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Chemistry: buffers and pH

by jaredmt
Tags: buffers, chemistry
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jaredmt
#1
Apr9-08, 03:04 PM
P: 122
1. The problem statement, all variables and given/known data
A mixture containing 500ml of .1 M HC3H5O2 (acetic acid) and 500 ml of .1 M NaC3H5O2

what is the pH?


2. Relevant equations
can someone walk me through this? i know the answer is 4.89 but im not sure how to get that. im not even entirely sure how to set up the equation


3. The attempt at a solution
i tried this to start, but not sure if its right:
notations:
wa = weak acid
sa = strong acid
wb = weak base
sb = strong base
n = neutral

-wa----sb-------wb-----n
HAc + NaOH <> NaAc + H20

even if this is right, im not sure where to go from here.
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chemisttree
#2
Apr9-08, 03:18 PM
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Have you studied the Henderson-Hasselbach equation?
PuzzledMe
#3
Apr9-08, 03:40 PM
P: 34
Just curious but are you provided with the Ka (acid dissociation constant) value for HC3H5O2?

For this case it's [salt] over [acid] (with Henderson-Hasselbach equation). Notice the acidic buffer there.

jaredmt
#4
Apr9-08, 04:01 PM
P: 122
Chemistry: buffers and pH

o sorry, the Ka = 1.3 x 10^-5
and yes we were taught the enderson-Hasselbach technique

i wish i could edit that into the 1st post but w/e
PuzzledMe
#5
Apr9-08, 04:07 PM
P: 34
Quote Quote by jaredmt View Post
o sorry, the Ka = 1.3 x 10^-5
and yes we were taught the enderson-Hasselbach technique

i wish i could edit that into the 1st post but w/e
Ooooo and I'm wondering why I can't find that vital piece of information. The Ka means everything. Use this equation:

pH = pKa + log10 [salt]/[acid]

pKa = -log10 (Ka)

It's that simple =) Try it out!

You don't even have to write out the equation. NaC3H5O2 is the salt and HC3H5O2 is the acid. What you do need to know is that this is an acidic buffer solution.
jaredmt
#6
Apr9-08, 11:48 PM
P: 122
ah ok that was easier than i thought, thanks


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