From the definition of ionic strength, we see that,

Now, from the Davies equation, we can estimate that g for both Ca²⁺ and SO₄^2- is equal to 0.947. The logarithm of the activity products on the right side of the mass action equation is thus,

This value is ever so slightly smaller than the logarithm of the equilibrium constant, -4.6, so we can conclude that gypsum is not necessarily favored to precipitate from solution. If we ignore the activity corrections, we would calculate that,

This value slightly exceeds the logarithm of the equilibrium constant, so we could conclude here that there are slightly favorable circumstances for precipitation of gypsum. At elevated concentrations, the activity coefficient factors become much more important. For example, if both ions are present at 0.1 mol/L, then I = 0.4 mol/L and g = 0.557.

From the mass action equations between the carbonate species, we see that,

Given the definition of pH, we can calculate that at pH 3, {HCO₃^-}/{CO₃^2-} = 2 x 10⁷ while {H₂CO₃}/{CO₃^2-} = 5 x 10¹⁰. Clearly, these activity ratios indicate that {H₂CO₃} >> {HCO₃^-} >> {CO₃^2-}. At pH 7, {HCO₃^-}/{CO₃^2-} = 2000 while {H₂CO₃}/{CO₃^2-} = 501. In this case, the ratios suggest that {HCO₃^-} is now the dominant species, although barely so, exceeding {H₂CO₃} by only a factor of 4 (you can see this by dividing the first ratio expression by the second); {CO₃^2-} is still quite insignificant. At pH 10, however, {HCO₃^-}/{CO₃^2-} = 2 and {H₂CO₃}/{CO₃^2-} = 5 x 10^-4, so that {CO₃^2-} ~ {HCO₃^-}, whereas {H₂CO₃} becomes negligible.

Contamination by atmospheric oxygen is a problem to be avoided when analyzing for ferrous iron. O₂ readily reacts with Fe²⁺ to form Fe³⁺, which combines with OH^- in solution to form Fe(OH)3, which is quite insoluble at neutral to high pH. This makes the iron unavailable for analysis in the aqueous phase. A usual precaution to avoid this problem is to make sure that there is no headspace (i.e., a pocket of air) trapped within the sample vial.

From the mass action expression, we see that,

which, upon some re-arrangement, gives us,

From the mass action relationship,

we see that,

At pH 7 and pE ñ4.5, the logarithm of the oxygen fugacity is ñ73.1. Clearly, 10^-73.1 is not a physically measurable quantity and is illustrative from a theoretical standpoint only. Also, from this expression, we can see that at {O₂(g)} = 0.21 and at pH 5, pE = 15.6, whereas at pH 8, pE = 12.6. If oxygen is the only oxidizing agent in the system, then the capacity of the system to oxidize reduced species is fixed. That is why oxygen fugacity is sometimes a more satisfying measure of the redox capacity of a system; it is absolute, whereas pE is always coupled with pH.

The presence of quartz in the modeled microcline dissolution problem would remove H₄SiO₄ from solution upon saturation if the equilibrium constant were very small (a large equilibrium constant would favor formation of the product as the reaction is written). This would interfere with the ability of kaolinite and muscovite to form.

If we construct a table of the atoms in a molecular unit of the clay mineral and the associated charges, we see that x must be given by the charge on cation exchange element M as,

so that the total charge on the molecular unit is equal to zero. Ignoring the contribution of element M, we can calculate the molecular weight of the clay mineral by multiplying the stiochiometry of eachspecies by its respective molecular weight, yielding 304.6 gm/mol. Since there are 0.8 mol of charge by exchangeable cation for each mol of the mineral, we can thus calculate that the cation exchange capacity is about 2.6 mol/kg.

To calculate this ration, letís first calculate the concentration of solute present if there is no adsorption. If we add a mass of solute M to a volume of a porous medium V with porosity n, then the resulting concentration without adsorption, C, is simply total mass over total void volume, or,

Now, if adsorption is taking place, things are more complicated. We know that mass must be conserved, so,

Now, letís assume that there is an equilibrium concentration in the aqueous phase, C^*, that we can use to expand this expression. We know that M_aq must be given by,

We also know from the definition of K_d (Eq.-1), that the mass sorbed onto the solid, M_s, is,

Substituting the expressions for M_aq and M_s into the mass balance expression and solving for C* gives us,

Finally, dividing the expression for C by that for C^*, and dividing out V and n, gives us,

QUESTION # 1

A classic example of disequilibrium chemistry in the Earthís atmosphere that is indicative of life is the co-existence of oxygen and methane, which react slowly with one another. The continuous supply of these materials to the atmosphere are indicative of biological processes.