## How does the OLI Software determine how much solid will form?

To begin this discussion we will assume we have a simple solution of CaCO_{3} and H_{2}O at 25 ^{o}C 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 CaCO_{3} in 1 Kg of H_{2}O 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 |

CaCO_{3} |
8.27E-03 |

CaHCO_{3}^{+1} |
6.75E-06 |

Ca^{+2} |
1.71E-03 |

Ca(OH)^{+1} |
9.62E-06 |

CO_{2} |
2.25E-08 |

CO_{3}^{-2} |
1.25E-03 |

HCO_{3}^{-1} |
4.74E-04 |

The scaling tendency for CaCO_{3} 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 K_{SP}. The K_{SP} is related the equilibrium equation. For CaCO_{3} we have:

CaCO_{3(s)} = Ca^{+2} + CO_{3}^{-2}

In short-hand form the Scaling Tendency would be

Scaling Tendency = [Ca^{+2}][CO_{3}^{-2}] / K_{SP}

The K_{SP} 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 CaCO_{3(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 CaCO_{3(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 CO_{3}^{-2} from the liquid phase to the solid phase to make 0.0099 moles of CaCO_{3} such that the scaling tendency of CaCO_{3} 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 CaCO_{3} and MgCO_{3}?

Let's change our example to have an additional 0.01 mol/Kg of MgCO_{3} to the solution. Our initial scaling tendency calculation is slightly different. Here we have the following values

MgCO_{3} = 1.17

CaCO_{3} = 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 CaCO_{3(s)}. The resulting scaling tendency of the other solid MgCO_{3(s)} is now 1.13. We now move the mass of Mg/CO_{3} 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 MgCO_{3} 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 a 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.