Calculate the pH of buffer solution
I can't calculate the $\mathrm{pH}$ using Henderson-Hasselbalch approximation, because the concentration of acid is 0.


Yes, as commented above, sodium acetate solution cannot be considered as buffer;the result at the equilibrium , will be the same as dissolving  $\pu{0.025 mol }$ of $\ce{CH3COONa}$ in $\pu{250 ml}$ water,so you have the salt solution of $\pu{0.1M}\ce{[CH3COONa]}(c_\mathrm{\text{salt}}=\ce{[CH3COONa]_\mathrm{I}}=\pu{0.1M})$


  I don't understand the last part, why is this so? what is...


The salt ($\ce{ CH3COONa}$) is  electrically neutral substance formed by cation $\ce{Na+}$ (an acid) and  anion $\ce{ CH3COO−}$ (a base), completely dissociated or ionized in an aqueous solution as :
  $$\ce{CH3COONa   -&gt; Na^+_\mathrm{(aq)}   + CH3COO^−_\mathrm{(aq)}}$$ 
a)  Since $\ce{Na+}$  is the conjugate acid of a strong base, it won't be strong enough to react with water; $\ce{Na+}$  actually spectator ion.

b)  Meanwhile, since $\ce{ CH3COO−}$ is the conjugate base of a weak acid, and  therefore strong enough to be able to hydrolyze and accept ions $\ce{ H+}$ from water, so water act as an acid leaving a hydroxide ion $\ce{OH−}$ as :
$$\ce{CH3COO^− +H2O &lt;=&gt;CH3COOH + OH^−}$$ $$(K_\mathrm{b(\ce{CH3COO-})}=\frac{K_\mathrm{w}}{K_\mathrm{a(\ce{CH3COOH})}}=\frac{10^{-14}}{1.76\times 10^{-5} }=5.68\times{10^{-10}})$$

c)The equilibrium equation of the hydrolysis of the conjugate base $\ce{CH3COO-}$:
$$ K_\mathrm{b} =\frac{[\ce{OH-}]_\text{equilibrium}[\ce{CH3COOH}]_\text{equilibrium}}{[\ce{CH3COO-}]_\text{equilibrium}}$$

Neglecting water autoionization and assuming :${\ce{[OH-]_\text{equilibrium}}=\ce{[CH3COOH]_\text{equilibrium}}}$, ${[\ce{CH3COO-}]_\text{equilibrium}=c_\mathrm{\text{salt}}=[\ce{CH3COONa}]_\mathrm{I}}=\pu{0.1 M}$,so:
$$ K_\mathrm{b} =\frac{[\ce{OH-}]_\text{equilibrium}^2}{c_\mathrm{\text{salt}}}$$

$$[\ce{OH-}]_\text{equilibrium}=\sqrt{  K_\mathrm{b}\cdot c_\mathrm{salt}}$$
You  can calculate $\ce{[H+]}$ -without neglecting the water autoionization  and without approximation- by substituting in the following cubic equation and solving bywolframalpha

$$\
 [\ce{H+}]^3 
+( K_\mathrm{a}+[\ce{CH3COONa}]_0) [\ce{H+}]^2 
  -K_\mathrm{w}[\ce{H+}] -    K_\mathrm{w} K_\mathrm{a}
  $$
$\ce{[H+]} =1.32658 ,\pu{pH}=8.877$