How much of a solid will form
How does the OLI Software determine how much solid will form?
To begin this discussion we will assume we have a simple solution of CaCO3 and H2O at 25 oC and 1 atmosphere. In addition we will ignore the activity coefficients to make the discussion simpler.
Let's start off with 0.01 mole/Kg of CaCO3 in 1 Kg of H2O at the aforementioned temperature and pressure. The OLI software initially assumes that no solids are present in the solution. We would have a solution concentration consisting of the following:
| Aqueous | |
| mol/Kg | |
| H+1 | 2.56E-11 |
| OH-1 | 4.71E-04 |
| CaCO3 | 8.27E-03 |
| CaHCO3+1 | 6.75E-06 |
| Ca+2 | 1.71E-03 |
| Ca(OH)+1 | 9.62E-06 |
| CO2 | 2.25E-08 |
| CO3-2 | 1.25E-03 |
| HCO3-1 | 4.74E-04 |
The scaling tendency for CaCO3 is calculated to be 428.4. The scaling tendency is the ratio of available ions in solution to the thermodynamic limit (also known as the KSP. The KSP is related the equilibrium equation. For CaCO3 we have:
CaCO3(s) = Ca+2 + CO3-2
In short-hand form the Scaling Tendency would be
Scaling Tendency = [Ca+2][CO3-2] / KSP
The KSP value is either obtained from thermodynamic values such as the Gibb's Free Energy of Reaction or from solubility experiments.
When the Scaling Tendency (ST) < 1.0 we say that the solid is under-saturated. When the ST > 1.0 the solid is said to be super-saturated. When the ST = 1.0 exactly, we say that the solid is saturated.
In this example, the ST = 428.4 which means that the solution is super-saturated with respect to CaCO3(s) The OLI numerical engine now performs a mathematical operation on the solution. We remove the mass of the calcium and carbonate ions from the liquid solution (maintaining the stoichiometry) and place the mass as a solid species CaCO3(s). We then recalculate the equilibrium to see the value of the new Scaling Tendency.
If the value is exactly 1.0 we know that we have moved the mass of the ions correctly from the liquid phase to the solid. If the value is still greater than 1.0 we repeat the process till the value is exactly 1.0 for the scaling tendency.
For this example we moved 0.0099 moles of both Ca+2 and CO3-2 from the liquid phase to the solid phase to make 0.0099 moles of CaCO3 such that the scaling tendency of CaCO3 is exactly 1.0.
What happens when we have more than one solid?
If the ions that make up the other solid phase are different than the first solid we solve the scaling tendency simultaneously for both solids. Each solid phase is independent of any other solid phases. A problem occurs when the solids have competing ions. For example, what happens if we have both CaCO3 and MgCO3?
Let's change our example to have an additional 0.01 mol/Kg of MgCO3 to the solution. Our initial scaling tendency calculation is slightly different. Here we have the following values
MgCO3 = 1.17
CaCO3 = 473.4
Now we first take the solid with the largest Scaling Tendency and make that 1.0 as above by moving its from the liquid to the solid phase. In this example we created 0.0099 moles of CaCO3(s). The resulting scaling tendency of the other solid MgCO3(s) is now 1.13. We now move the mass of Mg/CO3 from the liquid phase to the solid phase to make this solid's scaling tendency exactly 1.0.
The results in 0.0009 moles of MgCO3 being created.
Now both solids have formed such that they are at thermodynamic saturation.
Highly soluble solids.
Sometimes you will have a highly soluble solid which shares an common ion a less soluble solid. During this iterative process described above after the solid with the largest scaling tendency has been adjusted (to 1.0) the other solid may now have a scaling tendency less than 1.0. If this occurs, no further iterative calculations are performed.
