Calculating pH in titration of monoprotic weak acid with strong base
Your calculation is right.  Regarding your question to take into account the conjugate base's contribution to deprotonation of water, I'd say:

If you calculate the equilibrium constant of the neutralization reaction $$\ce{CH3COOH + OH- &lt;=&gt; CH3COO- + H2O}$$
$$K= \frac{K_a}{K_w}=\frac{10^{-4.75}}{10^{-14}}= 10^{9.25}$$

Now, if you write  of reaction of acetate ion with water:$$\ce{CH3COO- + H2O &lt;=&gt; CH3COOH + OH-}$$
It's the opposite of the neutralization reaction. Its  equilibrium constant  is  $10^{-9.25}$. 

You can see clearly that the neutralization reaction is the "predominant reaction". So, the pH is determined by this reaction as if it's the only one happening in the aqueous solution. So, you can neglect the reaction of of acetate ion with water.