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Can some kind person please show the full stochiometric equations for the reaction of MMS (NaClO2) and citric acid (C6H8O7)?
that is what is the full discription of this reaction in correct molar quantities ie with correct determination of the coefficients as described by sciencesoft.at/equation/index?
NaClO2 + C6H8O7 → ClO2 + NaCl + ?
I have search far an wide and still can't find a definative answer to this!!
As far as I can tell, citric acid has 3 acid COOH moieties and thus could donate a possible 3 Hydrogen atoms to the reaction.
ie please check the image of the citric acid molecular formula here: http://www.encyclopedia.com/topic/citric_acid.aspx
Then the molecular formular of citric acid could be rewiriten to show this as COH(CH2)2(COOH)3
TIA
Green
Thankyou for the link,
so far what I have found:
A) the dissociation of NaClO2 in water
1) 2NaClO2 + 2H2O = 2ClO2 + 2NaOH + H2 from en.wikipedia.org/wiki/Chlorine_dioxide
what is more likely to happen is that the NaOH would remain as Na+ and OH- as ions in solution.
2) 2NaClO2 + 2H2O = 2ClO2 + 2Na+ + 2OH- + H2
This implies that NaClO2 with water would yeild aalkaline solution , which is observed.
As for the generation of hydrogen, is this observed?
moving on....
B) the dissociation of citric acid in water, from en.wikipedia.org/wiki/Acid
3) COH(CH2)2(COOH)3 + 3H2O = (COH(CH2)2(COO)3)3- + 3(H3O-)
The hydronium ions from the citric acid reaction 3) would react with the hydroxyl ions in 2) to form water.
The removal of OH- from reaction 2) by citric acid would drive the reaction 2) to exhaust all NaClO2 to leave ClO2
Then combining these two equations and adding coefficients to balance:
COH(CH2)2(COOH)3 + 3H2O + 6NaClO2 + 6H2O = (COH(CH2)2(COO)3)3- + 3(H3O-) + 6ClO2 + 6Na+ + 6OH- + 3H2
and simplifing by adding all H2O and having 3 H3O+ and 3 OH- react to make 6H2O:
COH(CH2)2(COOH)3 + 6NaClO2 + 9H2O = (COH(CH2)2(COO)3)3- + 6H2O + 6ClO2 + 6Na+ + 3OH- + 3H2
removing excess 6 H2O from both sides of equation:
COH(CH2)2(COOH)3 + 6NaClO2 + 3H2O = (COH(CH2)2(COO)3)3- + 6ClO2 + 6Na+ + 3OH- + 3H2
finally rearranging to have NaClO2 and ClO2 as first terms:
6NaClO2 + COH(CH2)2(COOH)3 + 3H2O = 6ClO2 + (COH(CH2)2(COO)3)3- + 6Na+ + 3OH- + 3H2
or:
6NaClO2 + COH(CH2)2(COOH)3 + 3H2O = 6ClO2 + Na3COH(CH2)2(COO)3 + 3NaOH + 3H2
doubling the citric acid would eliminate the NaOH so:
6NaClO2 + 2(COH(CH2)2(COOH)3) + 3H2O = 6ClO2 + 2(Na3COH(CH2)2(COO)3) + 3H2O + 3H2
elimianting H2O
6NaClO2 + 2(COH(CH2)2(COOH)3) = 6ClO2 + 2(Na3COH(CH2)2(COO)3) + 3H2
checking for atomic and charge balance
atoms in/out : Na 6/6; Cl 6/6; O 12+14=26/12+14=26, C 12/12, H 16/10+6=16
charge 0/0
well it balances!!
and it works in http://sciencesoft.at/equation/index
which predicts the reaction of sodium chlorite and citric would produce chlorine dioxide, sodium citrate and hydrogen......
but is it correct?? all corrections and comments gratefully encouraged!!