In: Chemistry
CH3CO2- + H2O ------------------------> CH3COOH + OH-
0.800 0 0 -------------------> initial
-x x x ---------------------> changed
0.8-x x x -----------------------> equilibrium
Kb = [CH3COOH][OH-]/[CH3COO-]
Kw / Ka = x^2 / 0.8 -x
1.0 x 10^-14 / 1.8 x 10^-5 = x^2 / 0.8 -x
5.56 x 10^-10 = x^2 / 0.8 -x
x^2 + 5.56 x 10^-10 x - 4.45 x 10^-10 = 0
x = 2 .1 x 10^-5
[OH-] = 2 .1 x 10^-5 M
pOH = -log[OH-] = -log (2 .1 x 10^-5)
pOH = 4.68
pH + pOH = 14
pH = 9.32