maxcrummfan (maxcrummfan) wrote in chemistree,
maxcrummfan
maxcrummfan
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My version, versus, another version, interchangeable?

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 
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