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Re: Stoichiometry Anyone?
Jamie Johnson wrote:
> > A search of the archives turned up the following formula which seems
> > to be the basis for the CO2 charts on The Krib:
> > CO2 (mg/L) = 12.839 * KH * 10^(6.37 - pH)
[snip]
> > For the result to indicate dissolved CO2 in mg/L, shouldn't the
> > constant
> > be:
> > .3566 * 44.01 mg/mmole = 15.694 rather than 12.839?
> Todd, can't remember where I found it, but I've had it on my
> calculation sheet for a while:
>
> CO2 = 3 x dK x 10^(7.00-pH), where dK equals degrees alkalinity.
I don't know where that formula came from either. Carlos Munoz <still
reading Carlos?> pointed out that it is pretty much the same as the one
that George used and Todd is taking issue with. Do this little bit of
algebra:
CO2 = 12.839*KH*10^(6.37-pH)
= 12.839*KH*(10^-0.63)*10^(7.00-pH); 10^-0.63 = 0.2344 so
= 12.839*0.2344*KH*10^(7.00-pH)
= 3.0*KH*10^(7.00-pH)
It's like magic!
Roger Miller