Difference between revisions of "How much of a solid will form?"
| Line 26: | Line 26: | ||
|6.75E-06 | |6.75E-06 | ||
|---- | |---- | ||
| − | |Ca+2 | + | |Ca<sup>+2</sup> |
|1.71E-03 | |1.71E-03 | ||
|---- | |---- | ||
| − | | | + | |Ca(OH)<sup>+1</sup> |
|9.62E-06 | |9.62E-06 | ||
|---- | |---- | ||
Revision as of 10:37, 12 January 2015
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 |
