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b. na forum
http://www.winepress.us znalazłem jeszcze coś takiego
1. How do I make up a test solution of Kmeta to check out the system. If I want a 50 ppm solution in water, how many grams do I add to 100 ml of water? Seems we went over this but I can't find it.
PPM= (gram solute/gram solution)*1,000,000
1 liter of water = 1,000 grams
20 = (grams solute/1,000)*1,000,000
20 PPM = .02 gram/liter
Sodium Meta = 66g SO2/100g of powder ~ .66 factor.
(.02 gram/liter / .66 factor) =.0303 grams/liter for Na-Meta
Potassium Meta = 56.7g SO2/100g of powder ~ .567 factor.
(.02 gram/liter / .567 factor) = .0352 grams/liter for K-Meta
Potassium metabisulphite dissociates in water to potassium ions (K+) and singly ionised bisulphite, (HSO3)-. (Sodium metabisulphite dissociates in the same way.)
Metabisulphite dissociates in the following way to form these fractions:
K2S2O5 + H2O ===> 2K+ + 2(HSO3)-
Sulphur dioxide is a bifunctional acid, and dissociates into three fractions. The quantity of each of these fractions depends on the thermodynamic constants and the pH. The dissociation is almost instantaneous.
The three fractions are molecular SO2 (SO2), sulphite (SO3=), and bisulphite (HSO3-). Dissociation of the various fractions is almost immediate.
Since wine is acidic, hydrogen ions are present (H+) and the (HSO3)- can then transform into sulphur dioxide:
(HSO3)- + H+ <===> H2O + SO2
singly ionized bisulphite + hydrogen ion water + unionized (molecular) sulphur dioxide
Additionally,
(HSO3)- + H2O <===> H+ + SO3=
singly ionized bisulphite + water hydrogen ion + doubly ionized sulphite
Thus, the relationships of the forms of SO2 in wine are shown completely by:
H2O + SO2 <===> H+ + (HSO3)- <===> 2H+ + SO3=
water + molecular sulphur dioxide hydrogen ion + bisulphite hydrogen ion + sulphite