Understanding pH Calculations for a Buffer Solution

AI Thread Summary
To calculate the pH of a buffer solution containing 500 ml of 0.1 M acetic acid (HC3H5O2) and 500 ml of 0.1 M sodium acetate (NaC3H5O2), the Henderson-Hasselbalch equation is used. The pKa is derived from the acid dissociation constant (Ka = 1.3 x 10^-5), leading to pKa = -log10(Ka). The pH can then be calculated using the formula pH = pKa + log10[salt]/[acid], where the salt is NaC3H5O2 and the acid is HC3H5O2. This approach confirms that the pH of the solution is approximately 4.89. Understanding the relationship between the components of the buffer and their concentrations is crucial for accurate pH calculations.
jaredmt
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Homework Statement


A mixture containing 500ml of .1 M HC3H5O2 (acetic acid) and 500 ml of .1 M NaC3H5O2

what is the pH?

Homework Equations


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

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, I am not sure where to go from here.
 
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Have you studied the Henderson-Hasselbach equation?
 
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.
 
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
 
jaredmt said:
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.
 
Last edited:
ah ok that was easier than i thought, thanks
 
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TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...

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