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

0.4 0.4 0.4

INTIAL: 0.4 0.2 0

CHANGE: -x +x +x

EQUIL: 0.4-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

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 **