copyright by Philip A. Candela, 1997.

The equilibrium constant for the equilibrium:

CaCO_{3}^{calcite} = Ca^{2+} +
CO_{3}^{2-}

is K = a{Ca2+}a{CO3=}/a{CaCO_{3}^{calcite}}.

If the calcite is pure (i.e. one component), it is customary to write the equilibrium constant as a "solubility product":

1. Ksp (calcite) =a{Ca2+}a{CO3=} = J{Ca2+}m{Ca2+} J{CO3=}m{CO3=}

where J{i} is the activity coefficient of i.

The activity coefficients correct for long range, ion -ion interactions; i.e., in a 1M NaCl solution, the effective concentration (activity) of an a Ca2+ ion is reduced relative to a very dilute solution. This can be pictured as being due to a cloud of Cl- ions that surround the Ca2+ ion, and a cloud of Na+ ions that surround the CO3= ions. Note that the presence of the calcium and carbonate ions also lower the activities of the other, but in this case, their concentrtions are so low (limited by the sparingly soluble nature of calcite), that they are strongly outnumbered by Na and Cl in their own ionic "clouds".

We calculate the ionic strength of the the solution by using the equation

2. I= (1/2) SUM c(i){z(i)^2} ,

where c(i) is the conc of the ion (i), and z(i) is the formal charge on the ion.

The sum is over all the ions, including Na+, Cl-, Ca2+, CO3=, H+ and OH- included. If we think ions like HCO3-, or CaHCO3+ are present, then those are to be included in the ionic strength summation also. However, a few trials shows that most of these ions contribute little to the ionic strength of an aqueous solution saturated or near- saturated with respect to calcite, with the exception that HCO3- may be greater than CO3=. THE IONIC STRENGTH IS SIMPLY A MEASURE OF THE SALINITY OF THE SOLUTION. The ionic strength of a groundwater may be 10-4 M, whereas the ionic strength of seawater is about 0.7 M.

Activity coefficients can be determined experimentally (direct observation and calculation), which can be accurate, difficult and time consuming, or by calculation, (based on a theoretical model of how ions interact with each other and with water) , which can be innacurate, and mind numbingly simple and fast. Here, we will use the Debye - Huckel Equation:

3. log Ca2+ = A(z^2) [{I^1/2}/{1 + sB(I^1/2)}] A,B are constants that depend only on T & P, and s depends on the ion in question (Ca2+ in this example).

Note that as I--> 0, log Ca2+--> 0, and therefore, J( Ca2+)--> 1.

At high I, a significant proportion of the water molecules get tied up in the "hydration spheres" of the ions (oriented, polar water molecules love to crowd around charged ions in aqueous solution) and this drives the effective concentration (activity) of an ion UP (because the concentration of "free" water, the solute, has been driven DOWN); hence, the upward swing of the activity coefficients at high ionic strength. The D-H equation does not account for this!! This is a separate correction and will not be dealt with here. (see papers by Scott Wood for further details).

Given:

K1, K2, and K(CO2), with all concentrations replaced by activities, and with the activity coefficients for neutral complexes (e.g. H2CO3 ) set equal to one (PCO2 remains unchanged); and, the D-H equation for activity coefficients,;

4. and for charge balance: 2m{Ca2+} + m{(CaHCO3)+} +m{H+} +2m{Mg2+} = 2m{CO3=} + m{HCO3}- + m{OH-} +m{Cl-}

calculate the solubility of calcite (as moles of total Ca per Kg) in an aqueous solution with 0.001 molal MgCl2 with a pH of 10.3.

First, look for any approximations or "red flags": well, notice that the concentrations of carbonate and bicarbonate are going to be close, because of the pH...

From K2, we can say that a{CO3=} = a{HCO3-}, or...

5. m{HCO3-} = (J{CO3=}/ J{HCO3-}) mCO3=.

Also, the high pH means that the activity of H+ is low, and therefore, because of the hyperbolic relationship

6. Kw = aH+ aOH- the mOH- is high...

Calculate the activity of OH-, and note that aH+ << aOH- (you will need this for the charge balance equation).

In the charge balance, eliminate H+, because it is insignificant relative OH-, and we know the concentrations of Mg and Cl.

Using the relationship in 5., we can eliminate HCO3- from the charge balance; then, the first two terms on the left hand side of 4. can be combined by using the distributive property. 7.

2mCa2+ + m(CaHCO3)+ + 2mMg2+ = 2m{CO3=} [1+ ( CO3=/2 HCO3-)] + mOH-+mCl-. You have a numerical value for mOH-,. Given the complexation constant:

8. Kcomp = (aCaHCO3+)/aCa++ aHCO3- = 10+1.1 and substituting 5., solving for mCaHCO3+ , and replacing mCaHCO3+ in the charge balance yields:

2mCa2+ +2m{Mg2+} [Kcomp [ J{Ca2+}/J {Ca(HCO3)+}] [J{CO3=} m{CO3= } m{Ca2+}] + 0 = [2m{CO3=}] [1+ ( J{CO3=}/2J {HCO3-})] +mOH-

Because the solution is saturated with calcite, we can replace mCO3= by Ksp/(mCa J{CO3=} J{Ca}): 9: 2mCa2+ +2mMg2+ +{Kcomp (J CaHCO3+) Ksp} = 2Ksp/(mCa++) (J {CO3= }J Ca) {1+ ( J{CO3=}/2J {HCO3-})} +mOH- +mCl-

Notice that we have reduced the charge balance equation to a polynomial in one variable, mCa++.

from here, calculate activity coefficients using an ionic strength due to MgCl2 alone, substitute the activity coefficients into 9., multiply 9 by m{Ca2+}, and solve for m{Ca2+} using the quadratic formula (remember that all the other terms in the equation are constants that you should know the value of...).Try it.