[lablackey]

Somehow I think this should be simple, but true understanding eludes me.  I need to figure out what portion of sodium bisulfite and sodium metabisulfite ends up as sulfur dioxide in solution.  I've seen statements that it's 66%, but how is that calculated?  I'm tweaking a sulfur dioxide titration method designed for corn syrup for steep water and I can't assess the effectiveness of the method if I don't know what the results should be for any standards I make up.  Any help is much appreciated.
Thanks.
-LL

[Borek]

You must find out chemistry behind - try to write reaction equations.

[lablackey]

OKay here's what I have.

Sodium metabisulfite: Na₂S₂O₅ + H₂O = 2NaHSO₃ (sodium bisulfite)

And sodium bisulfite: NaHSO₃ + H₂O = H₂SO₃ (sulfurous acid)

Meanwhile, in water sulfur dioxide: SO₂ + H₂O =
H₂SO₃

The formula weight of sodium bisulfite is 104.17; sodium metabisulfite is 190.1; and sulfur dioxide is 64

If i start out with 10grams of sodium bisulfite that's 0.096 moles and 1 mole of bisulfite produces 1 mole of sulfurous acid, same as sulfur dioxide, which is 64g/mole so the equivalent of 6.14g SO₂.
Which is 61.44% of the bisulfite I weighed out. 
Yes?

Following the same calculations, 10 grams of metabisulfite is 0.053 moles, but one mole of metabisulfite produces two moles of bisulfite which goes one to one with sulfurous acid, equivalent to 1 mole SO₂ = 6.733 grams or 67.33% of the original weight.

Does this make sense? 

-LL

[Borek]

Using EBAS I got 61.56% and 67.40% respectively, so almost identical results.