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Why does [HA]=[A-] halfway thru a titration?

  1. May 14, 2013 #1
    I read somewhere that:

    "During the course of a titration, an acid in solution reacts with a base in solution, and
    when a strong acid or strong base is involved, this reaction always goes to completion...."

    HA(aq) + OH- (aq) → A- (aq) + H2O (1)"

    "At the equivalence point (which is the place where moles of acid = moles of base) the
    titration is complete, no HA remains; and the only substances in the beaker are water, the
    conjugate base, A-, and a spectator ion. The end point is the place at which the indicator
    changes color. The endpoint and the equivalence point are not always identical, but they
    are always very close."

    "Halfway to the end point, half of the HA has reacted to become its conjugate base A- and
    water. At that point, the concentrations of HA and A- are equal. When these
    concentrations are equal, log [A-]/[HA] is zero and pH = pKa (see equation 4). It is clear
    then that pKa can be read directly from the titration curve as the pH at the half-way point
    of a titration."

    So I understand most all of this, but one thing is bugging me:
    Why does [HA]=[A-] halfway to the end point? Shouldn't the concentrations be equal to one another the whole time b/c they are both in the same volume of container and dissociate with the same molar ratio 1:1?

    I must be totally misunderstanding something b/c that makes no sense to me.
  2. jcsd
  3. May 14, 2013 #2


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    Staff: Mentor

    It is hard to pinpoint where you are making the mistake - but yes, you are wrong.

    You start with just HA, so initially you have A- only from dissociation - so the concentrations are not equal as you suggest even before we start the titration. Then they mostly follow the stoichiometry - base reacts with the acid, so the amount of produced A- is more of less equal to amount of base added (with some minor differences caused by dissociation/hydrolysis), while concentration of HA is that of HA not yet neutralized. when you are halfway amount of base added is half of the amount of acid, so from stoichiometry [A-]=[HA].
  4. May 14, 2013 #3
    Oh my gosh. Thank You. It is crystal clear now.

    So, and correct me if I'm wrong please,

    [added base]=[conjugate base] throughout most of the reaction b/c the base combining with the acid leads to the formation of water and A-. And the minor differences in concentration are caused by the HA dissociating on its own, so technically the [A-] is a little higher than [OH-].

    At the halfway point, the [HA]=[OH-], but since the [OH-] is roughly equal to [A-], then

  5. May 14, 2013 #4


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    Staff: Mentor

    Yes, you got it. One remark:

    It actually depends on how strong the acid is. If it is very weak (say HCN, with pKa=9.31), conjugate base produced during neutralization will be a relatively strong Bronsted-Lowry base, reacting with water to produce HA and OH-, so it may happen that the concentration of HA will be actually higher than expected. If the acid is relatively strong (say trichloroacetic, pKa=0.7) concentration of HA is not even approximately equal to the concentration of base at midpoint, for typical titration concentrations it will be twice larger. But these are rather extreme cases.
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