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Annonn Annonn
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Posts: 116
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A month ago



1. Find the Molar concentration of HC2H3O2 (M)
-I'm not sure if each buffers affect the molar concentration, but I searched up the normal molar concentration for HC2H3O2 and found that it is 0.8527 M. Can anyone confirm that or steer me in the right direction on how to solve for the molar concentration?


2. Using Excel, prepare two charts, each with two series, from your data.
Create a separate chart for your titrations using NaOH with a series plotting your
titration with Buffer A and a series plotting your titration with Buffer B.
Do the same with your titrations using HCl.
-I think I understand how to make the graphs: there will be 4 separate graphs, Buffer A - NaOH, Buffer A - HCl, Buffer B - NaOH & Buffer B - HCl. Please give me any advice if my thought process is on the right track.


3. Write reaction equations to explain how your acetic acid-acetate buffer reacts with an acid and reacts with a base.

4. Buffer capacity has a rather loose definition, yet it is an important property of buffers. A commonly seen definition of buffer capacity is: “The amount of H+ or OH– that can be neutralized before the pH changes to a significant degree.”
We will consider 2 pH units to be significant. Use your data and your charts to determine the buffer capacity of Buffer A and Buffer B.

-I don't expect anyone to do the work for me (especially for the graphs), just please give me advice on where to start, any formulas that may help, etc.  Confounded Face

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wrote...
Educator
A month ago

Answer verified by a subject expert
bio_manbio_man
wrote...
Educator
Top Poster
Posts: 31040
A month ago
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See if this helps for question #4



Buffer A.

Moles of NaC2H3O2 (a base component of buffer) = 0.031 g/(82 g/mol) = 3.78*10-4 mol

Moles of HC2H3O2 (an acid component of buffer) = 0.1 mol/L * (100/1000) mL = 0.01 mol

Now, pH = pKa + Log([base]/[acid])

The buffer capacity range is (4.74-1) to (4.74+1) = 3.74 to 5.74

By the addition of acid, pH decreases, i.e. the max. pH for the buffer capacity = 3.74

i.e. 3.74 = 4.74 + Log{(3.78*10-4 - x)/(0.01+x)}

i.e. (3.78*10-4 - x)/(0.01+x) = 0.1

i.e. 3.78*10-4 - x = 0.001 + 0.1x

i.e. 1.1x = 6.22*10-4

i.e. The buffer capacity = 5.654*10-4 mol

Buffer B.

Moles of NaC2H3O2 (a base component of buffer) = 0.314 g/(82 g/mol) = 3.83*10-3 mol

Moles of HC2H3O2 (an acid component of buffer) = 1 mol/L * (100/1000) mL = 0.1 mol

Now, pH = pKa + Log([base]/[acid])

The buffer capacity range is (4.74-1) to (4.74+1) = 3.74 to 5.74

By the addition of acid, pH decreases, i.e. the max. pH for the buffer capacity = 3.74

i.e. 3.74 = 4.74 + Log{(3.83*10-3 - x)/(0.1+x)}

i.e. (3.83*10-3 - x)/(0.1+x) = 0.1

i.e. 3.83*10-3 - x = 0.01 + 0.1x

i.e. 1.1x = 6.17*10-3

i.e. The buffer capacity = 5.61*10-3 mol
This verified answer contains over 280 words.
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wrote...
Educator
A month ago
Also, I found a lab that contains many of the same elements as your question.
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wrote...
A month ago
First of all, I'd like to thank you so much for helping me and finding a similar lab!!
I'd like to ask one more thing about finding the buffer capacity. So, as I was filling out my website I have to use in order to submit my work is asking extra questions that were not in my lab manual:

I did the calculations based on the advice you gave me and ended up with Buffer A= 4.72*10-4 and Buffer B= 4.63*10-3. But I am not sure if those calculations would be for the NaOH or HCl treatment if that makes sense.
wrote...
Educator
A month ago
@Annonn, what did you get for the molar concentration? I will need that to verify your answers
wrote...
A month ago
Buffer A= 0.1 and Buffer B= 1.0
wrote...
Educator
A month ago
Ok, let's assume your data looks like this:



The calculations would look like this:



I can provide you with another example if you like, but this should sum it up
wrote...
A month ago
Thank you, that helps a lot! Slight Smile
wrote...
Educator
A month ago
Will mark the topic solved for now 👍
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