Dissolution of amorphous silica in alkaline solutions

[Aksenova]

Hello
I need your advice.
Please tell me what errors are in my code?
My goal is to plot the solubility of amorphous silica in a 10% mass sodium hydroxide  (NaOH) solution as a function of temperature.
I'm entering the code below.

The program gives me values, but I don't agree with the density value for this solution.
I took 10 ml of a 10% mass sodium hydroxide (NaOH) solution.

Code: [Select]

SOLUTION 1 NaOH 10%
        Units   mol/L
        pH      13.6 charge
        temp    25.0
        Na      2.5
        Density 1.1089
        water   0.01         
EQUILIBRIUM_PHASES 1
        SiO2(a)       0.0     0.017
REACTION_TEMPERATURE 1
        25.0 95.0 in 51 steps
SELECTED_OUTPUT
        -file   ex2.sel
        -temperature
        -si     SiO2(a)
USER_GRAPH 1 Example 2
        -headings Temperature SiO2(a)
        -chart_title "SiO2(a) Stability"
        -axis_scale x_axis 25 95 0
        -axis_scale y_axis auto 0.05 0.1
        -axis_titles "Temperature, in degrees celsius" "Saturation index"
        -initial_solutions false
  -start
  10 graph_x TC
  20 graph_y SI("SiO2(a)")
  -end
END

[dlparkhurst]

Yes, the calculated density is a little low: 1.096 rather than 1.107. However, you should let PHREEQC calculate the density so that PHREEQC calculates the correct molarity for your calculation. The corresponding molality is what is important for the calculation, and it should be estimated adequately. In any case, the differences in concentration are relatively small.

Code: [Select]

SOLUTION
pH 7 charge
-units mol/L
density 1.1 calc
Na 2.5
USER_PRINT
10 PRINT "Density:  ", RHO
20 PRINT "Molarity: ", TOTMOL("Na") / SOLN_VOL
30 PRINT "Molality: ", TOT("Na")
END
Provided you have sufficient SiO2 in EQUILIBRIUM_PHASES, it should dissolve to equilibrium, which will be SI = 0. If you want to see how much SiO2 dissolves, plot TOT("Si").

[Aksenova]

Thank you very much for your answer
I figured out the sodium concentration. I took 900 g of distilled water and 100 g of pure anhydrous sodium hydroxide. Accordingly, 1 liter requires 2.5 mol of sodium hydroxide (100 g / 40 g/mol).
I specified the "charge" function to automatically change the pH.
The solution was prepared at 25?C, then I added 0.017 mol of amorphous silica to 10 ml of the sodium hydroxide solution.
The program calculates that everything has dissolved, according to the data in the "Phase assembly" block, meaning the final amount of amorphous silica is 0.
The graph shows that the SiO2(a) saturation index increases with increasing temperature.

Code: [Select]

SOLUTION 1 NaOH 10%
        Units   mol/L
        pH      13.6 charge
        temp    25.0
        Na      2.5
        water   0.01         
EQUILIBRIUM_PHASES 1
        SiO2(a)       0.0     0.017
REACTION_TEMPERATURE 1
        25.0 120.0 in 5 steps
SELECTED_OUTPUT
        -file   ex2.sel
        -temperature
        -si     SiO2(a)
USER_GRAPH 1 Example 2
        -headings Temperature SiO2(a)
        -chart_title "SiO2(a) Stability"
        -axis_scale x_axis 25 120 0
        -axis_scale y_axis auto 0.005 0.1
        -axis_titles "Temperature, in degrees celsius" "Saturation index"
        -initial_solutions false
  -start
  10 graph_x TC
  20 graph_y TOT ("Si")
  -end
ENDHowever, if you take the alkali concentration to be two times less, you get a completely opposite graph
My idea was to find out how much of the 0.017 mol was not dissolved by my 10 ml solution

[dlparkhurst]

Here is a script that follows your recipe for the NaOH solution. I think it differs slightly from your definition.

Here is another way to do the calculation, where SiO2 is added until saturation is reached. Note that the solubility of your solid may be different than the solubility represented by SiO2(a) in the database.

