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  • Sodium Chlorite

    What formula what one use to convert a 75% purity Sodium Chlorite powder to 25% solution?

    Thanks.

  • #2
    Originally posted by Alex View Post
    What formula what one use to convert a 75% purity Sodium Chlorite powder to 25% solution?

    Thanks.
    The first question is what kind of percent? Is the basis mass/mass or mass over volume?

    Also, it is fairly nasty stuff. One of the makers has a pretty useful handbook. Please review the handling precautions.
    https://www.oxy.com/OurBusinesses/Ch...20Handbook.pdf

    Page 19 covers your question assuming mass/mass percentage (note they assumed 80% purity)
    If you want mass/volume, then you need 250 g of sodium chlorite per liter of solution, but if 75% pure, you need 250 g/0.75 = 333.3 g/L

    Comment


    • #3
      Originally posted by JohnS View Post

      The first question is what kind of percent? Is the basis mass/mass or mass over volume?

      Also, it is fairly nasty stuff. One of the makers has a pretty useful handbook. Please review the handling precautions.
      https://www.oxy.com/OurBusinesses/Ch...20Handbook.pdf

      Page 19 covers your question assuming mass/mass percentage (note they assumed 80% purity)
      If you want mass/volume, then you need 250 g of sodium chlorite per liter of solution, but if 75% pure, you need 250 g/0.75 = 333.3 g/L


      Thanks for your expert response and the link.


      Answer : 1. Mass/ volume percentage.


      A few questions:

      1. What formula did you use to come up with 250 g of Sodium Chlorite, if it's 75% purity to come up with a final solution of 25% in 1000 mL?

      2. I am being offered only 70, 73 and 75% purity, not 80%.

      3. What formula should I use to come up with such purties, lesser in grams and volume (e.g 70% purity, ?? grams in 125 mL) for a final as a 25% net mass/ vol solution?

      For example, for 50 grams 70% purity in 125 mL.......??

      4. Aim is to prepare a final 25% solution of sodium chlorite and use 4% HCL both in 125 mL to create 0.3% (ppm) chlorine dioxide for my aquarium.

      Hoping to hear from you,

      Thanks again!

      Comment


      • #4
        Response from a some company........

        80% tech grade sodium chlorite flakes (made in U.S.A.). has 6.4% inert stabilizers.

        You can never get 90-100% sodium chlorite solution. We do not offer a 25% solution. 22.4% is the standard. That is all we offer.

        Comment


        • #5
          Is this the correct formula as per pg 19 of the PDF?


          1.19 grams/ mL.


          125 mL X 1.19 = 148.75 mL


          148.75 mL X .25 / .73 = 50.94 grams.


          125 mL X 1.19 = 148.75 grams


          148.75 mL X .25 = 37.18 grams


          37.18 /.73 = 50.94 grams



          To make a 25% sodium chlorite sol, purity 73%, in 125 mL need to dissolve 50.94 grams??

          Thanks again!


          Comment


          • #6
            Originally posted by Alex View Post
            Is this the correct formula as per pg 19 of the PDF?


            1.19 grams/ mL.


            125 mL X 1.19 g/mL= 148.75 g


            148.75 g X .25 / .73 = 50.94 grams.


            125 mL X 1.19 g/mL= 148.75 grams


            148.75 g X .25 = 37.18 grams


            37.18 /.73 = 50.94 grams



            To make a 25% sodium chlorite sol, purity 73%, in 125 mL need to dissolve 50.94 grams??

            Thanks again!

            Yes, basically. I made a couple of edits in what you wrote to correct units, but the numbers are correct.

            Comment


            • #7
              Thanks highly appreciated!

              Comment


              • #8
                Corrections?

                10.27 lbs/gal = 1.23 grams/mL and not 1.19.



                125 mL x 1.23 = 153.75 grams.

                153.75 grams x .25% = 38.75 grams.

                38.75/.80= 48.04 grams.

                153.75 - 48.04= 105.71 mL of water.....Missed this step from page 19.

                105.71 /.99= 105 mL water required to make a 25% solution in 125 mL of 80% sodium Chlorite powder.

                How much of the powder in grams should I add to 105 mL?





                OR do it simply....

                50 grams/ 0.388 = 128 mL water ??
                Last edited by Alex; 07-02-2020, 03:27 AM.

                Comment


                • #9
                  Example: How much OxyChem Technical Sodium
                  Chlorite dry product and water is required to prepare
                  50 gallons of 25% sodium chlorite solution?
                  50 gals. X 10.27 lbs./gal. = 513.5 lbs.
                  513.5 lbs. X 0.25 = 128.38 lbs. of sodium
                  chlorite required to prepare 50 gallons of 25% active
                  sodium chlorite solution.
                  128.38 lbs. NaClO 2 / 0.8 = 160.48 lbs. of OxyChem
                  Technical Sodium Chlorite dry product required to
                  prepare the desired volume of 25% sodium chlorite
                  solution.
                  513.5 lbs. - 160.48 lbs. = 353.02 lbs. of water is
                  required to prepare 50 gallons of 25% sodium
                  chlorite solution.
                  353.02 lbs / 8.34 = 42.33 gallons of water is required
                  to prepare the 50 gallons of 25% sodium chloriteExample: How much OxyChem Technical Sodium
                  Chlorite dry product and water is required to prepare
                  50 gallons of 25% sodium chlorite solution?
                  50 gals. X 10.27 lbs./gal. = 513.5 lbs.
                  513.5 lbs. X 0.25 = 128.38 lbs. of sodium
                  chlorite required to prepare 50 gallons of 25% active
                  sodium chlorite solution.
                  128.38 lbs. NaClO 2 / 0.8 = 160.48 lbs. of OxyChem
                  Technical Sodium Chlorite dry product required to
                  prepare the desired volume of 25% sodium chlorite
                  solution.
                  513.5 lbs. - 160.48 lbs. = 353.02 lbs. of water is
                  required to prepare 50 gallons of 25% sodium
                  chlorite solution.
                  353.02 lbs / 8.34 = 42.33 gallons of water is required
                  to prepare the 50 gallons of 25% sodium chlorite

