You can put this solution on YOUR website! how much pure acid should be mixed with 8 gallons of 50% acid solution to get an 80% solution?
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Let x = amount of pure acid added
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"amt of acid added" + "amt of acid in 50%" = "desired amt of acid in mix"
x + .50(8) = .80(x+8)
x + 4 = .80x + 6.4
.20x = 2.4
x = 12 gallons