Example 1, U 3.3 ppb N(5)/N(-3), why?

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Taylor_211126:
this is a part of ex1 input file codes:

TITLE Example 1.--Add uranium and speciate seawater.
SOLUTION 1 SEAWATER FROM NORDSTROM AND OTHERS (1979)
units ppm
pH 8.22
pe 8.451
density 1.023
temp 25.0
redox O(0)/O(-2)
Ca 412.3
Mg 1291.8
Na 10768.0
K 399.1
Fe 0.002
Mn 0.0002 pe
Si 4.28
Cl 19353.0
Alkalinity 141.682 as HCO3
S(6) 2712.0
N(5) 0.29 gfw 62.0
N(-3) 0.03 as NH4
U 3.3 ppb N(5)/N(-3)
O(0) 1.0 O2(g) -0.7
----------------------------------------------------

I cannot understand why N(5)/N(-3) was chosen rather than O(0)/O(-2) to speciate uranium among its valence states, even though N(5)/N(-3) was inputted.

And besides,

Fe 0.002
Mn 0.0002 pe

I know ex1 is primarily to show something about how to define an input file, but I am still interested in knowing what to do if there are indeed some redox couples in the solution.

dlparkhurst:
The choice of redox only affects the distribution of U, Fe, and Mn in this example. Given the data, you have three options for calculating the pe that is used for the speciation of these three elements: pe, O(0)/O(-2), and N(5)/N(-3).

The choices in this example are completely arbitrary and are used simply for demonstration.

In general, there is no guarantee that one redox couple applies to a redox element better than another. In my opinion the best you can do is to use O2 if it is present (oxic), NO3/NH3 if there is no O2 and no H2S (nitric), SO4/H2S if there is no O2 and no NO3 (sulfidic), and CO2/CH4 if only methane is present (methanic). Of course you need the appropriate analyses to make these different calculations.

Taylor_211126:

--- Quote from: dlparkhurst on 29/05/24 20:36 ---The choice of redox only affects the distribution of U, Fe, and Mn in this example. Given the data, you have three options for calculating the pe that is used for the speciation of these three elements: pe, O(0)/O(-2), and N(5)/N(-3).

The choices in this example are completely arbitrary and are used simply for demonstration.

In general, there is no guarantee that one redox couple applies to a redox element better than another. In my opinion the best you can do is to use O2 if it is present (oxic), NO3/NH3 if there is no O2 and no H2S (nitric), SO4/H2S if there is no O2 and no NO3 (sulfidic), and CO2/CH4 if only methane is present (methanic). Of course you need the appropriate analyses to make these different calculations.

--- End quote ---

Thanks a lot.

What I've learned is that it depends on the results of chemical analyses and the capacities of redox couples, as per your response.

In my research, as exemplified below, the pe value is -5.11767, yet I am uncertain which redox couple governs the redox conditions. In the field, H2S smell was strongly. Therefore, could I hypothesize the presence of the SO4/H2S redox couple, and how can I determine the H2S concentration using Phreeqc? Thanks.
-------------------------------------------
SOLUTION 111
temp 81.3
pH 4.8 CO2(g) -0.215043599
pe -5.11767
redox pe
units mg/l
density 1
Ca 23.3
Cl(-1) 49.9
F 4.46
K 58.4
Mg 9.79
Na 470
S(6) 10.4
Si 50.6 as SiO2
Alkalinity 1418 as HCO3
-water 1 # kg
-----------------------------------------------------

dlparkhurst:
Yes, I think you can assume that you are in the sulfidic redox environment.

However, I don't think you have any way to estimate the sulfide concentration. You can measure it with electrode or chemical methods. If you had an iron analysis, you might consider what sulfide concentration would put you in equilibrium with, say, mackinawite.

Taylor_211126:

--- Quote from: dlparkhurst on 30/05/24 04:00 ---Yes, I think you can assume that you are in the sulfidic redox environment.

However, I don't think you have any way to estimate the sulfide concentration. You can measure it with electrode or chemical methods. If you had an iron analysis, you might consider what sulfide concentration would put you in equilibrium with, say, mackinawite.

--- End quote ---

Thanks a lot again.

Actrually, there are indeed some priate ore deposits around. So I do some changes as you said like this:
-------------------------------------------------
SOLUTION 113
temp 73
pH 8
pe -4.45896
redox pe
units mg/l
density 1
C(4) 1521
Ca 25.9
Cl(-1) 50.8
F 4.55
Fe 0.168
K 63
Mg 10.5
Na 502
S(6) 10.1
Si 132 as SiO2
S(-2) 0.01 Pyrite 0 # I set HS- in equilibrium with Pyrite
-water 1 # kg
-----------------------------------------------------

The result shows:

----------------------------Description of solution----------------------------

pH = 8.000
pe = -4.459
Activity of water = 0.999
Ionic strength (mol/kgw) = 2.656e-02
Mass of water (kg) = 1.000e+00
Total alkalinity (eq/kg) = 2.535e-02
Total CO2 (mol/kg) = 2.499e-02
Temperature (癈) = 73.00
Electrical balance (eq) = -1.571e-03
Percent error, 100*(Cat-|An|)/(Cat+|An|) = -3.08
Iterations = 4
Total H = 1.110749e+02
Total O = 5.560492e+01

---------------------------------Redox couples---------------------------------

Redox couple pe Eh (volts)

S(-2)/S(6) -4.7096 -0.3235
-----------------------------------------------------------------------------------

I think the result is good. The pe value for the S(-2)/S(6) redox couple is -4.7096, which is quite close to the pe value of the solution at -4.459. Additionally, the concentration of S(-2) is 8.418e-06. However, I am puzzled by the H(0) concentration, which is 1.248e-05.

I've realized that relying on insufficient analysis to determine the presence of species in a solution is hazardous. The results calculated by PHREEQC should be considered as a reference, not a definitive conclusion.

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