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  • Converting PPM to equivalents

    Hello all. I am interested in converting PPM of some cations such as Calcium, Magnesium, Potassium to equivalents and then in turn milliequivalents. I need to know this for some fertilizer requirements. I would also like to know the process of finding equivalents for anions such as the amount of Nitrogen milliequivalents in the nitrate NO3 form. For example I use a certain amount of Calcium Nitrate and get 150 PPM Calcium ion and 120 PPM Nitrate. Soils have a cation exchange capacity represented in milliequivalents/100g of soil and I would like to fulfill my fertilizer bases on a milliequivalent balance. I hope I explained my question well. Thank you.

  • #2
    Originally posted by KosherOnly View Post
    Hello all. I am interested in converting PPM of some cations such as Calcium, Magnesium, Potassium to equivalents and then in turn milliequivalents. I need to know this for some fertilizer requirements. I would also like to know the process of finding equivalents for anions such as the amount of Nitrogen milliequivalents in the nitrate NO3 form. For example I use a certain amount of Calcium Nitrate and get 150 PPM Calcium ion and 120 PPM Nitrate. Soils have a cation exchange capacity represented in milliequivalents/100g of soil and I would like to fulfill my fertilizer bases on a milliequivalent balance. I hope I explained my question well. Thank you.
    Ppm is an ambiguous term unless the basis of computing it is given; it can be based on mass, volume, or molar ratios or even mass/volume. In this case, is probably only adds confusion and I recommend going direct from mass to amount of substance (moles) then to equivalents. Calcium Nitrate is Ca(NO3)2. It has a molar mass of 164.088 grams per mole, computed by adding up the atomic weights of the constituent atoms. There is negligible error in using atomic weights rounded to integers and calling it 164 g/mol.

    When dissolved and ionized, that is 40 g (1 mol) of Ca++ ions, which will be 2 equivalents of calcium because of the double charge and 124 g ( 2 mol) of NO3- ions, or two equivalents of NO3. Note for ionized materials, equivalents are the number of moles multiplied by the charge of the ion.

    168 g of Ca(NO3)2 is 2 eqv of Ca++ and 2 eqv of NO3- added to whatever amount of water you dilute it in or soil you add it to. Divide by 1000 and 0.168 g per 100 g of soil would add 2 meqv each of calcium and nitrate to 100 g of soil.

    I believe following this method and avoiding ppm will simplify the calculations for you. Work out the molar masses of the other fertilizers in a similar manner. If you are forced to deal with ppms, can you determine the basis on which they are computed? The calculations to undo that depend on the basis.

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    • #3
      Originally posted by JohnS View Post

      Ppm is an ambiguous term unless the basis of computing it is given; it can be based on mass, volume, or molar ratios or even mass/volume. In this case, is probably only adds confusion and I recommend going direct from mass to amount of substance (moles) then to equivalents. Calcium Nitrate is Ca(NO3)2. It has a molar mass of 164.088 grams per mole, computed by adding up the atomic weights of the constituent atoms. There is negligible error in using atomic weights rounded to integers and calling it 164 g/mol.

      When dissolved and ionized, that is 40 g (1 mol) of Ca++ ions, which will be 2 equivalents of calcium because of the double charge and 124 g ( 2 mol) of NO3- ions, or two equivalents of NO3. Note for ionized materials, equivalents are the number of moles multiplied by the charge of the ion.

      168 g of Ca(NO3)2 is 2 eqv of Ca++ and 2 eqv of NO3- added to whatever amount of water you dilute it in or soil you add it to. Divide by 1000 and 0.168 g per 100 g of soil would add 2 meqv each of calcium and nitrate to 100 g of soil.

      I believe following this method and avoiding ppm will simplify the calculations for you. Work out the molar masses of the other fertilizers in a similar manner. If you are forced to deal with ppms, can you determine the basis on which they are computed? The calculations to undo that depend on the basis.
      Thank you for your reply JohnS! I believe the PPM I am working with is mass/volume. How I learned to use Ionic salts for fertilizer in terms of PPM was to use the formula mass of solute/mass of solvent x 10^6. I'm going to use Magnesium Sulfate as a example instead of Calcium Nitrate. If I were to make a 1 liter solution, my mass of solvent would be 1000g and if I add 1 gram of MgSO4, I would get 1000 PPM total. The MgSO4 has a Mg % of 9.8 and Sulfur has a S % of 13. I would then multiply the 1000 PPM by each element % to find out the actual elemental PPM in the solution. This would give me about 130PPM S and about 98 PPM Mg in this 1 liter solution.

