Question 597441: How much pure acid should be mixed with 6 gallons of a 20% acid solution in order to get a 90% acid solution? Answer by edjones(8007) (Show Source):
You can put this solution on YOUR website! 6*.2=1.2 total gal of acid.
Let x=gal of pure acid required.
x*1.00=x total gal of acid
(6+x)*0.9=.9(6+x) total gal acid
.
1.2+x=.9(6+x)
1.2+x=5.4+.9x
.1x=4.2
x=42 gallons
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Ed
