# Mixture Problems

How many ounces of pure water must be added to 80oz of a 20% acid solution to make a solution that is 12% acid? (hint: Pure water is 0% Acid.)

 s_cianci Posts: 5,481, Reputation: 4046 Über Member #2 Oct 31, 2006, 07:01 PM
Let x = # of ounces of pure water
You have 80 oz. Of 20% acid (before adding the water)
You have x + 80 oz. Of 12% acid after adding the water
The amount of acid is the same before the water is added and after the water is added
So: .2(80) = .12(x + 80)
Solve the above equation for x

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