Beginners > PHREEQC manual examples

Example 6 Changing K feldspar reaction pathway

**jt2020**:

Hello,

I am a first year PhD student at Imperial College London and am trying to reproduce a the reaction pathway for pyrite. I have been using example 6 in the User Manual to familiarise myself with the programme before I then produce a phase diagram of my own.

Example 6 is showing a reaction pathway for K-feldspar (KAlSi3O8). The one thing that is troubling me is under the USER_GRAPH section. I am struggling to understand what K 1, Al 1 and Si 1 next to each mineral represent (and why one of the three are assigned to a mineral in each solution). I also dont understand why a pH value of 11 has been chosen in Solution 1. I know that those elements make up K-keldspar but im not sure what part they play here (example below).

USER_GRAPH

-initial_solutions true

10 PLOT_XY LA("H4SiO4"), SI("K_H"), color = Black, symbol = None

SOLUTION 1

pH 11; K 2 K_H 8; Al 1 Gibbsite; Si 1 K-mica

SOLUTION 2

pH 7; K 1 K-mica; Al 1 Gibbsite; Si 1 Kaolinite

SOLUTION 3

pH 7; K 1 K-mica; Al 1 K-feldspar; Si 1 Kaolinite

SOLUTION 4

pH 7; K 1 K_H -1; Al 1 Kaolinite; Si 1 K-feldspar

END

USER_GRAPH

10 PLOT_XY LA("H4SiO4"), SI("K_H"), color = Black, symbol = None

SOLUTION 1

pH 11; K 1 K_H 8; Al 1 K-feldspar; Si 1 K-mica

SOLUTION 2

pH 7; K 1 K-mica; Al 1 K-feldspar; Si 1 Kaolinite

SOLUTION 3

pH 7; K 1 K-mica; Al 1 Gibbsite; Si 1 Kaolinite

SOLUTION 4

pH 7; K 1 K_H -1; Al 1 Gibbsite; Si 1 Kaolinite

END

I am recreating a reaction pathway but from mackinawite to greigite to pyrite, therefore, have only been considering Fe and S as appose to the 3 elements here (K, Al, Si). How would I be able to construct a graph with just the two elements as I suppose you need 3 to triangulate a point?

If anyone has any suggestions on what those points mean, that would be a great help.

Thanks

Jack

**dlparkhurst**:

To make the diagram, one of the axes is log K+/H+. A fictitious phase was defined that has a reaction that has SI equal to log(K+) - log(H+) (log K is 0.0 by default).

--- Code: ---PHASES

K_H; KH = K+ - H+; -no_check

--- End code ---

The following definition creates a solution of pH 11. The K concentration is adjusted to obtain a log(K+/H+) activity ratio of 8. The Al concentration is adjusted to obtain equilibrium with K-feldspar, and the Si concentration is adjusted to obtain equilibrium with K-mica.

So, this solution produces the point on the graph at log(K+/H+) 8 at the phase boundary between K-mica and K-feldspar. I don't think the pH of 11 is critical, but the pH must be high enough that a log ratio of 8 produces reasonable concentrations of K.

--- Code: ---SOLUTION 1

pH 11;

K 1 K_H 8;

Al 1 K-feldspar;

Si 1 K-mica

--- End code ---

**jt2020**:

Hi,

Thanks for replying. So it is essentially a case of trial and error until you a saturation index of 0.0 for all three phases.

I am curious on why K, Al and Si have been placed corresponding to each mineral though. For example, why in solution 1 is Gibbsite placed next to Al and not K-mica?

Also, I have a problem with my model. I am able to create a line for the boundary between mackinawite and greigite. I am, however, unable to get one for greigite-pyrite or even a triple point between mackinawite-greigite-pyrite. I have varied pH levels but it doesnt seem to want to work. Mackinawite is undersaturated, with the two others being heavily oversaturated. Is this a problem with the logK values from the database, and because of that, that point at which the three coexist, is impossible?

Below is my input.

TITLE Simulation 6A.--React to phase boundaries.

