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  • dosing xml in yL how ppm i get?

    hi I would like to have some information

    dosing 5ml of IRON in 180L of water to how much ppm I get?

    dosing 5ml of POTASSIUM in 180L of water to how much ppm of K I get?

    dosing 5ml of NITROGEN in 180L of water to how much ppm of N I get?

    dosing 5ml of PHOSPHORUS in 180L of water to how much ppm of P I get?

    dosing 5ml of flourish in 180L of water to how much ppm of Total Nitrogen, Available Phosphate, Soluble Potash, Calcium, Magnesium, Sulphur, Boron, Chlorine, Cobalt, Copper, Iron, Manganesem,Sodium, Zinc, Molybdenum I get?

    dosing 5g of equilibrium in 30L of water to how much ppm of Potassium, Calcium, Magnesium, Iron, Manganese I get?

    dosing 5ml of trace in 180L of water to how much ppm of Boron, Cobalt, Copper, Manganese, Molybdenum, Zinc, Rubidium, Nickel, Vanadium I get?

    thanks

  • #2
    Re: dosing xml in yL how ppm i get?

    I am going to cover this based on the recommended dosages to avoid any confusion for any one else. There is a formula that can help you do different dosages following this list.

    Flourish has 0.014 mg/L (ppm) of Nitrogen per 5 mL in 250 L
    Flourish has 0.002 mg/L (ppm) of Phosphorus per 5 mL in 250 L
    Flourish has 0.074 mg/L (ppm) of Soluble Potash per 5 mL in 250 L
    Flourish has 0.028 mg/L (ppm) of Calcium per 5 mL in 250 L
    Flourish has 0.022 mg/L (ppm) of Magnesium per 5 mL in 250 L
    Flourish has 0.055 mg/L (ppm) of Sulfur per 5 mL in 250 L
    Flourish has 0.0018 mg/L (ppm) of Boron per 5 mL in 250 L
    Flourish has 0.23 mg/L (ppm) of Chlorine per 5 mL in 250 L
    Flourish has 0.00008 mg/L (ppm) of Cobalt per 5 mL in 250 L
    Flourish has 0.00002 mg/L (ppm) of Copper per 5 mL in 250 L
    Flourish has 0.0064 mg/L (ppm) of Iron per 5 mL in 250 L
    Flourish has 0.00236 mg/L (ppm) of Manganese per 5 mL in 250 L
    Flourish has 0.00018 mg/L (ppm) of Molybdenum per 5 mL in 250 L
    Flourish has 0.026 mg/L (ppm) of Sodium per 5 mL in 250 L
    Flourish has 0.00014 mg/L (ppm) of Zinc per 5 mL in 250 L

    Equilibrium has 39 mg/L (ppm) of Potassium per 16 g in 80 L
    Equilibrium has 16.12 mg/L (ppm) of Calcium per 16 g in 80 L
    Equilibrium has 4.82 mg/L (ppm) of Magnesium per 16 g in 80 L
    Equilibrium has 0.22 mg/L (ppm) of Iron per 16 g in 80 L
    Equilibrium has 0.12 mg/L (ppm) of Manganese per 16 g in 80 L

    Flourish Nitrogen has 0.0912 mg/L (ppm) of Nitrogen per 2.5 mL in 160 L
    Flourish Nitrogen has 0.1216 mg/L (ppm) of Potassium per 2.5 mL in 160 L

    Flourish Potassium has 2 mg/L (ppm) of Potassium per 5 mL in 120 L

    Flourish Phosphorus has 0.05 mg/L (ppm) of Phosphorus (0.15 phosphate) per 2.5 mL in 80 L
    Flourish Phosphorus has 0.0795 mg/L (ppm) of Potassium per 2.5 mL in 80 L

    Flourish Iron has 0.2 mg/L (ppm) of Iron per 5 mL in 200 L

    Flourish Trace has 0.00175 mg/L (ppm) of Boron per 5 mL in 80 L
    Flourish Trace has 0.0001875 mg/L (ppm) of Cobalt per 5 mL in 80 L
    Flourish Trace has 0.002 mg/L (ppm) of Copper per 5 mL in 80 L
    Flourish Trace has 0.0053125 mg/L (ppm) of Manganese per 5 mL in 80 L
    Flourish Trace has 0.0001875 mg/L (ppm) of Molybdenum per 5 mL in 80 L
    Flourish Trace has 0.0105625 mg/L (ppm) of Zinc per 5 mL in 80 L
    Flourish Trace has 0.000005 mg/L (ppm) of Rubidium per 5 mL in 80 L
    Flourish Trace has 0.0000018 mg/L (ppm) of Nickel per 5 mL in 80 L
    Flourish Trace has 0.0000012 mg/L (ppm) of Vanadium per 5 mL in 80 L

    1000vP ÷V = C
    where v = volume being dosed in mL
    P = % of that element in our product
    V=volume of water being added to
    C=final concentration of element after dosing
    You just need to pull "P" from each of our labels for each element of interest and plug it into the equation to solve for "C"

    Comment


    • #3
      Re: dosing xml in yL how ppm i get?

      I don't know if I have understood well.... but if I want to have 5ppm of N in 180L i must dose 154.2ml of nitrogen

      is it correct?

      Comment


      • #4
        Re: dosing xml in yL how ppm i get?

        That would be correct if you wanted your Nitrate to read 5 mg/L. I would not suggest keeping it that high with our product. When relying on your fish waste to provide all of the Nitrogen and there is no ammonium or Nitrite present keeping your nitrates between 5-10 is great. When using our product you want to keep in mind that it is not just Nitrate. It has some Nitrate and some ammonium-like Nitrogen sources. With our product I would suggest just maintaining a trace of Nitrate. Add enough just to have a slight reading on your test kit.

