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how much pure acid should be mixed with 5 gallons of a 20% acid solution in order to get a 50% acid solution?

1 Answer

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% acid=[amount of acid / (amount of solution (acid + water))]*100

Data:
We have 5 gallons of a 20% acid solution:
amount of acid=20% of 5 gallons=(20/100)(5 gallons)=1 gallons
water amount=80% of 5 gallons=(80/100)(5 gallons)=4 gallons.

x=amount of pure acid (gallons).
(5+x)=amount of solution.
(1+x)=amount of acid in 50% solution

We have this equation:

50%=[(1+x) / (5+x)]*100%
(1+x)/(5+x)=50%/100%
(1+x)/(5+x)=1/2
2(1+x)=(5+x)
2+2x=5+x
2x-x=5-2
x=3

Answer: we need 3 gallons of pure acid.
User Dimitry K
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