Code: [Select]

SOLUTION 1
pH 7
END
MIX 1
1 0.9
REACTION
NaOH 1
2.5
SAVE solution 2
END
MIX 2
2 0.01
EQUILIBRIUM_PHASES
SiO2(a) 0 0
REACTION
SiO2 1
0.02 in 10
USER_GRAPH 1
    -headings               rxn SI(SiO2(a))
    -axis_titles            "SiO2 dissolved, moles" "SI SiO2(a)" ""
    -initial_solutions      false
    -connect_simulations    true
    -plot_concentration_vs  x
  -start
10 GRAPH_X RXN
20 GRAPH_Y SI("SiO2(a)")
  -end
    -active                 true
END

[Aksenova]

Thank you very much for your reply.
I understand that the solubility of my solid may differ from the solubility of amorphous silica in the database, of course.
I looked at your code, but there's no temperature dependence.
And I don't quite understand the data provided in the code.
What does "SiO2(a) 0.02 in 10" mean?

I understand that the y-axis doesn't show the saturation index; it shows the molality.
The higher this value, the more silica is present in the solution and the better the dissolution? Or is my logic flawed?

Code: [Select]

SOLUTION 1 NaOH 10%
        Units   mol/L
        pH      14.1
        temp    25.0
        Na      2.5 charge
        water   0.01         
EQUILIBRIUM_PHASES 1
        SiO2(a)       0.0     0.017
REACTION_TEMPERATURE 1
        25.0 120.0 in 5 steps
SELECTED_OUTPUT
        -file   ex2.sel
        -temperature
        -si     SiO2(a)
USER_GRAPH 1 Example 2
        -headings Temperature SiO2(a)
        -chart_title "SiO2(a) Stability"
        -axis_scale x_axis 25 120 0
        -axis_scale y_axis auto 0.005 0.1
        -axis_titles "Temperature, in degrees celsius" "Saturation index"
        -initial_solutions false
  -start
  10 graph_x TC
  30 graph_y TOT ("Si")
  -end
END

[dlparkhurst]

You had plotted SI of zero for all temperatures, which was not very informative. I thought you might want to see how SI varied with added SiO2. I did it only for 25C, but you could change the temperature.

Input for the REACTION keyword is described in the manual.

I see another issue, and that is the volume of NaOH solution. According to your recipe, 900 g water + 100 g NaOH, the volume of solution calculated by PHREEQC is about 900 mL. The fraction of that solution (generated in PHREEQC) that equals 10 mL  would be .01/0.900 = 0.011. The script calculates it more exactly. If this logic does not apply to your experiments, then you will have to adjust the PHREEQC calculation.

Note that the volume of solution will change with temperature. The script assumes that a 10 mL aliquot of the 25 C solution is then heated with sufficient pressure to retain the liquid phase. Again, if that does not match experimental conditions, you will need to adjust the calculation.

Code: [Select]

SOLUTION 1
    temp      25
    pH        7
END
MIX 1
    1    0.9 # ~900 mL
REACTION 1
    NaOH       1
    2.5 moles
SAVE solution 2
USER_PRINT
-start
10 PRINT "Mixing fraction for 10 mL:", 0.01 / SOLN_VOL
-end
END
USER_PRINT
10 REM Remove print for next calculations
MIX 2
    2    0.011052
EQUILIBRIUM_PHASES 1
    SiO2(a)   0 10 # excess of SiO2(a) is available to react
REACTION_TEMPERATURE 1
    25 120 in 5 steps
USER_GRAPH 1
    -headings               rxn Molality(Si) Grams(Si)
    -axis_titles            "Temperature, Celsius" "Solubility of SiO2(a), mol/kgw" "Grams of SiO2 in heated solution"
    -initial_solutions      false
    -connect_simulations    true
    -plot_concentration_vs  x
  -start
10 GRAPH_X TC
20 GRAPH_Y TOT("Si")
30 GRAPH_SY TOT("Si")*TOT("Water")*GFW("SiO2")
  -end
END

[Aksenova]

Thank you very much.
Yes, you're right. I was thinking I entered the concentration incorrectly. After all, I have 2.5 mol per 1 kilogram of water?that changes the graph quite significantly.
I understand what you mean by sufficient pressure.

