Difference between revisions of "Alkalinity and OLI"
|Line 118:||Line 118:|
Remove the 100 mg/L -and add 100 mg/L OH-
Remove the 100 mg/L
The computed alkalinity is 361.3 mg/L as HCO3-1. The formula weight - is 17.01 g/mole compared with 61.02 g/mole -1, so 3.587 times more moles - is added compared -. Therefore,
61.020/17.010×100 (mg as -)/L=358.7 (mg as -)/L
The computed alkalinity is 361.3 mg/L as HCO3-1. The formula weight
- neutralization actually happens at about 10 pH, not 4.5 pH, which is the experimental endpoint.
61.020/17.010×100 (mg as
===Non-Carbonate Alkalinity Contributions===
===Non-Carbonate Alkalinity Contributions===
Revision as of 17:11, 2 June 2020
What does OLI mean when it discusses Alkalinity? Alkalinity is a frequently measured and reported quality of many waters. Stumm and Morgan define alkalinity as:
“Acidimetric or alkalimetric titrations of carbonate-bearing water to the appropriate end points represent operations procedures for determining alkalinity and acidity, that is, the equivalent sum of the bases that are titratable with strong acid and the equivalent sum of the acids that are titratable with strong base. Alkalinity and acidity are then the capacity factors that represent, respectively, the acid- and base-neutralizing capacities … of an aqueous system. For solutions that contain no protolysis system other than that of aqueous carbonate, alkalinity is a measure of the quantity of strong acid per liter required to attain a pH equal to that of a total concentration (molar) solution of H2CO3. Alternatively, acidity is a measure of the quantity per liter of strong base required to attain a pH equal to that of a total concentration (molar) solution of Na2CO3”1
The key to this statement involves the fact that many users think that the alkalinity is the concentration of various forms of carbonate ion. This would be true of other acid or base systems were not present in solution. Even simple ions such as sodium and magnesium may affect the free carbonate in solution and have markedly different alkalinities.
OLI considers alkalinity to be the total base capacity of the brine. We will us a titration to determine the alkalinity exactly. We will now show some examples featuring the OLI/LabAnalyzer™ program.
We will consider a simple brine with the following concentration
1“Aquatic Chemistry. An Introduction Emphasizing Chemical Equilibria in Natural Waters”. Werner Stumm and James J. Morgan. John-Wiley & Sons, New York. 1981 p 185
Figure 1 Brine composition.
The user would suspect that the alkalinity would be the same as the bicarbonate concentration of 375 mg/L. The electrically neutral pH of this brine is:
Figure 2 the pH is 7.68
We will use HCl to titrate the brine to the standard end point pH of 4.5.
The amount of HCl added to bring the pH down to 4.5 is:
Figure 3 215.4 mg/L of HCl required.
This amount of HCl needs to be converted to equivalents of bicarbonate ion for reporting purposes.
So there is actually less alkalinity than the input concentration of bicarbonate ion would indicate. The reason for this is that some of the bicarbonate ion is tied up in the form of a complex, NaHCO3o and is not available to the alkalinity titration.
What would happen if organics acids were present in the brine? The following brine concentration has 150 mg/L of acetic acid added:
Figure 4 Added acetic acid
The alkalinity was determined as before to an end point pH of 4.5.
Figure 5 164.8 mg/L of HCl required.
This corresponds to an alkalinity of 275.8 mg/L as HCO3-1. This is considerably less than the original bicarbonate ion concentration of 375 mg/L. The problem occurs in that the acetic acid is also reacting with the HCl. At a pH of 4.5 (the titration end point) not all of the acetic acid has reacted. The reason is that the dissociation constant for acetic acid is near the end point pH.
CH3COOH = CH3COO-1 + H+1 pKa ~ 4.7
Most procedures state that if organic acids are present the pH end point must be lowered to react both the carbonate and the acid. We used a lower pH end point of 3.0 and obtained this result:
Figure 6 263.6 mg/L of HCl to reach the end point pH of 3.0
This corresponds to an alkalinity of 441 mg/L as HCO3-1. This value is greater than the bicarbonate ion concentration that was entered. This means that the organic acid is also a source of alkalinity.
It is true that any organic or inorganic species that has significant acid/base chemistry near the end point of the alkalinity titration will contribute the alkalinity. Species such as acetic acid (as just seen) and formic acid contribute to alkalinity. Inorganic species such as hydrofluoric acid (pKa = 4.3) also contribute. Some boric acids also contribute.
