[jukebo]

Propionic acid, HC3H5O2, reacts with KOH. What considering the point in the titration where 19.10 mL of .100 M HC3H5O2 has been mixed with 21.05 mL of .875 M KOH. Find each value of:

1. [C3H5O2-]
2. [HC3H5O2]
3. [K+]
4. [OH-]
5. [H3O+]
6. pH

[Borek]

You have to show your attempts at solving the question to receive help. This is a forum policy.

[jukebo]

First, I tried to find the values of [HC3H5O2] and [KOH]. I decided that the volume to be (19.10 + 21.05)mL. And the values came out like this.  [HC3H5O2]=.0475716065M &  [KOH]=.4582814446M.

With it, I used the HendersonHassebalch formula to find the pH, which came to be 5.81, and I used it to find the value of [H3O+].

Then I found the pOH using 14 - 5.81 = pOH and it was [OH-]=6.457e-9

I went back to the equation HC3H5O2 + KOH ==> H2O + KC3H5O2, and found that in the equation, KOH would be an excess and I would have .0165 mol of KOH left over. With it I did KOH ==> K+   +   OH- and assumed there would be also .0165 moles of OH- would be in (19.10 + 21.05 mL) of solution, which gave me
[OH-] = .410958....

Was I taking the wrong approach? What should I do?

[Borek]

[HC3H5O2]=.0475716065M &  [KOH]=.4582814446M.

With it, I used the HendersonHassebalch formula to find the pH

Show what you plugged into HH equation.

I bet you assumed concentration of base to be 0.458 M? That's wrong base. HH equation contains ratio of acid and CONJUGATE base. Which you can't easily calculate here.

Hint: generally speaking this is a limiting reagent problem.

[jukebo]

For the Henderson-Hasselbalch equation,

pH = -log(1.3e-5) + log(.458/.0476)

Hmm, so would the conjugate base be KC3H5O2?

[Borek]

For the Henderson-Hasselbalch equation,

pH = -log(1.3e-5) + log(.458/.0476)

As explained earlier - that's wrong.

Hmm, so would the conjugate base be KC3H5O2?

Yes. But you won't be able to easily calcualte its concentration in this case.

As I wrote earlier: start the question thinking in terms of a limiting reagent. What is in excess?