Speciation plots

[AnnaJulIQ]

H

[dlparkhurst]

You made one calculation reacting pure water with three minerals. The resulting pH is 13.5 and the plot shows the distribution of species at that pH.

Attached is a calculation that adds KOH incrementally, so that you have multiple pH.

Code: [Select]

TITLE #MINERALS RECOVERY WITH WATER
SOLUTION_SPECIES
K+ + H2O  =  KOH + H+
        log_k           -14.46
PHASES
KOH(s)
KOH = K+ + OH-
log_k 2.18

SOLUTION 1
pH 4.5
WATER   9.27      #9.27 Kg of water in 10 kg of whey
END
USE solution 1
#AMOUNT OF ALKALINE BATTERY 100 g
EQUILIBRIUM_PHASES 1
Pyrolusite 0 4.74        #MnO2     "Mol"
Manganite  0 10.56       #MnOOH    "Mol"
#KOH(s)     0 4.48        #KOH      "Mol"
REACTION
KOH 1
1e-7 1e-6 1e-5 1e-4 1e-3 1e-2 1e-1 1 4.48
USER_GRAPH 1 #Example Mn speciation
   -headings        "Mn(OH)3-" "MnOH+" "Mn+2"
   -axis_titles      "pH" "Fraction of Mn(2)"
   -chart_title       "Mn(2) speciation (dissolved)"
   -axis_scale x_axis        auto
   -axis_scale y_axis        auto
  -start
10 graph_y  MOL("Mn(OH)3-")/TOT("Mn"), MOL("MnOH+")/TOT("Mn"), MOL("Mn+2")/TOT("Mn")
20 graph_x   -LA("H+")
end