what happens at half equivalence points and equivalence points?

[mrh125]

i've always mixed these two up. I think at half equivalence points HA=A- and that Ph=Pka and that at full equivalence points you have 100% A- (but [H+] = [OH]-?) and that formula ph=1/2(pka1+pka2) applies if it's polyprotic. is that right? are there any other properties of half-equivalence and equivalence points that you generally need to know?

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[brood910]

i've always mixed these two up. I think at half equivalence points HA=A- and that Ph=Pka and that at full equivalence points you have 100% A- (but [H+] = [OH]-?) and that formula ph=1/2(pka1+pka2) applies if it's polyprotic. is that right? are there any other properties of half-equivalence and equivalence points that you generally need to know?

It's not "100%" A- at eq.
A- predominates there when you titrate a weak acid with strong base.
HA predominates there when you titrate a weak base with strong acid
Remember that mv = mv at eq as well.

Also, from the fact that pH range in buffer is +-1 of pKa, you can actually estimate the pH very closely.
(Ex: in weak acid titration, pH below half point can be found by : pKa-1 < pH < pKa).

Also, pH = pKa1 + pKa2/2 works only if it is a pure substance.
Dont just memorize things. You must understand.

 

[Blueprint MCAT Tutor]

I was just discussing this with a tutoring student yesterday. Here's on of the slides from the stuff we were covering. Hopefully it helps a bit!

 

[Czarcasm]

I was just discussing this with a tutoring student yesterday. Here's on of the slides from the stuff we were covering. Hopefully it helps a bit!

So -- at equivalence point, since the (H+ of) weak acid that dissociates is reacting completely with the strong base, then would it be accurate to say we have 100% A- at equivalence point? From that perspective, I would think it'd keep re-establishing equilibrium but each time it produces more H+, the protons would continue reacting with strong base until it went to completion and no protons from the initial weak acid were left in solution.

But then, what happens to the A- floating in solution, wouldn't they react to an appreciable amount in the reverse direction to re-establish equilibrium? I don't mean to over complicate things, just trying to get a feel for this topic. Maybe it's better to keep it direct and just say moles [OH] added = moles of weak acid titrated at equivalence point and ignore the other side products.

 

[brood910]

So -- at equivalence point, since the (H+ of) weak acid that dissociates is reacting completely with the strong base, then would it be accurate to say we have 100% A- at equivalence point? From that perspective, I would think it'd keep re-establishing equilibrium but each time it produces more H+, the protons would continue reacting with strong base until it went to completion and no protons from the initial weak acid were left in solution.

But then, what happens to the A- floating in solution, wouldn't they react to an appreciable amount in the reverse direction to re-establish equilibrium? I don't mean to over complicate things, just trying to get a feel for this topic. Maybe it's better to keep it direct and just say moles [OH] = moles of weak acid at equivalence point and ignore the other side products.

It cannot be "100%". Even TBR says it is not.
They even uses pH = pKa + log1000/1 or 100/1 to show that A- predominates, but not exist all alone.

 

[Czarcasm]

It cannot be "100%". Even TBR says it is not.
They even uses pH = pKa + log1000/1 or 100/1 to show that A- predominates, but not exist all alone.

Yeah, I figured that was true but I wanted a deeper understanding as to why that was the case and basically I guess the answer is, that no reaction is static, it's an equilibrium and the conjugate base that is formed will then react in the reverse direction to some degree until equilibrium is re-established. But regardless, I would think at that instantaneous moment in time when we reach equilibrium point, conjugate base is theoretically 100%. Or I think it's simply more accurate to say that at equivalence point, the moles of HA that we had initially has reacted completely to produce 100% moles A -- which is a bit different than simply saying we have 100% A in the solution I guess.

I admit this was always something I struggled with lol. Acid-Base chemistry was never my forte.
@Jepstein30 @Schenker @Cmdr_Shepard Any ideas?

 

[Jepstein30]

Is that @ thing supposed to give me an alert? It didn't! I just happened to open this up and see that.

To the OP, I'd also focus on knowing how to calculate the pH at all points on the curve (pre-titration, buffer region, at equivalence, post-titration, etc.).

As for the question whether its 100% A- or not... equivalence by definition is the point where total mol of acid is equal to total mol of base.. and given that such a reaction would be heavily favored (as all titrations are, especially with a weak component), it would be ~100% conjugate base. I don't think you can say it's 100% conjugate base (when can you ever really say something is 100% in one species? even if heavily favored.. it's still at equilibrium) but it's probably pretty damn close.

I'd think of it as 100% converting AT equivalence.. but as usual, some will react back into the acid. I'm not sure if a practical application of assuming its not 100% conjugate base. At equivalence, you wouldn't use the H-H equation to calculate pH.. unless I'm misremembering.

Don't really know for sure though so would be interested in hearing from others more chemically inclined. Perhaps some actual equilibrium constants would make this easier to figure out.

 

[Czarcasm]

Is that @ thing supposed to give me an alert? It didn't! I just happened to open this up and see that.

To the OP, I'd also focus on knowing how to calculate the pH at all points on the curve (pre-titration, buffer region, at equivalence, post-titration, etc.).

As for the question whether its 100% A- or not... equivalence by definition is the point where total mol of acid is equal to total mol of base.. and given that such a reaction would be heavily favored (as all titrations are, especially with a weak component), it would be ~100% conjugate base. I don't think you can say it's 100% conjugate base (when can you ever really say something is 100% in one species? even if heavily favored.. it's still at equilibrium) but it's probably pretty damn close.

I'd think of it as 100% converting AT equivalence.. but as usual, some will react back into the acid. I'm not sure if a practical application of assuming its not 100% conjugate base. At equivalence, you wouldn't use the H-H equation to calculate pH.. unless I'm misremembering.

Don't really know for sure though so would be interested in hearing from others more chemically inclined. Perhaps some actual equilibrium constants would make this easier to figure out.

Great, thank you!

I'm glad you brought that up. I'm gonna make an educated guess here, so you or someone can correct me if I'm wrong. I think to calculate the pH at equivalence point, we'd have to basically realize the conjugate base produced will react in the reverse reaction to some degree, producing [HA] and [OH-] in the process (A- + H2O <--> HA + OH-). Assuming they'd likely give us pKa or Ka for the original weak acid, we can solve for Kb and then set up an equilibrium expression to solve for [OH-] concentration. I believe we take the moles of A- as equal to the moles of HA we had initially (since it reacts completely in the presence of strong base), but the concentration overall would be diluted since we reacted strong base in the solution. Also, the concentrations of both [HA] and [OH-] would be equal so we can express that as x^2. Then we basically solve for [OH-] concentration and then manipulate that to get pH. Seems a little time consuming so I hope they wouldn't ask us to make that calculation.

 

[Czarcasm]

Yay, I was right! I vaguely recall reading something from EK a few years ago (2011) about this and I just cracked open my book to them saying this: Finding the pH at the equivalence point is a good exercise, but you won't have to do it on the MCAT. Here are the steps:" The steps they showed are essentially what I said above I haven't looked at this stuff since then, so I'm impressed with myself lol.

 

[Jepstein30]

Yep, sounds right to me.

Good to be able to figure out the pH at each point on the curve.. it makes sure you know what's going on throughout the whole process.