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*To*: "Bernd Ehret" <bernd.ehret@xxxxxxxxxxxxx>*Subject*: Re: bernd 050902*From*: "David L Parkhurst" <dlpark@xxxxxxxx>*Date*: Thu, 5 Sep 2002 08:33:29 -0600*Bcc*: "David L Parkhurst" <dlpark@xxxxxxxx>*In-reply-to*: <006d01c254cd$e04ee190$88042b8d@hydrologie.tucottbus.de>

I am pretty sure that PHREEQC is doing the calculation correctly in all cases. The problems you are solving are not equivalent. I just looked at case 1 and case 3 for comparison. Initially solutions 0, 1, 2, and 3 have volumes of 1 L. In case 1, you are taking .1 x 1 L of 0, .15 x 1 L of 1, .25 x 1 L of 2, and .5 x 1 L of 3 to make 1 L of 1001. However, in case 3, you are taking .072 x 1 L of 0, .108 x 1 L of 1, .180 x 1 L of 2, and .360 x 1 L of 3 to make .72 L of 1001. If you look at the results, the solution compositions of the mixture are the same in terms of molality. However, the next step causes a difference because solutions 0, 1, and 2 have 1 L of solution for both cases, but solution 1001 has 1 L for case 1 and .72 L for case 3. For the second mix in case 1, you add .1 x 1 L of solution 0, .2 x 1 L of solution 1, .3 x 1L of solution 2, and .4 x 1 L of solution 1001 to make 1 L of 1002. For case 3, you add .072 x 1 L of solution 0, .144 x 1 L of solution 1, .216 x 1 L of solution 2, and (here is the difference) .288 x .72 L of solution 1001 to make .6394 L of solution 1002. The proportion of 1001 in 1002 is different between the two cases .4x1/1 = .4 for case 1 and .288x.72/.6394 = .32 for case 3. I don't know which you want. Case 3 will give a constantly diminishing volume of water; case 2 will eventually give a huge volume of water; so I think you probably want to scale things correctly to keep a constant volume of water. David David Parkhurst (dlpark@xxxxxxxx) U.S. Geological Survey Box 25046, MS 413 Denver Federal Center Denver, CO 80225 Project web page: https://wwwbrr.cr.usgs.gov/projects/GWC_coupled

**References**:**Re: bernd 050902-2***From:*Bernd Ehret

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