        Comment


        • #5
          Hi,
          The formula doesn't work for me.

          For example you said:
          "Flourish has 0.014 mg/L (ppm) of Nitrogen per 5 mL in 250 L"

          And:
          "1000vP ÷V = C"

          The label of Flourish said:
          "Total Nitrogen (N)............ 0.07%"
          "5 ml for each 250 L"

          In the formula: 1000 x 5 (ml) x 0.07 (%) / 250 (L) = 1,4. (But you said 0,014)

          What am I doing wrong?

          Comment


          • #6
            5 mL Flourish at 0.007% total soluble N (as dry mass; mg) in 250L = 0.14 mg/L. I think you got the % wrong by an order of magnitude (0.07 instead of 0.007)

            1000v: 1000 * 5 = 5000 mL
            1000v * P: 5000 * 0.007 [mg] = 35 mg total soluble N
            (1000v * P) / V): 35 mg / 250 L = 0.14 mg/L total soluble N

            I hope this helps.

            Regards,
            PE

            Comment


            • #7
              But the label of flourish says:
              "Total Nitrogen (N)...... 0,07%
              0,07% Water soluble Nitrogen" (Attach image)

              Why do you use 0,007 instead 0,07%?

              Comment


              • #8
                That's odd, the information I referenced has 0.007%. Thank you for bringing this to my attention; I'll forward it to the appropriate people. Going off the label you have, 1.4% would be correct.

                Regards,
                PE

                Comment


                • #9
                  But that is very serious.
                  You told people for 15 years that the concentration was 100 times less than the real!!!
                  Causing this overdose and many algae !!

                  Comment


                  • #10
                    Good day Juan,

                    The label has the correct concentration; 0.07% total soluble N. I have brought the changes that need to be made to our website to the attention of the appropriate parties for revision. When in doubt of a concentration or amount of an ingredient in our products, please refer to the label on the bottle as that is what we use to register products.

                    Please keep in mind that Flourish is not a significant source of nitrogen and we do not recommend using it to supply N. Flourish Nitrogen or Aquavitro synthesis are the products you will want to use to increase nitrogen in your aquariums. We always recommend using our products as directed on the label.

                    Regards,
                    PE
                    Last edited by Tech Support PE; 05-11-2020, 14:09.

                    Comment


                    • #11
                      That´s not the problem. The problem is the formula is wrong.

                      I think I have it!!!

                      A 0.07% solution indicates that you have 0.07 grams in 100 ml, which is equivalent to 70 mg in 100 ml.
                      This is solved with the formula: 10 * v * P = c (mg / ml)
                      v = volume that is dosed in ml
                      P =% of that element in the product
                      c= concentration of element en v

                      We divide this into the volume to be added in ml and it remains in mg / ml.
                      We need to multiply the result by 1000 to give it in mg / L, which would be the same as
                      ppm.

                      In the example of the N in Flourish it would be: (10 * 5 ml * 0.07% / 250,000 ml) * 1000 = 0.014 mg / L (ppm) of N in 250L.

                      The correct formula is: (10*v*P/V)*1000=C
                      v = volume that is dosed in ml
                      P =% of that element in the product
                      V=volume of water being added to (in ml, multiply liters per 1000)

                      C=final concentration of element after dosing
                      You just need to pull "P" from each of our labels for each element of interest and plug it into the equation to solve for "C"


                      His examples are correct, his formula is not correct.
                      Last edited by Juan Restrepo; 05-11-2020, 14:50.

                      Comment


                      • #12
                        Hello Juan,

                        I checked with our lab director and she confirmed the original formula is correct as long as the percent is used as a decimal, i.e. 0.07% as 0.07. In this case, the 5 mL dose will net 1.4 mg/L total soluble N in 250 L. We don't recommend modifying our equations as we've carefully calibrated them to make sure they're correct.

                        Regards,
                        PE

                        Comment


                        • #13
                          0,07 grs in 100ml (A solution to 0,07%) is same to 3,5 mg in 5 ml.
                          If you dilute this in 250 It you have 0,014 mg/L.
                          I don't understand your 1.4 result.

                          Comment


                          • #14
                            1000vP / V = C where 1000 is a non-unit constant to get mL into L so the units match with total volume, v = 5mL, P = 0.07%, V = 250L, C = concentration of X


                            Step 1: 1000 * 5mL = 5000mL (equivalent to 5L)

                            Step 2: Covert 0.07% into decimal form equalling 0.0007

                            Step 3: Multiply 5000 by 0.0007 = 3.5 mg total soluble N

                            Step 4: Divide 3.5mg by 250L = 0.014 mg/L total soluble N when dosing 5mL into 250L.

                            Total soluble N from a 5mL dose into 250L = 0.014 mg/L

                            I had to go back and recalculate it as 1.4 mg/L didn't make sense and seemed like a much higher concentration than it should have been for a 5mL dose into 250L. I've got a decent chemistry background, but I misunderstood our lab director and continued with the 0.07 as the numerical value for P rather than properly converting it into 0.0007.

                            Regards,
                            PE
                            Last edited by Tech Support PE; 05-18-2020, 16:36.

                            Comment


                            • #15
                              We agree right now.

                              Just change in step 3: 0.007 for 0.0007.

                              The correct formula is: 1000*v*(P/100)/V=C

                              Comment

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