 8.09, consistent with observations.

Atmospheric carbon dioxide levels have varied over the Earth’s history and Table 6.2 shows the pH value expected for freshwater and marine waters in equilibrium with calcite and the atmosphere at 25 °C for glacial, pre-industrial, modern, Eocene, and projected levels in 2100 for two global warming scenarios (2 °C and 4 °C).

Table 6.2 Freshwater and ocean pH in equilibrium with calcite

Table 6.2 clearly shows that pH declines with increasing $P_{{{\text{CO}}_{2} }}$ levels. This ocean acidification is also known as the other CO₂ problem. The above calculations are based on full equilibration of seawater with both the atmosphere and calcium carbonate in the form of calcite. However, in the ocean, carbonate minerals are at the bottom of the ocean (depth of km’s) while gas exchange, and thus CO₂ invasion, occurs in surface waters. Ocean acidification for surface waters can better be calculated using the assumption of equilibrium gas exchange and constant alkalinity, i.e., no equilibration with calcite. In this case (last column), pH declines more steeply without this buffering effect of carbonate minerals, i.e., the surface ocean is more sensitive to atmospheric composition changes on timescales shorter than ocean turnover (needed to equilibrate with sedimentary carbonates).

Case 4 The pH of a soda lake

Evaporative lakes are usually alkaline and contain soda minerals such as natron (${\text{Na}}_{2} {\text{CO}}_{3} \cdot 10{\text{H}}_{2} {\text{O}})$. The equilibrium reaction

$$ {\text{Na}}_{2} {\text{CO}}_{3} \cdot 10{\text{H}}_{2} {\text{O}} \left( s \right) \leftrightarrow 2{\text{Na}}^{ + } + {\text{CO}}_{3}^{2 - } + 10{\text{H}}_{2} {\text{O}} $$

constrains the carbonate ion and sodium concentration of the lake. Calculate the pH of the alkaline lake knowing that the dissolved sodium concentration ($[{\text{Na}}^{ + } ])$ is 2.3 g kg⁻¹, that the lake water has a temperature of 25 °C and assuming equilibrium with the atmosphere, and a water density of 1 kg L⁻¹.

To solve this problem, we first identify the six unknowns species concentrations: $\left[ {{\text{H}}^{ + } } \right]$, $\left[ {{\text{OH}}^{ - } } \right]$, $\left[ {{\text{H}}_{2} {\text{CO}}_{3} } \right]$, $\left[ {{\text{HCO}}_{3}^{ - } } \right]$, $\left[ {{\text{CO}}_{3}^{2 - } } \right]$ and $\left[ {{\text{Na}}^{ + } } \right].$ Consequently, we need six equations to solve the problem. The first four are the self-ionization of water (Eq. 6.6), the first and second acid-base equilibria (Eqs. 6.13 and 6.14) and Henry’s Law combined with the $P_{{{\text{CO}}_{2} }}$(Eq. 6.10). The dissolved sodium concentration $\left[ {{\text{Na}}^{ + } } \right]$ is 0.1 mol L⁻¹ and provides the fifth equation (mass balance). To close the system, we derive the charge balance:

$$ \left[ {{\text{H}}^{ + } } \right] + \left[ {{\text{Na}}^{ + } } \right] = \left[ {{\text{OH}}^{ - } } \right] + \left[ {{\text{HCO}}_{3}^{ - } } \right] + 2 \left[ {{\text{CO}}_{3}^{2 - } } \right] $$

(6.35)

For an alkaline lake, $\left[ {{\text{Na}}^{ + } } \right] \gg \left[ {{\text{H}}^{ + } } \right]$, and we can thus modify the charge balance and link it to the mass balance:

$$ \left[ {{\text{Na}}^{ + } } \right] = \left[ {{\text{OH}}^{ - } } \right] + \left[ {{\text{HCO}}_{3}^{ - } } \right] + 2 \left[ {{\text{CO}}_{3}^{2 - } } \right] = 0.1 $$

(6.36)

Since it is an open system in equilibrium with the atmosphere, we follow the approach for the previous case and use Eqs. 6.6, 6.26, and 6.27 to rewrite the charge balance (Eq. 6.36):

$$ \frac{{K_{w} { }}}{{\left[ {{\text{H}}^{ + } } \right]}} + \frac{{K_{1} K_{H} \cdot P_{{{\text{CO}}_{{2}} }} }}{{\left[ {{\text{H}}^{ + } } \right]}} + 2\frac{{K_{1} K_{2} K_{H} \cdot P_{{{\text{CO}}_{2} }} }}{{\left[ {{\text{H}}^{ + } } \right]^{2} }} = 0.1 $$

