Question 220587: What quantity of a 80% acid solution must be mixed with a 30% solution to produce 600 ml of a 40% solution? Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Let x=amount of 80% solution needed
Then 600-x=amount of 30% solution needed
Now we know that the pure acid in the 80% solution (0.80x) plus the amount of pure acid in the 30% solution ((0.30)(600-x)) has to equal the amount of pure acid in the final mixture(0.40*600), so our equation to solve is:
0.80x+0.30(600-x)=0.40*600 get rid of parens
0.80x+180-0.30x=240 subtract 180 from each side
0.80x+180-180-0.30x=240-180 collect like terms
0.50x=60 divide each side by 0.50
x=120 ml---------------------amount of 80% solution needed
600-x=600-120=480 ml---------------amount of 30% solution needed
CK
120*0.80+480*0.30=0.40*600
96+144=240
240=240
Hope this helps---ptaylor
