I think of metabite as sulfur dioxide in solid form. Add water and you get the gas (which you can certainly smell): N2S2O5 + H2O --> 2Na+ + 2SO2 much of which stays dissolved in the water as the bisulfite ion: SO2 + H2O --> HSO3- + H+. The bisulfite ion is a reducing agent. That means the sulfur atom offers up its unshared electron pair to something that is electron grabbier than sulfur. In most cases this is oxygen. If, for example, we are using it to neutralize chlorine in water we get the reaction
HOCl + HSO3- --> Cl- + SO4-- + 2H+
in which the oxygen from the hypochlorite ion lets go of the electrons it is sharing with hydrogen and chlorine (they are reduced) in preference for the sulfur's unshared electrons (it is oxidized).
In the case of using it to neutralize chloramine we have
H2NCl + HSO3- + H2O --> NH4+ +Cl- + SO4-- + H+
the situation is a little dicier as it is water that is reduced (the sulfur atoms electrons go to its oxygen atom) releasing two hydrogens which in turn reduce the nitrogen and chlorine. But as in the chlorine case the oxygen from the water oxidizes the sulfur in the bisulfite ion to sulfate (sulfur's oxidation number increases from 4 to 6).
Were the bisulfite ion 'used' to reduce a carboxylic acid to an aldehyde:
HSO3- + RCOOH --> RCHO + SO4-- + H+
the terminal oxygen returns the electrons that it was sharing with the carbon and hydrogen to the carbon atom which is reduced from oxidation state +4 to +2 and, as in the other examples, takes up the sulfur's electrons instead so that the sulfur is, as in the other examples, oxidized.
The end effect for all the examples is the same: The bisulfite ion is oxidized to sulfate. Thus the stoichimetry is very simple:
•Each mole of bisulfite oxidized produces 1 mole of sulfate
•Each mole of bisulfite oxidized came from 1/2 mole of metabite which contained a mole of sodium
Thus if you add 25 mg of Na2S2O5 with molecular weight 190.11 mg/mmol you have added 25/190.11 = 0.131503 mmol of metabite and 2*25/190.11 = 0.2630 mmol of sodium which is, at 23 mg/mmol, 6.049 mg, irrespective of how much of the addition winds up as sulfate. This checks with your calculator.
Now 0.1315 mmol metabite also produces 0.2630 mmol of bisulfite and if that is all oxidized it results in 0.2630 mole of sulfate which, with its molecular weight of 96.08 mg/mmol corresponds to 25.2696 mg of sulfate ion. This too is close to what your calculator came up with so it seems the calculator is OK if you assume that all the bisulfite is in fact oxidized. If it is not there will be some residual bisulfite left in solution and this will either eventually find something to reduce in which case the sulfate level will eventually be correct or, if it can't find anything to reduce, be driven off as SO2 in the mash, sparge or boil.
