Show that the pH at the endpoint in the titration of $\mathrm{0.100\ mol\ dm^{-3}}$ ethanoic acid with $\mathrm{0.100\ mol\ dm^{-3}}$ sodium hydroxide is 8.72. The dissociation constant Ka of ethanoic acid is $\mathrm{1.80\ x\ 10^{-5}\ mol\ dm^{-3}}$.
I managed to obtain the value of Kb using the formula (Ka x Kb = Kw = $\mathrm{10^{-14}}$).
I worked out the chemical equation at the endpoint which is:
$$\ce{CH3COO- + H2O <--> CH3COOH + OH- }$$
However, I am unsure as to what the concentration of $\ce{CH3COO-}$ should be.
I understand that if I find [$\ce{OH-}$] then I would be able to find pOH and by extension pH.