SOLUTION 1 PURE WATER

pH 7.0 charge

temp 25.0

PHASES

Mackinawite

FeS + H+ = Fe+2 + HS-

log_k -4.648

delta_h -0 kcal

Greigite

Fe3S4 + 4H+ = 2Fe+3 + Fe+2 + 4HS-

log_k -45.035

delta_h -0 kcal

Pyrite

FeS2 + 2H+ + 2e- = Fe+2 + 2HS-

log_k -18.479

delta_h 11.3 kcal

SELECTED_OUTPUT

-file Phase test.sel

-activities Fe+2 H+ HS- Fe+3

-si Mackinawite Greigite Pyrite

-equilibrium Mackinawite Greigite Pyrite

END

PRINT; -user_print false

# --Plot the phase boundaries with USER_GRAPH..

PHASES

Fe_H; FeH = Fe+2 - H+; -no_check

USER_GRAPH

-initial_solutions true

10 PLOT_XY LA("S-2"), SI("Fe_H"), color = Black, symbol = None

SOLUTION 1

pH 11 ; Fe(2) 1 Fe_H 5; Fe(3) 1 Mackinawite; S(-2) 1 Greigite

SOLUTION 2

pH 4; Fe(2) 1 Greigite; Fe(3) 1 Mackinawite; S(-2) 1 Pyrite

SOLUTION 3

pH 6; Fe(2) 1 Fe_H -1; Fe(3) 1 Greigite; S(-2) 1 Mackinawite

END

USER_GRAPH

10 PLOT_XY LA("S-2"), SI("Fe_H"), color = Black, symbol = None

SOLUTION 1

pH 7; Fe(2) 1 Fe_H 5; Fe(3) 1 Greigite; S(-2) 1 Pyrite

SOLUTION 2

pH 7; Fe(2) 1 Greigite; Fe(3) 1 Mackinawite; S(-2) 1 Pyrite

SOLUTION 3

pH 7; Fe(2) 1 Fe_H -1; Fe(3) 1 Mackinawite; S(-2) 1 Pyrite

Thanks for helping,

Jack

**dlparkhurst**:

First, I think you should be using the ratio Fe+2/H+^2 rather than Fe+2/H+.

I think the trick is finding the right range of of Fe+2/H+^2. You can also run reactions to determine the characteristics of equilibrium between two phases.

--- Code: ---USE solution 1

EQUILIBRIUM_PHASES

Mackinawite 0 10

Greigite 0 10

--- End code ---

I suspect that the triple point among the three minerals is not physically feasible.

All that said, I think PhreePlot may be a better choice for generating a stability diagram for this system.

**dlparkhurst**:

So, yes, a lot of trial and error gives the following script and attached graph.

I think one set of plausible axes for the diagram is H2 and Fe+2/H+^2.

With these minerals, you easily flirt with the edge of the stability field for water, so, if you fix pH and pe, you may be outside the field, or require many moles of Fe or S. I let the pH and pe float to produce the required Fe/H^2 relation.

The edge of the stability field of water is in the neighborhood of log H2 equal zero (not H2O as in previous post). The intersection of the two lines would be well beyond log H2 = 0, so you would never have all three minerals in equilibrium.

--- Code: ---PHASES

Mackinawite

FeS + H+ = Fe+2 + HS-

log_k -4.648

delta_h -0 kcal

Greigite

Fe3S4 + 4H+ = 2Fe+3 + Fe+2 + 4HS-

log_k -45.035

delta_h -0 kcal

Pyrite

FeS2 + 2H+ + 2e- = Fe+2 + 2HS-

log_k -18.479

delta_h 11.3 kcal

SELECTED_OUTPUT

-file Phase test.sel

-activities Fe+2 H+ HS- Fe+3

-si Mackinawite Greigite Pyrite

-equilibrium Mackinawite Greigite Pyrite

END

PHASES

Fe_H2; FeH2 = Fe+2 - 2H+; -no_check

END

USER_GRAPH 1

-headings Greigite/Mackinawite

-initial_solutions true

-axis_titles "Log activity of H2(aq)" "Log activity (Fe+2/H+^2)" ""

-start

10 PLOT_XY LA("H2"), SI("Fe_H2"), color = Black, symbol = None

-end

SOLUTION 1

pH 11 ; Fe(2) 1 Fe_H2 8; Fe(3) 1 Mackinawite; S(-2) 1 Greigite

SOLUTION 3

pH 6; Fe(2) 1 Fe_H2 1; Fe(3) 1 Greigite; S(-2) 1 Mackinawite

END

USER_GRAPH

-headings Greigite/Pyrite

10 PLOT_XY LA("H2"), SI("Fe_H2"), color = Red, symbol = None

SOLUTION 2

pH 10 charge ; pe -8 Fe_H2 20; Fe 1 Pyrite; S 1 Greigite

SOLUTION 4

pH 8 charge; pe -8 Fe_H2 14; Fe 1 Pyrite; S 1 Greigite

--- End code ---

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