So for the Ka of acetic acid I have used the value which is well used in many websites = 1.8*10^-5
From the ICE table we have
HA + H2O -> H3O+ + A-
I 0.032 0M 0M
C -x +x +x
E 0.032 - x +x +x
Now using the equilibrium formula:
Ka = [H3O+] * [A-]/[HA]
1.8*10^-5 = x^2/0.032 - x
Since this x will give a small number we can "ignore" it, therefore,
1.8*10^-5 = x^2/0.032
x^2 = 5.76*10^7
x = 7.59*10^-4, this is the concentration of H+
Now let's use our pH formula
pH = -log [7.59*10^-4]
pH = 3.12
Now we have the pH, the pOH will be = 14 - 3.12
pOH = 10.88
If we check letter B
pOH = -log [1.32*10^-11]
pOH will be equal to 10.88 exactly
Therefore, letter B is the correct one
