Question 697704: how much of a 25% acid solution should be mixed with 5 L of a 15% acid solution to obtain a 20% acid solution? Found 2 solutions by Alan3354, Edwin McCravy:Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! how much of a 25% acid solution should be mixed with 5 L of a 15% acid solution to obtain a 20% acid solution?
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20 is the average of 15 & 25, so it's equal amounts.
5 liters.
The other tutor's answer is correct, but his method will work ONLY
when the percentage desired is exactly half-way between the two
percentages. Here is a more general way to do the problem, which will
work in ALL mixture problems:
Make this chart:
amt of liquid % acid amt of pure acid contained
1st solution
2nd solution
final solution
Let x be the answer, so fill that in first
amt of liquid % acid amt of pure acid contained
1st solution x
2nd solution
final solution
We are told that there are 5 L of the second solution, so
amt of liquid % acid amt of pure acid contained
1st solution x
2nd solution 5
final solution
Now we add x PLUS 5 to get the amt of final liquid, so we fill
x+5 in
amt of liquid % acid amt of pure acid contained
1st solution x
2nd solution 5
final solution x+5
Next we fill in the three percentages as decimals:
amt of liquid % acid amt of pure acid contained
1st solution x .25
2nd solution 5 .15
final solution x+5 .20
Then we multiply the three amts of liquid by the percentages expressed
as decimals to get the amt of pure acid in each amount:
amt of liquid % acid amt of pure acid contained
1st solution x .25 .23x
2nd solution 5 .15 .15(5)
final solution x+5 .20 .20(x+5)
The equation comes from:
.25x + .15(5) = .20(x+5)
Multiply through by 100
25x + 15(5) = 20(x+5)
25x + 75 = 20x + 100
5x = 25
x = 5
Edwin
