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My version, versus, another version, interchangeable? - CHS Chem Help Online

About My version, versus, another version, interchangeable?

Previous Entry My version, versus, another version, interchangeable? Oct. 24th, 2007 @ 10:35 pm
 

Hello All,
I'm in a college Chem 2 class and I've just recently taken a quiz.  The problem is, is that- I ulitmately got the correct answer, however, the approach of solving this problem is incorrect.  I was marked off points in the structure of solving this problem, however, my prof. didn't have time to go over the quiz with me and I want to be sure that my version, is theoretically and quantitavely, correct, and somehow my version can be interchangeable with the prof. version.  Any help, explanation, would be great.  Thank you for your time:

The Question:

A solution is prepared to be .20M HC2H3O2 and .4M KC2H3O2.  What is the PH of this solution? (Ka= 1.7 x 10^ -5 for acetic acid)

My version of solving this problem:

KC2H3O2 >>>>>  K  +   C2H3O2                                                                C2H3O2 + H20>>>>>HC2H3O2 +  OH
      0.4                    0.4          0.4                                                                          0.4            ----                  0.2               0

INTIAL:                           0.4          0.2              0
CHANGE:                      -x             +x              +x
EQUIL:                        0.4-x        0.2+x            x   


Kb=     Kw/Ka  =    1.00 x 10^ -14/ 1.7x 10^-5  =   5.88 x 10^ -10
Kb=    [HC2H3O2] [OH] /  [C2H3O2]

5.88 x 10^-10 = (0.2 + x) (x) /  (0.4 - x)

2.35 x 10 ^-10 = 0.2x

x= 1.18 x 10^ -9

-log(1.18x 10- -9) =   8.93  =  POH           PH= 14- 8.93 = 5.07


TEACHER VERSION:

HC2H3O2 + H2O >>>>>>  H3O + C2H3O2
     0.2             ----                      0            0.4


INTIAL:               0.2         0          0.4
CHANGE:           -x          +x          +x 
EQUIL:            0.2-x         x            0.4+x

Ka=    [H3O] [C2H3O2] / [HC2H3O2]

1.7 x 10^ -5 =   (x) (0.4 + x) / (0.2 -x)

8.5 x 10^-6 = x = [H]

-log(8.5 x 10^ -6) = 5.07 = PH 
 

Leave a comment
From:(Anonymous)
Date:March 28th, 2010 07:53 pm (UTC)
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What happen to the potassium in the teacher's version? Is it not needed?
From:(Anonymous)
Date:April 4th, 2010 03:17 am (UTC)
(Link)
hey max, so the only difference between your teachers and yours is that your teacher showed the reaction reacting as an acid, whereas you did it as a base. So, you discovered the pOH and subtracted that from 14, and she just took the negative log of the concentration of H30+, both of which are the same. hope that helps! and im also wondering about the potassium, does it just kind of go away?
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