Please check, I've changed the code.
I have another question.
Look, if I enter the code below, I still plot the change in molality as a function of temperature.
Am I correct in understanding that the lower the molality, the more silica has dissolved and entered the solution? Is that the wrong way to think?

Code: [Select]

SOLUTION 1 NaOH 10%
        Units   mol/kgw
        pH      14.1 charge
        temp    25.0
        Na      2.5
        water   0.01         
EQUILIBRIUM_PHASES 1
        SiO2(a)       0.0     0.017
REACTION_TEMPERATURE 1
        25.0 120.0 in 5 steps
SELECTED_OUTPUT
        -file   ex2.sel
        -temperature
        -si     SiO2(a)
USER_GRAPH 1 Example 2
        -headings Temperature SiO2(a)
        -chart_title "SiO2(a) Stability"
        -axis_scale x_axis 25 120 0
        -axis_scale y_axis auto 0.005 0.1
        -axis_titles "Temperature, in degrees celsius" "Saturation index"
        -initial_solutions false
  -start
  10 graph_x TC
  30 graph_y TOT ("Si")
  -end
END

[dlparkhurst]

"After all, I have 2.5 mol per 1 kilogram of water"

You said earlier that you used 900 g water and 100 g NaOH (2.5 mol). Is it 1 kg or 900 g of water? As stated, you made 1000 g of solution and about 900 mL of solution. You'll have to decide what is right.

You are wearing me down. Here is a script with results for a maximum of 0.017 mol of SiO2(a) available to react. Below 55 C, 0.017 mol is not sufficient to dissolve to equilibrium: all of the SiO2(a) dissolves, and the SI is less than zero. Above 55 C, enough SiO2 dissolves to reach equilibrium, and the SI is zero.

Code: [Select]

SOLUTION 1 NaOH 10%
        Units   mol/kgw
        pH      14.1 charge
        temp    25.0
        Na      2.5
        water   0.01         
EQUILIBRIUM_PHASES 1
        SiO2(a)       0.0    0.017
REACTION_TEMPERATURE 1
        25.0 120.0 in 20 steps
SELECTED_OUTPUT
        -file   ex2.sel
        -temperature
        -si     SiO2(a)
USER_GRAPH 1 Example 2
        -headings Temperature SI(SiO2(a)) Si,molality
        -chart_title "SiO2(a) Stability"
        -axis_scale x_axis 25 120 0
        -axis_titles "Temperature, in degrees celsius" "SI" "Si, molality"
        -initial_solutions false
  -start
  10 graph_x TC
  20 graph_y SI("SiO2(a)")
  30 graph_sy TOT("Si")
  -end
END

[Aksenova]

Yes, you're right.
Excuse me.
I made a mistake
I used 100 g of NaOH (2.5 moles)  and 900 g of water.

[Aksenova]

Could you please tell me why, when I specify 2.5 moles of sodium hydroxide per kilogram of solution, there's some kind of conversion? After all, I get 2.5 moles per 1 kg of solution (900 grams of water and 100 grams of sodium hydroxide).

Code: [Select]

SOLUTION 1 NaOH 10%
        Units   mol/kgs
        pH      14.1 charge
        temp    25.0
        Na      2.5
        water   0.01         
EQUILIBRIUM_PHASES 1
        SiO2(a)       0.0    0.017
REACTION_TEMPERATURE 1
        25.0 120.0 in 20 steps
SELECTED_OUTPUT
        -file   ex2.sel
        -temperature
        -si     SiO2(a)
USER_GRAPH 1 Example 2
        -headings Temperature SI(SiO2(a)) Si,molality
        -chart_title "SiO2(a) Stability"
        -axis_scale x_axis 25 120 0
        -axis_titles "Temperature, in degrees celsius" "SI" "Si, molality"
        -initial_solutions false
  -start
  10 graph_x TC
  20 graph_y SI("SiO2(a)")
  30 graph_sy TOT("Si")
  -end
END



[dlparkhurst]

PHREEQC converts mol/kgs to mol/kgw. All calculations are done with molality.