Guest guest Posted July 4, 2010 Report Share Posted July 4, 2010 Hi Tom I am trying to establish the ratios of NaClO2 to available chlorine dioxide and free ClO2. Activated and non-activated. You Wrote: 0.56 ml of 22.4% sodium chlorite and add 0.56 ml of 10% citric acid. In 500cc This solution has 150 PPM available NaClO2 and about 15 PPM free chlorine dioxide. Is the NaClO2 amount =0.56*.224= 1.254 gm concentration = 1.254/500 grams , ~251 PPM in a total of 500 mL? If correct; when activated, Is the amount of available ClO2, 60% of 250 ppm or equal to 150 ppm ? And is the amount of free ClO2 = 6% of NaClO2 in the solution ? Do the percentages mentioned change when not activated ? Thank You for your Patience and the help You have provided. R Quote Link to comment Share on other sites More sharing options...
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