Trying to figure out pH when 200 mL of .2M Acetic acid mixed with 200 mL of .2M KOH?

Question

Ka of Acetic acid is 1.82 x 10^-5 for buffers if it applies here (which I dont know as there is a strong base and not a conjugate base)

Answer 1

You have
mole CH3COOH= Ca*Va= 0.2*0.2=0.04
mole KOH= Cb*Vb= 0.2*0.2= 0.04

CH3COOH + KOH -> CH3COOK + H2O

The stoichiometry is 1:1 and you have equal mole of acid and base therefore they will react completely and you will have only CH3COOK in the final solution
However this is the salt of a weak acid with a strong base so it will hydrolyze according to

.. .. .. .. .. CH3COO- + H2O <=> CH3COOH + OH-
Initial .. .. .. .. C
React .. .. .. ..x
Produce .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. x .. .. .. .. .. . x
At Equil. .. .. C-x .. .. .. .. .. .. .. .. .. .. .. x .. .. .. . .. .. x

Kb=[CH3COOH][OH-] / [CH3COO-] = x^2/(C-x)

But Kb=Kw/Ka= 10^-14/(1.82*10^-5)
C=moleCH3COOK/Vfinal = Ca*Va/(Va+Vb)=0.04/0.4
substitute Kb,C and find x (you can assume that C>>x and C-x=C. this simplifies the equation to x^2/C=Kb. If you calculate x and it is truly much smaller than C the assumption is valid. Otherwise you have to solve the quadratic equation)

pOH=-logx and pH=14-pOH