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Concentration Problem

  1. Sep 30, 2009 #1
    Hello all,

    I am trying to revise over acids and bases before my exams, but there is one question I am stuck on, and I am not sure how to approach it. Any hints would be great ^^

    "100mL of HCL of pH of 4.0 is mixed with 100mL of HCL of PH of 5.0. Whats the resulting pH of the solution formed".

    Its probably really easy...

    This is not homework, just revision I decided to attempt.

    Cheers,
    Adrian
     
  2. jcsd
  3. Sep 30, 2009 #2

    CompuChip

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    You might be tempted to say (4.0 + 5.0) / 2 = 4.5, it's not that easy. It is straightforward though: just calculate the concentration of H+.

    What is the relation between that and pH?
     
  4. Sep 30, 2009 #3
    Thanks for the reply ^^

    lol if only...

    Hows this:

    [H+]1 = 10-4
    [H+]2 = 10-5

    pH = -log[10-4 + 10-5]

    Close? ^^
     
  5. Sep 30, 2009 #4

    CompuChip

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    Closer, yes :)
    Can you include the units of [H+] ?
    It's not moles, as your calculation now suggests.
     
  6. Sep 30, 2009 #5
    Remember that [H+] is concentration, which is logarithmic. You do seem to be getting closer though.
     
  7. Sep 30, 2009 #6

    Borek

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    No, pH is logarithmic. Concentration would be antilogarithmic :wink:

    --
     
  8. Sep 30, 2009 #7
    Thanks for the replies everyone.

    Right, so the units would be moles per litre or M

    Apart from that, would the addition of the two concentrations and then taking the negative log yield the correct answer? ^^
     
  9. Oct 1, 2009 #8

    Borek

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    No, you can't just add concentrations. It would violate one of the most basic laws that governs physics and chemistry.

    See diluting and mixing solutions for details of the correct approach.

    --
     
  10. Oct 1, 2009 #9

    CompuChip

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    Don't guess at the answer, work it out. You have a concentration with 10-4 mol/l and you take 100 ml. Then you add to that 100 ml of a concentration with 10-5 mol/l. You get how many with what concentration?
     
  11. Oct 1, 2009 #10
  12. Oct 1, 2009 #11
    Ok, hows this:

    n = cV so

    n1 = 0.1 x 10-4 = 10-5

    n2 = 0.1 x 10-5 = 10-6

    Total moles = n1 + n2
    so total concentration = total moles / .200Litres.

    C = (10-5 + 10-6) / 0.200

    pH = -log(C)

    pH = 4.26 ?
     
  13. Oct 2, 2009 #12

    Borek

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    Not checking math - approach is perfect :smile:

    --
    methods
     
  14. Oct 2, 2009 #13

    CompuChip

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    And the number sounds familiar from when I checked it a few days ago when you first posted the question.
    It is also what you would expect: mixing equal amounts of pH 4 and pH 5 solutions gives something with pH between 4 and 5, but lower than 4.5 due to the logarithmic scale.
     
  15. Oct 4, 2009 #14
    Cool, thanks everyone for helping :)

    Thanks!
    Adrian
     
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