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

  1. Jun 22, 2016 #1
    Hello!
    1. The problem statement, all variables and given/known data

    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?

    2. Relevant equations
    3. The attempt at a solution

    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!
     
  2. jcsd
  3. Jun 22, 2016 #2

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  4. Jun 22, 2016 #3
    Thank you! )
     
  5. Jun 22, 2016 #4

    Mark44

    Staff: Mentor

    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.
     
  6. Jun 23, 2016 #5
    I agree ) Thank you.
     
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