Greg Cook ) wrote:
>On 5/7/04 10:06 AM, in article , "David C
>Breeden" > wrote:
>> seb ) wrote:
>>> (Ben Rotter) wrote in message
>>> . com>...
>>>
>>>> It means you'll get ~58% *total* SO2. Some of that will then become
>>>> free, and some will become bound (total = bound + free).
>>
>>> Ben, here a quote from the winemakermag :
>>
>>> "The key question now is: How much free SO2 is really added for any
>>> given amount of sulfite? Potassium metabisulfite and Campden tablets
>>> consist of approximately 57 percent and 48 percent SO2, respectively.
>>> This means that roughly half of the sulfite actually becomes free SO2
>>> when a solution is prepared and added to wine."
>>
>>> Thanks for your link, I had already read it and many more about SO2
>>> but I read different things about the 57% explanation. That's the
>>> only part i can't be sure of.
>>
>>> Séb
>>
>> Hi,
>>
>> It's easy enough to calculate for yourself. Meta is just K2S2O5,
>> and the molecular weights of K, S, and O are 39, 32, and 16,
>> repsectively. Find the total weight, and divide it into the weight
>> of the 2 SO2 molecules you get from each meta molecule.
>>
>> Voila! 57%.
>>
>> :-)
>>
>> Dave
>> ************************************************** **************************
>> Dave Breeden
>That's exactly right. But that would be TOTAL SO2 - not free SO2. Some will
>still be bound as sulfite. The free SO2 will be dependant on the pH of the
>wine. So I guess 57% would be more like a "potential" SO2 rather than the
>actual free SO2 in solution.
>--
>Greg Cook
Depends on what your solution is. :-)
In water, it's really (theoretically) 57%.
But yeah, the figure I gave was just the theoretical chemical
yield.
Dave
************************************************** **************************
Dave Breeden