Guest guest Posted July 5, 2010 Report Share Posted July 5, 2010 Hello , The concentration of the NaClO2 is directly related to the available chlorine dioxide. For example, 5% NaClO2 has 50000 PPM available chlorine dioxide (5% times 1 million gives us 50000). The amount of free chlorine dioxide produced upon activation is related to how far the PH of the solution is reduced. There is no easy formula for this, so it is usually derived through experimentation. In general, when you use 10% citric acid in a 1:1 ratio with NaClO2, you end up with about 10% free chlorine dioxide. All of this depends upon the method of activation, so there are some variations. For example, the food industry is interested in chlorous acid and views chlorine dioxide as a waste product. They use a different method of activation that results in only about 3 PPM of the available chlorine dioxide being released as free chlorine dioxide. Tom > 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...
Recommended Posts
Join the conversation
You are posting as a guest. If you have an account, sign in now to post with your account.
Note: Your post will require moderator approval before it will be visible.