exception in "solubility increasses with increasing temperature"

[pia said]

In general, solubility increases with increasing temperature, but there are some exceptions such as CaCO3, CaSO4 and FePO4. My question is: Do PHREEQC considere the this exceptions ?

[dlparkhurst]

PHREEQC considers log K as a function of temperature, when it is defined, either by an analytical expression or by using the enthalpy and the Van't Hoff expression. So log K may increase or decrease as as temperature increases.

However, be careful of what you assume. In addition to log K for the mineral, the solubility depends on speciation and  activity coefficients, which are also functions of temperature.

You may be surprised to learn that calcite in a closed system increases in solubility with temperature. In equilibrium with air (uncomment CO2(g)), the solubility decreases.

Code: [Select]

SOLUTION
END
USE solution 1
EQUILIBRIUM_PHASES
Calcite 0 10
#CO2(g) -3.4 10
REACTION_TEMPERATURE
10 50 in 6 steps
USER_GRAPH 1
    -headings               TC TOT_C CO3 HCO3
    -axis_titles            "Temperature, C" "Molality"
    -initial_solutions      false
    -connect_simulations    true
    -plot_concentration_vs  x
  -start
10 GRAPH_X TC
20 GRAPH_Y TOT("Ca"), MOL("CO3-2"), MOL("HCO3-")
  -end
    -active                 true
END


[pia said]

Thank you!

[pia said]

Hi again, I'm modeling the Fe precipitation with the addition of calcite to increase pH. I need your help because I couldn't understand why in my system (open in contact with the air) the calcite dissolution increases with temperature. This is my code:



Beforehand thank you very much

[dlparkhurst]

I think there are a couple of issues. First, when you change the temperature of the initial solution, you do not have the same starting condition. The amount of sulfate that balances the charge is different depending on the temperature, probably because activity coefficients are different for H+. The activity coefficient converts pH to molality of H+. Also, there may be a difference in the K for HSO4- at different temperature.

I have simplified your input file to consider only equilibrium conditions (something you might have considered) for the two starting conditions. In each case, the solubility of calcite is slightly less at the higher temperature.

I would also change your rate equation for Fe(OH)3(a) to remove line 20. This line prevents Fe(OH)3(a) from attaining complete equilibrium, and further, the ultimate saturation state at long time can vary as parameters in the input file are varied.

Code: [Select]

SOLUTION 1
    temp      20
    pH        2 #2
    pe        6.33 O2(g) -0.68
    redox     pe
    units     ppm
    density   1
    Al        60.02
    As        3
    B         4
    Ca        228 #265
    Cl        957
    Fe(3)     56 #56 60.2
    K         70.3
    Mg        125
    Mn        10
    Na        351
    Pb        1
    S(6)      1950 charge #1883
    Zn        10
    -water    0.24 # kg

EQUILIBRIUM_PHASES 1
    CO2(g)    -1.54 10
    O2(g)     -0.68 10
    Fe(OH)3(a) 0 10
    Calcite    0 10
REACTION_TEMPERATURE
20 30
END


[pia said]

Ok thanks you !!