Register to reply

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

by JeweliaHeart
Tags: halfway, titration
Share this thread:
JeweliaHeart
#1
May14-13, 11:48 AM
P: 41
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.
Phys.Org News Partner Chemistry news on Phys.org
Team pioneers strategy for creating new materials
Plug n' Play protein crystals
Intracellular imaging gets interactive
Borek
#2
May14-13, 01:18 PM
Admin
Borek's Avatar
P: 23,584
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].
JeweliaHeart
#3
May14-13, 03:05 PM
P: 41
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
[HA]=[A-].

?

Borek
#4
May14-13, 03:26 PM
Admin
Borek's Avatar
P: 23,584
Why does [HA]=[A-] halfway thru a titration?

Yes, you got it. One remark:

Quote Quote by JeweliaHeart View Post
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-].
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.


Register to reply

Related Discussions
What is the pH at halfway to the equivalence point? Biology, Chemistry & Other Homework 2
Stuck with Differentiation halfway Calculus & Beyond Homework 4
Electric field halfway between dipole isn't 0? Introductory Physics Homework 4
At 0k, why is fermi level halfway in bandgap? General Physics 3
Halfway measures Introductory Physics Homework 6