Difference between revisions of "How much of a solid will form?"

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Let's start off with 0.01 mole/Kg of CaCO<sub>3</sub> in 1 Kg of H<sub>2</sub>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:
 
Let's start off with 0.01 mole/Kg of CaCO<sub>3</sub> in 1 Kg of H<sub>2</sub>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<sup>+1</sup>
 +
|2.56E-11
 +
|----
 +
|OH<sup>-1/<sup>
 +
|4.71E-04
 +
|----
 +
|CaCO<sub>3</sub>
 +
|8.27E-03
 +
|----
 +
|CaHCO<sub>3</sub><sup>+1</sup>
 +
|6.75E-06
 +
|----
 +
|Ca+2
 +
|1.71E-03
 +
|----
 +
|CaOH+1
 +
|9.62E-06
 +
|----
 +
|CO2
 +
|2.25E-08
 +
|----
 +
|CO3-2
 +
|1.25E-03
 +
|----
 +
|HCO3-1
 +
|4.74E-04
 +
|----
 +
|}

Revision as of 10:36, 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
CaOH+1 9.62E-06
CO2 2.25E-08
CO3-2 1.25E-03
HCO3-1 4.74E-04
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