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How much of a 10 liter 30% acid solution must be replaced

  • #1
86
0
Hello!
1. Homework Statement

Please, take a look at the problem and at my solution. Have I done it correctly and is my logic correct?
How much of a 10 liter 30% acid solution must be replaced with pure acid to obtain 10 liters

of a 50% solution?

Homework Equations


3. The Attempt at a Solution [/B]
I need to get 10 liters of final solution that contains 5 liters of acid.
Let X be the volume of 30% solution, and Y the volume of pure acid.

Than:
X + Y = 10
0.3 X + Y = 5 I take 0.3X of 30% solution and add Y of acid to get 5 liters of acid.

Y = 2.857

Thank you!
 

Answers and Replies

  • #2
CWatters
Science Advisor
Homework Helper
Gold Member
10,529
2,295
Correct. As a check...

10L of 30%
take out 2.857L gives you
7.143L of 30%.

of which
2.143L is acid
5.000L is Water

add 2.857L of acid gives you

5L acid
5L Water.
 
  • #3
86
0
Correct. As a check...

10L of 30%
take out 2.857L gives you
7.143L of 30%.

of which
2.143L is acid
5.000L is Water

add 2.857L of acid gives you

5L acid
5L Water.
Thank you! )
 
  • #4
33,270
4,975
Please, take a look at the problem and at my solution. Have I done it correctly and is my logic correct?
How much of a 10 liter 30% acid solution must be replaced with pure acid to obtain 10 liters of a 50% solution?
Since you have found a solution, you should get into the habit of checking it by substituting the values for X and Y into the original equation. This is what @CWatters did in his reply. If the values you found for the variables are correct, substituting them into your equations should result in true statements.
 
  • #5
86
0
Since you have found a solution, you should get into the habit of checking it by substituting the values for X and Y into the original equation. This is what @CWatters did in his reply. If the values you found for the variables are correct, substituting them into your equations should result in true statements.
I agree ) Thank you.
 

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