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Hawkeye3
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Anyone good at General Chemistry (Buffered solutions, yes, that's right)
#5497381 - 04/09/06 06:27 PM (17 years, 9 months ago) |
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Doing some buffered solutions problems. They're not difficult, I just have a question...
In this problem:
Calculate pH of the following solution: .100 M propanoic acid (HC3H5O2, Ka=1.3x10^-5) with .020 mol HCl (1.00 L of solution)
I know HC3H5O2 dissociates to H+ and C3H5O2-, but I'm having trouble understanding where the HCl comes in. From what I get, the HCl dissociates into H+ and Cl-. The Cl has a low affinity for protons and is thus not a major specie.
Which leaves: HC3H5O2, H2O, H+, C3H5O2-
So
The H+ from the HCl re-bonds with the C3H5O2-, right? Or am I missing something here? If someone could verify my conclusions thus far and point me in the right direction...
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zappaisgod
horrid asshole

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Re: Anyone good at General Chemistry (Buffered solutions, yes, that's right) [Re: Hawkeye3]
#5497461 - 04/09/06 06:46 PM (17 years, 9 months ago) |
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If the propanoic acid gives up the H+ why would you expect it to reassociate with H+ from HCl. Seems to me that you should just count the H+ions in the volume of water.
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Hawkeye3
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Re: Anyone good at General Chemistry (Buffered solutions, yes, that's right) [Re: zappaisgod]
#5497518 - 04/09/06 06:56 PM (17 years, 9 months ago) |
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I figured as much because the point of a buffered solution is to absorb an acid without changing pH. When HCl is added the propane re-associates with the H+ to give little change in pH. Unless the H+ is bonding with the water to give hydronium...
Edited by Hawkeye3 (04/09/06 06:57 PM)
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ChuangTzu
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Re: Anyone good at General Chemistry (Buffered solutions, yes, that's right) [Re: Hawkeye3]
#5498774 - 04/10/06 02:30 AM (17 years, 9 months ago) |
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Ka = [H3O+][A-]/[AH], that is the concentration of hydronium ions times the concentration of ionized acid, divided by the concentration of unionized acid. So, without any HCl in solution:
Code:
1.3*10^-5 = [H3O+][A-]/[AH], //this is the definition of Ka [H3O+] = [A-], //since H+'s don't appear out of nowhere [A-]+[AH] = .1M //total amount of A is given by the problem
You have 3 equations in 3 variables. You can solve this system of equations any way you know how. I used matlab to find: [H3O+] = {A-] = 0.0011M and [AH] = 0.099M. As you can see, propionic acid is a weak acid and is mostly nondissociated at this point.
Now if we throw in some HCl, we get the following system of equations: Code:
1.3*10^-5 = [H3O+][A-]/[AH], //this doesn't change [H3O+] = [A-] + [Cl-], //H+ ions now come from 2 sources [A-] + [AH] = .1M //this doesn't change either
Since HCl dissociates fully in water, [Cl-] = 0.02M (0.02mol/1L). Solve this system of 3 equations in 3 unknowns just like you did before. I got: [H3O+] = .20M, [A-] = 0.00M, [AH]=.10M. Now, just about none of the propionic acid is dissociated. The problem asks to calculate the pH = -log[H+}. Since [H+] = [H3O+], pH = -log(0.2) = 1.6. That's it.
(Does anyone know if there is a "verbatim"-like UBB tag other than "code"?)
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ChuangTzu
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Re: Anyone good at General Chemistry (Buffered solutions, yes, that's right) [Re: Hawkeye3]
#5498782 - 04/10/06 02:40 AM (17 years, 9 months ago) |
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Quote:
Hawkeye3 said: I figured as much because the point of a buffered solution is to absorb an acid without changing pH. When HCl is added the propane re-associates with the H+ to give little change in pH. Unless the H+ is bonding with the water to give hydronium...
A solution of propionic acid in water isn't a buffer. A buffer typically consists of a solution of a weak acid and its conjugate base (propionate in this case). Excess propionate (in the form of sodium propionate for example) was not added so the pH will swing quite wildly with the addition of strong acid/base. For example, the solution of propionic acid before addition of HCl will have a pH of about 3.
When talking about aqeuous systems, H+ and H3O+ are interchangable.
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