11.3: Forming minerals

Biomineralization is the process by which living organisms cause minerals to form (Benzerara et al., 2011). Biomineralization impacts the composition of the environment, by affecting the distribution of elements among different phases in geochemical systems. Biomineralization, like biofilm formation, also impacts the physical properties of environments, but creating solid phases that fill voids.

There are two main mechanisms by which microorganisms cause minerals to form (Fig. \(11.2\)). Microorganisms can provide surfaces that help nucleate minerals. They can also alter the chemistry of the environment in such a way that mineral precipitation becomes thermodynamically favorable. In other words, their activities can alter the saturation state of a mineral and can cause it to become supersaturated (Box \(11.1\)).

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Nucleation, in the context of this discussion, is the formation of a new solid phase and a basic requirement for mineral formation. Homogenous nucleation occurs when a nucleus forms from a bulk solution whereas heterogeneous nucleation occurs when a nucleus forms in contact with a surface (Drever, 1997). Microbial biomass, including cells and extracellular polymeric substances (EPS), can provide surfaces for heterogenous mineral nucleation. Specifically, the reactive functional groups on biomass such as carboxyl, phosphate, and amino groups can form complexes with ions from an adjacent solution (Fig. \(11.2\)). Once an ion has complexed, it can then serve as a nucleation site for mineral growth (Beveridge, 1989; Schultze-Lam et al., 1996). This reaction pathway can enhance precipitation by helping overcome some of the factors that can inhibit precipitation directly from solution, such as ion pairing and the presence of hydration spheres (Braissant et al., 2007).

Supersaturation is also a basic requirement for mineral growth. A solution becomes supersaturated if ingredients needed to form the mineral have higher concentrations than they would if they mineral were in equilibrium with the solution. Microbes can cause this situation to occur by generating those ingredients or otherwise changing the chemistry of the environment in such a way that precipitation becomes more thermodynamically favorable (Fig. \(11.2\)).

As an example, microbial iron reduction can create conditions that are more thermodynamically favorable for precipitation of the mineral siderite \(\left(\text{FeCO}_{3}\right)\) (Fredrickson et al., 1998). Siderite precipitation consumes ferrous iron and bicarbonate and releases a hydrogen ion: \[\text{Fe}^{2+} + \text{HCO}_{3}^{-} \longleftrightarrow \text{FeCO}_{3} \ (s) + \text{H}^{+}\]

Therefore, the reaction becomes more thermodynamically favorable as ferrous iron and bicarbonate activities increase and hydrogen ion activity decreases (or in other words pH increases). Microbial iron reduction can cause all three of these changes to occur, as shown in the following example reaction with acetate as the electron acceptor and goethite \(\left(\text{FeOOH}\right)\) as the source of ferric iron: \[\text{CH}_{3} \text{COO}^{-} + 15 \ \text{H}^{+} + 8 \ \text{FeOOH} \longleftrightarrow 2 \ \text{HCO}_{3}^{-} + 12 \ \text{H}_{2} \text{O} + 8 \ \text{Fe}^{2+}\]The reaction consumes a large number of hydrogen ions, which works to increase pH. It also produces ferrous iron and bicarbonate, and thus can cause concentrations of those ions to increase.

Regardless of how, why would they do it? In many cases, microorganisms are not intentionally causing minerals to form. We refer to this scenario as biologically induced mineralization (Lowenstam, 1981). They may, for example, just happen to produce ingredients needed for the mineral or provide a favorable surface for precipitation, which generally occurs outside of the cell. In contrast, some microorganisms intentionally precipitate minerals for some physiological purpose (Konhauser, 2007). We refer to this scenario as biologically controlled mineralization. During controlled mineralization, microorganisms often produce vesicles for mineral growth. Supersaturation may be induced in some cases through active accumulation of an ingredient. Often, a template for mineral growth is provided by the cell and as a consequence, the morphology of the mineral product is not random.

