Proper stoichiometric mixing of 22.4% MMS solution with 50% citric acid solution by Stephan2 ..... Miracle Mineral Solution MMS Forum
Date: 11/17/2009 8:48:27 PM ( 12 years ago ago)
PLEASE NOTE THAT THIS POST IS INCORRECT! I assumed that the precentages of the MMS and acid solutions were by volume but they are by WEIGHT! I have therefore posted a table with (hopefully) correct number of activator drops for a given number of MMS drops. I regret that I made a mistake and I hope you accept my appologies. Thanks also to SilverFox who put me on the right track. I'll leave the material still here for review and maybe in a month or so I will delete it. So please don't follow the instructions below and skip to my post on February 1st, 2010, in this thread.
Accurate Preparation of Chlorine Dioxide Solution
Presented is an analysis of the amount of chemical substances for the MMS solution and citric acid needed to get a stoichiometric accurate mix to produce chlorine dioxide. Wikipedia has for the definition of stoichiometry:
Stoichiometry (sometimes called reaction stoichiometry to distinguish it from composition stoichiometry) is the calculation of quantitative (measurable) relationships of the reactants and products in a balanced chemical reaction (chemicals). It can be used to calculate quantities such as the amount of products that can be produced with the given reactants and percent yield. First we look at the molecular masses of sodium chlorite and citric acid (looked up at Wikipedia). A molar mass Mmol is defined as the mass of one mole of substance in units of gm/mol; one mole has 6.022*1023 of molecules (or atoms), which is called the Avogadro constant NA. More precisely, NA is defined as the number of atoms in exactly 12 grams of the pure isotope 12C. The molar mass takes into account the fractions of natural isotopes found in nature.
Sodium Chlorite: NaClO2 single valence
Molar Mass: MNaClO2 = (23+35.5+32) g/mol , MNaClO2 = 90.5 g/mol
Citric Acid: C6H8O7 triple valence
MCitricAcid = 210.14 g/mol
Ratio: 210.14/(3*90.5) = 0.774
The molar mass is for the monohydrate form, which means each citric acid molecule has a single molecule of water added to it.
Sodium Citrate: C6H5Na3O7 triple valence
MSodiumCitrate = 258.07 g/mol
3 NaClO2 + C6H8O7 --> C6H5Na3O7 + 3 ClO2
Begin Edit 11/21/09: Silverfox pointed out the formation of chlorous acid, which can form when chlorine dioxide captures an electron to form a negative chlorite ion:
ClO2 + e- --> ClO2-, which then forms chlorous acid:
H+ + ClO2- --> HClO2
Chlorous acid is unstable (Wikipedia) and will form hypochlorous acid and chloric acid:
2HClO2(aq) → HClO(aq) + HClO3(aq)
When ingested, the hypchlorous acid can react with the stomach hydrochloric acid (Wikipedia) and form chlorine gas (Wikipedia shows it the other way around, however the chemical reaction can reverse):
Cl2 + H2O HClO + HCl
or, hypochloric acid can form chloric acid (Wikipedia):
3HClO → HClO3 + 2 HCl
The chemical reaction chain of activated MMS is not trivial and one has to keep in mind that other, less disirable byproducts may be generated when ingesting activated MMS. (End edit 11/21/09)
By molar mass, one needs therefore three times as much sodium chlorite than citric acid. That means by actual weight, for each gram of sodium chlorite one would need about 0.77 grams of citric acid.
The density of sodium chlorite is 2.5 g/cm3 and the density of citric acid is 1.665 g/cm3.
By volume for each cm3 of sodium chlorite powder one would need 1.156 cm3 of
citric acid powder. So almost equal portions for each substance by volume.
MMS is made as a 22.4% solution by volume in water, the concentrated citric acid is made as
a 50% solution by volume in water. Therefore this is a bit more citric acid than needed. The
proper stoichimetric amount would be two drops of MMS solution for each drop of 50% citric
acid. Since the MMS solution is 22.4% and not 25%, there is a little more citric acid by volume.
The ratio 25%/22.4% is 1.12 vs. 1.156 for the optimum.
Within this accuracy two drops of MMS for each drop of concentrated (50%) citric acid should
Note that for the 10% citric acid solution one would need to mix two drops of MMS solution with 5 drops of 10% citric acid solution!
According to Wikipedia, natural lemon juice contains 47 g/liter or 0.047 g/cm3 of citric acid. Bottled lemon juice is more concentrated. I think it is twice as concentrated as in the juice of a lemon, that is what I found on the web, so it should have about 0.1 g/cm3 citric acid.
I think that the recommendations of 5 times as much bottled lemon juice for each drop of MMS solution is on the low side. To calculate, we take the ratio of the concentration of MMS with that of the bottled lemon juice and multiply the result by the stoichiometric weight ratio of 1.3:
(0.224 * 2.5)/0.1 = 7.28
You need 7-8 drops of bottled lemon juice per drop of MMS solution.
Quick check for the 50% citric acid solution:
MMS is 22.4% in water, so the concentration is 0.224*2.5g/cm3 = 0.56 g/cm3.
50% citric acid has 0.5*1.665 g/cm3 = 0.833 g/cm3.
The weight ratio of two parts of 22.4% MMS to one part of 50% citric acid is therefore:
2*0.56 g/0.833 g = 1.345
The ideal weight ratio was 1/0.774 = 1.3 - close enough!
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