Question 877862: I have one gallon of acid 30% solution . I need to reduce it to 5% acid. How much water should I add?
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! "I have one gallon of acid 30% solution"
So you have 1*0.30 = 0.30 gallons of pure acid.
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This means
(amount of pure acid)/(total amount of solution) = (0.30)/(1) = 0.30
which points back to the 30% acid solution
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Let x = amount of water to add (amount in gallons)
If you add x gallons of water to the solution, you get 1+x total gallons (acid + water)
so we now have
(amount of pure acid)/(total amount of solution) = (0.30)/(1+x)
You then take (0.30)/(1+x) and set that equal to 0.05 because you want to force the final solution to be 5% acid.
So,
(0.30)/(1+x) = 0.05
Now solve for x
(0.30)/(1+x) = 0.05
0.30 = 0.05(1+x)
0.30 = 0.05+0.05x
0.30-0.05 = 0.05x
0.25 = 0.05x
0.25/0.05 = x
5 = x
x = 5
You need to add 5 gallons of pure water to dilute the 30% acid solution to a 5% acid solution.
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