(6.37)

and obtain an equation with all terms known except for $\left[ {{\text{H}}^{ + } } \right].$

Substituting the appropriate values for the constants,

$$ \frac{{10^{ - 14} { }}}{{\left[ {{\text{H}}^{ + } } \right]}} + \frac{{10^{ - 11.19} }}{{\left[ {{\text{H}}^{ + } } \right]}} + 2\frac{{ 10^{ - 21.53} }}{{\left[ {{\text{H}}^{ + } } \right]^{2} }} = 0.1 $$

and multiplying with $\left[ {H^{ + } } \right]^{2}$ we arrive at

$$ 10^{ - 14} \cdot \left[ {{\text{H}}^{ + } } \right] + 10^{ - 11.19} \cdot \left[ {{\text{H}}^{ + } } \right] + 10^{ - 21.23} = 10^{ - 1} \left[ {{\text{H}}^{ + } } \right]^{2} $$

Which after re-arrangement turns out to be a quadratic equation

$$ \left[ {{\text{H}}^{ + } } \right]^{2} - 10^{ - 10.19} \cdot \left[ {{\text{H}}^{ + } } \right] - 10^{ - 20.23} = 0 $$

and results in a $\left[ {H^{ + } } \right]$ of 1.1 × 10⁻¹⁰ and thus a pH of 9.9. This estimate is approximate because of our neglect of non-ideal solution effects, i.e., using concentrations rather than activities.

Fig. 6.2

Conceptual figure of karst system with dissolution at surface and precipitation in caves

Box 4 The Physical Chemistry of Karst

Rainwater entering soils becomes enriched in carbon dioxide from respiration of the primary producers (trees and plants), and the microbial degradation of soil organic matter. These higher $P_{{{\text{CO}}_{2} }}$ levels cause dissolution of carbonate bed rocks, and consequently dolines and sinkholes are formed, and streams and rivers may go underground. Soil waters will eventually recharge subsurface water reservoirs. When the groundwater drains into stream or rivers exposed to air, carbon dioxide will escape from the water to the air to re-establish equilibrium. Consequently, the water becomes supersaturated with respect to calcium carbonates and travertine is formed. Similarly, when groundwater enters caves with lower $P_{{{\text{CO}}_{2} }}$ levels, carbon dioxide will be transferred to the gas phase and stalagmites and stalactites may slowly precipitate (Fig. 6.2).

Consider a typical soil $P_{{{\text{CO}}_{2} }}$ level of 2000 ppm (because of extensive root and microbial respiration). Using equation (6.34) we can estimate that soil water in equilibrium with calcite (carbonate bedrock) would have a pH of 7.73. To calculate the dissolved calcium concentration in the soil, we must re-arrange Eq. (6.33) to isolate $\left[ {{\text{Ca}}^{2 + } } \right]$ on the left-hand side.

$$ \left[ {{\text{Ca}}^{2 + } } \right]^{3} = P_{{{\text{CO}}_{2} }} \left[ {\frac{{K_{1} K_{sp} K_{H} { } }}{{4K_{2} }}} \right] $$

(6.38)

Soil water recharging the groundwater would have a $\left[ {{\text{Ca}}^{2 + } } \right]$ of 0.81 mM (32.5 mg L⁻¹) and TA and DIC values of 1.63 mM each. If we assume that $\left[ {{\text{Ca}}^{2 + } } \right],$ TA and DIC are transported conservatively, i.e., they do not change during transport, and that downstream cave/stream waters re-equilibrate with air with a $P_{{{\text{CO}}_{2} }}$ level of 420 ppm, we can calculate the full carbon dioxide system in equilibrium. Cave and stream re-equilibrate with air and loose about 0.66 mM DIC. The equilibrium pH, TA and $\left[ {{\text{Ca}}^{2 + } } \right]$ would be 8.18, 0.98 mM and 0.48 mM (19.3 mg L⁻¹), respectively, because of calcium carbonate precipitation induced by the carbon dioxide transfer from water to air.

	Soil water ($P_{{{\text{CO}}_{2} }}$ = 2000 ppm)	Cave/stream water ($P_{{{\text{CO}}_{2} }}$ = 420 ppm)
pH	7.73	8.18
$\left[ {{\text{Ca}}^{2 + } } \right]$ (mM)	0.81	0.48
DIC (mM)	1.63	0.97
TA (mM)	1.63	0.98
$\left[ {{\text{HCO}}_{3}^{ - } } \right]$ (mM)	1.62	0.97
$\left[ {{\text{CO}}_{3}^{2 - } } \right]$ (μM)	4.1	6.8