      I chose MgSO4 because the Calcium Nitrate that I use isn't just calcium nitrate, I believe it is Calcium Ammonium Nitrate and has a molar mass of about 1000 or so, so I think MgSO4 is a better example for this understanding. So if I use 1 gram MgSO4 (it is the hydrated version with a Molar mass of ~246) and convert that to moles I get 0.004065 moles. I'm not to sure how to convert this to equivalents and then milliequivalents? When Ionized Mg has a double + charge and SO4 has a double - charge, do I just multiple the moles by the charge and that would be my eqv? Thanks again.

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      • #4
        Yes. The relationship of equivalents to moles is based on a mole of electrons exchanged, either donated or accepted. For many elements (and ions), there is only one ionized form, for example Mg++. There are exceptions such as iron which can form both ferrous and ferric compounds and you have to look at the particular reaction which is considered "defining." In those cases, I would have to refresh my memory from a chemistry text. So you would have 0.00813 eqv or 8.13 meqv of Mg++ and SO4--.

        I prefer to replace ppm with units six orders of magnitude apart so I can use unit cancellation, g/g for mass/mass, L/L for volume/volume, mg/L for mass/volume, etc. Your calculations are correct, but you might find this a useful way to look at it. 1000 ppm MgSO4 is 1000 mg/L. Using the molar mass of 246 g/mol
        1000 mg/L x 1 mol/246 g = 4.065 mmol/L MgSO4. Since it ionizes is is 4.065 mmol/L each of Mg++ and SO4--. Due to the double charge, double that for milliequivalents. There is no nbeed to work through the elemental ppms.

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        • #5
          Originally posted by JohnS View Post
          Yes. The relationship of equivalents to moles is based on a mole of electrons exchanged, either donated or accepted. For many elements (and ions), there is only one ionized form, for example Mg++. There are exceptions such as iron which can form both ferrous and ferric compounds and you have to look at the particular reaction which is considered "defining." In those cases, I would have to refresh my memory from a chemistry text. So you would have 0.00813 eqv or 8.13 meqv of Mg++ and SO4--.

          I prefer to replace ppm with units six orders of magnitude apart so I can use unit cancellation, g/g for mass/mass, L/L for volume/volume, mg/L for mass/volume, etc. Your calculations are correct, but you might find this a useful way to look at it. 1000 ppm MgSO4 is 1000 mg/L. Using the molar mass of 246 g/mol
          1000 mg/L x 1 mol/246 g = 4.065 mmol/L MgSO4. Since it ionizes is is 4.065 mmol/L each of Mg++ and SO4--. Due to the double charge, double that for milliequivalents. There is no nbeed to work through the elemental ppms.
          Hello JohnS thanks for your help. I am still trying to get used to the conversion of these units (mg/L, mmol, meq etc) but I think I got some of the calculations down. I started looking through my notes to see what was the last nutrient ratios I used for my macro nutrients. These are Calcium, Nitrogen, Magnesium, Potassium, Phosphorous and Sulfur.

          Starting with MgSO4, I would use 1.2g per gallon of water. My calculation starts with 1200mg/3.785L = 317 PPM = 317 mg/L x 1/264g = 1.201mmol/L = 1.201mEq x 2 because of the double charges on Sulfate and Mg which equals 2.402mEq Mg and SO4

          Monopotassium Phosphate - KH2PO4, using 1.000g per gallon of water. 1000mg/3.785L = 264 PPM = 264mg/1L = 1mol/136g = 1.941mmol/L = 1.941 mEq K+ and 1.941 mEq H2PO4-. I believe the Potassium mEq is correct but not sure about the counter anion, dihydrogen phosphate anion. Is the H2PO4- also 1.941 mEq because its overall charge is -?

          Potassium Sulfate - K2SO4, using 0.4g per gallon of water. 400mg/3.785L = 105.7 PPM = 105.7mg/L = 1mol/174.2g = 0.6066mmol/L = 0.6066mEq K+ and 1.213mEq SO4^2- because of the double negative charge.

          I think I got those right but the when I get to the Calcium Ammonium Nitrate (CAN) ionic salt it's kind of tricky. The CAN I use has the formula, 5Ca(NO3)2 NH4NO3 10H2O which has a Molar Mass of 1080.71g/mol. I use 3.000g per gallon of water. 3000mg/3.785L = 792PPM = 792mg/1L x 1mol/1080g = 0.733mmol/L = 0.733mEq. Ca^2+ is divalent so 0.733x2 = 1.466mEq Ca ions.
          Not sure about this one because its not a two part ionic salt like MgSO4.

          I appreciate all the help JohnS. Would you be able to double check my calculations to confirm if I am on the right track? Thank you!

