Question Regarding the MIX command

[jlucas91]

Greetings,

I am working to develop an understanding of the PHREEQC computer program and I have a question regarding the mix command and how it arrives at the final system volume. Below is my input file.

SOLUTION 1
    temp      25
    pH        7
    pe        4
    redox     pe
    units     mmol/kgw
    density   1
    -water    0.08 # kg
REACTION 1
    Na2CO3     1
    0.39968 millimoles in 1 steps

SAVE solution 1
END

SOLUTION 2
    temp      25
    pH        7
    pe        4
    redox     pe
    units     mmol/kgw
    density   1
    -water    1 # kg
REACTION 2
    HCl        1
    0.926 moles in 1 steps
SAVE solution 2

END
MIX
    1 1
    2 1e-4
END


I am attempting to model a titration using the mix command I want the entire 0.08kg sodium carbonate solution and to that I want to add .1 mL of the HCl acid solution. So if I assign a mixing fraction of 1 to the carbonate solution then the total volume and concentration should be taken. If I assign the HCl solution a mixing fraction of 1e-4 then 0.1 mL of the acid solution should be taken for the mix. This should yield a total volume of 80.10 mL however the model indicates a final volume of 80.34 mL. Having read the manual I have come up with no reason as to why this difference is observed. Could anyone here offer me some insight as to why my final result is what it is?

Thanks,
John

[dlparkhurst]

MIX doesn't mix on volume, it mixes on the number of moles of each element in solution. So, in your case,

Na, moles in final solution = 0.39968 * 1
Cl, moles in final solution = 0.926 * 1e-4
H and O are also mixed by moles.

Note that H2O is not necessarily conservative in the mixing because of hydrolysis reactions.

If you want to mix by volume, you need to look at the volumes calculated for each solution (only phreeqc.dat, Amm.dat, and pitzer.dat can be used to calculate solution volume):

Na2CO3 solution:
Volume (L)  =   0.08023

HCl solution:
Volume (L)  =   1.02048

To get 0.08 L of the Na2CO3 solution, you would need a mixing fraction of 0.08/0.08023; and for 1e-4 L of the HCl solution 1e-4/1.02048. Note that the volume of the resulting solution might not be 0.0801 L because the speciation changes in the mix and different species have different molar volumes.

[jlucas91]

Thanks Dr. Parkhurst. That clears up my misunderstanding.