                  Comment


                  • #10
                    Alex, I am sorry. I have been a little confused by this thread, and just figured it out. Because I click on the "latest post" option, I never noticed three posts in quick succession, and saw only post 5, not #3, and #4, so I was missing what you were really asking.

                    There are fundamental differences in mass/volume and mass/mass calculations. The pdf I posted focused on mass/mass concentrations and we went off on that track when it is not what you want.

                    Mass/volume: The units aren't "real" percents because numerator and denominator are not the same units, but they are used a lot. 1% is defined as 10 g in 1 L of solution (it essentially assumes the 1 L weighs 1 kg regardless of whether it does or doesn't). So 25% is 250 g of sodium chlorite per liter of solution. Note that it is per liter of solution; if you add 1 L of water, the powder occupies some volume, so you will get more than a liter, and it will be the wrong concentration.

                    Since it is not pure, you have to divide by the purity to get the correct mass of powder to be 250 g of sodium chlorite. 250 g/ 0.75 = 333.3 g of the impure powder

                    Whatever amount you wish to make up should be in these proportions. If you only want 100 mL, use 33.3 g of powder and dilute to 100 mL.

                    Add the solute to a beaker, add enough water to reach about 80% (not critical, but less than the full volume) of the final volume, stir until fully dissolved, then add remaining water, to the full final volume.

                    Comment


                    • #11
                      Let me try to put it simply.

                      I have 500 grams of 80% Sodium Chlorite powder.

                      How much of it would I require to make a 25% solution in 100 mL.

                      Care to share the formula.

                      Thanks.



                      500 grams/. 80%= 625 grams.

                      How do you come up with the amount for 100 mL of 25%??

                      For 100 mL of 80%, 62.5 grams be dissolved in 100mL to get a net 25% solution.... Correct??



                      What does this formula imply

                      Amount of Sodium Chlorite 80% in grams/ Volume of water in mL = . 388

                      To calculate the amount of water for 62.5 grams

                      62.5 * 0.388= 24.25 mL of water and not 100 mL?

                      What exactly is 0.388?

                      Why is there an ambiguity in the 2.... 62.5 grams of 80% as per your formula for 100 mL.

                      And 24.25 mL of water in the other?

                      Appreciate if you'd clarify... Which to go with.

                      Again require a 25% net solution in 100 mL.... What amount of 500 grams of 80% Sodium Chlorite would be needed?

                      Thanks.
                      Last edited by Alex; 07-02-2020, 07:31 PM.

                      Comment


                      • #12
                        Originally posted by JohnS View Post
                        Alex, I am sorry. I have been a little confused by this thread, and just figured it out. Because I click on the "latest post" option, I never noticed three posts in quick succession, and saw only post 5, not #3, and #4, so I was missing what you were really asking.

                        There are fundamental differences in mass/volume and mass/mass calculations. The pdf I posted focused on mass/mass concentrations and we went off on that track when it is not what you want.

                        Mass/volume: The units aren't "real" percents because numerator and denominator are not the same units, but they are used a lot. 1% is defined as 10 g in 1 L of solution (it essentially assumes the 1 L weighs 1 kg regardless of whether it does or doesn't). So 25% is 250 g of sodium chlorite per liter of solution. Note that it is per liter of solution; if you add 1 L of water, the powder occupies some volume, so you will get more than a liter, and it will be the wrong concentration.

                        Since it is not pure, you have to divide by the purity to get the correct mass of powder to be 250 g of sodium chlorite. 250 g/ 0.75 = 333.3 g of the impure powder

                        Whatever amount you wish to make up should be in these proportions. If you only want 100 mL, use 33.3 g of powder and dilute to 100 mL.

                        Add the solute to a beaker, add enough water to reach about 80% (not critical, but less than the full volume) of the final volume, stir until fully dissolved, then add remaining water, to the full final volume.
                        Whatever amount you wish to make up should be in these proportions. If you only want 100 mL, use 33.3 g of powder and dilute to 100 mL.

                        Dissolving 33.3(75%)/ 62.5 (80%) g in 100 mL give a net 25% solution?

                        Thanks.
                        Last edited by Alex; 07-03-2020, 12:04 AM.

                        Comment


                        • #13
                          For 75% purity, divide 25 g by 0.75, for 80% purity, divide 25 g/0.8, or 31.1 g. In either case, dilute to final volume of 100 mL. That will be a bit less than 100 mL of water but mass/volume is always net grams of solute divided by liters of solution.
                          Last edited by JohnS; 07-03-2020, 07:12 AM.

                          Comment


                          • #14
                            Thanks for the prompt expert clarification. Appreciate it.

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