Why does dissolving a salt in water get a high pH?
It's a little more complicated than that.. When you dissolved $\ce{Na2CO3}$ in water you had a solution of ions: $\ce{Na+}$, $\ce{CO3^2-}$, $\ce{H3O+}$ and $\ce{OH-}$ (the later are present in water even when there is no salt, since it dissociates too, however they are in equal quantity in pure water so the pH is 7).

Now when you mix all those ions, the equilibrium isn't only the equilibrium of water dissociation ($\ce{H2O + H2O &lt;=&gt; H3O+ + OH-}$). Now you have a whole series of equilibria occurring in the solution simultaneously:

1) Water dissociation: 
$\ce{H2O + H2O &lt;=&gt; H3O+ + OH-}$

2) Salt dissociation:
$\ce{Na2CO3 &lt;=&gt; 2Na+ + CO3^2-}$

3) Base equilibrium*:
$\ce{NaOH &lt;=&gt; Na+ + OH-}$

4) Acid equilibrium*:
$\ce{CO3^2- + 2 H3O+ &lt;=&gt; H2CO3 + 2 H2O}$

The salt dissociation doesn't affect the pH directly. We have two equilibria though that do unbalance the proportion of $\ce{H3O+}$ and $\ce{OH-}$ ions, which are the "base equilibrium" and "acid equilibrium". We don't think about them at first sight because we didn't add any $\ce{NaOH}$ or $\ce{H2CO3}$ to the solution, but everything needed to form them is right there.

$\ce{NaOH}$ is a strong base, however $\ce{H2CO3}$ is a weak acid. So it's likely (and you witnessed it in experience) that more $\ce{H3O+}$ will be consumed in comparison with the $\ce{OH-}$, making the solution more alkaline.