Why is the hydrolysis of the conjugate base of a weak acid neglected in buffer solutions?
$\mathrm{p}K_\mathrm{a} = 4.75$, therefore $K_\mathrm{a} = 10^{-4.75}$. Thus,

$$K_\mathrm{a} = \frac{[\ce{AcO^-}][\ce{H+}]}{[\ce{AcOH}]}$$

So, let's assume $[\ce{AcOH}] = \pu{1.0 M}$. Just to get a feel, how far the acid is dissociated:

$$K_\mathrm{a} = \frac{x \cdot x}{\pu{1.0 M}} \to x = \sqrt{10^{-4.75}}~\pu{M} = 0.00422 \approx 0.4\%$$

where we have assumed that the contribution from the dissociation of water is negligible, and therefore $[\ce{H+}] = [\ce{AcO-}] = x$.

Now, what happens if I add $\pu{1.0 M}$ $\ce{AcO-}$? Then, effectively, we have:

$$K_\mathrm{a} = \frac{[\ce{AcO-}][\ce{H+}]}{[\ce{AcOH}]} = \frac{\pu{1.0 M}
 \cdot [\ce{H+}]}{\pu{1.0 M}}$$

Therefore, we have:

$$[\ce{H+}] = K_\mathrm{a} \frac{\pu{1.0 M}}{\pu{1.0 M}} = \pu{10^{-4.75} M}$$

If we compare the two cases, before addition of acetate, we had a $\mathrm{pH}$ of

$$[\ce{H+}] = \sqrt{10^{-4.75}} \to \mathrm{pH} = 2.375$$

and after addition of acetate, we have

$$[\ce{H+}] = 10^{-4.75} \to \mathrm{pH} = 4.75$$

Meaning that the solution got more basic after the addition of acetate, as excpected.