The OLI code does not make any direct calculation of alkalinity since we do not know apriori what ions may appear in solution. We calculate the alkalinity via a titration. The OLI/ScaleChem program also uses this philosophy. When ions exist in the brine (e.g., acetate, formate, fluoride, borate, etc.) it is left to the user to determine what end point pH is appropriate for the alkalinity titrations.
This is a quick example of how OLI calculates the alkalinity values, and how different species contribute to the alkalinity value.
The simple definition for carbonate alkalinity is:
Carbonate Alkalinity= Total H+ required to convert all CO3-2, HCO3-,and OH- to CO2
Carbonate Alkalinity = H+ (CO3-2) + H+ (from HCO3- and neutralized CO3-2) + H+ (from OH-) = total acid added to bring the pH to <4.5
Carbonate Alkalinity = =2(CO3-2)+ HCO3-+ OH-
However, there are some other non-carbonate alkalinity contributors. These contributors will affect the direct use of the alkalinity value in the carbonate equation. Contributing species include borate, phosphate, formate, acetate, propanate, bisulfide, and any anion that will accept H+ ion above 4.5 pH.
The calculations below show how the alkalinity calculations are performed, for both the carbonate and non-carbonate alkalinity contributions. All of these analysis have been done using the Brine Analysis object.
Carbonate Alkalinity Contributions
1. Find HCO3-1 in the Anions grid and give it a value of 100 mg/L. Sodium ions (Na+) are selected for the electroneutrality valance
2. Select the Reconcile tab. Leave the default reconciliation option (Concentration Data Only)
3. Click calculate and document the pH and alkalinity
The calculated pH is 8.27. The calculated alkalinity is 100.7 mg/L and is entirely the bicarbonate alkalinity. The 0.7 mg/L addition to the alkalinity value is due to the pH endpoint and minor activity coefficient effects.
1. Remove the 100 mg/L HCO3-1 and add 100 mg/L CO3-2
2. Select the Reconcile tab and recalculate
The alkalinity is 204.1 mg/L as HCO3-1 twice the CO3-2 concentration added, and is consistent with the above equation. The HCO3-1 formula weight is 61.02 g/mole and CO3-2 is 60.01 g/mole – a mole ratio of 1.7%. Two factors contribute to the alkalinity increase.
The first is that CO3-2 accepts two H+ ions one above 8.5 pH and one above 4.5 pH. The second is that 1.7% additional moles of carbonate are added because of the weight difference. Thus, the theoretical alkalinity of 100 mg/L alkalinity as HCO3-1*2*1.7% = 203.4 mg/L alkalinity as HCO3-1. The computed value is 204.1.
Note the slight bias forming in the results. Instead of exactly 203.4 mg/L, the value is 0.70 mg/L higher. This difference is due to non-ideal effects and the precise pH choice. The equations provided above are for ideal conditions. However, we are well aware that there are non-ideal effects that must be considered.
1. Remove the 100 mg/L CO3-2 and add 100 mg/L OH-
2. Select the Reconcile tab and recalculate
The computed alkalinity is 361.3 mg/L as HCO3-1. The formula weight of OH- is 17.01 g/mole compared with 61.02 g/mole for HCO3-1, so 3.587 times more moles of OH- is added compared with HCO3-1. Therefore, 61.020/17.010×100 (mg as HCO3-1)/L=358.7 (mg as HCO3-1)/L
Complete OH- neutralization actually happens at about 10 pH, not 4.5 pH, which is the experimental endpoint.
Non-Carbonate Alkalinity Contributions
Zero out any existing concentrations and add 100 mg/L acetate C2H3O2-1
Select the Reconcile tab and recalculate
The computed alkalinity is 68.9mg/L as 〖HCO〗_3^-. The acetate formula weight is 59.05 g/mole, close to〖HCO〗_3^- (61.02 g/mole). So, if all the acetate contributes to alkalinity, then 100 mg/L acetate should be close to that for 100 mg/L〖HCO〗_3^- as shown in the following equation: (61020 (mg 〖HCO〗_3^-)/mol)/(59050 (mg 〖Acetate〗^-)/mol)×100 (mg 〖Acetate〗^-)/L=103.3 mg/L as 〖HCO〗_3^- The theoretical alkalinity contributed by acetate is 103.3mg/l as 〖HCO〗_3^-. This means that 68.95⁄103.3=67% of the acetate ion accepted anH^+ at 4.5 pH (33% remained as the C_2 H_3 O_2^- ion). Acetate concentrations therefore have significant impact on total alkalinity.