# 1. Acid-Base Chemistry of Natural Aquatic Systems

Natural waters contain a wide variety of solutes that act together to determine the pH, which typically ranges from 6 to 9. Some of the major processes that affect the acid-base balance of natural systems are:

• Contact with atmospheric carbon dioxide
• Input of acidic gases from volcanic and industrial emissions
• Contact with minerals, rocks, and clays
• Presence of buffer systems such as carbonate, phosphate, silicate, and borate
• Presence of acidic anions, such as $$Fe(H_2O)^{3+}_6$$
• Input and removal of $$CO_2$$ through respiration and photosynthesis
• Other biological processes, such as oxidation ($$O_2+ H^+ +e^- \rightarrow H_2O$$), nitrification, denitrification, and sulfate reduction.

In this chapter and also in the next one which deals specifically with the carbonate system, we will consider acid-base equilibria as they apply to natural waters. We will assume that you are already familiar with such fundamentals as the Arrhenius and Brfinsted concepts of acids and bases and the pH scale. You should also have some familiarity with the concepts of free energy and activity. The treatment of equilibrium calculations will likely extend somewhat beyond what you encountered in your General Chemistry course, and considerable emphasis will be placed on graphical methods of estimating equilibrium concentrations of various species.

### 1 Proton donor-acceptor Equilibria

#### 1.1 Acid-base strengths

The tendency of an acid or a base to donate or accept a proton cannot be measured for individual species separately; the best we can do is compare two donor-acceptor systems. One of these is commonly the solvent, normally water. Thus the proton exchange

$HA + H_2O \rightleftharpoons H_3O^+ + A^- \tag{M1.1}$

with equilibrium constant $$K_1$$

is the sum of two reactions

$HA \rightleftharpoons H^+ + A^- \tag{M1.2}$

with equilibrium constant $$K_2$$ and

$H_2O + H^+ \rightleftharpoons H_3O^+ \tag{M1.3}$

with equilibrium constant $$K_3=1$$.
The unity value of $$K_3$$ stems from the defined value of $$\Delta G^o = 0$$ for this reaction, and assumes that the activity of the $$H_2O$$ is unity. Combining these equilibrium constants, we have

Formally, equilibrium constants for reactions in ionic solutions are defined in terms of activities, in which the reference state is a hypothetical one in which individual ion activities are unity but there are 2 no ion-ion interactions in the solution of the ion in pure water.

$K=\dfrac{fH+gfA°gf}{HAgfH2Og} \tag{M1.6}$

There has been considerable debate about the $$K_a$$ values of water and of the hydronium ion. The conventional value of 10° 14 shown for $$H_2O$$ in Table 1 is very commonly used, but it does not reflect the observed relative acid strength of $$H_2O$$ when it is compared with other very weak acids. When such comparisons are carried out in media in which $$H_2O$$ and the other acid are present in comparable concentrations, water behaves as a much weaker acid with

Kaº10°16

To understand the discrepancy, we must recall that acids are usually treated as solutes,so we must consider a proton-donor $$H_2O$$ molecule in this context. Although the fraction of $$H_2O$$ molecules that will lose a proton is extremely small (hence the designation of those that do so as solute molecules), virtually any $$H_2O$$ molecule is capable of accepting the proton, so these would most realistically be regarded as solvent molecules.

The equation that defines the acid strength of water is

$H_2O (solute)+H2O(solvent)°!H3O++OH°$

whose equilibrium constant is

$K=(fH+g=1)(fA°g=1)(fH2Og=1)(fH2Og=55:5) \tag{M1.7}$

in which the standard states are shown explicitly. Do you see the problem? The standard state of a solute is normally taken as unit molality, so we do not usually show it (or even think about it!) in most equilibrium expressions for substances in solution. For a solute, however, the standard state is the pure liquid, which for water corresponds to a molality of 55.5.

The value of $$K_a=10^{14}$$ for water refers to the reaction H2O°!H++OH°in which $$H_2O$$ is treated only as the solvent. Using Equation M1.7, the corresponding $$K_a$$ has the value $$55.5/10^{14} = 1:8 \times 10^{16}$$, which is close to the observed acid strength noted above.

What difference does all this make? In the context of Table 1 or Fig. 2, the exact $$pK_a$$ of wateris of little significance since no other species having similar $$pK_a$$s are shown. On the other hand, if one were considering the acid-base reaction between glycerol (pKa=14.2) and water, the prediction of equilibrium concentrations (and in this case, the direction of the net reaction) would depend on which value of the water p $$K_a$$ is used.

Instead of using pure water as the reference state, an alternative convention is to use a solution of some arbitrary constant ionic strength in which the species of interest is \infinitely dilute". In practice, this means a concentration of less than about one-tenth of the total ionic concentration. This convention,which is widely used in chemical oceanography, incorporates the equilibrium quotient into the equilibrium  constant:

$^cK=\dfrac{[H^+][A^-]}{[HA]} \tag{M1.8}$

A third alternative is to use a "mixed acidity constant" in which the hydrogen ion concentration is expressed on the activity scale (which corresponds to the values obtained by experimental pH measurements), but the acid and base amounts are expressed in concentrations.

$K'=\dfrac{\{H^+\}[A^-]}{[HA]} \tag{M1.8}$

The value of K0 can be estimated from the Güntelberg approximation for single-ion activities:

$pK' = pK+ \dfrac{0.5(z^2_{acid}-z^2_{base}\sqrt{I}}{1+\sqrt{I}} \tag{M1.10}$

in which

• $$I$$ is the ionic strength, and
• $$z_{acid}$$ and $$z_{base}$$ are the ionic charges of the acid and base species, respectively.

### Contributors

Stephen Lower, Professor Emeritus (Simon Fraser U.) Chem1 Virtual Textbook