          Comment


          • #6
            Yes. Three of the four are "tricky," and make me wonder if equivalents are the right approach for gardening. Equivalents were mostly used to equate power of acids and bases to neutralize each other and form salts. For gardening, you may be more concerned with available elemental nutrients. The MgSO4 is correct.

            On the K2SO4, you have 0.6066 mmol/L of K2SO4, but there are two atoms of potassium in the formula, hence 1.213 mmol/L of elemental (or singly-charged ion) of potassium, so 1.213 meqv/L.

            The last two are even trickier and involve what's important in gardening vs acids and bases neutralizing each other. I am not entirely sure of the correct answer; take with a grain of salt if you'll pardon the pun. On KH2PO4, it appears you are correct per Wikipedia. On the 5Ca(NO3)2 NH4NO3 10H2O, I am unsure how to handle it as both ammonia and nitrate are used as fertilizers and offer "available nitrogen" for plant growth. Per Wikipedia, it is a mixture of calcium nitrate and ammonium nitrate that crystallizes as a "mixed crystal". Using the chemical formula, it appears that 1 mole of it ionizes to 10 mol Ca++ (20 eqv due to double charge), 1 mol NH4+, and 11 mol NO3-. I am unsure how to consider the "available nitrogen" from the combination of ammonia and nitrate.

            Comment


            • #7
              Originally posted by JohnS View Post
              Yes. Three of the four are "tricky," and make me wonder if equivalents are the right approach for gardening. Equivalents were mostly used to equate power of acids and bases to neutralize each other and form salts. For gardening, you may be more concerned with available elemental nutrients. The MgSO4 is correct.

              On the K2SO4, you have 0.6066 mmol/L of K2SO4, but there are two atoms of potassium in the formula, hence 1.213 mmol/L of elemental (or singly-charged ion) of potassium, so 1.213 meqv/L.

              The last two are even trickier and involve what's important in gardening vs acids and bases neutralizing each other. I am not entirely sure of the correct answer; take with a grain of salt if you'll pardon the pun. On KH2PO4, it appears you are correct per Wikipedia. On the 5Ca(NO3)2 NH4NO3 10H2O, I am unsure how to handle it as both ammonia and nitrate are used as fertilizers and offer "available nitrogen" for plant growth. Per Wikipedia, it is a mixture of calcium nitrate and ammonium nitrate that crystallizes as a "mixed crystal". Using the chemical formula, it appears that 1 mole of it ionizes to 10 mol Ca++ (20 eqv due to double charge), 1 mol NH4+, and 11 mol NO3-. I am unsure how to consider the "available nitrogen" from the combination of ammonia and nitrate.
              HI JohnS. What honestly raised this question for me was when I was discussing different element nutrient ratios for multiple substrates. Another member of a forum stated that the ratio I was formulating was not balanced when converted to meqv and was cation heavy. Unfortunately the member didn't elaborate any further and left me stumped. I stated the following values in PPM
              Calcium : 150
              Magnesium : 30
              Potassium : 120
              Phosphorous : 60
              Nitrogen : 120
              Sodium : 10
              Sulfur : 75

              This is what led me to ask the question of this post. What does it mean to be cation heavy in milliequivalents? Understanding this term and conversion will help me tremendously with agriculture as knowing how to push a balanced fertilizer based on a substrates cation exchange capacity. Thank you!

              Comment


              • #8
                The total charge must balance electrically. However, I do not see how he could have determined that from the data you provided. Just as an example, available nitrogen can come from both NH4+ and NO3-, both cation and anion. He may have made assumptions about the actual form of the above nutrients and calculated an imbalance. Also the salts used may include other elements not significant to plant growth in either the cations or anions and not be reflected in the list. However, when fully accounted for, if there really is a charge imbalance, then there was likely an error in calculations. NOTE: I have not worked through the numbers above to estimate an imbalance. I am just claiming that I have a shot at fudging the numbers with two sources of available nitrogen.

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                • #9
                  Please
                  Let know this

                  Converting 5.1 mmol/l Cd to mg/kg

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                  • #10
                    Originally posted by Younger Dryas View Post
                    Please
                    Let know this

                    Converting 5.1 mmol/l Cd to mg/kg
                    First step: The molar mass of Cd is 112.4 g/mol so 5.1 mmol/L x 112.4 g/mol = 573 mg/L.

                    The second step depends on what the denominator represents. If it is simply a solution of a cadmium salt in a solvent, you need the density of the mixture to convert from volume of the mixture to mass of the mixture. On the other hand, if this is a dried soil sample that has been mixed with water to extract solubles, you need the ratio of liters of water to kilograms of soil. Since you didn't provide the context, I can't do that step.

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