PPM solutions
Chris,
Thanks for the offer to help. I do have a few questions, as I am not an expert on chemistry.
First off, whenever I have made solutions that were more than one or two percent I have often run into insolubility problems. The stuff would just precipitate out over night and turn to rock. Most of the time for aquatic uses I don't need to use concentrations higher than say 2%.
How much error is there in a using 9.0 grams of KH2PO4 in 250 ml of water to make a 2.51% PO4 solution?
This site:
http://www.epin.ncsu.edu/apti/ol_2000/module4/concent4/concent4.htm
states that for using water as a solvent, mg/l = ppm.
It also suggests that for low amounts of solutes the volume of the solute can be ignored.
Also, even though I see your point, your example of 150 grams of KNO3 in 250 ml seems a bit extreme. The Merck site has the solubility of KNO3 as 320 grams/liter. I would think by this info that you can't mix that much KNO3 into 250 ml of water. You example seems to be about a 60% solution.
OK, now I understand most of your example. What I don't understand yet is the conversion of grams of KNO3 to millilters. It seems the Merck Chemdat site gives the density of KNO3 as 2.11 g/cc. Is this the density you are using? This density seems to be fairly consistant. It is obviously not a dry bulk density with air voids between the crystals. So I assume now, that it is the wet density of KNO3, or what density it would be in a solution. It this correct? Probably not, since this would only give me 71 ml of volume in the solution, and you say its 165 ml.
Also if ppm is really a mass comparison, and for water ml should equal approximately mg, why not just add the masses together: 150000 mg x (1000)/(150+250)? (OK, I think I now realize that we would have to weigh it out, instead of measuring a volume). So we need a good value for the density so we can add the volume of the solute to the volume of the solvent.
I guess what I need to know is how you got 165 ml for 150 grams of KNO3 as that would give a density of .91 gm/cc.
All of us here at AquaticPlantCentral do appreciate your input. We really do want to make this calculator an accurate tool. Once we get the theory down we'll correct the calculations.
I don't want to change the description to "total volume of solution" as I think the average hobbyist, like myself, would be further confused.
Thanks in advance,
Steve Pituch
Chris,
Thanks for the offer to help. I do have a few questions, as I am not an expert on chemistry.
First off, whenever I have made solutions that were more than one or two percent I have often run into insolubility problems. The stuff would just precipitate out over night and turn to rock. Most of the time for aquatic uses I don't need to use concentrations higher than say 2%.
How much error is there in a using 9.0 grams of KH2PO4 in 250 ml of water to make a 2.51% PO4 solution?
This site:
http://www.epin.ncsu.edu/apti/ol_2000/module4/concent4/concent4.htm
states that for using water as a solvent, mg/l = ppm.
It also suggests that for low amounts of solutes the volume of the solute can be ignored.
Also, even though I see your point, your example of 150 grams of KNO3 in 250 ml seems a bit extreme. The Merck site has the solubility of KNO3 as 320 grams/liter. I would think by this info that you can't mix that much KNO3 into 250 ml of water. You example seems to be about a 60% solution.
OK, now I understand most of your example. What I don't understand yet is the conversion of grams of KNO3 to millilters. It seems the Merck Chemdat site gives the density of KNO3 as 2.11 g/cc. Is this the density you are using? This density seems to be fairly consistant. It is obviously not a dry bulk density with air voids between the crystals. So I assume now, that it is the wet density of KNO3, or what density it would be in a solution. It this correct? Probably not, since this would only give me 71 ml of volume in the solution, and you say its 165 ml.
Also if ppm is really a mass comparison, and for water ml should equal approximately mg, why not just add the masses together: 150000 mg x (1000)/(150+250)? (OK, I think I now realize that we would have to weigh it out, instead of measuring a volume). So we need a good value for the density so we can add the volume of the solute to the volume of the solvent.
I guess what I need to know is how you got 165 ml for 150 grams of KNO3 as that would give a density of .91 gm/cc.
All of us here at AquaticPlantCentral do appreciate your input. We really do want to make this calculator an accurate tool. Once we get the theory down we'll correct the calculations.
I don't want to change the description to "total volume of solution" as I think the average hobbyist, like myself, would be further confused.
Thanks in advance,
Steve Pituch
Unless I'm missing something?
