Question 134502: How many milliliters of a 40% acid solution must be added to 150 milliliters of a 65% acid solution to obtain a 50% acid solution? Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Let x=number of ml of 40% acid solution needed
Now we know that the amount of pure acid in the 40% solution (0.40x) plus the amount of pure acid in the 65% solution(0.65*150) has to equal the amount of pure acid in the final mixture (0.50(150+x)). So our equation to solve is:
0.40x+0.65*150=0.50(150+x) simplify
0.40x+97.5=75-0.50x subtract 75 and also 0.40x from each side
0.40x-0.40x+97.5-75=75-75+0.50x-0.40x collect like terms
22.5=0.10x divide both sides by o.10
x=225 ml------------------amount of 40% solution needed
CK
0.40*225+0.65*150=0.50(225+150)
90+97.5=187.5
187.5=187.5
