Predicting the pH of a weak acid and weak base solution
Note: The expected $\text{pH}$ values refer to 1:1 mixtures as suggested in the question



One major feature of weak acids and bases is that they do not dissociate completely in water, producing less free oxonium or hydroxide ions than strong acids and bases do (Brønsted). For example in Sodiumacetate, a weak base, there is an equilibrium where the reactions left to right and right to left occur at the same speed, hence cancelling each other. 
\begin{aligned}\ce{
   2NaOAc_{(aq)} + H2O &amp;&lt;=&gt; 2Na+_{(aq)} + {}^{-}OAc_{(aq)} + HOAc_{(aq)} + {}^{-}OH_{(aq)}} &amp;(1)
\end{aligned}

The same also applies to weak bases like ammonium hydroxide $\ce{NH4OH}$. Note, that ammonium itself is not a Brønsted base as it lacks to donate hydroxide ions.
\begin{aligned}\ce{
   NH4OH_{(aq)} &amp;&lt;=&gt; NH4+_{(aq)} + {}^{-}OH_{(aq)}} &amp;(2)
\end{aligned}

Weak acids can be expanded in the same way, example benzoic acid:
\begin{aligned}\ce{
   Ph-COOH_{(aq)} + 2H2O&amp;&lt;=&gt; Ph-COO^{-}_{(aq)} + H+3O_{(aq)}} &amp;(3)
\end{aligned}

In the Lewis definition, weak acids are not able to complex the hydroxide ions of water's autoprotolysis.

\begin{aligned}\ce{
   2H2O &amp;&lt;=&gt; H+3O_{(aq)} + {}^{-}OH_{(aq)} &amp;\\
   BF3_{(aq)} + 3H2O &amp;&lt;=&gt; [HOBF3]^{-}_{(aq)} + 2H+3O_{(aq)} + {}^{-}OH_{(aq)} &amp;(4)
}\end{aligned}

These equations are also in equilibrium, meaning all species are present in aqueous solution.

In the Lewis definition ammonia $\ce{NH3}$ is a base and it reacts analogous to the above. 

\begin{aligned}\ce{
   2NH3_{(aq)} + 2 H2O &amp;&lt;=&gt; NH4OH_{(aq)} + NH4+_{(aq)} + {}^{-}OH_{(aq)}} &amp;(5)
\end{aligned}

Now the rate of dissociation is strongly dependent on the weakest bond. In $(1)$ that is $\ce{Na-O}$ and successively $\ce{H-O}$. Overall the reaction will produce excess hydroxide ions, hence the solution will have a $\text{pH}&gt;7~(@25~^\circ{}\!C)$.
The same applies analogously to $(2)$ and $(5)$

In $(3)$ the opposite is the case. The reaction will produce excess oxonium ions, hence having a $\text{pH}&lt;7~(@25~^\circ{}\!C)$. The same applies to $(4)$.

So when adding a weak base to a weak acid, one has to evaluate the whole equilibrium:
\begin{aligned}\ce{
   NH4OH_{(aq)} + HOAc_{(aq)} + H2O 
   &lt;=&gt;&amp;\\ [(NH4)(OAC)]_{(aq)} + NH4+_{(aq)} &amp;+ {}^{-}OH_{(aq)} +  {}^{-}OAc_{(aq)} + H+3O_{(aq)}
} &amp;(6)
\end{aligned}

It is therefore very much dependent on the chemical properties of all compounds involved in the solution. These systems are also very well known as buffer solutions.

So a good start would be comparing $\text{p}K_a$ and $\text{p}K_b$ values of the involved components, giving you a hint whether it will be more acidic or more basic. But you also have to pay attention to the amount of substance in particular in these cases. Dilution also plays an important role in this kind of chemistry.