SOLUTION: how much of 80% acid must be mixed with 25% acid to produce 396ml of 50% solution?

Question 638776: how much of 80% acid must be mixed with 25% acid to produce 396ml of 50% solution?
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Let x=amount of 80% acid needed
Now we know that the amount of pure acid that exists before the mixture takes place (0.80x+0.25(396-x))has to equal the amount of pure acid that exists after the mixture takes place (0.50*396). Sooooo:
0.80x+0.25(396-x)=0.50*396 simplify
0.80x+99-0.25x=198 subtract 99 from each side and collect like terms
0.55x=99
x=180 ml----amount of 80% acid needed
CK
0.80*180+0.25*216=198
144+54=198
198=198

Hope this helps---ptaylor

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