the lab keeps two acid solutions on hand. one is 20% and the other is 35%. how much 20% and how much 35% acid solution should be used to prepare 25 liters of a 26% solution?
Can someone help me solve this?
volume of the 20% stuff ---- x L
volume of the 35% stuff ---- 25-x L
solve:
.2x + .35(25-x) = .26(25)
I suggest multiplying each term by 100 to clear the decimals.
carry on ...
To determine how much of each solution is needed to prepare a 25-liter solution with 26% acidity, we can set up a system of equations. Let's assume x represents the volume (in liters) of the 20% acid solution and y represents the volume (in liters) of the 35% acid solution.
The first equation represents the volume of acid in the final mixture:
0.20x + 0.35y = 0.26 * 25
The second equation represents the total volume of the mixture:
x + y = 25
This gives us a system of two equations:
0.20x + 0.35y = 6.5
x + y = 25
To solve this system of equations, we can use various methods such as substitution or elimination. Let's use the substitution method here.
Rearranging the second equation, we get:
x = 25 - y
Substituting this value of x in the first equation:
0.20(25 - y) + 0.35y = 6.5
Now, we can solve for y:
5 - 0.20y + 0.35y = 6.5
0.15y = 1.5
y = 1.5 / 0.15
y = 10
Substituting the value of y back into the second equation:
x + 10 = 25
x = 25 - 10
x = 15
Therefore, 15 liters of the 20% acid solution and 10 liters of the 35% acid solution should be mixed to prepare 25 liters of a 26% solution.