Calculating pH: Weak Acid, Strong Base
At the equivalence point, you have an aqueous solution of weak base $\ce{Na}A$ at the concentration of $c$. The pH of the solution is given by the equation: $$\mathrm{pH}=\frac{1}{2}(\mathrm{p}K_a +\mathrm{p}K_w -\mathrm{p}c)$$
In your case: $\mathrm{pH}=\frac{1}{2}(4.0 +14 -2.0)=8.0$
Now, if you are not familiar with the above equation, you can write the reaction that could happen in the solution; the reaction of  $\ce{Na}A$, completely dissociated, with water:
$$A^- +\ce{ H2O &lt;=&gt; OH- +H}A$$
The constant of this equilibrium is: $$K=\frac{K_w}{K_a}= 10^{-10}$$

On the other hand,  you have:


$$K=\frac{x^2}{c-x}=\frac{x^2}{1.0\times10^{-2}-x}\approx\frac{x^2}{1.0\times10^{-2}} =1.0\times10^{-10}$$
Where $x$ is the concentration of ion hydroxide and the weak acid $\ce{H}A$.

We have, $x=1.0\times10^{-6}\mathrm{M}$.

$$[\ce{H+_\mathrm{(aq)}}]=\frac{10^{-14}}{1.0\times10^{-6}}\mathrm{M}$$
 $$\mathrm{pH}=8.0$$