Racking and Sulphite

I've got a batch of Carignane wine aging in the carboy right now and have a question about how much sulphite to add at each racking. Here's what I've done so far:

50 ppm sulphite at crush, 50 ppm sulphite at the first racking (after ML fermentation was finished in November). My plan is to do a 2nd racking, adding oak chips, then leave it till the Fall - do a 3rd racking then bottle a week or so later.

Given the sulphite I've already added, how much should I add at the 2nd and 3rd rackings? pH is 3.55-3.60.

Thanks!

Reply to
Trav77
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The better and safer way would be to measure the free SO2 before racking and do the addition based on that. If you want to ballpark it, then I'd suggest 20-40ppm at the 2nd racking, probably closer to the end of the range to be safe. I'd definitely do the free SO2 check on the next racking if you're planning to bottle right after, aiming for

30ppm or so at bottling.

Pp

Reply to
pp

How does one calculate how to add 20-40ppm of SO2?

Reply to
Pino

Potassium metabisulphite salt contains 57% free SO2 (by weight). For x liters, to get y ppm (parts per million) = y mg/l (y l)/ (0.57) = grams of potassium metabisulphite salt.

For 23 liters, to get 40 ppm (parts per million) (40 mg/l x 23 l)/(0.57) = 0.99 grams of potassium metabisulphite salt. HTH, Guy

Reply to
guy

Guy's answer gives you the exact amount. If you don't have a precision scale, a reasonable compromise is: 1/4 teaspoon of potassium meta gives about 50ppm of free SO2 for 5 US gals, so dilute this in a known amount of water - say 20-25 ml - and then calculate how much of the solution you need for your volume.

Ex., to get 20ppm for 6 US gals: the whole amount of 20ml would give (50ppm x 5 gals) / 6 glas ~ 42 ppm for the 6 gals, so to get 20ppm, use (20ml x 20) / 42 ~ 10ml of the solution.

This is not as precise but good enough as long as the 1/4 tsp measurement is reasonably correct - Iverson recommends getting the spoon form a pharmacy instead of using the usual kitchen tool.

Pp

Reply to
pp

Wrong pp! I gave the right solution but the wrong answer. For 23 liters, to get 40 ppm (parts per million) (40 mg/l x 23 l)/(0.57) = 1614 mg or 1,61 grams of potassium metabisulphite salt.

Easier solution is 40g x 23 l /570 = 1,61 g

Guy

Reply to
guy

Reply to
Joe Sallustio

I will let Guy answer but I think he was / is correct. Approximately half (0.57) will be free the rest will be bound so if you multiply by any number less than one you get a smaller number. You need almost twice as much total to get approx. half of it as free. Multiplying will give you only about 1/4 of what you need. I have always been mathematically challenged - Ray Calvert - where are you, math tutor. Are all those beers we consumed tutoring me 35 years ago for naught?

Reply to
Paul E. Lehmann

I'm an idiot, you are right. No idea what I was thinking there. Guy had it right. Sounds like Ray has to keep an eye on two of us now.

Joe

Reply to
Joe Sallustio

I don't blame you Joe, that / after the 40mg is confusing. Besides, I made a mistake in my first post so....

Cheers, Guy

Reply to
guy

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