SOLUTION: How many liters of pure acid must be added to 3 liters of a 50% acid solution to obtain a 75% acid solution?

Question 232371: How many liters of pure acid must be added to 3 liters of a 50% acid solution to obtain a 75% acid solution?
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Let x=amount of pure acid needed
Now we know that the amount of pure acid in the liters that are added (x) plus the amount of pure acid in the 3 liters of 50% solution (0.50*3) has to equal the amount of pure acid in the final mixture (0.75(3+x)). So our equation to solve is:
x+0.50*3=0.75(3+x) get rid of parens and simplify
x+1.5=2.25+0.75x subtract 0.75x and also 1.5 from each side
x-0.75x+1.5-1.5=2.25-1.5+0.75-0.75 collect like terms
0.25x=0.75 divide each side by 0.25
x=3 liters-------amount of pure acid needed
CK
3+1.5=0.75*6
4.5=4.5
Hope this helps----ptaylor
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