Just looking for a dilution confirmation for 98% w/w sulfuric to 25% w/w sulfuric for liquor treatment (alkalinity and SO4 addition).
No advisements against using for safety or "not good for brewing - build up from RO" please.
98% w/w H2SO4 = 18.35 M
25% w/w H2SO4 = 3.18 M
To get 1 L of 25%/3.18 M sulfuric acid, by c1v1 = c2v2,
v1 = (3.18 * 1000)/18.35 = 173.3 ml sulfuric acid, with enough chilled DI water to make up 1 L, roughly 826.7 ml cold DI water, acid to water in small increments, to ease the exothermic reaction.
Thanks.
No advisements against using for safety or "not good for brewing - build up from RO" please.
98% w/w H2SO4 = 18.35 M
25% w/w H2SO4 = 3.18 M
To get 1 L of 25%/3.18 M sulfuric acid, by c1v1 = c2v2,
v1 = (3.18 * 1000)/18.35 = 173.3 ml sulfuric acid, with enough chilled DI water to make up 1 L, roughly 826.7 ml cold DI water, acid to water in small increments, to ease the exothermic reaction.
Thanks.
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