Box \(11.1\): Saturation Index

The saturation index of a mineral tells us whether a solution is in equilibrium with a mineral phase and if not, whether dissolution or precipitation of the mineral is thermodynamically favorable. We calculate the saturation index of a mineral in a solution based on the reaction quotient \((Q)\) for the mineral’s dissolution reaction and the equilibrium constant for the reaction \((K)\). Consider the following reaction, which describes dissolution of the mineral siderite \(\left(\text{FeCO}_{3}\right)\): \[\text{FeCO}_{3} \ (s) + \text{H}^{+} \longleftrightarrow \text{Fe}^{2+} + \text{HCO}_{3}^{-} \nonumber\]

This reaction is simply reaction \(\PageIndex{1}\) written as a mineral dissolution reaction (i.e., the mineral dissolves as the reaction moves forward). The quotient for this reaction calculated as: \[Q = \frac{\left[\text{Fe}^{2+}\right] \left[\text{HCO}_{3}^{-}\right]}{\left[\text{H}^{+}\right] \left[\text{FeCO}_{3} \ (s)\right]} \nonumber\]

Assuming an activity of one for siderite, the equation simplifies to \[Q = \frac{\left[\text{Fe}^{2+}\right] \left[\text{HCO}_{3}^{-}\right]}{\left[\text{H}^{+}\right]} \nonumber\]

Putting the equation into log form, we get: \[\log Q = \log \left[\text{Fe}^{2+}\right] + \log \left[\text{HCO}_{3}^{-}\right] - \log \left[\text{H}^{+}\right] \nonumber\]

This is the same as \[\log Q = \log \left[\text{Fe}^{2+}\right] + \log \left[\text{HCO}_{3}^{-}\right] - pH \nonumber\]

To calculate the saturation index (SI) for siderite, we subtract the \(\log K\) of the mineral dissolution reaction from \(\log Q\): \[SI = \log Q - \log K \nonumber\]

If the result is zero, the solution is equilibrated with the mineral, or in other words, the solution is saturated with respect to the mineral. If the result is a negative number \((\log Q - \log K < 0)\), the solution is undersaturated and dissolution of the mineral is possible. If the result is positive \((\log Q - \log K > 0)\), then the solution is supersaturated, and mineral precipitation is possible. Be aware that the interpretation of these results will differ if the mineral reaction is not written in terms of dissolution (mineral on the reactant side of the reaction) and if the reaction quotient and equilibrium constants are not in log form.

For an example calculation, imagine that a solution has a temperature of \(15^{\circ}\text{C}\) and composition equal to that of Solution A in the table below. What is the saturation index for siderite in the solution? First, calculate the reaction quotient value: \[\log Q = (-3.812) + (-2.888) + 6 = -0.700\]

Next, calculate the saturation index, given that the \(\log K\) value for siderite dissolution is \(-0.0339\) at \(15^{\circ} \text{C}\): \[SI = (-0.700) - (-0.0339) = -0.666\]

This result indicates that it is not thermodynamically favorable for siderite to precipitate from the solution. In other words, siderite is undersaturated. Note, this calculation should typically be carried out using chemical activities rather than concentrations. Refer to Chapter 8, Box \(8.1\), for more information on chemical activities and how to calculate them with geochemical modeling software.

Box \(11.1.1\) Practice

Repeat the saturation index calculation but increase the pH and log activities of ferrous iron and bicarbonate, as indicated for Solution B in the table below, to simulate impacts of microbial iron reduction.

Table \(11.1\): Hypothetical solutions used for Box \(11.1.1\) example and practice calculations

	Solution A		Solution B
Parameter	Concentration/value	Log activity*	Concentration/value	Log activity*
\(\text{Fe}^{2+}\)	\(200 \ \mu \text{M}\)	\(-3.812\)	\(1 \ \text{mM}\)	\(-3.16\)
\(\text{HCO}_{3}^{-}\)	\(5 \ \text{mM}\)	\(-2.888\)	\(6 \ \text{mM}\)	\(-2.523\)
\(\text{H}^{+}\)	\(pH \ 6.00\)	\(-6.00\)	\(pH \ 6.50\)	\(-6.50\)
\(\text{Na}^{+}\)	\(2 \ \text{mM}\)	\(-2.732\)	\(2 \ \text{mM}\)	\(-2.741\)
\(\text{Cl}^{-}\)	\(1 \ \text{mM}\)	\(-3.032\)	\(1 \ \text{mM}\)	\(-3.042\)
\(T\)	\(15^{\circ}\text{C}\)		\(15^{\circ}\text{C}\)
*Activities are calculated using The Geochemist’s Workbench program SpecE8.

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