OK - having read some of the 2/2 posts, I'm now getting (baking soda on an anhydrous basis)...
4dKh = 1.43mEQ/L = 0.715mM CO3¯ ¯
1L*0.715mM*84mg/mmole=60.06mg baking soda per Liter 4dKH solution
6g/5L=28.57meq or 80dKH, dilute 10mL to 200mL for 4dKH (again, all assuming anhydrous baking soda).
I guess I'll have to do a Karl Fischer titration on a sample of the baking soda I have first, divide my intended weight of baking soda by my % baking soda (100%-%water).
Maybe this explains why I've always seen either green or yellow in my drop checkers.
Helpful reference:
http://ozreef.org/library/tables/alk...onversion.html