Jennifer Templeton - 7-5-2010 at 16:57
Hello everyone. 
I hope it's okay to ask a question like this here. So, I'm
going through my old chemistry textbook and trying to brush up on some stuff and i'm having a little trouble with this particular problem. Just
wondering if someone could show me how it's done: 
Calculate the pH and pOH of a solution that contains 0.0100M (molarity) acetic acid (CH3COOH and 0.100M (molarity) sodium acetate (NaCH3COO). Acetic
acid has a K(subscript)a = 1.8 X 10^-5 at 25 degrees celcius
[Edited on 8-5-2010 by Jennifer Templeton]
entropy51 - 7-5-2010 at 17:13
Henderson-Hasselbalch equation What's the problem? Your Google broken? Can we get a moderator over here to delete the homework problems? This
forum is going to hell in a heating mantle.
Jennifer Templeton - 7-5-2010 at 18:00
C'mon is that really neccesary? I just needed to see someone solve one of the many problems i will be working out of this chapter. I am not trying to
clutter your board.
DJF90 - 8-5-2010 at 01:59
The material needed to answer problems at the end of a chapter is usually contained within that chapter.
Jennifer Templeton - 8-5-2010 at 09:54
you guys so aren't invited to any of my birthday parties
entropy51 - 8-5-2010 at 11:50
There are forums whose principal purpose is to answer chemistry homework questions.
I'm sure those guys would love to come to your parties.
blogfast25 - 8-5-2010 at 13:31
I think some here are being a little hard on young Jennifer here. The derivation of [H3O+] and [OH-] isn't so easy from first principles and Henderson
- Hasselbalch is only a good approximation, because in Jenni's problem the concentrations are nominal and not actual (in reality actual and nominal
will be close).
It involves setting up a system of four equations in [H3O+], [OH-], [CH3COO-] and [CH3COOH] and solving that system (usually approximating a little)
for [H3O+].
The four equations are:
1) [H3O+].[OH-] = 10^-14
2) Acid-base equilibrium for CH3COOH in water (Ka)
3) Electrical Neutrality equation
4) Material balance equation
[Edited on 8-5-2010 by blogfast25]
blogfast25 - 9-5-2010 at 08:07
Full development (FYE only, Jenni ):
Call the nominal concentration of acetic acid, Ca (= 0.01 M) and the actual equilibrium concentration, [HA].
Call the nominal concentration of sodium acetate, Cb (= 0.1 M) and the actual equilibrium concentration, [A-].
The equations:
Water constant: (eq. 1) [H3O+].[OH-] = Kw
Acid/base equilibrium for Bronsted acids in water: (eq. 2) [H3O].[A-]/[HA] = Ka
Electrical neutrality requirement: (eq. 3) [A-] + [OH-] = [Na+] + [H3O+]
Of course [Na+] = Cb, so [A-] + [OH-] = Cb + [H3O+]
Mass balance requirement: (eq. 4) [HA] + [A-] = Ca + Cb
So we have 4 unknowns ([OH-], [H3O+], [HA] and [A-]) and 4 equations: perfect! To solve, start from the assumption that [H3O+] << Cb: after all
there's more base than acid in solution and the pH is likely to be higher than 7, so [H3O+] < 10^-7
(eq. 3) can thus be approximated as [A-] + [OH-] = Cb, or (eq. 5) [A-] = Cb - [OH-]
and by substituting it into (eq. 4) we get (eq. 6) [HA] = Ca + [OH-]
Combining (eq, 1) and (eq. 2) we get:
Kw.[A-]/([HA].[OH-]) = Ka
Or [OH-] = (Kw/Ka)/([A-]/[HA])
We know that Kw = Ka.Kb and so Kw/Ka = Kb
Or [OH-] = Kb.([A-]/[HA]) and with (eq. 5) and (eq. 6):
(eq. 7) [OH-] = Kb.{(Cb - [OH-])/(Ca + [OH-])}
Two cases now arise:
a. For relatively concentrated solutions, Cb >> [OH-] and Ca >> [OH-] and the equation is then reduced to the classic buffer
equation for a weak acid/base system:
(eq. 8) [OH-] = Kb. (Cb/Ca), here pOH = - log [OH-] = 8.26
And because pH + pOH = 14, pH = 5.74. So the pH wasn't higher than 7 but the assumption [H3O+] << Cb still holds.
Using (eq.1) and Kw = Ka.Kb, (eq. 8) can also be rewritten as:
[H3O+] = Ka.(Ca/Cb) and pH = pKa - log(Ca/Cb) = - log (1.8.10^-5) - log (0.01/0.1) = - (log 1.8 + log 10^-5) - log 0.1 = -0.255 +5 +1 = 5.74
b. For very dilute systems, (eq. 7) has to be made explicit in [OH-] and the resulting quadratic equation solved by the usual means.
Enjoy!
[Edited on 9-5-2010 by blogfast25]