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  • H2O2 oxygen content confusion!

    According to photo metric measurements, 1 drop of activated Sodium Chlorite ( 0.05 mL) produces about 0.4 mg ClO₂ in a solution; that is the equivalent of about 0.2 mg of H202.

    If a 3% solution is used, this 0.2 mg of H202 would be contained in 0.007 mL, i.e. about 1/7th of a drop (3% x 0.2 mg / 100 mL = approx .0006 mL). That's fine.

    But than who does this calculation is done(though similar to above), am failing to understand......


    The calculation is as follows:


    100 mL 3% H202 solution = 3000 milligrams H202

    x ml = 0.2 milligrams H202 ??

    x – 0.007 ml ??



    Thanks.

    Last edited by Alex; 09-21-2020, 08:48 PM.

  • #2
    Originally posted by Alex View Post
    According to photo metric measurements, 1 drop of activated Sodium Chlorite ( 0.05 mL) produces about 0.4 mg ClO₂ in a solution; that is the equivalent of about 0.2 mg of H202.

    If a 3% solution is used, this 0.2 mg of H202 would be contained in 0.007 mL, i.e. about 1/7th of a drop (3% x 0.2 mg / 100 mL = approx .0006 mL). That's fine.

    But than who does this calculation is done(though similar to above), am failing to understand......


    The calculation is as follows:


    100 mL 3% H202 solution = 3000 milligrams H202

    x ml = 0.2 milligrams H202 ??

    x – 0.007 ml ??



    Thanks.
    If I accept that the drop of sodium chlorite solution is equivalent to 0.2 mg H2O2 in 0.05 mL, that is an equivalent H2O2 concentration pf 0.2 mg/0.05 mL = 4 mg/mL or 0.4g/100 mL, aka 0.4% (w/v).

    That is considerably weaker than your 3% H2O2 solution( 3g/100 mL), so you expect a much smaller volume (2/15) of 3% H2O2 to contain 0.2 mg H2O2,

    0.0002 g x 100 mL/3 g = 0.00667 mL or 6.67 L

    Comment


    • #3
      Originally posted by JohnS View Post

      If I accept that the drop of sodium chlorite solution is equivalent to 0.2 mg H2O2 in 0.05 mL, that is an equivalent H2O2 concentration pf 0.2 mg/0.05 mL = 4 mg/mL or 0.4g/100 mL, aka 0.4% (w/v).

      That is considerably weaker than your 3% H2O2 solution( 3g/100 mL), so you expect a much smaller volume (2/15) of 3% H2O2 to contain 0.2 mg H2O2,

      0.0002 g x 100 mL/3 g = 0.00667 mL or 6.67 L


      Thanks for your expert reply.

      1. 100 mL 3% H202 solution = 3000 milligrams H202.....Fine 100 mL would contain 3 grams.

      2. x ml = 0.2 milligrams H202 ?................How did 0.2 mg come? What was multiplied with certain mL?

      3. x = 0.007 ml ?........same how did 0.0007 mL come up?

      Comment


      • #4
        Originally posted by Alex View Post



        Thanks for your expert reply.

        1. 100 mL 3% H202 solution = 3000 milligrams H202.....Fine 100 mL would contain 3 grams.

        2. x ml = 0.2 milligrams H202 ?................How did 0.2 mg come? What was multiplied with certain mL?

        3. x = 0.007 ml ?........same how did 0.0007 mL come up?
        The 0.2 mg came directly from your "If I accept" statement. The numbers you threw out seem logical, although the 0.007 ml is more like 0,0067. I was just trying to show they make sense, based on your "givens" but I don't know where your givens came from.

        Comment


        • #5
          Originally posted by JohnS View Post

          The 0.2 mg came directly from your "If I accept" statement. The numbers you threw out seem logical, although the 0.007 ml is more like 0,0067. I was just trying to show they make sense, based on your "givens" but I don't know where your givens came from.

          True 0.2 mg came from me.
          I agree totally it should be 0.0067 vs 0.007 mL.
          My only query is how did the following formula brought it to 0.0067/ 0.007 mL?



          100 mL 3% H202 solution = 3000 milligrams H202.....Fine 100 mL would contain 3 grams.... Correct.


          x ml = 0.2 milligrams H202 ?............... What was multiplied by certain mL to come up with 0.2 mg?

          x = 0.007 ml ?........And again at step 3, what was multiplied to,0.2 mg to come up with 0.0069/0.007 mL?

          I forget the formula for that.
          Apologize for my ignorance!
          Thanks.

          Comment


          • #6
            Originally posted by Alex View Post


            True 0.2 mg came from me.
            I agree totally it should be 0.0067 vs 0.007 mL.
            My only query is how did the following formula brought it to 0.0067/ 0.007 mL?



            100 mL 3% H202 solution = 3000 milligrams H202.....Fine 100 mL would contain 3 grams.... Correct.


            x ml = 0.2 milligrams H202 ?............... What was multiplied by certain mL to come up with 0.2 mg?

            x = 0.007 ml ?........And again at step 3, what was multiplied to,0.2 mg to come up with 0.0069/0.007 mL?

            I forget the formula for that.
            Apologize for my ignorance!
            Thanks.
            Review my answer. You have to divide the 0.2 mg by 3g/100 mL. That is equivalent to multiplying by the reciprocal (100 mL/3 g). It is also easier to change the 0.2 mg to 0.0002 g so the unwanted units cancel directly without prefix manipulation. As to "why" it came about, I assumed you wanted the "drop" of 3% H2O2 equivalent to the drop of sodium chlorite.

            Comment


            • #7
              Originally posted by JohnS View Post
              Review my answer. You have to divide the 0.2 mg by 3g/100 mL. That is equivalent to multiplying by the reciprocal (100 mL/3 g). It is also easier to change the 0.2 mg to 0.0002 g so the unwanted units cancel directly without prefix manipulation. As to "why" it came about, I assumed you wanted the "drop" of 3% H2O2 equivalent to the drop of sodium chlorite.

              Thanks again for explaining.
              Best regards!

              Comment

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