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Titration Problem Chemistry pH

  1. Sep 10, 2013 #1
    1. The problem statement, all variables and given/known data

    Ok, so, I've been thinking about this for a while now.

    When I was doing titration back in high school, we had to put water (so that the solution would have a larger volume to work with) in the acid in which we were going to pour base from a buret.

    Now, my question is this:

    Why doesn't the water change the pH of the acid? You add more water which dilutes the concentration of H+. So, then, there is the equilibrium formula:

    K=[H+][A-]/[HA]

    Since you dilute the solution (let's assume by doubling the volume of the solution), the K changes:
    [0.5H+][0.5A-]/[0.5HA]

    So, equilibrium constant is not the same anymore. Shouldn't this cause changes in the concentration of the acid and thereby false the titration?

    I feel there is something that doesn't click here.
     
  2. jcsd
  3. Sep 11, 2013 #2

    Borek

    User Avatar

    Staff: Mentor

    It does.

    It also dilutes all other things present in the solution.

    No, K doesn't change. Equilibrium shifts till K has exactly the same value it had before.

    K didn't change, but even if it would, it would not change the titration result. Amount of acid (note: amount, not concentration!) is calculated from the neutralization stoichiometry, not from pH.
     
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