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Re: DO input concentration




It's a little confusing, but PHREEQC wants to know the number of moles of
oxygen in the zero valence state. To get that, you need to take the total
number of milligrams of oxygen and divide by 16.  It really does not matter
whether the total milligrams is specified to be O or O2; the number of
milligrams of O is equal to the number of milligrams of O2.

If you use "as O2", you will divide by 32, but this gives the number of
moles of O2, whereas PHREEQC wants the number of moles of just plain O.

David


David Parkhurst (dlpark@xxxxxxxx)
U.S. Geological Survey
Box 25046, MS 413
Denver Federal Center
Denver, CO 80225

Project web page: https://wwwbrr.cr.usgs.gov/projects/GWC_coupled



                                                                                                                 
                      "Bruno Haerens"                                                                            
                      <bruno.haerens@xxxxxxxx        To:       <dlpark@xxxxxxxx>                                 
                      uven.ac.be>                    cc:                                                         
                                                     Subject:  DO input concentration                            
                      10/28/2003 09:48 AM                                                                        
                                                                                                                 
                                                                                                                 




David,


I'm a bit confused about the input concentration for DO in mg/l.
If I have 4mgO2/L, should I than enter 4mg/L for O(0)?
I tried to find out, and this gave me the correct result of 1.25e-5molO2/L
and 2.5e-5molO(0)/L
is this correct?


This is maybe a stupid question, but I got confused why not entering conc
"as O2", but then I get 6.284e-005molO2/l as a result, which is 2mgO2/l.


best regards,
bruno


------------------------------------
Reading input data for simulation 1.
------------------------------------


        DATABASE C:\Program Files\Phreeqc\Phreeqc.dat
        SOLUTION 1
            temp      25
            pH        7.92
            pe        4
            redox     pe
            units     mg/l
            density   1
            O(0)      4.02
            Na        19.8
            K         4.9
            Ca        73.3
            Mg        33.5
            Cl        78
            S(6)      95 as SO4
            N(5)      55 as NO3
            Alkalinity 140
            water    1 # kg
-------------------------------------------
Beginning of initial solution calculations.
-------------------------------------------


Initial solution 1.


-----------------------------Solution
composition------------------------------


        Elements           Molality       Moles


        Alkalinity       2.799e-003  2.799e-003
        Ca               1.830e-003  1.830e-003
        Cl               2.201e-003  2.201e-003
        K                1.254e-004  1.254e-004
        Mg               1.379e-003  1.379e-003
        N(5)             8.874e-004  8.874e-004
        Na               8.617e-004  8.617e-004
        O(0)             2.514e-004  2.514e-004
        S(6)             9.894e-004  9.894e-004


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


                                       pH  =   7.920
                                       pe  =   4.000
                        Activity of water  =   1.000
                           Ionic strength  =  1.079e-002
                       Mass of water (kg)  =  1.000e+000
                    Total carbon (mol/kg)  =  2.823e-003
                       Total CO2 (mol/kg)  =  2.823e-003
                      Temperature (deg C)  =  25.000
                  Electrical balance (eq)  = -4.625e-004
 Percent error, 100*(Cat-|An|)/(Cat+|An|)  =  -3.26
                               Iterations  =   7
                                  Total H  = 1.110152e+002
                                  Total O  = 5.552149e+001


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


        Redox couple             pe  Eh (volts)


        O(-2)/O(0)          12.6251      0.7469


----------------------------Distribution of
species----------------------------


                                                   Log       Log
Log
        Species            Molality    Activity  Molality  Activity
Gamma


        OH-              9.282e-007  8.325e-007    -6.032    -6.080
-0.047
        H+               1.320e-008  1.202e-008    -7.880    -7.920
-0.040
        H2O              5.551e+001  9.998e-001     1.744    -0.000
0.000
C(4)            2.823e-003
        HCO3-            2.656e-003  2.396e-003    -2.576    -2.621
-0.045
        CO2              6.462e-005  6.478e-005    -4.190    -4.189
0.001
        CaHCO3+          3.731e-005  3.367e-005    -4.428    -4.473
-0.045
        MgHCO3+          2.590e-005  2.329e-005    -4.587    -4.633
-0.046
        CaCO3            1.725e-005  1.729e-005    -4.763    -4.762
0.001
        CO3-2            1.410e-005  9.346e-006    -4.851    -5.029
-0.179
        MgCO3            7.389e-006  7.407e-006    -5.131    -5.130
0.001
        NaHCO3           1.038e-006  1.041e-006    -5.984    -5.983
0.001
        NaCO3-           1.496e-007  1.344e-007    -6.825    -6.871
-0.046
Ca              1.830e-003
        Ca+2             1.663e-003  1.101e-003    -2.779    -2.958
-0.179
        CaSO4            1.120e-004  1.123e-004    -3.951    -3.950
0.001
        CaHCO3+          3.731e-005  3.367e-005    -4.428    -4.473
-0.045
        CaCO3            1.725e-005  1.729e-005    -4.763    -4.762
0.001
        CaOH+            1.691e-008  1.520e-008    -7.772    -7.818
-0.046
Cl              2.201e-003
        Cl-              2.201e-003  1.975e-003    -2.657    -2.704
-0.047
H(0)            2.042e-027