Question 688822: How many gallons of a 30% acid solution must be mixed with 20 gallons of a 19% solution to obtain a solution that is 25% acid? Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Let x=amount of 30% solution needed
Now we know that the amount of pure acid that exists before the mixture takes place (0.30x+0.19*20) has to equal the amount of pure acid that exists after the mixture takes place(0.25(20+x)). Sooooooo
0.30x+0.19*20=0.25(20+x)
0.30x+3.8=5+0.25x subtract 0.25x and also 3.8 from each side
0.05x=1.2
x=24 gal---amount of 30% acid solution needed
CK
0.30*24+3.8=0.25*44
7.2+3.8=11
11=11
Hope this helps----ptaylor
