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Topic: % dissociation of C6H5COOH in a solution please help  (Read 13001 times)

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Offline noapologies01

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% dissociation of C6H5COOH in a solution please help
« on: August 03, 2008, 08:43:30 PM »
Calculate the % dissociation of C6H5COOH in a solution which is 0.0150M C6H5COOH (aq) and 8.33M HCl (aq).

Ka for C6H5COOH = 6.3 x 10-5

A. 1.26 x 10-3
B. 2.27 x 10-3
C. 1.76 x 10-3%
D. 4.35 x 10-4
E. 7.56 x 10-4

i ended up here and cant figure why i cant get E as an answer
(0.0150)(6.3 x 10-5) = 8.33

Offline enahs

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Re: % dissociation of C6H5COOH in a solution please help
« Reply #1 on: August 03, 2008, 08:52:09 PM »
Show how you are setting up your I.C.E. diagram (Initial. Change. Equilibrium).

Offline noapologies01

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Re: % dissociation of C6H5COOH in a solution please help
« Reply #2 on: August 03, 2008, 09:48:01 PM »
HCl Ice table i got H+ = 8.33
      
Solve for H+

[H+] = 8.33

d: Draw and ICE Table for C6H5COOH and plug (8.33) in for H+

   
   
C6H5COO   

I   
0.0150      
      
8.33

C   
- x      
+ x   
   
+ x

E
   
(0.0150 – x)      
x      
(8.33 + x)

Solve for x

(0.0150)(6.3 x 10-5) = 8.33


Offline enahs

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Re: % dissociation of C6H5COOH in a solution please help
« Reply #3 on: August 03, 2008, 10:35:11 PM »
Always start with your equation. Also, think about the setup. The initial concentration of H+ is set by the HCl. You are solving for the dissociation of the benzoic acid, so you need to use your equation for that!


C6H5COOH    -->   C6H5COO-  +   H+        =        Ka


I  0.015                      0             8.33
C  -x                        +x             +x
E   0.015+x                +x            8.33 + x        =   Ka

Yes? No? Understand?

So now you have the equilibrium conditions for the reactions, can you setup the proper form to solve for x?


Review the material on this site:
http://dbhs.wvusd.k12.ca.us/webdocs/Equilibrium/Calc-equib-from-init-cond.html


Offline noapologies01

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Re: % dissociation of C6H5COOH in a solution please help
« Reply #4 on: August 04, 2008, 12:03:26 AM »
i got 8.32 +/- 4.16 = 12.5 or 4.16 but i have no idea what im doing, i have a mental block, so can u please just solve this problem so i dont have to spend the rest of my night on it

Offline AWK

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Re: % dissociation of C6H5COOH in a solution please help
« Reply #5 on: August 04, 2008, 05:00:33 AM »
H3O+ from benzoic acid can be safely neglected, then  the dissociation degree is
alpha=[C6H5COO-]/[C6H5COOH]≈Ka/[H3O+]
then % od diss. =  100 x Ka/[H3O+]

The serious error (though rather difficult to estimate) comes in this case from a difference between the thermodynamic Ka and the concentration Ka. Thus the real % of diss. will be probably about two times lower than calculated one.
AWK

Offline enahs

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Re: % dissociation of C6H5COOH in a solution please help
« Reply #6 on: August 04, 2008, 08:27:57 AM »
That is not a valid option in this case AWK. While I and most people would say the answers D and E are essentially the same given the scale and science behind it, the teacher would consider one correct and incorrect most likely. So one can not estimate the order of magnitude (which is the more important part) but the precise number from the basic formula.

Offline Borek

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Re: % dissociation of C6H5COOH in a solution please help
« Reply #7 on: August 04, 2008, 09:45:22 AM »
That is not a valid option in this case AWK. While I and most people would say the answers D and E are essentially the same given the scale and science behind it, the teacher would consider one correct and incorrect most likely. So one can not estimate the order of magnitude (which is the more important part) but the precise number from the basic formula.

AWK wrote about two things. One - you may safely ignore ICE table and calculate dissociation percentage assuming known concentration of H+ (that's how I would do this question, note that ignoring second acid H+ you are making error lower than 0.2%, that's just from comparison of concentrations, doesn't matter how strong or weak second acid is). Two - result will be completely off. Please elaborate - what is not a valid option, as it is unclear to me.

I don't think AWK suggested to use his estimate as the final answer, he just pointed to the fact that the calculated answer is off.
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Offline enahs

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Re: % dissociation of C6H5COOH in a solution please help
« Reply #8 on: August 04, 2008, 11:33:38 AM »
Quote
Please elaborate - what is not a valid option, as it is unclear to me.
I turned by calculator back on (luckily it has a memory of the previous X number of calculations) and see that when I was doing the question my self I was typing in 8.88 and not 8.33 and 6E-5 not 6.5E-5 when I originally did the shortcut method, which gave a result in between answer D and E.

My bad.
I honestly think I might be developing dyslexia. My yearly eye exam is soon, I better ask my doctor to check that. I seem to be making these kinds of mistakes more and more.

Offline Borek

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Re: % dissociation of C6H5COOH in a solution please help
« Reply #9 on: August 04, 2008, 12:23:13 PM »
My bad.

Happens, I do it all the time.

Quote
I honestly think I might be developing dyslexia. My yearly eye exam is soon, I better ask my doctor to check that. I seem to be making these kinds of mistakes more and more.

You may need glasses, or you my be tired and overworked, but it is not dyslexia. You don't develop dyslexia in later age, you have dyslexia from the very beginning.

Over half of my income is still from dyslexia therapy software (see www.bpp.com.pl - be warned, it is in Polish), I know a little bit about the subject :)

And no, Junior was not dyslexic, it was never designed to help him.
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