[Fish200398]

0,5 M KOH + 4,5 CH3COOH
At equlibrium = CH3COOH left 4M
Produce CH3COOK 0,5 M
H2O 0,5M
Ka = 2x10-5

1. [H] = Ka [CH3COOH] : [CH3COOK]

2. The left CH3COOH 4M react with water
CH3COOH + H2O = CH3COO- + H3O+
In the end we get, 4M CH3COOH and x mol CH3COO- and H3O+
x = [H]

• ^2 = Ka . [4M]

3. salt CH3COOK react with water
CH3COOK + H2O = CH3COO- + K+
CH3COO- + H2O = CH3COOH + OH-
We left with 0,5 M CH3COO- and y mol of CH3COOH and OH
y = [OH]

Kb = [CH3COOH][OH] / [CH3COO-]
[y]^2 = Kb.[CH3COO-] = Kw/Ka . 0,5 M

pH = 14-pOH

From 3 methods, i get 3 different pH.. what is wrong with my calculations?

[AWK]

Forget equations 2 and 3. Just insert your concentrations and Ka into equation 1 (Henderson-Hasselbach equation). You will get pH below 4.

[Fish200398]

But why 2 and 3 is wrong?

[mjc123]

2. You don't have x M acetate. You have (4-x) M AcOH and (0.5+x) M AcO^-. Since x is small, this is equivalent to method 1.
You can't ignore the presence of the KOAc and try and work it out for the AcOH alone.
3 is wrong for the same reason.

[AWK]

But why 2 and 3 is wrong?

There are three basic types for calculating acid-base equilibria: 1) acid or base, 2) buffer solution and 3) salt hydrolysis. Before you start calculating you must decide (correctly) which system you are dealing with.

[Borek]

To add to what mjc123 said: each of these methods is based on some simplifying assumptions. You need to understand where these assumptions come from and when they hold. If they don't hold, method produces wrong results.

Compare http://www.chembuddy.com/?left=pH-calculation&right=general-pH-calculation (and derivations of specific cases on other site pages).