Can I combine an Acid dissociation with autoionization of H2O?

[Hammad Shahid]

TL;DR

Can I do this?

HCH3COO + H2O <<-> H3O+ + CH3COO- ; K= 1.8*10^-5

H3O+ + OH- <->>>>> 2H2O ; K= 10^14

Net Equation:
HCH3COO + OH- <->>> H2O + CH3COO-
K= (1.8*10^-5) * (10^14) = 1.8*10^9 = extremely large

The problem states that I'm adding a certain volume of a known [KOH] to a certain volume of a known [HCH3COO].

The goal is to calculate the final pH.

Since I don't know the K value of the rxn of HA w/ OH-, I set up 2 equations and combined their K values to derive the K value.

Since the new K value is extremely large, I assume the reaction goes fully to completely and that all OH- (limiting reagent) reacts with the HCH3COO to provide an equivalent amount of conjugate base and an excess HCH3COO.

Next, since I find the concentrations of the acid and the conjugate base. Since they are of appreciable amounts, I can assume that the initial volumes of both change negligibly.

Finally, I use the Henderson-Hasselbach equation to find the pH.*My question is: is this a legit way to solve this problem?

Thank you.

 

[Borek]

Since the new K value is extremely large

Actually it is not as large as you think.

I assume the reaction goes fully to completely and that all OH- (limiting reagent) reacts with the HCH3COO to provide an equivalent amount of conjugate base and an excess HCH3COO.

(...)

Finally, I use the Henderson-Hasselbach equation to find the pH.

That's a reasonable approach

Note however, that the K value you have derived is just a reciprocal of the base dissociation constant of acetate. While acetate is a weak base, it is still strong enough for the acetate solutions to be slightly alkaline. Assumption that the neutralization went to completion works OK for a typical buffer preparation, but will fail for more exotic cases (high dilution, or attempting to neutralize more than - say - 95% of the acid).

 

[Hammad Shahid]

Actually it is not as large as you think.
That's a reasonable approach

Note however, that the K value you have derived is just a reciprocal of the base dissociation constant of acetate. While acetate is a weak base, it is still strong enough for the acetate solutions to be slightly alkaline. Assumption that the neutralization went to completion works OK for a typical buffer preparation, but will fail for more exotic cases (high dilution, or attempting to neutralize more than - say - 95% of the acid).

In my class, we're told to assume the rxn goes to completion if K is 10^10 or greater, and since this is pretty close to that, I put in that assumption.

Oh wow, I just realized that the net reaction is the acetate with water rxn in reverse. Thanks, that'll save me quite some time on the actual tests.

Alright, so I understand why for very dilute solutions the approximation with Henderson-Hesselbach equation becomes less accurate, but why for attempting to neutralise more than 95%?

 

[Borek]

why for attempting to neutralise more than 95%?

Don't treat this number too seriously, just a rule of thumb - if neutralization goes close to completion it is better to check if the assumption was valid.

 

[Hammad Shahid]

Don't treat this number too seriously, just a rule of thumb - if neutralization goes close to completion it is better to check if the assumption was valid.

Ok I was actually concerned on why the assumption may not hold for neutralizing more acid?

 

[Borek]

The close we get to the 100% neutralization the more of the conjugate base gets hydrolized. If you assume 99% of the acid is neutralized and ignore the fact 1% of the conjugate base reacts with water, concentration of the undissociated acid estimated this way is wrong by 100%.

 

[Hammad Shahid]

The close we get to the 100% neutralization the more of the conjugate base gets hydrolized. If you assume 99% of the acid is neutralized and ignore the fact 1% of the conjugate base reacts with water, concentration of the undissociated acid estimated this way is wrong by 100%.

I'm sorry, but I don't quite understand this. Is there an example of this that could be given.