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Homework Help: Strong base weak acid titration

  1. Aug 9, 2008 #1
    Hi All

    I have a question that I require an answer to. For the above titration, concentration of A-(conjugate base of acid) is the same as concentration of HA at half equivalence. It is at this point that pka=ph of the solution. I agree to this conclusion, but why is the concentration of A- equals to concentration of HA at this point? Any help will be deeply appreciated. Thanks.
     
  2. jcsd
  3. Aug 9, 2008 #2

    Borek

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    Staff: Mentor

    Write reaction equation, think in terms of the neutralization stoichiometry. This is a weak acid, so it dissociates very slightly on its own - so slightly, that its own dissociation can be ignored.
     
  4. Aug 9, 2008 #3
    so, let HA be a weak acid

    let x be the amount of acid dissociated. so starting with 1 mol of HA

    ------------HA <-> H+ + A-
    conc--------1-x ----x ---x

    in order to achieve the required concentration, conc of HA = conc of A-

    1-x=x
    x=0.5

    0.5 mol of H+ is displaced which requires 0.5 mol of Naoh which is also half of total naoh required, am i right?
     
  5. Aug 9, 2008 #4

    Borek

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    Staff: Mentor

    You are mostly right, but you are using ICE instead of following simple stoichiometry.

    HA + NaOH -> NaA + H2O
     
  6. Aug 9, 2008 #5
    You can see this pretty easy from the Henderson Hasslebach equation:

    pH = pKa + log ([A-]/[HA])

    When [A-] is equal to [HA], you get log 1 in the above equation which is equal to 0. So pH = pKa. The Henderson Hasslebach equation is derived from the standard Ka equilibrium equation, and you can see that derivation if you want a deeper understanding.
     
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