Question 575062: What quantity of a 50% acid solution must be mixed with a 20% solution to produce 150 mL of a 40% solution?
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! Mixture problems are solved by setting up two equations: one for the total mixture amount balance, and another for the amount balance for the compound of interest.
In algebra, little attention is paid to the science, or the units, and any volume changes due to mixing are ignored.
Chemists are truly a dying breed
http://www.chemicalprocessing.com/articles/2011/cartoon_caption_62.html
So I'll pretend I'm not a chemist.
Let x be the volume of 50% solution to be used.
Let y be the volume of 20% solution to be used.
Total solution balance
 (all quantities are in mL, and volumes are additive)
Amount of acid balance:
Let's assume that the percentages are in v/v (volume to volume basis), so that a 50% solution contains 50 mL of acid (whatever is meant by that) in a total volume of 100 mL.
The amount (in mL) of acid in x mL of 50% solution is (0.50)x, and the amount (in mL) of acid in y mL of 20% solution is (0.20)y.
The total amount of acid in x mL of 50% solution plus y mL of 20% solution is:
We want that to be the amount of acid in 150 mL of 40% solution, which is (0.40)150, and
 , so
We need to solve the system of equations
There are several ways to solve the system.
USING ELIMINATION:
You could multiply the second equation times 2 to get
 and then subtract it from the first equation to get
 -->  -->  -->  -->  -->  --> 
And then, substituting the value found for y into , you would get
 , which you could solve for x to find  .
You need to mix  mL of 50% solution with  mL of 20% solution to get 150 mL of 40% solution.
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