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Titration - Calculating the endpoint

  1. Oct 21, 2006 #1
    Hello,

    I have titrated 25 ml of NaOH with 25 ml of HCl. Both are 0.1M.
    I need to calculate the expected endpoint for the titration of the strong base with the strong acid.

    I know that:
    0.025 L x 0.1M = 2.5 x 10^-3 moles of NaOH and HCl each.
    I know that I need the same volume of both to do the titration.

    But what I don't understand is: how do I calculate the endpoint? What is it supposed to be, a mole amount, a M amount, a volume, a pH?...

    Thank you,

    J.
     
  2. jcsd
  3. Oct 21, 2006 #2

    Borek

    User Avatar

    Staff: Mentor

    Question is ambiguous for me. You are asked either about volume of titrant or about pH.
     
  4. Oct 21, 2006 #3
    That's what I thought too...

    So, if it is the volume of titrant, since the 2 chemicals have the same molarity, the volume of the initial solution would be equal to the volume of the titrant. So, if they ask for the volume, I can do it.

    What if I have to calculate the pH at the endpoint? I know how to predict whether it will be above at or below 7.0 depending on what we start with and what the titrant is, but I am not sure what the formula is to find the exact pH...

    Any idea?

    Thank you!
     
  5. Oct 21, 2006 #4
    Is this an actuall reaction you are doing or just on paper? In the past all titrations ive done have been volume based.
     
  6. Oct 22, 2006 #5

    Borek

    User Avatar

    Staff: Mentor

    Acid/base titration curve calculation. Don't worry about the beginning (description of the general approach). Scroll to read the text below the equations.
     
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