SOLUTION: a chemist wants to mix a 22 acid solution with a 50 acid solution to get 28 l of a 30 acid solution. how many liters of the 22% solution and how many liters of the 50% solution sho

Question 697274: a chemist wants to mix a 22 acid solution with a 50 acid solution to get 28 l of a 30 acid solution. how many liters of the 22% solution and how many liters of the 50% solution should be mixed?
Found 2 solutions by KMST, lwsshak3:
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
If a chemist (like me) wanted to mix a 22% acid solution with a a 50% acid solution to make a 22% acid solution, she/he would first want to know what kind of acid we are dealing with, and how accurate we need to be.

However, in math problems (unlike in real life), when you mix volumes of different solutions the final volume is the sum of the original volumes (and nothing dangerous can happen, either).

IF YOU HAVE STUDIED systems of linear equations:
Volume balance:
= volume of 22% acid solution to be used (in L)
= volume of 50% acid solution to be used (in L)
final volume = L = 28 L (the abbreviation, as per the SI, should be a capital L)
So <--> is our first equation.

Acid amount balance:
Whatever the units are (grams, liters, or whatever)
L of 22% solution contain units of acid
L of 50% solution contain units of acid
L of 30% solution should contain units of acid
So
Multiplying both sides of the equal sign times , we get

If you have studied systems of linear equations, you would say that you have the system

Then, you could substitute the expression for y, into and get to solve for .

--> --> --> --> -->

IF YOU HAVE NOT STUDIED systems of linear equations:
Volume balance:
= volume of 22% acid solution to be used (in L)
= volume of 20% acid solution to be used (in L)
The amount of acid in the final mixture would be

and that must equal ,
so our equation is

and that equation can be solved as shown avbove.
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
a chemist wants to mix a 22 acid solution with a 50 acid solution to get 28 l of a 30 acid solution. how many liters of the 22% solution and how many liters of the 50% solution should be mixed?
**
let x=amt of 22% acid solution to mix
(28-x)=amt of 50% acid solution to mix
22%x+50%(28-x)=30%*28
.22x+14-.5x=8.4
.28x=5.6
x=20
28-x=8
amt of 22% acid solution to mix=20 liters
amt of 50% acid solution to mix=8 liters
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