Question 966256: how much pure acid do you have to add to 250 mL of an 8% acid solution to get a 20% acid solution Found 2 solutions by lwsshak3, ikleyn:Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! how much pure acid do you have to add to 250 mL of an 8% acid solution to get a 20% acid solution.
let x=amt of pure acid to add
8%(250)+100%x=20%(250+x)
20+x=50+.2x
0.8x=30
x=37.5
how much pure acid do you have to add? 37.5 mL
You can put this solution on YOUR website!
Note that the concentrations in this problem are Volume-to-Volume concentrations, and they have the dimension of (i.e., actually, are non-dimensional).
The amount (the volume) of pure acid in 250 mL of 8% acid solution is 0.08*250 mL = 20 mL.
Let x be the amount (the volume) of pure acid we have to add to 250 mL of 8% acid solution to get 20% solution.
Then the volume of the obtained solution will be (250 + x) mL and the volume of the pure acid in it will be (20 + x) mL.
The concentration of the obtained solution is , and it should be 20%.
It gives an equation
= 0.2.
To solve it, multiply both sides by . You will get
20 + x = 0.2*(250 + x).
Open the parentheses and simplify step by step:
20 + x = 50 + 0.2*x
0.8*x = 50 - 20 = 30.
Hence, x = = 37.5 mL.
Answer. 37.5 mL of pure acid should be added to 250 mL of 8% acid solution to get 20% acid solution.
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